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ion Tower, Dallas, 582 Show Caves, Texas Hill Country, Sonora, 141 Southwestern University, Georgetown, 448 Texas Coins, 741 Titan, Arlington, 449 Problem-Solving Plan, xxviii Problem-Solving Strategies, S40–S49 Draw a Diagram, S40 Find a Pattern, S44 Guess and Test, S42 Make a Model, S41 Make a Table, S48 Make an Organized List, S47 Solve a Simpler Problem, S49 Use a Venn Diagram, S45 Use Logical Reasoning, S46 Work Backward, S43 Problems measuring to solve, 736–737 reading to solve, 745 solving simpler, S49 Proof, 228, 312, 338, 391, 397, 404, 405, 411–415, 425, 427, 434, 523, 753, 758, 762, 778–779, 783, 788–789 of angle-angle-side (AAS) congruence, 254 of the Chord-Chord Product Theorem, 797 of the Circumcenter Theorem, 308 by contradiction, 332 of the Converse of the Hinge Theorem, 340 coordinate, see Coordinate proof of the Distance Formula, 354 hands-on, of the Pythagorean Theorem, 347 indirect, see Indirect proof of the Inscribed Angle Theorem, 772, 778 of the Isosceles Triangle Theorem, 273 of the Law of Cosines, 557 of the Perpendicular Bisector Theorem, 300 of the Pythagorean Theorem, 347–348, 798 of the Secant-Secant Product Theorem, 793 of the Secant-Tangent Product Theorem, 797 of the Triangle Inequality Theorem, 338 of the Triangle Midsegment Theorem, 326 of the Triangle Sum Theorem, 222, 223 The Proof Process, 112 Proofs, 104, 178 algebraic, 104–107 design plans for, 117 flowchart, 118–119, 122, 123, 168 of the Common Segments Theorem, 118 of the Converse of the Alternate Interior Angles Theorem, 168 of the Converse of the Common Segments Theorem, 118 geometric, 110–113 paragraph, 120–124, 159, 168, 169, 173, 187, 305, 306, 345, 354, 362, 396, 404, 424, 467 of the Converse of the Same-Side Interior Angles Theorem, 168 of the Perpendicular Transversal Theorem, 173 of the Same-Side Interior Angles Theorem, 159 of the Vertical Angles Theorem, 120 Index S171S171 point-slope form of a line, 190 two-column, 111–125, 159, 164, 173, 175, 187, 305, 306, 318, 342, 343, 344, 394, 395, 396, 403, 413, 423, 424, 435, 472 of the Alternate Exterior Angles Theorem, 159 of the Common Segments Theorem, 118 of the Congruent Complements Theorem, 112 of the Congruent Supplements Theorem, 111 of the Linear Pair Theorem, 111 of the Right Angle Congruence Theorem, 112 of the Vertical Angles Theorem, 120 Properties of congruence, 106 of equality, 104 of exponents, S54 of inequality, 330 of kites, 427 of linear inequalities, S60 of parallelograms, 391, 392 of real numbers, S51 of rectangles, 408 of rhombuses, 409 of squares, 410 of trapezoids, 429–431 Proportion(s), 454–457 properties of, 455 solving, 313 Proportional Perimeters and Areas Theorem, 490 Proportional relationships, 488–490 Protractor, 20, F47 using, 21 Protractor Postulate, 20 Prove statement, 111 Pyramid Arena, 695 Pyramid of Cheops, 531 Pyramids, 654 altitude of, 689 drawing, 653 frustum of, 696 regular, see Regular pyramids surface area of, 689–692 vertex of, 689 volume of, 705–708 Pythagorean Identity, 531–532 Pythagorean Inequalities Theorem, 351 Pythagorean Theorem, 45, 220, 348–352, 461, 522, 707, 749 Converse of the, 350 deriving the, 522 hands-on proof of the, 347 Math Builders, xxvi–xxvii proof of the, 347–348, 798 solving quadratic equations using, 760, 761, 770 using, in three dimensions, 671 Pythagorean triple, 349 Q Quadrants, 42, S56 Quadratic equations, see Equations Quadratic formula, using, 272, 279, S66 Quadrilaterals, 98, 382 developing formulas for, 589–593 opposite angles, 391 opposite sides, 391 polygons and, 376–449 regular, 380 special, 391 Quartiles, S80 Question type, any check with a different method, 372–373 estimate, 578–579 highlight main ideas, 890–891 interpret a diagram, 510–511 measure to solve problems, 736–737 use a formula sheet, 646–647 R Racing, 392 Radicals, simplifying, 44, 346, 519–521 Radius, 37, 747 of a sphere, 714 Rainforest Pyramid, 705, 706 Range, 41, 345, 389, 533, 597 Rate of change, 182, see also Slope Ratio(s), 33, 454–457, 754 area, 490 perimeter, 490 rate, S70 in similar polygons, 462–464 similarity, 463, 490 trigonometric, 524, 525–528 Rational numbers, 80, S53 Rattler, 233 Rays, 7 Reading and Writing Math, 5, 73, 145, 215, 299, 397, 453, 517, 587, 653, 745, 823, see also Reading Strategies; Study Strategies; Writing Strategies Reading Math, 273, 300, 455, 456, 534, 570, 670, 748 Reading Strategies, see also Reading and Writing Math Learn Math Vocabulary, 299 Read and Interpret a Diagram, 73 Read and Understand the Problem, 453 Read Geometry Symbols, 215 Read to Solve Problems, 745 Read to Understand, 517 Use Your Book for Success, 5 Ready to Go On?, 35, 59, 103, 127, 181, 201, 239, 281, 329, 365, 407, 437, 479, 503, 543, 569, 615, 639, 679, 725, 771, 807, 855, 881, see also Assessment Real Estate, 486 Real numbers classifying, S53 operations with, S50 Reasonableness, 66–67, 332, 428, 445, 453, 578–579, 632, 749, S52 Reasoning deductive, see Deductive reasoning direct, 332 inductive, see Inductive reasoning logical, S46 spatial, 650–741 Reciprocals, opposite, 184 Recreation, 15, 92, 108, 271, 476, 564, 636, 673–674, 828, 850 Rectangle, 36 properties of, 408 proof of, 408 Rectangular prism, right, diagonal of a, 671 Reduction, 495, 873 Reference angle, 570 Reflection symmetry, glide, 863 Reflections, 50, 824–826 constructing, 829 in the coordinate plane, 826 describing transformations in terms of, 850 of figures, constructing, 824 glide, 848, 851 of parent functions, 838 Reflexive Property, 168, 176 of congruence, 106 of equality, 104 of similarity, 473 Regression, 494, see also Lines of best fit Regular polygons, 380–382, 818–819 area of, 601 center of, 601 central angles of, 601 constructing, 380–381 developing formulas for, 600–602 Regular polyhedrons, 669 Regular pyramids, 689 lateral area of, 689 slant height of, 689 surface area of, 689 Regular tessellations, 864 Related conditionals, 83 Relations, 389, S61 Relationships functional, in formulas, 713, S63 proportional, 488–490 Relative error, S73 Reliant Stadium, 140 Remember!, 36, 82, 104, 106, 129, 182, 191, 217, 242, 260, 269, 275, 282, 283, 309, 348, 351, 358, 382, 383, 384, 393, 398, 420, 429, 454, 489, 552, 560, 562, 589, 590, 592, 602, 617, 630, 682, 698, 765, 801, 831, 841 Remote interior angles, 225 Repeat unit, 868 Representations of three- dimensional figures, 661–664 S172 S172 Index Resultant vectors, 561 Reunion Tower, 2, 582 Rhombus(es), 409 area of, 591 conditions for, 419 constructing, 415 properties, 409 proof, 409 Rhombus method, 170 Richter scale, S74 Right angle, 21 Right Angle Congruence Theorem, 112 proof of the, 112 Right cone, 690 lateral area of, 690 slant height of, 690 surface area of, 690 Right cylinder, 681 lateral area of, 681 modeling, 688 surface area of, 681 Right prism, 680 lateral area of, 680 surface area of, 680 Right rectangular prism, diagonal of a, 671 Right triangle(s), 216 constructing, 258 similarity in, 518–520 solving, 534–537 special, 526, 529, 530 trigonometry and, 512–583 Rigid motions, 824 Rigidity, triangle, 242 Rio Grande river, 186 Rise, 182 Roberval, Gilles Personne de, 404 Roller coasters, 92, 233, 449 Roman numerals, 42 Rotational symmetry, 857 angle of, 857 order of, 857 Rotations, 50, 74, 839–841 constructing, 844 in the coordinate plane, 840 of figures, constructing, 839 Ruler, F47 Ruler Postulate, 13 Run, 182 S 7A Ranch, 70 Safety, 158, 349, 353, 386, 395, 530 Sailing, 245 Salinon, 768 Same-side interior angles, 147 Same-Side Interior Angles Theorem, 156 Converse of the, 163 proof of the, 168 proof of the, 159 Sample space, 628 San Jacinto Monument, 514 SAS (side-angle-side) congruence, 243 applying, 242–245 exploring, 240–241 SAS (side-angle-side) similarity, 471 Satellite, 797 Scalar multiplication, 566 Scale, 489 Scale drawing, 489 Scale factor, 495, 872 Scalene triangle(s), 217 constructing, 248, 313 Scatter plots, 198, S79 Scavenger Hunt, xxii Science, 92, 786 Scientific notation, S54 Scoring Rubric, 208 Secant, 531, 746 Secant-Secant Product Theorem, 793 proof of the, 793 Secant segment, 793 Secant-Tangent Product Theorem, 794 proof of the, 797 Seconds (in degrees), 27 Sector of a circle, 764 area of, 764–766 Segment(s), 7 of a circle, 765 area of a, 765 congruent to a given segment, 14 constructing, 14 of given length, 18 constructing, 18 measuring and constructing, 13–16 secant, 793 tangent, 794 that intersect circles, 746 Segment Addition Postulate, 14 Segment bisectors, 16 constructing, 16 Segment relationships in circles, 790–795 Seismograph, 804 Selected Answers, S88–S114 Self-similar figures, 882 Semicircle, 756 Semiregular tessellations, 864 Sequences, 558 Shen Kua, 493 Shipping, 395 Short Response, 26, 69, 101, 139, 161, 211, 293, 306, 345, 375, 405, 414, 447, 459, 467, 511, 513, 549, 579, 581, 636, 649, 667, 739, 798, 819, 878, 893 Write Short Responses, 208–209 Shuffleboard, 305 Shuffleboard Link, 305 Side-angle-side (SAS) congruence, 243 Side-angle-side (SAS) similarity, 471 Side lengths, triangle classification by, 217 Side-side-side (SSS) congruence, 242 Side-side-side (SSS) similarity, 470 Sides corresponding, 231 opposite, of quadrilaterals, 391 of polygons, 382 of triangles, included, 252 Sierpinski tetrahedron, 883 Sierpinski triangle, 882 Significant digits, S73 Similar, 462 Similar polygons defined, 462 ratios in, 462–464 Similar triangles angle-angle (AA), 470 applying properties of, 481–484 in the coordinate plane, 495–497 properties of, 473 in right triangles, 518–520 side-angle-side (SAS), 471 side-side-side (SSS), 470 Similarity ratio, 463, 490 Similarity transformations, 873 Simplest radical form, 346 Simplifying expressions, see Expressions Sine, 525, 841 Sines, Law of, 551–554 ambiguous case of the, 556 Sketch, 14, 17 Skew lines, 146 Skew rays, 146 Skew segments, 146 Skills Bank, S50–S81 Slant height, 818 of regular pyramids, 689 of right cones, 690 Slides, see Translations Slope(s), 182–185, 188, 322, 324, 539 formula, 182, 183, 185, 186, 199 finding, 279
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through two points, 182, 183, 185, 186, 558 of parallel lines, 184–186, 188, 306 of perpendicular lines, 184–186, 189, 306, 617 point-slope form, 303, 305 of vertical lines, 182 Slope-intercept form, 188, 190, 191, 194 proof of, 196 Social Studies, 403 Solids, 654, see also Three-dimensional figures Platonic, 669 Solving compound inequalities, 330 equations, see Equations inequalities, see Inequalities Southwestern University, 448 Space, 671 Space Exploration, 354, 491, 492, 751 Space Shuttle, 548 Index S173S173 Special parallelograms, 410 conditions for, 416–421 properties of, 408–411 Special points in triangles, 321 Special quadrilaterals, 391 Sphere, 714–717, 805, 819 center of a, 714 circumference of a great circle of a, 769 defined, 714 drawing, 653 radius of a, 714 surface area of a, 716 volume of a, 769 Spherical geometry, 726–729 Spherical Geometry Parallel Postulate, 726 Spherical Triangle Sum Theorem, 726 Spherical triangles, area of, 727 Sports, 17, 19, 40, 46, 149, 165, 175, 259, 458, 492, 530, 562, 603, 635, 720, 729, 761, 851 Spreadsheet, 541 Springboks, 271 Square, 36, 410 properties of, 410 proof of, 410 Square base in origami, 594 Square roots, S55 simplifying, S55 Square units, 36 Square window on calculator screen, 189 SSS (side-side-side) congruence, 242 applying, 242–245 exploring, 240–241 SSS (side-side-side) similarity, 470 Standard normal curve, 860 Standardized Test Prep, 69, 139, 211, 293, 375, 447, 513, 581, 649, 739, 819, 893, see also Assessment Statements biconditional, 96–98 compound, 128 conditional, 81–84 logically equivalent, 83 Statue of Liberty, 466 Stonehenge, 787 Straight angles, 21 Straightedge, 14, F47, see also Construction(s), using compass and straightedge Strategies for positioning figures in the coordinate plane, 267 Student to Student, 21, 121, 157, 233, 359, 411, 463, 535, 632, 662, 800, 865 Study Guide: Preview, see Assessment Study Guide: Review, see Assessment Study Strategies, see also Reading and Writing Math Memorize Formulas, 587 Prepare for Your Final Exam, 822 Take Effective Notes, 145 Substitution, solving systems of equations by, 316–318, 396, S67 Substitution Property of Equality, 104 Subtend, 772 Subtraction Property of Equality, 104 Subtraction Property of Inequality, 330 Summarizing, 791 Supplementary angles, 29 Supplements, 29 Surface, lateral, of a cylinder, 681 Surface area of cones, 689–692 of cubes, 680 of cylinders, 680–683 of prisms, 680–683 of pyramids, 689–692 of regular pyramids, 689 of right cones, 690 of right cylinders, 681 of right prisms, 680 of spheres, 716 of three-dimensional figures, 680 and volume, comparing, 722–723 Surveying, 25, 224, 256, 257, 263, 276, 353, 474, 547, 556 Swing bridges, 895 Syllogism, Law of, 89 Symbolic logic, 128–129 Symbols, see back cover Symmetric Property, 168, 176 Symmetric Property of Congruence, 106 Symmetric Property of Equality, 104 Symmetric Property of Similarity, 473 Symmetry, 856–858 about an axis, 858 axis of, 362, 856, S65 defined, 856 glide reflection, 863 identifying, in three dimensions, 858 line, 856 line of, 318, 856 plane, 858 rotational, see Rotational symmetry translation, 863 Systems of equations, see Equations T Tables making, 230, 763, S48, S78 using, 19, 40 Taffrail log, 278 TAKS Practice, S4–S39 TAKS Prep, 68–69, 138–139, 210–211, 292–293, 374–375, 446–447, 512–513, 580–581, 648–649, 738–739, 818–819, 892–893, see also Assessment TAKS Tackler, see also Assessment Any Question Type Check with a Different Method, 372–373 Estimate, 578–579 Highlight Main Ideas, 890–891 Identify Key Words and Context Clues, 290–291 Interpret Coordinate Graphs, 208–209 Interpret a Diagram, 510–511 Measure to Solve Problems, 736–737 Use a Formula Sheet, 646–647 Gridded Response: Record Your Answer, 136–137 Multiple Choice: Eliminate Answer Choices, 444–445 Recognize Distracters 816–817 Work Backward, 66–67 TAKS Tip, 67, 69, 137, 139, 209, 210, 291, 293, 373, 375, 445, 447, 511, 513, 579, 581, 647, 649, 737, 739, 817, 819, 891, 893 Tangency, point of, 746 Tangent, 525, 746, 805 to a circle at a point, constructing, 748 to a circle from an exterior point, constructing, 779 common, 748 Tangent circles, 747 Tangent segment, 794 Tangram, 589 Technology, 92, 809 using drawing figures in one- and two-point perspectives, 668 geometry software, 249 graphing calculator, 879 segment and angle bisectors, 27 spreadsheet, 541 Technology Lab, see also Geometry lab Compare Surface Areas and Volumes, 722–723 Explore Angle Relationships in Circles, 780–781 Explore the Golden Ratio, 460–461 Explore Isosceles Trapezoids, 426 Explore Parallel and Perpendicular Lines, 188–189 Explore Parallel Lines and Transversals, 154 Explore Properties Associated with Points, 12 Explore Segment Relationships in Circles, 790–791 Explore Transformations, 56–57 Explore Transformations with Matrices, 846–847 Explore Trigonometric Ratios, 524 Investigate Angle Bisectors of a Triangle, 480 Predict Conditions for Special Parallelograms, 416–417 Predict Other Triangle Congruence Relationships, 250–251 Predict Triangle Similarity Relationships, 468–469 Special Points in Triangles, 321 S174 S174 Index TEKS, TX28–TX35 Three-dimensional figures, 650–741 TEKS correlation charts are found in every chapter. Some examples: 4, 72, 144, 214, 298 Temperature, 105 Terminal side of reference angle, 570 Terms, undefined, 6 Tessellations, 863–866 dual of, 868 regular, 864 semiregular, 864 using transformations to create, 864 using transformations to extend, 870–871 Test Prep, see also Assessment Test Prep questions are found in every exercise set. Some examples are: 11, 19, 26, 33, 40 Test-taking strategy, see TAKS Tackler Test-taking tips, see TAKS Tip and Hot Tip Tetrahedron, 669 Texas flag, 245 Texas Link, xix Archaeology, Hunt, 787 Bicycles, Austin, 337 Bird Watching, Llano, 401 Engineering Bluff Dale, 115 San Antonio, 233 Entertainment, 683 Austin, 833 Dallas, 803 Geography, Big Bend National Park, 626 History, Katy, 48 Kites, Austin, 428 Landscaping, Austin, 607 Marine Biology, Corpus Christi, 698 Meteorology, 476 Oceanography, Brownsville, 174 Pets, Garland, 361 Racing, Fort Worth, 392 Recreation, New Braunfels, 673 Space Shuttle, Houston, 548 Sports, Austin, 530 Transportation, Dallas, 183 Texas Motor Speedway, 392 Texas Star Ferris wheel, 841 Texas State Aquarium, 698 Texas State Capitol, 742 Texas state gemstone, 752 Textiles, 125 Theater, 246 Theorems, 110, 748–749, 757–758, 774–775, 782–784, 848–850 For a complete list, see pages S82–S87 proofs of, 753, 758, 762, 779, 783, 788–789 Theoretical probability, 628 Third Angles Theorem, 226 30°-60°-90° triangle, 358 drawing, 653 representations of, 661–664 surface area of, 680 Three dimensions Distance Formula in, 672 formulas in, 670–673 identifying symmetry in, 858 Midpoint Formula in, 672 using Pythagorean Theorem in, 671 Tick marks, 13 Tiling, 863, see also Tessellations Titan, 449 Tolerance, S72 Too much information, 209 Tools of Geometry, xxi Transformations, 50, 79 compositions of, see Compositions of transformations congruence, 824, 854 in the coordinate plane, 50–52 describing, in terms of reflections, 850 exploring, with geometry software, 56–57 of functions, 838 Math Builders, xxiv–xxv with matrices, 846–847 of parent functions, 838, S63 similarity, 873 using to create tessellations, 864 to extend tessellations, 870–871 Transit (tool), 20 Transitive Property, 168, 176 Transitive Property of Congruence, 106 Transitive Property of Equality, 104 Transitive Property of Inequality, 330 Transitive Property of Similarity, 473 Translation symmetry, 863 Translations, 50, 327, 831–833 constructing, 836 in the coordinate plane, 832 of figures, constructing, 831 general, in the coordinate plane, 832 horizontal, see Horizontal translations vertical, see Vertical translations Transportation, 183, 194, 360, 386, 620, 631, 633, 866 Transversals, 147 parallel lines and, angles formed by, 154–157 Trapezoid, 51, 426, 429, 819 area of, 590 base angles of, 429 bases of, 429 isosceles, see Isosceles trapezoids legs of, 429 midsegment of, 431 properties of, 429–431 proof of, 435 Tree rings, 604 Trend lines, S79 Trefoil shape, 313 Triangle(s), 36, 98, 216, 382 acute, 216 altitudes of, 314–317 defined, 316 angle bisectors of, 480 angle relationships in, 223–226 angle-side relationships in, 333 area of, 590 bisectors of, 307–310 centroid of, 314 constructing, 314 circumcenter of, 307 constructing, 307 circumscribe a circle about, 778 classifying, 216–219, 230 congruent, see Congruent triangles developing formulas for, 589–593 equiangular, 216 equilateral, see Equilateral triangles incenter of a, 309 isosceles, see Isosceles triangles medians of, 314–317 midsegment, 322 constructing, 327 musical, 218 obtuse, 216 orthocenter of a, 316 constructing, 320 right, see Right triangle(s) scalene, 217 solving, 535 special points in, 321 spherical, see Spherical triangles two, inequalities in, 340–342 Triangle Angle Bisector Theorem, 483 Triangle classification by angle measures, 216 by side lengths, 217 Triangle congruence, 212–295 applying ASA, AAS, and HL, 252–255 applying SSS and SAS, 242–245 CPCTC, 260–262 exploring SSS and SAS, 240–241 predicting other relationships, 250–251 Triangle inequalities, exploring, 331 Triangle Inequality Theorem, 334 proof of the, 338 Triangle Midsegment Theorem, 322–324 proof of the, 326 Triangle Proportionality Theorem, 481 Converse of the, 482 Triangle rigidity, 242 Triangle similarity AA, SSS, and SAS, 470–473 predicting, relationships, 468–469 Triangle Sum Theorem, 223 Trapezoid Midsegment Theorem, 431 Travel, 17, 54, 84, 335, 458, 484 Treadmill, 539 developing the, 222 proof of the, 223 Triangulation, 223 Index S175S175 Trigonometric functions, inverse, 533, Vertex, 20 534 Trigonometric ratios, 524–528 Trigonometry indirect measurement using, 550 right triangles and, 524–583 unit circle and, 570–571 Trisecting angles, 25 Triskelion, 861 Trundle wheel, 605 Truth table, 128 Truth value, 82 T
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urns, see Rotations Two-column proofs, see Proofs, two-column Two-point perspective, 662 drawing figures in, 668 Two-Transversal Proportionality Corollary, 482 U Undefined terms, 6 Unit circle, 570 trigonometry and the, 570–571 University of Texas Longhorn Band, 833 Urban legends, 88 Use more than one method, 45 V Vanishing point, 662 Variation constant of, 501 direct, 501 Vasarely, Victor, 860 Vector(s), 559–563 addition of, 561–562 equal, 561 negation of a, 566 parallel, 561 resultant, 561 Venn diagrams, 80, S45, S81 of a cone, 690 of a polygon, 382 of a pyramid, 689 of a three-dimensional figure, 654 Vertex angles, 273 Vertical angles, 30 Vertical Angles Theorem, 120–122 proof of the, 120 Vertical line, equation of a, 190 Vertical translations in the coordinate plane, 832 of parent functions, 838 Vocabulary, 9, 17, 24, 31, 38, 47, 53, 77, 84, 91, 99, 107, 113, 122, 148, 185, 194, 219, 227, 234, 245, 256, 262, 270, 276, 304, 311, 317, 324, 336, 352, 386, 395, 412, 432, 457, 465, 491, 498, 521, 529, 547, 563, 603, 609, 633, 657, 665, 674, 684, 693, 701, 709, 718, 751, 760, 767, 776, 795, 827, 851, 859, 866, 875 math, learning, 299 Vocabulary Connections, 4, 72, 144, 214, 298, 378, 452, 516, 586, 652, 744, 822 Volume, 697, 789 of cones, 705–708 of cylinders, 697–700 of prisms, 697–700 of pyramids, 705–708 of spheres, 769 surface area and, comparing, 722–723 Vortex train, 867 W Washington, George, 18 What if...?, 26, 30, 108, 165, 183, 193, 230, 253, 275, 323, 349, 357, 359, 385, 424, 428, 456, 473, 495, 539, 545, 554, 556, 562, 673, 692, 698, 706, 793, 801, 825, 833, 849, 873 Whole numbers, 80 Working backward, 889, S43 Write About It Write About It questions appear in every exercise set. Some examples: 10, 18, 26, 33, 40 Write and solve an equation linear, 31–34, 38–41 literal, 41 Write Short Responses, 208–209 Writing equations, see Equations, writing Writing Math, 81, 96, 111, 463, 525, 756 Writing Strategies, see also Reading and Writing Math Draw Three-Dimensional Figures, 653 Write a Convincing Argument, 379 X x-axis, 42, S56 reflections across the, 826 x-coordinate, finding, 841 x-intercept, see Intercepts Y y-axis, 42, S56 reflections across the, 826 y-coordinate, finding, 841 y-intercept, see Intercepts Z ZDecimal, 189 ZSquare, 189 ZStandard, 189 Zilker Kite Festival, 428 Zulu people, 834 S176 S176 Index Credits Credits Abbreviations used: (t) top, (c) center, (b) bottom, (l) left, (r) right, (bkgd) background rights reserved.; 193 (tr), Alamy Images; 195 Sipa Photos/Newscom.com; 196 (tl), Comstock/Fotosearch; 200 (tl), Comstock/Fotosearch; 200 (tr), Pierre Vivant Staff Credit Bruce Albrecht, Angela Beckmann, Nancy Behrens, Lorraine Cooper, Marc Cooper, Lana Cox, Jennifer Craycraft, Martize Cross, Nina Degollado, Lydia Doty, Sam Dudgeon, Kelli R. Flanagan, Mary Fraser, Stephanie Friedman, Jeff Galvez, Pam Garner, Diannia Green, Tracie Harris, Tessa Henry, Liz Huckestein, Jevara Jackson, Kadonna Knape, Cathy Kuhles, Jill M. Lawson, Peter Leighton, Christine MacInnis, Jonathan Martindill, Susan Mussey, Kim Nguyen, Matthew Osment, Sara Phelps, Manda Reid, Patrick Ricci, Michael Rinella, Michelle Rumpf-Dike, Beth Sample, Annette Saunders, Kay Selke, Robyn Setzen, Patricia Sinnott, Victoria Smith, Jeannie Taylor, Ken Whiteside, Sherri Whitmarsh, Aimee F. Wiley, David W. Wynn Photo All images HRW Photo unless otherwise noted. Master Icons—teens, authors (all), Sam Dudgeon/HRW. Front Matter: vi (l), ©Joseph Sohm; ChromoSohm Inc./CORBIS; vii (r), Courtesy of 7A Ranch; viii (l), Laurence Parent; ix (r), ©Richard Cummins/CORBIS; x (l), Courtesy of Texas Highways Magazine; xi (r), ©Donne Bryant/Art Resource; xii (l) Laurence Parent; xiii (r), Painet, Inc.; xiv (l), Jim Wark; xv (r), Doug Hopfer/HRW; xvi (l), Victoria Smith/HRW; xvii (r), ©Royalty-Free/Corbis; xviii, (tc), Corbis images; (cr), ©Royalty Free/CORBIS; (bc), HRW Photo; xix (tc), Mark Sykes/Alamy Images; xix (cl), Photo by Walt Disney Studios/ZUMA Press ©1998 by Courtesy of Walt Disney Studios; xix (c), ©Owaki-Kulla/CORBIS; xix (cr), The Image Bank/Getty Images; xx Victoria Smith/HRW Photo; xxi (cl) Sam Dudgeon/HRW; TX2 (stamp), on-page credit; (border), Photodisc; (icon), Photodisc Front Cover: johnleepbs/Dreamstime.com Chapter 1: 2–3 Joseph Sohm/ChromoSohm Inc./CORBIS; 6 (tr, br), Sam Dudgeon/ HRW; 10 (tl), Photodisc Royalty Free; 10 (b)(inset), Reunion des Musee Nationaux/Art Resource, NY; 13 (tr), Tony Freeman/Photo Edit; 16 (tl) (tc) (tr), Andy Christiansen/ HRW Photo; 18 (tl), Photodisc Royalty Free; 19 (c), Gabriel Bouys/AFP/Getty Images; 20 (tr), Gary Conner/Photo Edit; 21 (tl), HRW Photo; 26 (tl), Photodisc Royalty Free; 28 (t), Jon Feingersh/CORBIS; 32 (bl), Photodisc Royalty Free; 34 (tl), Photodisc Royalty Free; 34 (br), Gibson Stock Photography; 38 (c-triangle), Victoria Smith/HRW Photo; 39 (bl), HRW Photo; 40 (l), Alamy Images; 41 (l), The Bridgeman Art Library/ Getty Images; 43 (tl), HRW Photo; 43 (tr), Russell Andorka; 46 (tl), Duomo/CORBIS; 48 (cl), Sam Dudgeon/HRW; 48 (bl), HRW Photo by Sam Dudgeon; 54 (tl), BrandX/ Fotosearch.com; 54 (bl), HRW Photo; 58 (tl), HRW Photo by Sam Dudgeon; 58 (c), Marty Granger/HRW Photo Chapter 2: 70–71 Courtesy of 7A Ranch; 74 (tr), Stephen Frink/CORBIS; 76 NASA Images; 78 (tl), Index Stock/Picturequest; 78 (bl), Victoria Smith/HRW Photo; 78 (br), The Granger Collection; 81 (tr), BARRY RUNK/STAN/Grant Heilman Photo; 83 (cl), Digital Vision/Getty; 83 (cr), Picturequest Images; 85 (bl), Victoria Smith/HRW Photo; 85 (br), Corbis; 86 Mark Sykes/Alamy Images; 87 (bl), Michael Newman/PhotoEdit; 88 (tr), Alamy Images; 88 (br), Taxi/Getty Images; 92 (cl), Daniel Miller/AP Wide World; 92 (bl), Victoria Smith/HRW Photo; 95 (br), Sam Dudgeon/HRW Photo; 96 Darrell Gulin/CORBIS; 99 Getty/Stone; 100 (tl), Nibsc/Photo Researchers, Inc.; 100 (bl) Victoria Smith/HRW; 100 (br), The Granger Collection; 102 (tl), HRW/Victoria Smith; 102 (c-Alice and Chesire Cat), Corbis Images; 102 (br), (tr), The Pierpont Morgan Library/Art Resource, NY; 104 Getty Images; 109 (tl), Photo-objects/Fotosearch; 109 (cr), Paul A. Souders/CORBIS; 110 (tr), REAL LIFE ADVENTURES ©2004 GarLanco. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved.; 115 (cl, cr), Courtesy of the Texas Department of Transportation; 115 (bl), Photoobjects/Fotosearch; 118 (tr), HRW Photo; 120 Alamy Images; 121 (tl), Danilo Donadoni/Bruce Coleman Inc.; 124 (bl), Photo-objects/Fotosearch; 125 (all) Victoria Smith/HRW; 126 (tl), Photo-objects/Fotosearch; 126 (tc), David Buffington/Photodisc/ Getty; 140 (cr), Gordon McGregor; 141(bl, tr), Laurence Parent; 141 (br), ©John Elk III Chapter 3: 142–143 Laurence Parent; 146 (tr), Cameron Cambell Integrated Studio; 149 (bl), Mark McKenna Photography & Design; 150 (tl), Rick Davis/Darkhouse and Fun Enthusiast; 155 (tr), Alamy Photos; 157 (tl), Doug Menuez/Photodisc/Getty; 159 (cl), B.S.P.I./Corbis Images; 162 (tr), Ken Hawkins/Mira.com; 166 Alamy Images; 167 (cr), The Art Archive/Victoria and Albert Museum London/Sally Chappell; 168 (tl), Rick Davis/Darkhouse and Fun Enthusiast; 171 (tl)(tr)(cl)(tr)(bc), Andy Christiansen/HRW Photo; 172 (tr), Alamy Images; 174 (tl), CORBIS Images; 175 Michael Melford/The Image Bank/Getty Images; 176 (tr), Klaus Lahnstein/Stone/Getty Images; 176 (bl), Rick Davis/Darkhouse and Fun Enthusiast; 177 Terraserver.com; 180 (tl) (br), Rick Davis/Darkhouse and Fun Enthusiast; 182 (tr), Craig Cameron Olsen/Stone/Getty; 183 (bl), Courtesy of Texas Highways Magazine; 186 (cl), David Muench/CORBIS Images; 186 (bl),Comstock/Fotosearch; 190 (tr), CLOSE TO HOME ©1996 John McPherson. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All Chapter 4: 212 (bkgd), Richard Cummins/CORBIS; 212 (inset-teen), HRW Photo; 216 (t), Philip Gould/CORBIS; 218 (t), Digital Image ©2007 PhotoDisc; 219 (cl), Sam Dudgeon/HRW; 220 (cl), Alamy Photo; 220 (bl), Sam Dudgeon/HRW; 222 (c, b, t), Andy Christiansen/HRW; 223 (tr), ©Library of Congress/CORBIS; 227 (tr), Eckhard Slawik/Photo Researchers; 229 (bl), Sam Dudgeon/HRW; 231 (br, tr), NASA; 233 (tl), FotoSearch.com; 233 (cl), Courtesy of Six Flags Fiesta Texas; 233 (cr), John Foxx/ ImageState; 234 (br), Courtesy of Texas Highways Magazine; 236 (tl), Sam Dudgeon/ HRW; 237 (bl), FotoSearch/COMSTOCK, Inc.; 238 (tl), Sam Dudgeon/HRW; 238 (all), Andy Christiansen/HRW; 240 (t, b), Sam Dudgeon/HRW; 241 (t, b), Sam Dudgeon/ HRW; 242 (tr), The Image Bank/Getty Images; 247 (bl), Fotosearch; 248 (tl), Jack Fields/Photo Researchers, Inc.; 252 (tr), ©Steve Skjold/Alamy Photos; 252 (cr), Stockbyte Royalty-Free Images/HRW Library; 257 (tl), Mary Evans Picture Library; 258 (tl), Fotosearch/Photodisc; 259 (cr), Victoria Smith/HRW; 260 (tr) Chris Lisle/ CORBIS; 262 (cr), Kevin Taylor/Alamy Photos; 264 (tl) Fotosearch/Photodisc; 267 (tr), Eric Vandeville, Rome; 271 (cl), Photodisc/Getty; 271 (bl), Fotosearch/Photodisc; 273 (tr), Jason T. Ware/Photo Researchers, Inc.; 273 Jason T. Ware/Photo Researchers, Inc.; 278 (tl), Fotosearch/Photodisc; 278 (bl), Sam Dudgeon/HRW;280 (tl), Fotosearch/ Photodisc; 280 (cr), ©Chuck Smith/Southern Stock/PictureQuest; 294 (cr), ©Cavanaugh Flight Museum 2005; 294 (tr, br), ©Jim Wilson Photography; 295 (tr), ©Longview Chamber of Commerce Chapter 5: 296–297 Courtesy of Texas Highways Magazine; 299 Sam Dudgeon/ HRW Photo; 300 (tr), The Image Bank/Getty Images; 302 (br), ©Gunter Marx Photography/CORBIS; 305 (bl), Creatas/Punchstock.com; 305 (cr), Scott McDermott/ IPN; 305 (cl), Lake Country Museum/CORBIS; 307 (tr), Firefly Productions/CORBIS; 307 (cl, c, cr), Sam Dudgeon/HRW Photo; 312 (bl), Creatas/Punchstock.com; 313 (cl), Corbis Images; 314 (tr), Calder, Alexander (1898–1976) ©ARS, NY Ordinary, 1969, 580 x 600 x 580 cm. ©Copyright ARS, NY. Painted Steel. Private Collection Photo Credit: Art Resource, NYART127373; 318 (cl), Corbis Images; 319 (tl), Creatas/ Punchstock.com; 320 (cl) (c) (cr), Sam Dudgeon/HRW Photo; 320 (bl), Corbis Images; 322
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(tr), Getty Images; 324 (cr), Imagebroker/Alamy Images; 326 (tl), Creatas/ Punchstock.com; 328 (tl), Creatas/Punchstock.com; 328 (c), Photodisc Red/RF/Getty Images; 331 (tr, br), Sam Dudgeon/HRW Photo; 332 (tr), Real Life Adventures by Gary Wise and Lance Aldrich; 337 (tl), Stefano Rellandini/Reuters/Corbis; 338 (tl), Photodisc Red/RF/Getty Images; 340 (tr), Stone/Getty; 343 (cl, cr),Victoria Smith/HRW Photo; 344 (tl)(tr), Alamy Images; 344 (cr), Victoria Smith/HRW Photo; 344 (bl), Transtock Inc./Alamy; 347 (cr), Sam Dudgeon/HRW Photo; 348 (tr), Danny Lehman/ CORBIS; 353 (tr), Peter Van Steen/HRW Photo; 353 (cl), Erich Lessing/Art Resource; 354 (bl), Transtock Inc./Alamy Images; 356 (tr), Taxi/Getty Images; 357 (cr), HRW Photo; 359 (tl), Stockbyte/Getty Images; 359 (cr), Sam Dudgeon/HRW Photo; 360 (cr), Corbis Images/Punchstock.com; 360 (br), Sam Dudgeon/HRW Photo; 361 (tl), PhotoStockFile/Alamy; 361 (cr), HRW Photo; 361 (bl), Transtock Inc./Alamy Images; 364 (tl), Transtock Inc./Alamy Images; 364 (b), Paul Doyle/Alamy Chapter 6: 376–377 Donne Bryant/Art Resource; 382 (tr), AP Photo/Pat Sullivan; 385 (cr), Custom Medical Stock; 386 (tl), Ingram Image/Picturequest; 386 (tcl), Punchstock; 386 (tr), Comstock/Punchstock; 386 (tcr), Alamy Images; 386 (cr), Alamy Images; 387 (tr), Peter Van Steen/HRW Photo; 387 (bl), Sam Dudgeon/HRW Photo; 387 (br), HRW Photo by Peter Van Steen; 390 (t, c, b), HRW Photo; 391 (tr), George D. Lepp/CORBIS; 392 (cl), Michael Kim/Corbis; 392 (br), Art Reference: BasketballHoops Unlimited; 395 (tr), Robert Harding World Imagery/Getty Images; 395 (cr), Photo Edit Inc.; 396 (tr), Photo Courtesy Mohawk Lifts; 396 (bl), HRW Photo; 396 (br), HRW Photo by Peter Van Steen; 398 (tr), Sam Dudgeon/HRW; 401 (cl), Cottonwood Photography; 401 (cr), Sam Dudgeon/HRW; 402 (cr-parallel rule) Victoria Smith/HRW; 402 (cr-map), Alamy Images; 403 (tr)(inset), Sam Dudgeon/HRW; 404 (cl), Historical Picture Archive/CORBIS; 404 (cr), Robervals’s balance/Varnished wood, brass French, Late 18th century/700 mm x 360 mm; diam. 160 mm Inventory number 1983.26.36, Stewart Museum, Montreal, Canada; 404 (bl), Sam Dudgeon/HRW Photo; 404 (br), HRW Photo by Peter Van Steen; 405 (cr), Stinar Corporation; 406 (bl), Guinea REUTERS/Corbis Images; 406 (tl), HRW Photo; 407 (tr-Flourite) (cl-Rhodochrosite) (inset), Chip Taylor/NMNH/Smithsonian Institution; 407 (cr) Amazonite, HRW Photo; 407 (cr), HRW Photo; 408 (tr), Courtesy of Wimberley Stain Glass/HRW Photo by Peter Van Steen; 408 (stained glass, br), Jill Stephenson/Alamy; 411 (tl), Gareth Brown/CORBIS; 412 (tr), CORBIS; 412 (cr), Tony Freemman/Photo Edit Inc.; 413 (cl), Roger Ressmeyer/CORBIS; 413 (cr), Paul S. Calter; 414 (tl), BrandX/Fotosearch; 418 (tr), Digital Vision/Picturequest; 422 (cr), Peter Van Steen/HRW Photo; 424 (tl), BrandX/fotosearch; 425 (cr), HRW Photo; 427 (tr), CALVIN AND HOBBES ©1995 Watterson. Dist. By UNIVERSAL PRESS SYNDICATE. Reprinted with permission. All rights reserved.; 427 (tr), Spencer Grant/Photo Edit Inc.; 428 (cl), Sam Dudgeon/HRW; 428 (tr), Victoria Smith/HRW Photo; 432 (r), HRW Photo by Sam Dudgeon; 434 (tl), BrandX/Fotosearch; 434 (cl), The Granger Collection; 434 (cr), Photo Edit Inc; 436 (tl), Punchstock; 436 (b), Punchstock; 437 (br), Royalty-Free/Corbis/Fotosearch; 442 (br), Royalty-Free/Fotosearch; 448 (all), Sam Dudgeon/HRW; 449 (br, tc), Courtesy of Six Flags Over Texas Credits S177 S177 Chapter 7: 450–451 Laurence Parent; 453 (cr), Custom Medical Stock; 454 (tr), Photofest; 456 (cr), Photofest; 456 (br), Everett Collection OR Everett/CSU Archives; 457 (br), Joseph Sohm; ChromoSohm Inc./CORBIS; 458 (cl), ©John Elk III; 458 (cr), ©Zaw Min Yu/Lonely Planet Images; 458 (bl), ©Hemera Technologies/Alamy Photos; 461 (cr), George Post/Photo Researchers, Inc.; 461 (br), Wei Yan/Masterfile; 461 (cl), Age Fotostock/Morales; 461 (bl), Corbis Images; 462 (tr), Jens Meyer/AP/Wide World Photos; 463 (bl), ©Dennis Boissavy/Getty Images; 464 (t), ©Nathan Keay/HRW; 465 (cr), ©Cameron Cross; 466 (tr), ©Nathan Keay/HRW Photo; 466 (cl), OwakiKulla/CORBIS; 466 (bl), ©Hemera Technologies/Alamy Photos; 470 (tr), RoyaltyFree/CORBIS; 472 (b), PhotoDisc/Getty Images; 476 (cl), AFP PHOTO/NOAA/NewsCom; 476 (tl), ©Hemera Technologies/Alamy Photos; 478 (tl), ©Hemera Technologies/ Alamy Photos; 478 (cr), ©David Young-Wolff/PhotoEdit; 481 (tr), ©Christie’s Images/ CORBIS; 486 (tl), Royalty-Free/Comstock; 488 (tr), roadsideamerica.com, Kirby, Smith & Wilkins; 489 (br), ©Peter Gridley/Getty Images; 491 (br), Courtesy NASA/JPL/ASU; 492 (cl), ©Reuters/CORBIS; 492 (bl), Royalty-Free/Comstock; 493 (tl), ©Sygma/ CORBIS; 493 (tr), ©Tom Bean/CORBIS; 494 (tr), ©Jack Van Dusen; 494 (bl), ©Mark Richards/PhotoEdit; 495 (tr), David Neuse Photography; 495 (cr), ©Photodisc/Getty Images; 495 (br), Darrell Gulin/CORBIS; 499 (bl), Royalty-Free/Comstock; 502 (tr), Comstock/HRW Photo Research Library; 502 (cr), ©Jeff Cadge/Getty Images Chapter 8: 514–515 (bkgd), Painet Inc.; 518 (tr), Twisted Texas/Wesley Treat; 522 (bl), Fotosearch; 525 (tr), Alamy Images; 529 (br), Alamy Images; 530 (tl), Phil Gustafson; 530 (cr), Kevin Fleming/Corbis; 530 (bl), Fotosearch; 531 (tl), Alamy Images; 534 (tr), Photo Edit Inc.; 535 (bl), PhotoEdit; 536 (c), Getty Images; 538 (tr), Getty Images Sport/Bobby Julich; 539 (tl), Fotosearch; 539 (cl), Photo Edit Inc.; 542 (tl), Fotosearch; 542 (b), Superstock; 544 (tr), Stone/Getty Images; 548 (tl), Scott Berner/Index Stock Imagery, Inc.; 548 (bl), Fotosearch; 550 (tr), HRW Photo; 550 (cr), HRW Photo; 551 (tr), Alamy Images; 556 (tl), Brad Smith/News & Observer/AP/Wide World Photos; 557 (tl), Fotosearch; 559 (tr), Stone/Getty; 565 (tl), Fotosearch; 566 (t), FOXTROT ©1999 Bill Amend. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved.; 566 (cl), Photodisc/Picturequest; 568 (tr), Fotosearch; 568 (b), Stone/Getty Images; 582 (b), Chad Ehlers/PictureQuest; 582 (tr), Jim Olive/ Stockyard.com; 583 (tc), Yoichi R. Okamoto/LBJ Library Photo; 583 (bl), Steve Warble/ Mountain Magic Photography; 583 (cr), National Park Service Chapter 9: 584 Jim Wark; 589 (tr), Victoria Smith/HRW Photo; 594 (c), HRW Photo by Sam Dudgeon; 594 (cr), HRW Photo by Sam Dudgeon; 595 (cl), The Granger Collection, New York; 595 (tl), Photodisc/Getty Images; 596 (cr), HRW Photo; 598 (all) Sam Dudgeon/HRW Photo; 600 (tr), ©gkphotography/Alamy Photos; 604 (cl), ©Royalty-Free/CORBIS; 604 (bl), ©Photodisc/Getty Images; 606 (tr), ©Rose/Zefa/ Masterfile; 607 (bl), Courtesy of Texas Highways Magazine; 610 (bl), ©Photodisc/ Getty Images; 612 (bl), ©Royalty Free/CORBIS; 614 (tr), Thinkstock/PictureQuest; 614 (c), ©Otto Rogge/CORBIS; 620 (bl), ©Royalty-Free/CORBIS; 626 (cl), ©Patrick Ray Dunn/Alamy Photos; 626 (bl), ©Royalty-Free/CORBIS; 628 (c), Peter Van Steen/ HRW; 629 (t), Peter Van Steen/HRW; 630 (tr), AP Photo/Reed Saxon; 628 (cr), Peter Van Steen/HRW; 632 (tl), Warren Morgan/CORBIS; 635 (tl), Romeo Gacad/AFP/Getty Images; 635 (bl), Royalty-Free/CORBIS; 637 (cr), HRW Photo by Sam Dudgeon; 638 (tl), Royalty-Free/CORBIS; 638 (b), Photofusion Picture Library/Alamy Photos; 638 (cr), Dennis MacDonald/PhotoEdit Chapter 10: 650–651 Doug Hopfer/HRW; 654 (tr), AFP PHOTO/JIJI PRESS/Newscom; 655 (tl), Fotosearch; 655 (tr), David Young-Wolff/Photo Edit; 655 (cl), HRW Photo; 655 (cr), HRW Photo; 656 (cr), Newscom; 657 (tl) Bonillo/Photo Edit; 657 (tc), Photodisc/RF/Fotosearch; 657 (tr), HRW Photo; 658 (bl), David Young-Wolff/Photo Edit; 661 (cr, br), HRW Photo; 662 (tr, cr, bc), HRW Photo; 662 (bl), Corbis Images; 664 (tr, cr), Victoria Smith/HRW Photo; 665 (tl) Victoria Smith/HRW Photo; 665 (all cubes), Victoria Smith/HRW Photo; 666 (tl, tc, tr, cr), Victoria Smith/HRW Photo; 666 (bl), David Young-Wolff/Photo Edit; 667 (tr)–poster, Duomo/CORBIS; 670 (tr), Jeff Hunter/The Image Bank/Getty Images; 673 (cl), Peter Essick/Getty Images; 675 (bl), David Young-Wolff/Photo Edit; 675 (tl), Stone/Getty Images; 678 (tl), David Young Wolf/Photoedit; 678 (b), Courtesy of Texas Highways Magazine; 679 (cr), Victoria Smith/HRW Photo; 680 (tr), Robert Harding World Imagery/Getty Images; 683 (cl), Creative Ice Carvings; 686 (bl) HRW Photo; 687 (bl)(bc)(br), HRW Photo; 688 (all), HRW Photo; 692 (cl), Marc Golub/HRW Photo; 695 (tl), Victoria Smith/HRW; 695 (cl), Dennis MacDonald/Photo Edit; 697 (tr), Jeff Greenberg/Photoedit Inc.; 697 (tc), HRW Photo; 698 (cl), ©John Elk III; 699 (tc)(tr), HRW Photo; 701 (cr), AFP/TIMOTHY A. CLARY/Getty Images; 703 (tl), Victoria Smith/HRW Photo; 703 (cl), AKG Images; 703 (cr), HRW Photo; 705 (tr), ©Mark Gibson Photography; 706 (cr), ©Lyndol Descant/LyndolDotCom; 711 (bl), HRW Photo; 718 (cl)(cr), Victoria Smith/HRW Photo; 719 (tr), ©Susan Van Etten/Photo Edit; 720 (Chart-Golf Ball), Martin Paul Ltd., Inc./ Picturequest; 720 (Chart-Cricket ball), Photodisc/RF/Fotosearch; 720 (Chart-Tennis ball), Stockdisc/RF/Getty Images; 720 (Chart-Petanque ball), Dk Images/RF/Getty Images; 720 (cl), Ralph White/CORBIS; 720 (bl), HRW Photo; 721 (cr), Lyndol Descant/ LyndolDotCom; 724 (tl), HRW Photo; 724 (b), Picturequest; 726 (bl), HRW Photo; 729 (tcl) (tcr), ©Dorling Kindersley/Getty Images; 729 (cr), Royalty-Free/Alamy Images; 729 (br), Stockbyte/RF/Picturequest; 731 (tl, tr, c), Victoria Smith/HRW Photo; 734 (tc), HRW Photo; 740 (tr), ©Space Imaging; 740 (bl), Alberto Tamargo/Getty Images; 740 (br), ©Cut and Deal Ltd/Alamy; 741 (tc), United States Mint image; 741 (cr), Lyle Engleson/Goldberg Coins; 741 (cl, bc), Jerry Adams Chapter 11: 742–743 Victoria Smith/HRW; 746 (tr), ©NASA/Roger Ressmeyer/ CORBIS; 749 (tr), ©Alan Kearney/Getty Images; 749 (br), Gamma; 752 (cl), ©CORBIS; 752 (bc), Courtesy of Texas Highways Magazine; 753 (bl), Photolibrary.com.pty. ltd./Index Stock Imagery, Inc.; 756 (tr), ©Brand X Pictures/PunchStock; 762 (bl), ©photolibrary.com.pty.ltd./Index Stock Imagery, Inc.; 762 (c), Victoria Smith/HRW Photo; 764 (tr), AP/Wide World Photos; 764 (tr), Jim Wark/Airphoto; 767
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(br), ©Christer Fredriksson/Lonely Planet Images; 768 (tl) ©Photolibrary.com.pty.ltd./Index Stock Imagery, Inc.; 768 (b), ©Tony Freeman/PhotoEdit; 768 (cl), Scala/Art Resource, NY; 768 (bl), ©Photolibrary.com.pty.ltd./Index Stock Imagery, Inc.; 768 (br), Photo by Eisenmann, N.Y./Library of Congress; 772 (tr), Victoria Smith/HRW Photo; 773 (br), Victoria Smith/HRW Photo; 776 (tc), Victoria Smith/HRW Photo; 777 (bl), Victoria Smith/HRW Photo; 777 (cr), H. Armstrong Roberts/RobertStock.com; 778 (tr), ©Archivo Iconografico, S.A./CORBIS; 787 (b) Linda Owen; 788 (bl) Victoria Smith/ HRW Photo; 792 (tr), ©Joathan Blair/CORBIS; 795 (br), ©Chris Lisle/CORBIS; 796 (cr), ©Michael T. Sedam/CORBIS; 797 (tl), NASA Marshall Space Flight Center (NASAMSFC); 797 (bl), Victoria Smith/HRW Photo; 799 (tr), Cartoon copyrighted by Mark Parisi, printed with permission.; 800 (bl), ©RuberBall/Alamy Photos; 802 (br) ©Alamy Photos; 803 (tl), www.lonestarthrills.com; 803 (bl), Victoria Smith/HRW Photo; 804 (tl), The Granger Collection, New York; 805 (bl), Sam Dudgeon/HRW; 806 (tr), Victoria Smith/HRW Photo Chapter 12: 820–821 ©Royalty-Free/Corbis; 823 (tl), Sam Dudgeon/HRW; 823 (tcl), Sam Dudgeon/HRW; 823 (bcl), Sam Dudgeon/HRW; 823 (bl), Sam Dudgeon/HRW; 824 (tr), Scott Teven/photohouston; 824 (cr), Sam Dudgeon/HRW; 824 (c), Sam Dudgeon/HRW; 824 (cl), Sam Dudgeon/HRW; 824 (br), Sam Dudgeon/HRW; 828 (cr), Musee d’Orsay, Paris, France/Erich Lessing/Art Resource, NY; 829 (tl), ©Brian Hagiwara/Brand X Pictures/Getty Images; 831 (tr), ©Steve Boyle/NewSport/CORBIS; 831 (cl), Sam Dudgeon/HRW; 831 (c), Sam Dudgeon/HRW; 831 (cr), Sam Dudgeon/ HRW; 831 (br), Sam Dudgeon/HRW; 833 (cl), ©Kelly-Mooney Photography/CORBIS; 834 (cr), Bonhams, London, UK/The Bridgeman Art Library; 835 (tl), Photo by Walt Disney Studios/ZUMA Press ©Copyright 1998 by Courtesy of Walt Disney Studios; 835 (bl), ©Brian Hagiwara/Brand X Pictures/Getty Images; 836 (tr), Victoria Smith/ HRW; 839 (tr), ©Roger Ressmeyer/CORBIS; 839 (cl, c, cr, br), Sam Dudgeon/HRW; 841 (cl), ©L. Clarke/CORBIS; 843 (bl), ©Brian Hagiwara/Brand X Pictures/Getty Images; 844 (tr), ©Roger Ressmeyer/CORBIS; 848 (tr), Scott Teven/photohouston; 853 (tl), ©Brian Hagiwara/Brand X Pictures/Getty Images; 854 (tl), ©Brian Hagiwara/ Brand X Pictures/Getty Images; 854 (br), ©Brand X Pictures/Getty Images; 856 (tr), Jan Hinsch/Photo Researchers, Inc.; 856 (br), ©One Mile Up, Inc; 857 (c-purple diatoms), Alfred Pasieka/Photo Researchers, Inc.; 857 (bc), Eric Grave/Photo Researchers, Inc.; 857 (br), John Burbidge/Photo Researchers, Inc.; 859 (cr), spaceimaging.com/Getty Images; 859 (br), 859 (br), ©Brand X Pictures/Alamy Photos; 859 (tr), M. C. Escher’s “Circle Limit III” ©2005 The M.C. Escher CompanyHolland. All rights reserved. www.mcescher.com; 860 (tr), (c) ARS, NY/Art Resource, NY; 861 (tl), M. C. Escher’s Wooden Ball with Fish ©2005 The M.C. Escher CompanyHolland. All rights reserved. www.mcescher.com; 861 (cr), Comstock Images/ PictureQuest; 861 (cr), ©bildagentur-online.com/de/Alamy Photos; 863 (tr), ©Russell Gordon/DanitaDelimont.com; 863 (tcl), ©Anna Zuckerman-Vdovenko/PhotoEdit; 863 (tcr), ©Danita Delimont/Alamy Photos; 863 (cl) ©Paul Souders/WorldFoto/ Alamy Photos; 863 (cr), ©Jeff Greenberg/Alamy Photos; 863 (cbl)(cbr), ©Danita Delimont/Alamy Photos; 865 (tl), ©Comstock Images/Alamy Photos; 865 (cl), Darren Matthews/Photographer’s Direct; 865 (c), ©G. Schuster/Photo-AG/CORBIS; 865 (cr), Brand X Pictures/PictureQuest; 866 (bc), ©M. Angelo/CORBIS; 866 (br), ©Paul Almasy/CORBIS; 866 (bl), Wolfgang Kaehler Photography; 867 (all), Sam Dudgeon/HRW; 868 (tl), M. C. Escher’s Wooden Ball with Fish ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher.com; 868 (tr); M. C. Escher’s “Symmetry Drawing E103” ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher.com; 868 (tcl), M. C. Escher’s “Verbum” ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher.com; 868 (tcr), M. C. Escher’s “Symmetry Design E38” ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher.com; 868 (cl, br), Sam Dudgeon/HRW; 870 (tr, cr, br), Sam Dudgeon/HRW; 871 (all), Sam Dudgeon/HRW; 872 (tr), Mark Lennihan/AP/Wide World; 876 (tl), Texas State Library & Archives Commission; 876 (bl), M. C. Escher’s Wooden Ball with Fish ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher.com; 876 (br), M. C. Escher’s “Drawing Hands” ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher.com; 876 (tl), AP Photo/ The Truth, Jennifer Shephard; 877 (cl)(cr), Adam Hart-Davis/Photo Researchers, Inc.; 880 (tcl), M. C. Escher’s “Symmetry Drawing E93” ©2005 The M.C. Escher CompanyHolland. All rights reserved. www.mcescher.com; 880 (c), ©Alamy Photos; 881 (tl), M. C. Escher’s Wooden Ball with Fish ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher.com; 880 (tcr), M. C. Escher’s “Symmetry Drawing E91” ©2005 The M.C. Escher Company-Holland. All rights reserved. www.mcescher. com; 880 (bcl), M. C. Escher’s “Path of Life III” ©2005 The M.C. Escher CompanyHolland. All rights reserved. www.mcescher.com; 880 (bcr), M. C. Escher’s “Symmetry Drawing E69” ©2005 The M.C. Escher Company-Holland. All rights reserved. www. mcescher.com; 880 (b), M. C. Escher’s “Reptiles” ©2005 The M.C. Escher CompanyHolland. All rights reserved. www.mcescher.com; 882 (tr), GEORGE POST/Photo Researchers, Inc.; 883 (cr), ©George W. Hart; 894 (cr), ©Buddy Mays/CORBIS; 894 (bl), Courtesy of the Smithsonian Institution, NMAH/Transportation; 895 (bc), Stephanie Friedman/HRW; 895 (cr), Courtesy of Texas Department of Transportation Back Matter: S2 Don Couch/HRW; S3 John Langford/HRW S178 S178 Credits a207se_toc_vii-xix.sw.indd viii a207se_toc_vii-xix.sw.indd viii 11/29/05 4:47:45 PM 11/29/05 4:47:45 PM
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two outputs, 1 and and 9, also have more than one output, 3 26 1. is not a function Two other inputs, 4 1, 1 2 1 , , 0, 0 1, 1 , 1 51 Although 1 appears as an output twice, each input has one and only one output. is a function. 4, 2 9, 3 26 , , , , 2 9, 3 , 0, 0 4, 2 51 because each input corresponds to one and only one output. 9, 3 1, 1 9, 3 1, 1 4, 2 26 , , , , , is a function ■ Calculator Exploration Make a scatter plot of each set of ordered pairs in Example 4. Examine each scatter plot to determine if there is a graphical test that can be used to determine if each input produces one and only one output, that is, if the set represents a function. Example 5 Finding Function Values from a Graph The graph in Figure 1.1-8 defines a function whose rule is: For input x, the output is the unique number y such that (x, y) is on the graph. y 4 2 −2 −4 −2 x 2 4 Figure 1.1-8 a. Find the output for the input 4. b. Find the inputs whose output is 0. x 2. c. Find the y-value that corresponds to d. State the domain and range of the function. Solution a. From the graph, if x 4, then y 3. Therefore, 3 is the output corresponding to the input 4. Section 1.1 Real Numbers, Relations, and Functions b. When y 0, x 3 or x 0 or x 2. Therefore, 3, 0, and 2 are the inputs corresponding to the output 0. x 2 c. The y-value that corresponds to d. The domain is all real numbers between is 4 y 3. range is all real numbers between and 3, inclusive. 2 and 5, inclusive. The 9 ■ CAUTION The parentheses in d(t) do not denote multiplication. The entire symbol d(t) is part of the shorthand language that is convenient for representing a function, its input and its output; it is not the same as algebraic notation. Function Notation Because functions are used throughout mathematics, function notation is a convenient shorthand developed to make their use and analysis easier. Function notation is easily adapted to mathematical settings, in which the particulars of a relationship are not mentioned. Suppose a function is given. Let f denote a given function and let a denote a number in the domain of f. Then f(a) denotes the output of the function f produced by input a. For example, f(6) denotes the output of the function f that corresponds to the input 6. y is the output produced by input x according to the rule of the function f NOTE The choice of letters that represent the function and input may vary. is abbreviated which is read “y equals f of x.’’ y f x , 2 1 In actual practice, functions are seldom presented in the style of domainrule-range, as they have been here. Usually, a phrase, such as “the function of directions, as shown in the following diagram. 2x2 1, x f 2 1 ” will be given. It should be understood as a set Name of function Input number > > 2x2 1 f x ⎧⎨⎩ ⎧⎪⎨⎪⎩ 2 > 1 > Output number Directions that tell you what to do with input x in order to produce the corresponding output f(x), namely, “square it, add 1, and take the square root of the result.” For example, to find f(3), the output of the function f for input 3, simply replace x by 3 in the rule’s directions. 2x2 Similarly, replacing x by 29 1 210 1 5 2 and 0 produces the respective outputs. 202 1 1 2 1 226 and f 0 1 2 10 Chapter 1 Number Patterns Technology Tip is to enter 2 1 y f One way to evaluate a x function f its rule into the equation memory as and use TABLE or EVAL. See the Technology Appendix for more detailed information. x 1 2 NOTE Functions will be discussed in detail in Chapter 3. Example 6 Function Notation For h a. h A x 2 1 23 x2 x 2, find each of the following: B b. h 1 2 2 c. h 1 a 2 Solution h To find rule of h. A 23 B and h 2 1 2 , replace x by 23 and 2, respectively, in the a. b. h h 23 A 2 1 2 B 1 23 A 2 B 2 2 2 23 2 3 23 2 1 23 2 1 2 2 4 2 2 0 The values of the function h at any quantity, such as can be found by using the same procedure: replace x in the formula for h(x) by the quantity a a, and simplify. a 2 1 2 a2 a 2 ■ In Exercises 6–8, sketch a scatter plot of the given data. In each case, let the x-axis run from 0 to 10. 6. The maximum yearly contribution to an individual retirement account in 2003 is $3000. The table shows the maximum contribution in fixed 2003 dollars. Let correspond to 2000. x 0 Year 2003 2004 2005 2006 2007 2008 Amount 3000 2910 3764 3651 3541 4294 7. The table shows projected sales, in thousands, of x 0 personal digital video recorders. Let correspond to 2000. (Source: eBrain Market Research) Year 2000 2001 2002 2003 2004 2005 Sales 257 129 143 214 315 485 Exercises 1.1 1. Find the coordinates of points A–I In Exercises 2–5, find the coordinates of the point P. 2. P lies 4 units to the left of the y-axis and 5 units below the x-axis. 3. P lies 3 units above the x-axis and on the same . vertical line as 6, 7 1 2 4. P lies 2 units below the x-axis and its x-coordinate is three times its y-coordinate. 5. P lies 4 units to the right of the y-axis and its y-coordinate is half its x-coordinate. 8. The tuition and fees at public four-year colleges in the fall of each year are shown in the table. Let x 0 correspond to 1995. (Source: The College Board) Section 1.1 Real Numbers, Relations, and Functions 11 Tuition Year & fees Tuition Year & fees 1995 $2860 1998 $3247 1996 $2966 1999 $3356 1997 $3111 2000 $3510 9. The graph, which is based on data from the U.S. Department of Energy, shows approximate average gasoline prices (in cents per gallon) between 1985 and 1996, with corresponding to 1985. x 0 y 120 100 80 60 40 20 1 2 3 4 5 6 7 8 9 10 11 x a. Estimate the average price in 1987 and in 1995. b. What was the approximate percentage increase in the average price from 1987 to 1995? c. In what year(s) was the average price at least $1.10 per gallon? 10. The graph, which is based on data from the U.S. Department of Commerce, shows the approximate amount of personal savings as a percent of disposable income between 1960 and 1995, with x 0 corresponding to 1960. y 10 10 15 20 25 30 35 x a. In what years during this period were personal savings largest and smallest (as a percent of disposable income)? b. In what years were personal savings at least 7% of disposable income? 11. a. If the first coordinate of a point is greater than 3 and its second coordinate is negative, in what quadrant does it lie? b. What is the answer to part a if the first coordinate is less than 3? 12. In which possible quadrants does a point lie if the product of its coordinates is b. negative? a. positive? 13. a. Plot the points (3, 2), 4, 1 1 , 2 1 2, 3 , and 2 5, 4 . 2 1 b. Change the sign of the y-coordinate in each of the points in part a, and plot these new points. c. Explain how the points (a, b) and a, b are 2 1 related graphically. Hint: What are their relative positions with respect to the x-axis? 14. a. Plot the points (5, 3), 4, 2 1 , 2 1 1, 4 , and 2 3, 5 . 2 1 b. Change the sign of the x-coordinate in each of the points in part a, and plot these new points. c. Explain how the points (a, b) and a, b are 2 1 related graphically. Hint: What are their relative positions with respect to the y-axis? In Exercises 15 – 18, determine whether or not the given table could possibly be a table of values of a function. Give reasons for each answer. 15. 16. 17. 18. Input Output Input Output Input Output Input Output .5 0 0 3 4 2 ± 14 5 3 7 8 3 8 12 Chapter 1 Number Patterns Exercises 19 – 22 refer to the following state income tax table. Annual income Amount of tax Less than $2000 0 $2000–$6000 2% of income over $2000 More than $6000 $80 plus 5% of income over $6000 19. Find the output (tax amount) that is produced by each of the following inputs (incomes): $500 $6783 $1509 $12,500 $3754 $55,342 20. Find four different numbers in the domain of this function that produce the same output (number in the range). 21. Explain why your answer in Exercise 20 does not contradict the definition of a function. 22. Is it possible to do Exercise 20 if all four numbers in the domain are required to be greater than 2000? Why or why not? 23. The amount of postage required to mail a firstclass letter is determined by its weight. In this situation, is weight a function of postage? Or vice versa? Or both? 24. Could the following statement ever be the rule of a function? For input x, the output is the number whose square is x. Why or why not? If there is a function with this rule, what is its domain and range? Use the figure at the top of page 13 for Exercises 25–31. Each of the graphs in the figure defines a function. 25. State the domain and range of the function defined by graph a. 26. State the output (number in the range) that the function of Exercise 25 produces from the following inputs (numbers in the domain): 2, 1, 0, 1. 27. Do Exercise 26 for these numbers in the domain: 1 2 , 5 2 , 5 2 . 28. State the domain and range of the function defined by graph b. 29. State the output (number in the range) that the function of Exercise 28 produces from the following inputs (numbers in the domain): 2, 0, 1, 2.5, 1.5. 30. State the domain and range of the function defined by graph c. 31. State the output (number in the range) that the function of Exercise 30 produces from the following inputs (numbers in the domain): 2, 1, 0, , 1. 1 2 32. Find the indicated values of the function by hand and by using the table feature of a calculator. 2x 4 2 x g 2 1 2 a. g d. g(5) 1 2 b. g(0) e. g(12) c. g(4) 33. The rule of the function f is given by the graph. Find a. the domain of f b. the range of f c. d. e. f4 −3 −2 −1 1 2 3 4 −2 −3 −4 34. The rule of the function g is given by the graph. Find a. the domain of g b. the range of g c. g d. g e. g f4 −3 −2 −1 1 2 3 4 −2 −3 − Section 1.2 Mathematical Patterns 13 y y y 4 3 2 1 −1 −1 −2 −3 −4 −3 −2 x 1 2 3 −3 −2 4 3 2 1 −1 −1 −2 −3 −4 x 1 2 3 4 −3 −2 4 3 2 1 −1 −1 −2 −3 −4 x 1 2 3 a. b. c. 1.2 Mathematical Patterns Objectives • Define key terms: sequence sequence notation recursive functions • Create a graph of a sequence • Apply sequences to real- world situatio
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ns Definition of a Sequence Visual patterns exist all around us, and many inventions and discoveries began as ideas sparked by noticing patterns. Consider the following lists of numbers. 4, 1, 2, 5, 8, ? 1, 10, 3, 73, ? Analyzing the lists above, many people would say that the next number in the list on the left is 11 because the pattern appears to be “add 3 to the previous term.’’ In the list on the right, the next number is uncertain because there is no obvious pattern. Sequences may help in the visualization and understanding of patterns. A sequence is an ordered list of numbers. Each number in the list is called a term of the sequence. An infinite sequence is a sequence with an infinite number of terms. Examples of infinite sequences are shown below. 2, 4, 6, 8, 10, 12, p 1, 3, 5, 7, 9, 11, 13, p 2 5 , p 2, 1, 2 3 2 7 2 6 2 4 , , , , The three dots, or points of ellipsis, at the end of a sequence indicate that the same pattern continues for an infinite number of terms. A special notation is used to represent a sequence. 14 Chapter 1 Number Patterns Sequence Notation The following notation denotes specific terms of a sequence: • The first term of a sequence is denoted u2. • The second term • The term in the nth position, called the nth term, is u1. denoted by un. • The term before un is un1. NOTE Any letter can be used to represent the terms of a sequence. Example 1 Terms of a Sequence Make observations about the pattern suggested by the diagrams below. Continue the pattern by drawing the next two diagrams, and write a sequence that represents the number of circles in each diagram. Diagram 1 1 circle Diagram 2 3 circles Diagram 3 5 circles Solution Adding two additional circles to the previous diagram forms each new diagram. If the pattern continues, then the number of circles in Diagram 4 will be two more than the number of circles in Diagram 3, and the number of circles in Diagram 5 will be two more than the number in Diagram 4. Diagram 4 7 circles Diagram 5 9 circles The number of circles in the diagrams is represented by the sequence 1, 3, 5, 7, 9, p , 5 6 Technology Tip If needed, review how to create a scatter plot in the Technology Appendix. which can be expressed using sequence notation. u1 1 u2 3 u3 5 u4 7 u5 9 p un un1 2 ■ Graphs of Sequences A sequence is a function, because each input corresponds to exactly one output. • The domain of a sequence is a subset of the integers. • The range is the set of terms of the sequence. Section 1.2 Mathematical Patterns 15 Because the domain of a sequence is discrete, the graph of a sequence consists of points and is a scatter plot. Example 2 Graph of a Sequence Graph the first five terms of the sequence 1, 3, 5, 7, 9, p 5 . 6 Solution The sequence can be written as a set of ordered pairs where the first coordinate is the position of the term in the sequence and the second coordinate is the term. 10 (1, 1) (2, 3) (3, 5) (4, 7) (5, 9) Figure 1.2-1 The graph of the sequence is shown in Figure 1.2-1. ■ Recursive Form of a Sequence In addition to being represented by a listing or a graph, a sequence can be denoted in recursive form. 10 0 0 Recursively Defined Sequence A sequence is defined recursively if the first term is given and there is a method of determining the nth term by using the terms that precede it. Example 3 Recursively Defined Sequence Define the sequence 7, 4, 1, 2, 5, p 5 6 recursively and graph it. Solution 10 0 10 10 Figure 1.2-2 The sequence can be expressed as 7 u2 4 u3 u1 1 u4 2 u5 5 The first term is given. The second term is obtained by adding 3 to the first term, and the third term is obtained by adding 3 to the second term. Therefore, the recursive form of the sequence is 7 n 2. 3 and for un1 un u1 The ordered pairs that denote the sequence are 1, 7 2, 4 3, 1 2 1 The graph is shown in Figure 1.2-2. 1 1 2 2 4, 2 1 2 5, 5 1 2 ■ 16 Chapter 1 Number Patterns Calculator Exploration An alternative way to think about the sequence in Example 3 is Each answer Preceding answer 3. • Type 7 into your calculator and press ENTER. This establishes the first answer. • To calculate the second answer, press to automatically place ANS at the beginning of the next line of the display. • Now press 3 and ENTER to display the second answer. • Pressing ENTER repeatedly will display subsequent answers. Figure 1.2-3 See Figure 1.2-3. Technology Tip Alternate Sequence Notation Sometimes it is more convenient to begin numbering the terms of a sequence with a number other than 1, such as 0 or 4. The sequence graphing mode can be found in the TI MODE menu or the RECUR submenu of the Casio main menu. On such calculators, recursively defined function may be entered into the sequence memory, or your instruction manual for the correct syntax and use. Y list. Check u0, u1, u2, p or b4, b5, b6, p Example 4 Using Alternate Sequence Notation A ball is dropped from a height of 9 feet. It hits the ground and bounces to a height of 6 feet. It continues to bounce, and on each rebound it rises to 2 3 the height of the previous bounce. a. Write a recursive formula for the sequence that represents the height of the ball on each bounce. b. Create a table and a graph showing the height of the ball on each bounce. c. Find the height of the ball on the fourth bounce. Solution a. The initial height, u0, is 9 feet. On the first bounce, the rebound height, u1, is 6 feet, which is 2 3 the initial height of 9 feet. The recursive form of the sequence is given by u0 9 and un 2 3 un1 for n 1 b. Set the mode of the calculator to Seq instead of Func and enter the function as shown on the next page in Figure 1.2-4a. Figure 1.2-4b displays the table of values of the function, and Figure 1.2-4c displays the graph of the function. Section 1.2 Mathematical Patterns 17 10 0 0 Figure 1.2-4c Figure 1.2-4a Figure 1.2-4b c. As shown in Figures 1.2-4b and 1.2-4c, the height on the fourth bounce is approximately 1.7778 feet. 10 ■ Applications using Sequences Example 5 Salary Raise Sequence If the starting salary for a job is $20,000 and a raise of $2000 is earned at the end of each year of work, what will the salary be at the end of the sixth year? Find a recursive function to represent this problem and use a table and a graph to find the solution. Solution is 20,000. The amount of money earned at the end of The initial term, u1 the first year, u0, , will be 2000 more than un1 20,000 and un u0 The recursive function u0. 2000 for n 1 will generate the sequence that represents the salaries for each year. As shown in Figures 1.2-5a and 1.2-5b, the salary at the end of the sixth year will be $32,000. 50,000 Figure 1.2-5a 0 0 Figure 1.2-5b 10 ■ In the previous examples, the recursive formulas were obtained by either adding a constant value to the previous term or by multiplying the previous term by a constant value. Recursive functions can also be obtained by adding different values that form a pattern. 18 Chapter 1 Number Patterns Example 6 Sequence Formed by Adding a Pattern of Values A chord is a line segment joining two points of a circle. The following diagram illustrates the maximum number of regions that can be formed by 1, 2, 3, and 4 chords, where the regions are not required to have equal areas. 1 Chord 2 Chords 3 Chords 4 Chords 2 Regions 4 Regions 7 Regions 11 Regions a. Find a recursive function to represent the maximum number of regions formed with n chords. b. Use a table to find the maximum number of regions formed with 20 chords. Solution 300 0 0 Figure 1.2-6a Let the initial number of regions occur with 1 chord, so . The maximum number of regions formed for each number or chords is shown in the following table. u1 2 Number of chords 1 2 3 4 p n Maximum number of regions u1 u2 u3 u4 un 2 4 u1 7 u2 11 u3 p un1 n 2 3 4 21 Figure 1.2-6b The recursive function is shown as the last entry in the listing above, and the table and graph, as shown in Figures 1.2-6a and 1.2-6b, identify the 20th term of the sequence as 211. Therefore, the maximum number of regions that can be formed with 20 chords is 211. ■ Example 7 Adding Chlorine to a Pool Dr. Miller starts with 3.4 gallons of chlorine in his pool. Each day he adds 0.25 gallons of chlorine and 15% evaporates. How much chlorine will be in his pool at the end of the sixth day? Solution The initial amount of chlorine is 3.4 gallons, so and each day 0.25 gallons of chlorine are added. Because 15% evaporates, 85% of the mixture remains. u0 3.4 Section 1.2 Mathematical Patterns 19 The amount of chlorine in the pool at the end of the first day is obtained by adding 0.25 to 3.4 and then multiplying the result by 0.85. 0.85 1 3.4 0.25 2 3.1025 The procedure is repeated to yield the amount of chlorine in the pool at the end of the second day. Continuing with the same pattern, the recursive form for the sequence is 0.85 1 3.1025 0.25 2.85 2 u0 3.4 and un 0.85 un1 0.25 for n 1. 1 As shown in Figures 1.2-7a and 1.2-7b, approximately 2.165 gallons of chlorine will be in the pool at the end of the sixth day. 2 ■ Figure 1.2-7a 10 Figure 1.2-7b 4 0 0 Exercises 1.2 In Exercises 1 – 4, graph the first four terms of the sequence. 11. u1 un 1, u2 un1 2, u3 un2 3, un3 and for n 4 1. 2. 3. 4. 2, 5, 8, 11, p . 6 3, 6, 12, 24, p . 6 4, 5, 8, 13, p . 6 4, 12, 36, 108, p 5 5 5 5 6 In Exercises 5–8, define the sequence recursively and graph the sequence. 5 5 5 5. 6. 7. 8. 6, 4, 2, 0, 2, p 6 4, 8, 16, 32, 64, p 6 6, 11, 16, 21, 26, p 8, 4, 2, 1 12. u0 1, u1 1 and un nun1 for n 2 13. A really big rubber ball will rebound 80% of its height from which it is dropped. If the ball is dropped from 400 centimeters, how high will it bounce after the sixth bounce? 14. A tree in the Amazon rain forest grows an average of 2.3 cm per week. Write a sequence that represents the weekly height of the tree over the course of 1 year if it is 7 meters tall today. Write a recursive formula for the sequence and graph the sequence. 15. If two rays have a common endpoint, one angle is formed. If a third r
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ay is added, three angles are formed. See the figure below. In Exercises 9–12, find the first five terms of the given sequence. 2 3 1 9. u1 4 and un 2un1 3 for n 2 10. u1 5 and un 1 3 un1 4 for n 2 Write a recursive formula for the number of angles formed with n rays if the same pattern continues. Graph the sequence. Use the formula to find the number of angles formed by 25 rays. 20 Chapter 1 Number Patterns 16. Swimming pool manufacturers recommend that the concentration of chlorine be kept between 1 and 2 parts per million (ppm). They also warn that if the concentration exceeds 3 ppm, swimmers experience burning eyes. If the concentration drops below 1 ppm, the water will become cloudy. If it drops below 0.5 ppm, algae will begin to grow. During a period of one day 15% of the chlorine present in the pool dissipates, mainly due to evaporation. a. If the chlorine content is currently 2.5 ppm and no additional chlorine is added, how long will it be before the water becomes cloudy? b. If the chlorine content is currently 2.5 ppm and 0.5 ppm of chlorine is added daily, what will the concentration eventually become? c. If the chlorine content is currently 2.5 ppm and 0.1 ppm of chlorine is added daily, what will the concentration eventually become? d. How much chlorine must be added daily for the chlorine level to stabilize at 1.8 ppm? 17. An auditorium has 12 seats in the front row. Each successive row, moving towards the back of the auditorium, has 2 additional seats. The last row has 80 seats. Write a recursive formula for the number of seats in the nth row and use the formula to find the number of seats in the 30th row. 18. In 1991, the annual dividends per share of a stock were approximately $17.50. The dividends were increasing by $5.50 each year. What were the approximate dividends per share in 1993, 1995, and 1998? Write a recursive formula to represent this sequence. 19. A computer company offers you a job with a starting salary of $30,000 and promises a 6% raise each year. Find a recursive formula to represent the sequence, and find your salary ten years from now. Graph the sequence. new students. What will be the enrollment 8 years from now? 22. Suppose you want to buy a new car and finance it by borrowing $7,000. The 12-month loan has an annual interest rate of 13.25%. a. Write a recursive formula that provides the declining balances of the loan for a monthly payment of $200. b. Write out the first five terms of this sequence. c. What is the unpaid balance after 12 months? d. Make the necessary adjustments to the monthly payment so that the loan can be paid off in 12 equal payments. What monthly payment is needed? 23. Suppose a flower nursery manages 50,000 flowers and each year sells 10% of the flowers and plants 4,000 new ones. Determine the number of flowers after 20 years and 35 years. 24. Find the first ten terms of a sequence whose first u1 two terms are for n 3 term 1 terms. 2 and 1 and whose nth is the sum of the two preceding 1 u2 Exercises 25–29 deal with prime numbers. A positive integer greater than 1 is prime if its only positive integer factors are itself and 1. For example, 7 is prime because its only factors are 7 and 1, but 15 is not prime because it has factors other than 15 and 1, namely, 3 and 5. 25. Critical Thinking a. Let un6 be the sequence of 5 prime integers in their usual ordering. Verify that the first ten terms are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. b. Find u17, u18, u19, u20. In Exercises 26 – 29, find the first five terms of the sequence. 26. Critical Thinking un is the nth prime integer larger 20. Book sales in the United States (in billions of than 10. dollars) were approximated at 15.2 in the year 1990. The book sales increased by 0.6 billion each year. Find a sequence to represent the book sales for the next four years, and write a recursive formula to represent the sequence. Graph the sequence and predict the number of book sales in 2003. 21. The enrollment at Tennessee State University is currently 35,000. Each year, the school will graduate 25% of its students and will enroll 6,500 27. Critical Thinking un is the square of the nth prime integer. 28. Critical Thinking un integers less than n. is the number of prime 29. Critical Thinking un is the largest prime integer less than 5n. Exercises 30–34 deal with the Fibonacci sequence { which is defined as follows: un } 1, u2 u1 preceding terms, and for un 1, n 3, un is the sum of the two That is, un1 un2 u1 u2 uz u3 u4 and so on. u3 u4 u5 u1 u2 u3 30. Critical Thinking Leonardo Fibonacci discovered the sequence in the thirteenth century in connection with the following problem: A rabbit colony begins with one pair of adult rabbits, one male and one female. Each adult pair produces one pair of babies, one male and one female, every month. Each pair of baby rabbits becomes adult and produces its first offspring at age two months. Assuming that no rabbits die, how many Section 1.3 Arithmetic Sequences 21 n 1, 2, 3, p ? adult pairs of rabbits are in the colony at the end of n months, helpful to make up a chart listing for each month the number of adult pairs, the number of onemonth-old pairs, and the number of baby pairs. Hint: It may be 31. Critical Thinking List the first ten terms of the Fibonacci sequence. 32. Critical Thinking Verify that every positive integer less than or equal to 15 can be written as a Fibonacci number or as a sum of Fibonacci numbers, with none used more than once. 33. Critical Thinking Verify that perfect square for un2 n 1, 2, p , 10. 5 1 2 4 1 n 2 1 is a 34. Critical Thinking Verify that for n 2, 3, p , 10. 1 n1 1 2 2 un2 1 un1 un1 1.3 Arithmetic Sequences Objectives • Identify and graph an arithmetic sequence • Find a common difference • Write an arithmetic sequence recursively and explicitly • Use summation notation • Find the nth term and the nth partial sum of an arithmetic sequence An arithmetic sequence, which is sometimes called an arithmetic progression, is a sequence in which the difference between each term and the preceding term is always constant. Example 1 Arithmetic Sequence Are the following sequences arithmetic? If so, what is the difference between each term and the term preceding it? a. b. 5 5 14, 10, 6, 2,2, 6, 10, p 3, 5, 8, 12, 17, p 6 6 Solution a. The difference between each term and the preceding term is 4 . So this is an arithmetic sequence with a difference of 4 . b. The difference between the 1st and 2nd terms is 2 and the difference between the 2nd and 3rd terms is 3. The differences are not constant, therefore this is not an arithmetic sequence. ■ is an arithmetic sequence, then for each un un1 n 2, the term preceding is some constant—usually called un6 5 is If un un1 d. Therefore, un and the difference d. un1 22 Chapter 1 Number Patterns Recursive Form of an Arithmetic Sequence In an arithmetic sequence { un } un1 for some constant d and all n 2. un d The number d is called the common difference of the arithmetic sequence. Example 2 Graph of an Arithmetic Sequence is an arithmetic sequence with un6 If two terms, 5 u1 3 and u2 4.5 as its first 15 0 0 Figure 1.3-1a a. find the common difference. b. write the sequence as a recursive function. c. give the first seven terms of the sequence. d. graph the sequence. Solution a. The sequence is arithmetic and has a common difference of u1 4.5 3 1.5 u2 10 Figure 1.3-1b b. The recursive function that describes the sequence is n 2 1.5 3 and un1 for un u1 c. The first seven terms are 3, 4.5, 6, 7.5, 9, 10.5, and 12, as shown in Figure 1.3-1a. d. The graph of the sequence is shown in Figure 1.3-1b. ■ Explicit Form of an Arithmetic Sequence Example 2 illustrated an arithmetic sequence expressed in recursive form in which a term is found by using preceding terms. Arithmetic sequences can also be expressed in a form in which a term of the sequence can be found based on its position in the sequence. Example 3 Explicit Form of an Arithmetic Sequence Confirm that the sequence un expressed as 7 un n 1 1 2 un1 4. 4 with u1 7 can also be Solution Use the recursive function to find the first few terms of the sequence. u1 u2 7 7 4 3 Section 1.3 Arithmetic Sequences 23 2 u3 u4 u5 1 1 1 is 12 5 4 7 4 4 7 16 , which is the first term of the sequence with Notice that the common difference of 4 added twice. Also, which is is the first term of the sequence with the common difference of 4 added n 1 The three times. Because this pattern continues, 2 table in Figure 1.3-2b confirms the equality of the two functions. n 1 7 and un 4 with u1 7 3 4, 7 7 un1 4. 4 un un u3 u4 1 1 2 ■ Figure 1.3-2a As shown in Example 3, if un1 difference d, then for each tion in terms of n, the position of the term. un6 5 n 2, un is an arithmetic sequence with common can be written as a func- d Figure 1.3-2b Applying the procedure shown in Example 3 to the general case shows that u2 u3 u4 u5 u1 u2 u3 u4 d d d d u1 u1 u1 1 1 1 d 2 2d 3d d u1 d u1 d u1 2d 3d 4d 2 2 u5 Notice that 4d is added to u1 So ence, d, added un n 1 yields un. times. 1 2 u1 to obtain is the sum of n numbers: . In general, adding to and the common differ- d 1 2 u1 n 1 Explicit Form of an Arithmetic Sequence In an arithmetic sequence { un n 1 u1 un } with common difference d, d for every n 1. 1 2 u0, If the initial term of a sequence is denoted as arithmetic sequence with common difference d is the explicit form of an un u0 nd for every n 0. Example 4 Explicit Form of an Arithmetic Sequence Find the nth term of an arithmetic sequence with first term mon difference of 3. Sketch a graph of the sequence. 5 and com- Solution Because u1 10 5 and u1 d 3, n 1 1 2 un the formula in the box states that n 1 3 3n 8 d 5 1 2 Figure 1.3-3 The graph of the sequence is shown in Figure 1.3-3. ■ 30 0 10 24 Chapter 1 Number Patterns Example 5 Finding a Term of an Arithmetic Sequence What is the 45th term of the arithmetic sequence whose first three terms are 5, 9, and 13? Solution The first three terms show that d, is 4
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. Apply the formula with and that the common difference, u45 u1 1 45 1 2 44 4 2 21 1 181 ■ 5 u1 n 45 . d 5 Example 6 Finding Explicit and Recursive Formulas un6 If is an arithmetic sequence with 5 sive formula, and an explicit formula for u6 57 un. and u10 93, find u1 , a recur- Solution The sequence can be written as p , 57, —, —, —, 93, p u8 u9 u10 u7 { { { { { u6 The common difference, d, can be found by the difference between 93 and 57 divided by the number of times d must be added to 57 to produce 93 (i.e., the number of terms from 6 to 10). d 93 57 10 6 36 4 9 d 9 Note that is the difference of the output values (terms of the sequence) divided by the difference of the input values (position of the terms of the sequence), which represents the change in output per unit change in input. The value of equation u1 can be found by using n 6, u6 57 and d 9 in the n 1 6 1 d 2 9 2 57 5 9 57 45 12 1 1 u1 u6 57 u1 u1 d 9, and the recursive form of the arithmetic sequence u1 Because is given by 12 u1 12 and un un1 9, for n 2 The explicit form of the arithmetic sequence is given by un n 1 12 1 9n 3, for n 1. 9 2 ■ NOTE If u1, u2, u3, p is an arithmetic sequence, then an expression of u3 u1 the form (sometimes written as q u2 p Section 1.3 Arithmetic Sequences 25 Summation Notation It is sometimes necessary to find the sum of various terms in a sequence. For instance, we might want to find the sum of the first nine terms of the sequence Mathematicians often use the capital Greek letter sigma to abbreviate such a sum as follows. un6 5 . 1 2 un ) is called an a n1 arithmetic series. 9 a i1 ui u1 u2 u3 u4 u5 u6 u7 u8 u9 Similarly, for any positive integer m and numbers c1, c2, p , cm , Summation Notation m a k1 means c1 ck c2 c3 p cm Example 7 Sum of a Sequence Compute each sum. a. 5 a n1 1 7 3n 2 b. 4 a n1 3 3 n 1 1 4 2 4 Solution a. Substitute 1, 2, 3, 4, and 5 for n in the expression 7 3n and add the terms. 5 a n1 1 7 3n 10 12 1 2 7 15 2 b. Substitute 1, 2, 3, and 4 for n in the expression add the terms. 3 n 1 2 1 4, and 4 a n1 12 12 2 3 4 3 3 4 4 ■ 26 Chapter 1 Number Patterns Technology Tip SEQ is in the OPS submenu of the TI LIST menu and in the LIST submenu of the Casio OPTN menu. SUM is in the MATH submenu of the TI LIST menu. SUM is in the LIST submenu of the Casio OPTN menu. Using Calculators to Compute Sequences and Sums Calculators can aid in computing sequences and sums of sequences. The SEQ (or MAKELIST) feature on most calculators has the following syntax. SEQ(expression, variable, begin, end, increment) The last parameter, increment, is usually optional. When omitted, increment defaults to 1. Refer to the Technology Tip about which menus contain SEQ and SUM for different calculators. The syntax for the SUM (or LIST ) feature is SUM(list[, start, end]), where start and end are optional. When start and end are omitted, the sum of the entire list is given. Combining the two features of SUM and SEQ can produce sums of sequences. SUM(SEQ(expression, variable, begin, end)) Example 8 Calculator Computation of a Sum Use a calculator to display the first 8 terms of the sequence un 7 3n and to compute the sum 50 a n1 7 3n. Solution Using the Technology Tip, enter which produces Fig1 ure 1.3-4. Additional terms can be viewed by using the right arrow key to scroll the display, as shown at right below. , 2 SEQ 7 3n, n, 1, 8 Figure 1.3-4 Figure 1.3-5 fore, 1 22 7 3n 3475. 1 50 a n1 17. The first 8 terms of the sequence are To compute the sum of the first 50 terms of the sequence, enter SUM Figure 1.3-5 shows the resulting display. There- 4, 1, 2, 5, 8, 11, 14, 7 3n, n, 1, 50 SEQ and . Partial Sums Suppose k terms of the sequence is called the kth partial sum of the sequence. is a sequence and k is a positive integer. The sum of the first un6 5 ■ Section 1.3 Arithmetic Sequences 27 Calculator Exploration Write the sum of the first 100 counting numbers. Then find a pattern to help find the sum by developing a formula using the terms in the sequence. Partial Sums of an Arithmetic Sequence un If { } is an arithmetic sequence with common difference d, then for each positive integer k, the kth partial sum can be found by using either of the following formulas. 1. k a n1 k 2. a n1 un k 2 (u1 un ku1 uk) k(k 1) 2 d Sk represent the kth partial sum Proof Let terms of the arithmetic sequence in two ways. In the first representation of Write the u1 Sk, u2 p uk. repeatedly add d to the first term. u3 p uk2 u2 d 2d u1 u1 u1 u1 Sk uk1 uk p 3 4 In the second representation of uk uk uk1 uk Sk 3 3 , repeatedly subtract d from the kth term. 2 1 4 4 Sk u1 k 1 d uk2 p u3 d 2d uk u2 p u1 uk k 1 d 4 Sk 3 3 are added, the multiples of d add to zero 1 2 4 4 If the two representations of and the following representation of u1 uk d d u1 uk u1 uk Sk Sk 4 4 3 3 2Sk 2d 2d is obtained. p p 4 4 u1 3 uk terms ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩ 2Sk Sk u1 1 k 2 1 u1 1 uk2 uk2 uk2 u1 p u1 1 uk2 k u1 1 uk2 Divide by 2. The second formula is obtained by letting equation. uk u1 k 1 2 1 d in the last k 2 3 u1 u1 k 1 1 k 2 3 d 2 4 2u1 k 1 ■ d 2 4 1 Sk k 2 1 ku1 u1 uk2 k 1 k 1 2 2 d 28 Chapter 1 Number Patterns Example 9 Partial Sum of a Sequence Find the 12th partial sum of the arithmetic sequence below. 8, 3, 2, 7, p Solution First note that d, the common difference, is 5 and u1 8 . u12 u1 8 11 12 1 5 2 d 47 1 1 2 Using formula 1 from the box on page 27 yields the 12th partial sum. 12 a n1 un 12 2 1 8 47 234 2 ■ Example 10 Partial Sum of a Sequence Find the sum of all multiples of 3 from 3 to 333. Solution Note that the desired sum is the partial sum of the arithmetic sequence 3, 6, 9, 12, p . The sequence can be written in the form 3 1, 3 2, 3 3, 3 4, p , 333 3 111 is the 111th term. The 111th partial sum of the where sequence can be found by using formula 1 from the box on page 27 with k 111, 333. 3, and u111 u1 111 a n1 un 111 2 1 3 333 111 2 1 2 336 2 18,648 ■ Example 11 Application of Partial Sums If the starting salary for a job is $20,000 and you get a $2000 raise at the beginning of each subsequent year, how much will you earn during the first ten years? Solution The yearly salary rates form an arithmetic sequence. 20,000 22,000 24,000 26,000 p u1 The tenth-year salary is found using 10 1 u1 2 d 1 20,000 9 1 2000 u10 $38,000 2 20,000 and d 2000. Section 1.3 Arithmetic Sequences 29 The ten-year total earnings are the tenth partial sum of the sequence. 10 a n1 un 10 2 ˛ 1 u1 u102 20,000 38,000 10 2 ˛ 1 5˛1 58,000 2 $290,000 2 ■ Exercises 1.3 In Exercises 1–6, the first term, and the common difference, d, of an arithmetic sequence are given. Find the fifth term, the explicit form for the nth term, and sketch the graph of each sequence. u1, 1. u1 5, d 2 2. u1 4, d 5 3. u1 4, d 1 4 4. u1 6, d 2 3 5. u1 10, d 1 2 6. u1 p, d 1 5 In Exercises 7– 12, find the sum. 7. 9. 5 a i1 3i 16 a n1 1 2n 3 2 11. 36 a n151 2n 8 2 8. 4 a i1 1 2i 10. 12. 75 a n1 1 31 a n1 1 3n 1 2 300 n 1 1 2 ˛2 2 In Exercises 13–18, find the kth partial sum of the arithmetic sequence { } with common difference d. un 13. k 6, u1 2, d 5 14. k 8, u1 2 3 , d 4 3 15. k 7, u1 3 4 , d 1 2 16. k 9, u1 6, u9 24 17. k 6, u1 4, u6 14 18. k 10, u1 0, u10 30 In Exercises 19–24, show that the sequence is arithmetic and find its common difference. 19. 5 3 2n 6 21. 23. 24. 5 3n 2 e f c 2n 6 2b 3nc 5 5 20. 4 n e 3 f 22. p n 2 e f c constant 2 1 6 1 b, c constants 2 In Exercises 25–30, use the given information about the arithmetic sequence with common difference d to find un. and a formula for u5 25. u4 12, d 2 26. u7 8, d 3 27. u2 4, u6 32 28. u7 6, u12 4 29. u5 0, u9 6 30. u5 3, u9 18 In Exercises 31–34, find the sum. 31. 33. 20 a n1 1 3n 4 2 40 a n1 n 3 6 32. 34. 25 a n1 a n 4 5 b 30 a n1 4 6n 3 35. Find the sum of all the even integers from 2 to 100. 36. Find the sum of all the integer multiples of 7 from 7 to 700. 37. Find the sum of the first 200 positive integers. 38. Find the sum of the positive integers from 101 to 200 (inclusive). Hint: Recall the sum from 1 to 100. Use it and Exercise 37. 30 Chapter 1 Number Patterns 39. A business makes a $10,000 profit during its first year. If the yearly profit increases by $7500 in each subsequent year, what will the profit be in the tenth year? What will be the total profit for the first ten years? 40. If a man’s starting annual salary is $15,000 and he receives a $1000 increase to his annual salary every six months, what will he earn during the last six months of the sixth year? How much will he earn during the first six years? 41. A lecture hall has 6 seats in the first row, 8 in the 42. A monument is constructed by laying a row of 60 bricks at ground level. A second row, with two fewer bricks, is centered on that; a third row, with two fewer bricks, is centered on the second; and so on. The top row contains ten bricks. How many bricks are there in the monument? 43. A ladder with nine rungs is to be built, with the bottom rung 24 inches wide and the top rung 18 inches wide. If the lengths of the rungs decrease uniformly from bottom to top, how long should each of the seven intermediate rungs be? second, 10 in the third, and so on, through row 12. Rows 12 through 20 (the last row) all have the same number of seats. Find the number of seats in the lecture hall. 44. Find the first eight numbers in an arithmetic sequence in which the sum of the first and seventh terms is 40 and the product of the first and fourth terms is 160. 1.4 Lines Objectives • Find the slopes of lines, including parallel and perpendicular lines • Describe the connection between arithmetic sequences and lines • Graph lines • Write the equations of lines, including horizontal and vertical lines A graph is a set of points in a plane. Some graphs are based on data points, such as those shown in Section 1.3 where arithmetic sequences were graphed as scatter plots. Other graphs arise from equations. A solution of an equation in two variables, say x and y, is a pair of numbers such that the substitution of t
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he first number for x and the second for y produces a true statement. The graph of an equation in two variables is the set of points in a plane whose coordinates are solutions of the equation. Thus, the graph is a geometric picture of the solutions. Recall that an arithmetic sequence is a sequence in which the difference between each term and the preceding term is constant. For example, n 2 Position 1 un2 Term 1 3, 8, 13, 18, p 5 1 2 3 is an arithmetic sequence. p n 5 4 3 8 13 18 23 p 3 6 n 1 1 2 5 The graph of the sequence above has an infinite number of discrete points because the value of the sequence depends upon the term of the sequence, which must be a counting number. See Figure 1.4-1a. The graph of y 3 is a continuous line that contains the discrete points of the arithmetic sequence, as shown in Figures 1.4-1a and 1.4-1b on page 31. x 1 5 2 1 NOTE If needed, review graphing functions on a graphing calculator in the Technology Appendix. 25 25 Section 1.4 Lines 31 0 0 Figure 1.4-1a 6 0 0 Figure 1.4-1b 6 As the positions within a sequence increase by one, the value of the terms increases by 5. Notice on the graph that as the x-values move to the right a distance of one, the y-values move up by a distance of 5. This common difference represents one of the most identifiable characteristics of a line, its slope. Slope When you move from a point P to a point Q on a line, two numbers are involved, as illustrated in Figure 1.4-2. • The vertical distance you move is called the change in y, which is sometimes denoted and read “delta y.” ¢y • The horizontal distance you move is called the change in x, which is sometimes denoted and read “delta x.” ¢x Q Q Q Change in y = 4 Change in y = 4 P Change in x = 4 = 1 = 4 4 Change in y Change in x b. Figure 1.4-2 Change in x = 6 Change in y Change in x a. = 4 6 = 2 3 Change in y = 4 P Change in x = 1 Change in y Change in x c. = 4 1 = 4 y The fraction change in y change in x ¢y ¢x (x1, y1) pose P has coordinates Figure 1.4-3. measures the steepness of the line. Sup- and Q has coordinates (x2, y2), as shown in P y2 y1 • The change in y is the difference of the y-coordinates of P and Q. ¢y y2 y1 x • The change in x is the difference of the x-coordinates of P and Q. ¢x x2 x1 (x2, y2) Q y2 − y1 (x1, y1) P x2 − x1 x2 x1 Figure 1.4-3 32 Chapter 1 Number Patterns Consequently, slope is defined as follows. Slope of a Line y 1 1 2 3 4 −1 Figure 1.4-4 CAUTION When finding slopes, you must subtract the y-coordinates and the x-coordinates in the same order. With the points (3, 4) and (1, 8), for instance, if you use 8 4 in the numerator, you must write 1 – 3 in the denominator, not 3 1. and (x2, y2) (x1, y1) are points with If the line through these points is the ratio y2 x2 change in y change in x ¢y ¢x x1 y1 x1 x2 , then the slope of Example 1 Finding Slope Given Two Points Find the slope of the line that passes through ure 1.4-4. 1 0, 1 2 and (4, 1). See Fig- x Solution Apply the formula in the previous box with x2 4, y2 1. x1 0, y1 1 and Slope y2 x2 y1 x1 1 (1) 4 0 2 4 1 2 The order of the points makes no difference; if you use (4, 1) for and in Example 1, the result is the same. (0, 1) for (x2, y2) ■ x1, y12 1 Example 2 Finding Slope From a Graph Find the slope of each line shown in Figure 1.4-5. The lines shown are determined by the following points: and 1, 1 3, 5 1 L4: L2: 2 and 2 0, 2 1 3, 1 2 1 1 and 1 2 L5: 2, 4 L3: and 6, 2 1 2, 2 2 1, 0 1 2 1 2 and 3, 2 2 1 2 L1: 1 0, 2 2 Solution The slopes are as follows. 3 0 1 3 1 L1: 2 1 0 1 L2 L4: L3: L5: 2 2 1 2 y L1 L2 5 4 3 2 1 L3 x −6 −5 −4 −3 −2 −1 −1 −2 3 L4 1 2 L5 Figure 1.4-5 ■ Section 1.4 Lines 33 Example 2, page 32, illustrates how the slope measures the steepness of the line, summarized as follows. The lines referenced refer to Figure 1.4-5. Properties of Slope The slope of a nonvertical line is a number m that measures how steeply the line rises or falls. • If m 77 0, the line rises from left to right; the larger m is, the more steeply the line rises. [Lines and L2 ] , the line is horizontal. [Line L1 ] L3 • If • If m 0 m 66 0, the line falls from left to right; the larger is, m 0 0 the more steeply the line falls. [Lines L4 and ]L5 Slope-Intercept Form A nonvertical line intersects the y-axis at a point with coordinates (0, b), because every point on the y-axis has first coordinate 0. The number b is called the y-intercept of the line. For example, the line in Figure 1.4-4 has y-intercept because it crosses the y-axis at (0, 1). 1 y (x, y) y − b (0, b) x − 0 Let L be a nonvertical line with slope m and y-intercept b. Therefore, (0, b) is a point on L. Let (x, y) be any other point on L, as shown in Figure 1.4-6. Use the points (0, b) and (x, y) to compute the slope of L Multiply both sides of the equation by x, and solve for y. mx y b y mx b Figure 1.4-6 Thus, the coordinates of any point on L satisfy the equation which leads to the following. y mx b, Slope-Intercept Form The line with slope m and y-intercept b is the graph of the equation y mx b. Did you notice, as you recall your work with arithmetic sequences, that the explicit form of the sequence looks very similar to y mx b ? Example 3 Graphs of Arithmetic Sequences and Lines and 8. Use the The first three terms of an arithmetic sequence are explicit form of the sequence to express the nth term, and compare it to 2, 3, 34 Chapter 1 Number Patterns the slope-intercept form of the equation of a line that passes through the points on the graph of the sequence. Graph both the sequence and the corresponding line on the same set of axes. Solution 25 The common difference for the sequence is 5 and explicit form is u1 2. Therefore, the 3 5 10 Figure 1.4-7 Connection between Arithmetic Sequences and Lines un n 1 u1 2 ˛d 1 n 1 2 2 ˛5 2 5n 5 5n 7 1 The last equation, un has the form y mx b, where m d 5 b u0 d 7 and x corresponds to n. 5n 7, u1 Figure 1.4-7 shows graphs of the sequence and the line. ■ Below is an important summary connecting the explicit form of an arithmetic sequence to the slope-intercept form of a line. The connection between the explicit form of an arithmetic sequence, a line, (n 1)˛d , is as follows. u1 un y mx b , and the slope-intercept form of • The slope of the line corresponds to the common difference of the sequence, m d. • The y-intercept represents the value of the first term of the sequence minus the difference, b u1 d. A linear equation expressed in slope-intercept form defines a relation for all the ordered pairs (x, y) on the line. The equation represents the rule and each x represents an input. For every input x there is one and only one output y, so y is a function of x. The graph of a linear function f is the graph of mx b. x y f˛ 1 2 Most graphing calculators are called function graphers because they graph a relation only if it can be expressed as a function Thus, slopeintercept form is useful for graphing a line both on paper and with a graphing calculator. y f ˛1 x . 2 Example 4 Graphing a Line Sketch the graph of ing calculator. 2y 5x 2, and confirm your sketch with a graph- y 6 Solution Begin by solving the equation for y. 2y 5x 2 2y 5x 2 y f 1 5 2 x 2 Section 1.4 Lines 35 x 1 2 Figure 1.4-8a 8 x Therefore, the graph is a line with slope 5 2 , the coefficient of x, and y-intercept 1, the constant term. Because the y-intercept is 1, the point (0, 1) is on the line. When x 2, then , so (2, 6) is also on the line. Plotting and connecting the points (0, 1) and (2, 6) produces the line in Figure 1.4-8a. Figure 1.4-8b displays the same graph produced by a graphing calculator. ■ 3 5 Example 5 Linear Depreciation 1 Figure 1.4-8b An office buys a new computer system for $7000. Five years later its value is $800. Assume that the system depreciates linearly. a. Write the equation that represents value as a function of years. b. Find its value two years after it was purchased, that is, the y-value when x 2. c. Graph the equation. d. Find how many years before the system is worthless, that is, the x-value that corresponds to a y-value of 0. Solution a. Linear depreciation means that the equation that gives the value y of the computer system in year x has the form for some constants m and b. Because the system is worth $7000 new (when x 0 ) the y-intercept is 7000 and the equation can be written as y mx b y mx 7000 y mx b y 800 when Because the system is worth $800 after 5 years (i.e., x 5 , 2 y mx 7000 800 m 5 7000 6200 5m m 6200 5 1240 The depreciation equation is the value of the system after x years. y 1240x 7000, where y represents 36 Chapter 1 Number Patterns 10,000 0 0 10 Figure 1.4-9 b. The value of the system after two years 1 x 2 7000 is 2 y 1240 $4520. 2 1 2 c. The graph of y 1240x 7000 1240, is shown in Figure 1.4-9. Notice represents the depreciation of the that the slope of the line, system per year. That is, the value of the system decreases $1240 each year. d. The trace feature or the zero feature of a graphing calculator shows that the x-value corresponding to Figure 1.4-9. That is, the system will be worthless in 5 years and 0.6 12 7.2 as shown in months. is y 0 x 5.6, ■ The Point-Slope Form Suppose the line L passes through the point and has slope m. Let x1, y12 (x, y) be any other point on L. Use the fixed point x1, y12 and the variable 1 to compute the slope m of L, and it will generate another useful point x x1, form of a line called the point-slope form. For x, y 1 2 1 y y1 x x1 y y1 m slope of L x x12 m 1 Thus, the coordinates of every point on L satisfy the equation y y1 m . x x12 1 Point-Slope Form The line with slope m through the point of the equation (x1, y1) is the graph y y1 m (x x1). There are two interesting observations about point-slope form. 1. Although slope-intercept form and point-slope form can be used to write the equation of a line, the point-slope form is easier to use, unless you know the y-intercept. 2. The point-slope form can also be used to graph a line because any point of the line can be used as the
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initial point and the remaining points can be found by using the equation’s slope. The slope determines how to find a second point from the initial point by moving vertically an amount equal to the numerator of the slope, which represents and then moving horizontally an amount equal to the denominator of the slope, which represents ¢x. ¢y, y 0 1 −1 −2 −3 −4 −5 −6 −7 x 1 765432 2 1 Figure 1.4-10 Section 1.4 Lines 37 Example 6 Point-Slope Form of a Line Sketch the graph and find the equation of the line that passes through the point with slope 2. Write the equation in slope-intercept form. 1, 6 1 2 Solution and identify another point on To graph the line, start at the point 1 the line by moving 2 units vertically and 1 unit horizontally. The point 2, 4 is also on the line. Because a unique line is determined by two 1 points, connecting the points produces the line, as and shown in Figure 1.4-10. 2, 4 1, 6 2 2 1 1 2 2 1, 6 To find the equation of the line, substitute 2 for m and in the point-slope equation. m˛1 2 1 1 y 6 2x 2 y 2x 8 x x12 x 1 2 y y1 6 Point-slope form y Slope-intercept form 2 1, 6 1 2 for x1, y12 1 ■ Vertical and Horizontal Lines When a line has 0 slope, it is called a horizontal line, and it can be written as y 0x b b. Example 7 Equation of a Horizontal Line Describe and sketch the graph of the equation y 3. x Solution y 3 can be written as its graph is a line with slope Because 0 and y-intercept 3. This is sufficient information to obtain the graph shown in Figure 1.4-11. y 0x 3, ■ Vertical Lines The preceding discussion does not apply to vertical lines, whose equations have a different form than those examined earlier because a vertical line is not a function. Example 8 Equation of a Vertical Line x y 1 −1 −3 2 Figure 1.4-11 y 3 2 1 0 −1 1 3 4 Find the equation of the vertical line shown in Figure 1.4-12. −2 Solution Figure 1.4-12 Every point on the vertical line in Figure 1.4-12 has first coordinate 2. and the line is the graph Thus, every point on the line satisfies x 0y 2, 38 Chapter 1 Number Patterns of the equation x 2. (2, 1) and (2, 4), you obtain If you try to compute the slope of the line, say using 4 1 2 2 3 0 , which is not defined. ■ Parallel and Perpendicular Lines The slope of a line measures how steeply it rises or falls. Because parallel lines rise or fall equally steeply, their slopes are the same. Two lines that meet in a right angle, that is, a angle, are said to be perpendicular. There is a close relationship between the slopes of two perpendicular lines. 90° Parallel and Perpendicular Lines Two nonvertical lines are parallel when they have exactly the same slope. Two nonvertical lines are perpendicular when the product of their slopes is 1. Example 9 Parallel and Perpendicular Lines Given the line M whose equation is 2, 1 of the lines through the point 1 2 3x 2y 6 0, find the equation a. parallel to M. b. perpendicular to M. Solution First find the slope of M by rewriting its equation in slope-intercept form. 3x 2y 6 0 2y 3x 6 y 3 2 x 3 Therefore, M has slope 3 2 . a. The line parallel to M must have the same slope, and because is on the parallel line, use the point-slope form to find its 2, 1 1 equation. 2 y y y1 x12 x 2 2 1 x 3 x 4 Section 1.4 Lines 39 b. The line perpendicular to M must have slope 2 3 1 3 2 . Use point- slope form again to find the equation of the line perpendicular to M through 2 ■ The standard form of a line is where A, B, and C are integers, both 0. , and A and B are not Ax By C , A 0 equation shown in Example 9, for the perpendicular line, Any line, including vertical lines, can be written in this form. The last x 1 3 can be written in standard form by multiplying both sides by 3, the least common denominator of all the terms, and then adding 2x to both sides. The resulting equation is 2x 3y 1. y 2 3 , Standard Form of a Line NOTE The standard form of a line is sometimes called the general form of a line and may also be written as Ax By C 0. The following box summarizes the different forms of the equation of a line and when each form is best used. Forms of Linear Equations The forms of the equation of a line are Ax By C standard form y mx b slope-intercept form Graphing y y1 m(x x1) point-slope form Write equations A horizontal line has slope 0 and an equation of the form y b. A vertical line has undefined slope and an equation of the form x c. 40 Chapter 1 Number Patterns Exercises 1.4 1. For which of the line segments in the figure is the slope a. largest? b. smallest? c. largest in absolute value? d. closest to zero? y A In Exercises 7–10, find the slope of the line through the given points. 7. 9. 1, 2 ; 1 2 1 3. 10. 1 A 1, 2 ; 2 1 2, 1 2 22, 1 ; 1 B 2, 9 2 In Exercises 11–14, find a number t such that the line 2. passing through the two given points has slope B C D E x 11. 13. 1 1 0, t ; 9, 4 1 2 t 1, 5 2 ; 1 2 6, 3t 7 2 12. 14. 1 1 1, t ; 2 1 3, 5 2 t, t ; 1 2 5, 9 2 2. The doorsill of a campus building is 5 ft above ground level. To allow wheelchair access, the steps in front of the door are to be replaced by a straight ramp with constant slope 1 12 , as shown in the figure. How long must the ramp be? [The answer is not 60 ft.] 15. Let L be a nonvertical straight line through the origin. L intersects the vertical line through (1, 0) at a point P. Show that the second coordinate of P is the slope of L. 16. On one graph, sketch five line segments, not all meeting at a single point, whose slopes are five different positive numbers. In Exercises 17–20, find the equation of the line with slope m that passes through the given point. 17. 19. 3, 5 2 m 1; 1 m 1; 6, 2 2 1 18. m 2; 20. m 0; 2, 1 2 4, 5 1 1 2 In Exercises 21–24, find the equation of the line through the given points. 21. 0, 5 1 and 1 2 3, 2 2 22. 4, 3 2 1 and 2, 1 1 2 23. 4 3 a , 2 3b and 1 3 a , 3 b 24. (6, 7) and (6, 15) In Exercises 25–28, determine whether the line through P and Q is parallel or perpendicular to the line through R and S, or neither. 25. P , Q 2, 5 2 1 1 1, 1 2 and R , S 4, 2 1 2 6, 1 1 2 26. P 3 2b 0, a , Q 1, 1 1 2 and R , S 2, 7 1 2 3, 9 1 2 R a m p 5 In Exercises 3–6, find the slope and y-intercept of the line whose equation is given. 3. 2x y 5 0 4. 3x 4y 7 5. 3 6 27. P S P R 28. , Q 1, 1 1 2 and R 2, 0 , 2 1 1 3, 3b a 4, 2 a 3b 3, 3 2 2, 2 , Q 1 , S 3, 1 4, 5 2 1 2 2 1 1 and Section 1.4 Lines 41 In Exercises 29–31, determine whether the lines whose equations are given are parallel, perpendicular, or neither. 1 10 2 7 3 4 4 1 5 2 29. 2x y 2 0 and 4x 2y 18 0 30. 3x y 3 0 and 6x 2y 17 0 31. y 2x 4 and 0.5x y 3 47. The first three terms of an arithmetic sequence are 5. 7, 1, and Write the sequence’s equation in slope-intercept form. 32. Use slopes to show that the points , 1, 12 7, 0 and 2 1 2 4, 6 , 2 1 straight line. 1 all lie on the same 33. Use slopes to determine if (9, 6), , are the vertices of a right triangle. 1, 3 1 2 1, 2 and 1 2 34. Use slopes to show that the points 5, 2 1 1 of a parallelogram. 3, 1 , 2 1 2 3, 0 , and 5, 3 1 2 2 are the vertices 48. For a given arithmetic sequence, the common u1 difference is y-intercept of the graph of this sequence. Find the slope and 6. and 3 49. For a given arithmetic sequence, the common u1 difference is 8 and y-intercept of the graph of this sequence. Find the slope and 2. 50. Let L be a line that is neither vertical nor horizontal and which does not pass through the origin. Show that L is the graph of x a y b 1, In Exercises 35–42, find an equation for the line satisfying the given conditions. where a is the x-intercept and b is the y-intercept of L. 35. through 2, 1 1 2 with slope 3 36. y-intercept 7 and slope 1 37. through 38. through 1 1 2, 3 2 1, 2 and parallel to 3x 2y 5 and perpendicular to y 2x 3 2 39. x-intercept 5 and y-intercept 5 40. through 5, 2 2 (1, 2) and (4, 3) 1 and parallel to the line through 41. through 1, 3 through (0, 1) and (2, 3) 1 2 and perpendicular to the line 42. y-intercept 3 and perpendicular to 2x y 6 0 43. Find a real number k such that kx 2y 7 0. line 1 3, 2 2 is on the 44. Find a real number k such that the line 3x ky 2 0 has y-intercept 3. 45. Write the equation for the given arithmetic sequence in slope-intercept form. 1 2 2 2 3 6 4 10 5 14 46. Write the equation for the given arithmetic sequence in slope-intercept form. 51. Let A, B, C, and D be nonzero real numbers. Show that the lines Ax By D 0 are parallel. Ax By C 0 and 52. Sales of a software company increased linearly from $120,000 in 1996 to $180,000 in 1999 a. Find an equation that expresses the sales y in year x (where x 0 corresponds to 1996). b. Estimate the sales in 2001. 53. The poverty level income for a family of four was $9287 in 1981. Due to inflation and other factors, the poverty level income rose to approximately $18,267 in 2001. (Source: U.S. Census Bureau) a. Find a linear equation that approximates the x 0 poverty level income y in year x (with corresponding to 1981). b. Use the equation of part a to estimate the poverty level income in 1990 and 2005. 54. At sea level, water boils at 1100 ft, water boils at between boiling point and height is linear. a. Find an equation that gives the boiling point The relationship At a height of 210° F. 212° F. y of water at a height of x feet. Find the boiling point of water in each of the following cities (whose altitudes are given). b. Cincinnati, OH (550 ft) c. Springfield, MO (1300 ft) d. Billings, MT (3120 ft) e. Flagstaff, AZ (6900 ft) 42 Chapter 1 Number Patterns 55. A small plane costs $600,000 new. Ten years later, 61. A hat company has fixed costs of $50,000 and it is valued at $150,000. Assuming linear depreciation, find the value of the plane when it is 5 years old and when it is 12 years old. 56. In 1950, the age-adjusted death rate from heart disease was about 307.2 per 100,000 people. In 1998, the rate had decreased to 126.6 per 100,000. a. Assuming the rate decreased linearly, find an equation that gives the number y of deaths per x 0 100,000 from heart d
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isease in year x, with corresponding to 1950. Round the slope of the line to one decimal place. b. Use the equation in part a to estimate the death rate in 1995 and in 2005. 57. According to the Center of Science in the Public Interest, the maximum healthy weight for a person who is 5 ft 5 in. tall is 150 pounds and for someone 6 ft 3 in. tall is 200 pounds. The relationship between weight and height here is linear. a. Find a linear equation that gives the maximum healthy weight y for a person whose height is x x 0 inches over 5 ft 5 in. ( corresponds to 5 ft to 5 ft 7 in., etc.). 5 in., b. Use the equation of part a to estimate the x 2 maximum healthy weight for a person whose height is 5 ft and for a person whose height is 6 ft. 58. The profit p (in thousands of dollars) on x thousand units of a specialty item is p 0.6x 14.5. The cost c of manufacturing x c 0.8x 14.5. items is given by a. Find an equation that gives the revenue r from selling x items. b. How many items must be sold for the company to break even (i.e., for revenue to equal cost)? variable costs of $8.50 per hat. a. Find an equation that gives the total cost y of producing x hats. b. What is the average cost per hat when 20,000 are made? 50,000? 100,000? Use the graph and the following information for Exercises 62–64. Rocky is an “independent” ticket dealer who markets choice tickets for Los Angeles Lakers home games (California currently has no laws against scalping). Each graph shows how many tickets will be demanded by buyers at a particular price. For instance, when the Lakers play the Chicago Bulls, the graph shows that at a price of $160, no tickets are demanded. As the price (y-coordinate) gets lower, the number of tickets demanded (x-coordinate) increases. e c i r P 160 140 120 100 80 60 40 20 0 Bulls Suns Mavericks 10 20 30 40 Quantity 62. Write a linear equation expressing the quantity x of tickets demanded at price y when the Lakers play the indicated team. a. Dallas Mavericks b. Phoenix Suns c. Chicago Bulls Hint: In each case, use the points where the graph crosses the two axes to determine its slope. 59. A publisher has fixed costs of $110,000 for a 63. Use the equations from Exercise 62 to find the mathematics text. The variable costs are $50 per book. The book sells for $72. Find equations that give the required information. a. the cost c of making x books b. the revenue r from selling x books c. the profit p from selling x books d. the publisher’s break-even point (see Exercise 58b) number of tickets Rocky would sell at a price of $40 for a game against the indicated team. a. Mavericks b. Bulls 64. Suppose Rocky has 20 tickets to sell. At what price could he sell them all when the Lakers play the indicated team. a. Mavericks b. Suns 60. If the fixed costs of a manufacturer are $1000 and it costs $2000 to produce 40 items, find a linear equation that gives the total cost of making x items. Section 1.5 Linear Models 43 1.5 Linear Models Objectives • Algebraically fit a linear model People working in business, medicine, agriculture, and other fields frequently want to know the relationship between two quantities. For instance, • Calculate finite differences How does money spent on advertising affect sales? and use residuals to determine the model of best fit • Use a calculator to determine a linear model • Find and interpret the correlation coefficient for a model • Create and interpret a residual plot for a linear model What effect does a fertilizer have on crop yield? How much do large doses of certain vitamins lengthen life expectancy? In many such situations there is sufficient data available to construct a mathematical model, such as an equation or graph, which demonstrates the desired relationship or predicts the likely outcome in cases not included in the data. In this section applications are considered in which the data can be modeled by a linear equation. More complicated models will be considered in later sections. When you are given a set of data points, you should first determine whether a straight line would be a good model for the data. This can be done graphically by making a scatter plot of the data, as shown in Figures 1.5-1a and 1.5-1b. Visual inspection suggests that the data points in Figure 1.5-1a are approximately linear but that those in Figure 1.5-1b are not. So a line would be a good model for the data points in Figure 1.5-1a, but for those in Figure 1.5-1b a line is not a good model. y y x x Figure 1.5-1a Figure 1.5-1b You can also determine whether a line is a good model for a given set of data points, without graphing, by using finite differences. To understand y 3x 1, the idea, consider the equation whose graph is known to be a line. Consider the table of values shown on the next page and look at the difference between each y-entry and the preceding one. The differences are the same; all of them are equal to the slope of the line y 3x 1. This fact suggests that if the successive differences of the y-coordinates of the data points are approximately equal, then a line should be a good model for the data. 44 Chapter 1 Number Patterns x 1 2 3 4 5 y 3x 1 2 5 8 11 14 Difference 5 2 3 8 5 3 11 8 3 14 11 3 Example 1 Linear Data Estimated cash flows from a company over the five-year period 1988– 1992 are shown in the table. Year 1988 1989 1990 1991 1992 Cash flow per share ($) 2.38 2.79 3.23 3.64 4.06 Determine whether a line would be a good model for this data. Use two different methods. a. Calculate the finite differences for the data points. b. Draw a scatter plot of the data. Solution Year Cash Flow Differences a. Subtract each cash flow from the preceding one and record the difference, as shown in Figure 1.5-2a. 0.41 0.44 0.41 0.42 Because the differences are approximately equal, a line is a good model for this data. b. Let x 0 correspond to 1988. The scatter plot for the data points is shown in Figure 1.5-2b, where the points appear to be linear. Therefore, a line is a reasonable model. ■ 2.38 2.79 3.23 3.64 4.06 Figure 1.5-2a 1988 1989 1990 1991 1992 5 Once it has been determined that a line would be a good model for a set of data points, there are several ways to determine an appropriate model. The simplest way is to choose two of the data points and find the equation of the line that includes the points. This may require some experimenting to see which two points appear to produce a line that fits the data well. The number of data points above the line should balance with the number of data points below the line. Of course, there are many choices of two points and many possible lines that model the data. So there must be some way of determining which line fits the data best. Modeling Terminology Suppose (x, r) is a data point and that the corresponding point on the is called a residual. Residuals model is (x, y). Then the difference r y 0 0 Figure 1.5-2b 5 Section 1.5 Linear Models 45 are a measure of the error between the actual value of the data, r, and the value y given by the model. Graphically, the residual is the vertical distance between the data point (x, r) and the model point (x, y), as shown in Figure 1.5-3. y (x, r) Data point Residual r − y (x, y) Model point x Figure 1.5-3 The residual represents a directed distance that is positive when the data point is above the model point and negative when the data point is below the model point. When the sum of the residuals is 0, which indicates that the positive and negative errors cancel out each other, the model is probably a reasonable one. However, this is not always enough to determine which of several models is best because their residuals may all have the same sum. Consequently, to find which model among several fits the data best, use the sum of the squares of the residuals because this sum has no negative terms and no canceling. Using the sum of the squares as a measure of accuracy has the effect of emphasizing large errors, those with absolute value greater than 1, because the square is greater than the residual. It minimizes small errors, those with absolute value less than 1, because the square is less than the residual. Example 2 Modeling Data The data below shows the weekly amount spent on advertising and the weekly sales revenue of a small store over a seven-week period. Advertising Expenditure x (in hundreds of dollars) Sales revenue (in thousands of dollars Find two models for the data, each determined by a pair of data points. Then use residuals to see which model best fits the data. 46 Chapter 1 Number Patterns Solution Let the two models be denoted as A and B. For Model A, use the points (1, 2) and (3, 3). The slope of the line through these points is m 3 2 3 1 1 2 0.5 The equation of the line through (1, 2) and (3, 3) is y 2 0.5 x 1 2 1 y 0.5x 1.5 Model A is shown in Figure 1.5-4 and its residuals are shown in the table below. Notice that the sum of the squared residuals is 2. Data point (x, r) 0, 1 1, 2 2, 2 3, 3 4, 3 5, 5 6 Model point (x, y) 0, 1.5 1, 2 2 1 2, 2.5 3, 3 2 1 4, 3.5 5, 4 2 1 6, 4.5 1 1 1 1 2 2 2 2 Residual r y Squared residual (r y)2 0.5 0 0.5 0 0.5 1 0.5 Sums 0 0.25 0.25 0.25 0.25 0.25 1.25 0.25 2.00 For Model B, use the point (1, 2) and (6, 5). The slope of the line through these points is 5 2 6 1 3 5 y 2 0.6˛ 0.6 and its equation is x 1 1 2 or y 0.6 x 1.4 Figure 1.5-5 shows the graph of Model B and the table below shows its residuals. The sum of the squared residuals is 1.56. Data point (x, r) 0, 1 1, 2 2, 2 3, 3 4, 3 5, 5 6 Model point (x, y) Residual r y Squared residual (r y)2 1 1 1 1 1 0, 1.4 1, 2 2 1 2, 2.6 3, 3.2 4, 3.8 5, 4.4 6.4 0 0.6 0.2 0.8 0.6 0 1.4 Sums 0.16 0.16 0.36 .04 0.64 0.36 0.16 1.56 Because this sum is smaller that the sum for Model A, conclude that Model B is the better of the two models Advertising 6 Figure 1.5-4 2 4 Advertising 6 Figure 1.5-5 NOTE If needed, review how to compute linear regression equations in the Technology Appendix. Section 1.5 Linear Models 47 Least–Squares Reg
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ression Lines It can be proved that for any set of data there is one and only one line for which the sum of the squares of the residuals is as small as possible. Such a line is called the least–squares regression line, and the computational process for finding it is called linear regression. Most graphing calculators have the linear regression process built-in. For example, Figure 1.5-6 shows the approximate least–squares regression line for the data in Example 2. y 0.679x 0.964 The sum of the squared residuals for this model is approximately 1.107, slightly less than the corresponding sum for Model B in Example 2. The number r in Figure 1.5-6, which is called the correlation coefficient, is a statistical measure of how well the least–squares regression line fits 1 the data points. The value of r is always between and 1, the closer the r absolute value of r is to 1, the better the fit. When the fit is perfect: all the data points are on the regression line. Conversely, a correlation coefficient near 0 indicates a poor fit. 1, 0 0 r2 is called the coefficient of determination. It is the proportion of variation in y that can be attributed to a linear relationship between x and y in the data. Example 3 Modeling Data A circle can be circumscribed around any regular polygon. The lengths of the radii of the circumscribed circles around regular polygons whose sides have length of one unit are given as follows. Number of sides Radius 3 4 5 6 7 8 9 0.577 0.707 0.851 1.00 1.152 1.306 1.462 a. Draw a scatterplot. b. Calculate the finite differences for the data points. c. Find the model that best fits the data using the regression feature on a calculator. d. What does the correlation coefficient indicate about the data? Figure 1.5-6 Technology Tip Many calculators have a List command that ¢ can be entered into the label cell of a list. This will automatically calculate the differences between the items of the list specified. For example, List(L2) produces a list of differences between the items of list L2. ¢ 48 Chapter 1 Number Patterns Solution a. A scatter plot of the data is shown in Figure 1.5-7a. b. Subtract each radius from the preceding one and record the differences in a list. Notice that the differences are approximately equal, as shown in Figure 1.5-7b. 1.5 0 0 10 Figure 1.5-7a Figure 1.5-7b c. Use the linear regression feature to obtain Figure 1.5-7c, which shows that the least–squares regression line is approximately y 0.15x 0.12. d. The correlation coefficient, r 0.9996308197, is very close to 1, which indicates that this linear model is a very good fit for the data. ■ Although only linear models are constructed in this section, you should always allow for the possibility that a linear model may not be the best choice for certain data. The least–squares regression line may give a reasonable model, which is the line that fits the data best, but there may be a nonlinear equation that is an even better model for the data. Polynomial models, for example, are presented in Chapter 4. Figure 1.5-7c In addition to finding finite differences and a scatter plot of the data, another way to check that a linear model is appropriate is to construct a x, r y , scatter plot of the residuals. In other words, plot the points 2 x, y is the corresponding point on the where model. The general rule is two fold: is a data point and x, r 1 1 2 1 2 Use a linear model when the scatter plot of the residuals shows no obvious pattern, as shown in Figure 1.5-8. Use a nonlinear model when the scatter plot of the residuals has a pattern, as shown in Figure 1.5-9. r y x x Figure 1.5-8 Figure 1.5-9 Technology Tip The regression equation is stored in a variable that is usually called RegEQ each time the regression coefficients are calculated. The residuals are stored in a variable called RESID each time a regression is performed. See the Technology Appendix for specific instructions for graphing a regression line and residuals. Section 1.5 Linear Models 49 Example 4 Linear Regression and Residuals A local resident owns an espresso cart and has asked you to provide an analysis based on last summer’s data. To simplify things, only data for Mondays is provided. The data includes the amount the workers were paid each day, the number of cups sold, the cost of materials, and the total revenue for the day. The owner also must spend $40 each operating day on rent for her location and payment toward a business loan. Sales taxes have been removed from the data, so you need not consider them, and amounts have been rounded to the nearest dollar. Date Salaries ($) Cups sold Material cost ($) Total revenue ($) June 02 June 09 June 16 June 23 June 30 July 07 July 14 July 21 July 28 August 04 August 11 August 18 August 25 68 60 66 63 63 59 57 61 64 58 65 57 64 112 88 81 112 87 105 116 122 100 80 96 108 93 55 42 33 49 38 45 49 52 48 36 42 52 47 202 119 125 188 147 159 165 178 193 112 158 162 166 a. Find a linear regression model for the daily revenue as a function of the number of cups sold. b. Use a scatter plot of the residuals to determine if the linear model is a good fit for revenue. c. Find a linear regression model for the daily cost as a function of the number of cups sold. Be sure to include the pay for the workers, the fixed daily cost, and the cost of the material. d. Draw a scatter plot of the residuals to determine if the proposed model is a good fit for cost. e. Find the break-even point, that is, when revenue is equal to cost. 50 Chapter 1 Number Patterns Solution a. Enter two lists in your calculator, using the column labeled “Cups Sold” and “Total Revenue” in the chart. That is, the data points are (112, 202), (88, 119), and so on. Next, use the linear regression function on these lists to approximate the least–squares regression line. y 1.586x 0.895 Store its equation as 1.5-10a and 1.5-10b. y1 in the equation memory, as shown in Figure Figure 1.5-10a Figure 1.5-10b b. To obtain a scatter plot of the residuals for the least squares regression line for revenue, plot the points whose first coordinates are given by the CUPS list and whose second coordinates are given by the RESID, the variable that holds the residuals each time a regression is performed. 35 75 −35 125 Figure 1.5-11a Figure 1.5-11b c. First, create a new list that shows the total cost each day. For June 2, Salaries Material Cost Rent/Loan Costs Daily Cost $68 $55 $40 $163 Compute the total daily cost for each date, as shown in Figure 1.5-12a, where the salaries list is called PAY, the materials cost list is called COST, and TOTCO is the total daily cost list. To find a model for the total daily cost as a function of cups sold, use the data points given by the lists CUPS and TOTCO with the regression feature. Find the closest approximate least–squares regression line. y 0.395x 107.659 Technology Tip Placing RESID into Ylist will use the last computed regression’s residual values in a scatter plot. Figure 1.5-12a Technology Tip A formula can be placed in the upper cell of a list to perform the operation on all the elements of the list. Section 1.5 Linear Models 51 Store this equation as 1.5-12b. y2 in the equation memory, as shown in Figure Figure 1.5-12b d. Use the same procedure as in part b to obtain the scatter plot of the residuals for cost shown in Figure 1.5-12c. It shows no obvious pattern, which again indicates that a linear model is a good choice for this data. 30 80 30 Figure 1.5-12c 125 320 60 40 e. The break-even point occurs when revenue is equal to cost. Plot the revenue equation found in part a and the cost equation found in part c on the same screen, and find the x-coordinate of their intersection (shown in Figure 1.5-13). Since 89.6 cups cannot be sold, 90 cups a day must be sold to break even. ■ 180 Figure 1.5-13 Example 5 Prediction from a Model The total number of farm workers (in millions) in selected years is shown in the following table. Year Workers Year Workers Year Workers 1900 1920 1930 1940 29.030 42.206 48.686 51.742 1950 1960 1970 1980 59.230 67.990 79.802 105.060 1985 1990 1994 106.210 117.490 120.380 a. Use linear regression to find an equation that models the data. Use the equation to estimate the number of farm workers in 1975 and in 2000. 52 Chapter 1 Number Patterns b. According to the model, when will the number of workers be 150 million? c. Is a line the best model for the data? Solution x 0 Let correspond to 1900 and enter the data into two lists. Perform linear regression on the data, and display the scatter plot of the data together with the graph of the least–squares regression line. 100 y 1.0116x 18.3315 Figure 1.5-14 As suggested by Figure 1.5-14, the regression line provides a reasonable model for approximating the number of farm workers in a given year. a. If If x 75, x 100, then then y 1.0116 1 y 1.0116 1 18.3315 94.202 75 . 2 18.3315 119.492 . 100 2 Therefore, there were approximately 94,202,000 farm workers in 1975 and 119,492,000 in 2000. 130 0 0 15 −5 −15 Figure 1.5-15 b. To determine when the number of workers will be 150 million, solve the regression equation when y 150. 1.0116x 18.3315 150 100 1.0116x 131.6685 x 131.6685 1.0116 130.159 There will be 150 million farm workers in approximately 2030. c. As shown in Figure 1.5-15, there seems to be a pattern in the residuals, so there is a better, nonlinear model for the data. Nonlinear models are discussed in Section 4.3.A. ■ Correlation and Slope The correlation coefficient, r, always has the same sign as the slope of the least squares regression line. So when r is negative, the regression line slants downward from left to right. In other words, as x increases, y decreases. In such cases, we say that the data has a negative correlation. When r is positive, the regression line slopes upward from left to right, and the data is said to have a positive correlation. As x increases, y also increases. When r is close to 0 (regardless of sign), there is no correlation between the quanti
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ties. Exercises 1.5 1. a. In Example 2, find the equation of the line through the data points (1, 2) and (5, 5). b. Compute the sum of the squares of the errors for this line. Is it a better model than any of the models in the example? Why? 2. The linear model in Example 5 is the least squares regression line with coefficients rounded. Find the correlation coefficient for this model. 3. a. In Example 5, find the slope of the line through the data points for 1920 and 1994. b. Find the equation of the line through these two data points. c. Which model predicts the higher number of farm workers in 2010: the line in part b or the regression line found in Example 5? In Exercises 4–7, determine whether the given scatter plot of the data indicates that there is a positive correlation, negative correlation, or very little correlation. 4. y 5. y 6. y 7. y x x x x Section 1.5 Linear Models 53 8. The U.S. gross domestic product (GDP) is the total value of all goods and services produced in the United States. The table shows the GDP in billions x 0 of 1996 dollars. Let (Source: U.S. Bureau of Economic Analysis) a. Use a scatter plot to determine if the data correspond to 1990. appears to be linear. b. If so, is there a positive or negative correlation? Year 1990 1992 1994 1996 1998 2000 GDP $6707.9 $6880.0 $7347.7 $7813.2 $8495.7 $9318.5 In Exercises 9–13, construct a scatter plot for the data and answer these questions: a. What are the finite differences for the data? b. Do the finite differences confirm that the data is linear? If so, is there a positive or negative correlation? 9. The table shows the monthly premium (in dollars) for a term life insurance policy for a female nonsmoker. Let x represent age and y the amount of the premium. Age Premium 25 30 35 40 45 50 55 $11.57 $11.66 $11.83 $13.05 $16.18 $21.32 $29.58 10. The table shows the percent of persons in the United States below the U.S. poverty level in x 0 correspond to 1960. selected years. Let 54 Chapter 1 Number Patterns Year 1960 1965 1970 1975 1980 1985 1990 1992 1994 1996 1997 1998 1999 Percent below poverty level 22.2 17.3 12.6 12.3 13.0 14.0 13.5 14.8 14.5 13.7 13.3 12.7 11.8 11. The vapor pressure y of water depends on the temperature x, as given in the table. Temperature (C) Pressure (mm Hg) 0 10 20 30 40 50 60 70 80 90 100 4.6 9.2 17.5 31.8 55.3 92.5 149.4 233.7 355.1 525.8 760.0 12. The table shows the U.S. Census Bureau’s population data for St. Louis, Missouri in selected years. Let correspond to 1950. x 0 Year Population 1950 1970 1980 1990 2000 856,796 622,236 452,801 396,685 348,189 13. The table shows the U.S. disposable income (personal income less personal taxes) in billions of dollars. (Source: Bureau of Economic Analysis, U.S. Dept. of Commerce). Let 1990. correspond to x 0 Year 1990 1992 1994 1996 1998 1999 2000 Disposable personal income 4166.8 4613.7 5018.9 5534.7 6320.0 6618.0 7031.0 14. The table gives the annual U.S. consumption of beef and poultry, in million of pounds. (Source: U.S. Dept. of Agriculture) Year Beef Poultry 1990 24,031 22,151 1991 24,113 23,270 1992 24,261 24,394 1993 24,006 25,099 1994 25,125 25,754 1995 25,533 25,940 1996 25,875 26,614 Section 1.5 Linear Models 55 a. Make scatter plots for both beef and poultry consumption, using the actual years (1990, 1991, etc.) as x in each case. b. Without graphing, use your knowledge of slopes to determine which of the following equations models beef consumption and which one models poultry consumption. Confirm your answer by graphing. 717.46x 1,405,160 329.86x 632,699 y1 y2 15. The table at the bottom of the page gives the a. Find a linear model for this data, using x 0 to correspond to 1950. b. In the unlikely event that the linear model in part a remains valid far into the future, will there be a time when death from heart disease has been completely eliminated? If so, when would this occur? 17. The table shows the share of total U.S. household income earned by the poorest 20% of households and the share received by the wealthiest 5% of households. (Source: U.S. Census Bureau) median weekly earnings of full-time workers 25 years and older by their amount of education. (Source: U.S. Bureau of Labor Statistics) a. Make four scatter plots, one for each educational group, using to 1990. x 0 to correspond b. Four linear models are given below. Match each model with the appropriate data set. y1 y3 20.74x 392 y2 34.86x 543 y4 12.31x 238 15.17x 354 In Exercises 16–22, use the linear regression feature of your calculator to find the required model. Year Lowest 20% Top 5% 1985 1990 1995 1996 4 3.9 3.7 3.7 17.0 18.6 21.0 21.4 a. Find a linear model for the income share of the poorest 20% of households. b. Find a linear model for the income share of the wealthiest 5% of households. 16. The table shows the number of deaths per 100,000 c. What do the slopes of the two models suggest people from heart disease. of each? Year 1950 1960 1970 1980 1990 1999 Deaths 510.8 521.8 496.0 436.4 368.3 265.9 d. Assuming that these models remain accurate, will the income gap between the wealthy and the poor grow, stay about the same, or decline in the year 2000? Median Weekly Earnings By Amount of Education No High School Diploma High School Graduate Some College College Graduate $317 $321 $337 $346 $360 $378 $443 $461 $479 $490 $506 $520 $518 $535 $558 $580 $598 $621 $758 $779 $821 $860 $896 $924 Year 1996 1997 1998 1999 2000 2001 56 Chapter 1 Number Patterns 18. The table shows what percent of federal aid is given in the form of loans to students at a particular college in selected years. Year (in which school year begins) Loans (%) 1975 1978 1984 1987 1990 18 30 54 66 78 a. Find a linear model for this data, with x 0 corresponding to 1975. b. Interpret the meaning of the slope and the y-intercept. c. If the model remains accurate, what percentage of federal student aid were loans in 2000? 19. The table shows the percent of federal aid given in the form of grants or work-study programs to students at the college of Exercise 18. Year (in which school year begins) Grants and work-study (%) 1975 1978 1984 1987 1990 82 70 46 34 22 a. Find a linear model for this data, with x 0 corresponding to 1975. b. Graph the model from part a and the model from Exercise 18 on the same axes. What appears to be the trend in the federal share of financial aid to college students? c. In what year is the percent of federal aid the same for loans as for grants and work-study? b. Find a linear model for the data. c. According to the model, what was the average number of take-out meals purchased per person in 1993? in 2000? Average number of annual take-out meals per person 43 48 53 55 57 61 65 Year 1984 1986 1988 1990 1992 1994 1996 21. The table shows the median time, in months, for the Food and Drug Administration to approve a new drug after the application has been made. (Source: U.S. Food and Drug Administration) Year 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 Median time for approval 24.3 22.1 22.6 23.0 17.5 15.9 14.3 13.4 12.0 11.6 15.6 a. Make a scatter plot of the data, with x 0 20. The table gives the average number of takeout corresponding to 1990. meals per person purchased at restaurants in selected years. (Source: NPD Group’s Crest Service) a. Make a scatter plot of the data, with x 0 corresponding to 1980. b. Find a linear model for the data. c. What are the limitations of this model? Hint: What does it say about approval time in the year 2009? 22. The ordered pairs below give production (x) and consumption ( y) of primary energy in quadrillion BTUs for a sample of countries in 1995. Australia (7.29, 4.43) Brazil (4.55, 6.76) Canada (16.81, 11.72) China (35.49, 35.67) France (4.92, 9.43) Germany (5.42, 13.71) India (8.33, 10.50) Indonesia (6.65, 3.06) Iran (9.35, 3.90) Japan (3.98, 21.42) Mexico (8.15, 5.59) Poland (3.74, 3.75) Russia (39.1, 26.75) Saudi Arabia (20.34, 3.72) South Africa (6.08, 5.51) United States (69.1, 88.28) United Kingdom (10.57, 9.85) Venezuela (8.22, 2.53) a. Make a scatter plot of the data. b. Find a linear model for the data. Graph the model with the scatter plot. c. In 1995, what three countries were the world’s leading producers and consumers of energy? d. As a general trend, what does it mean if a country’s coordinates lie above the linear model? e. As a general trend, what does it mean if a f. country’s coordinates lie below the linear model? Identify any countries whose coordinates appear to differ dramatically from most of the others. 23. The table shows the winning times, in minutes, for men’s 1500-meter freestyle swimming at the Olympics in selected years. Year 1912 1924 1936 1948 1960 1972 1984 1996 Time 22.00 20.11 19.23 19.31 17.33 15.88 15.09 14.94 a. Find a linear model for this data, with corresponding to 1900. x 0 Section 1.5 Linear Models 57 e. Make a residual plot of the model. Is a linear model appropriate for the data? 24. The following table shows, for selected states, the percent of high school students in the class of 2001 who took the SAT and the average SAT math score. State Connecticut Delaware Georgia Idaho Indiana Iowa Montana Nevada New Jersey New Mexico North Dakota Ohio Pennsylvania South Carolina Washington Students who took SAT (%) Average math score 82 67 63 17 60 5 23 33 81 13 4 26 71 57 53 510 499 489 542 501 603 539 515 513 542 599 539 499 488 527 a. Make a scatter plot of the percent of students who took the SAT (x) versus the average SAT math score (y). b. Find a linear model for the data. c. What is the slope of your linear model? What does this mean in the context of the problem? d. Below is the data on four additional states. How well does the model match the actual figures for these states? State Students taking SAT (%) Average math score b. Kieren Perkins of Australia set the Olympic Oklahoma record of 14.72 minutes in 1992. How accurately did your model estimate his time? c. How long is this model likely to remain accurate? Why
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? d. Find the correlation coefficient for the model. Arizona Alaska Hawaii 8 34 51 52 561 525 510 515 58 Chapter 1 Number Patterns 1.6 Geometric Sequences Objectives • Recognize a geometric sequence • Find a common ratio • Graph a geometric sequence • Write a geometric sequence recursively and explicitly • Find partial sums of a geometric sequence Recall that in an arithmetic sequence, each term is obtained from the preceding term by adding a constant, d. A geometric sequence, which is sometimes called a geometric progression, is a sequence in which terms are found by multiplying a preceding term by a nonzero constant. Like an arithmetic sequence, where the difference between consecutive terms is the constant d, the quotient of consecutive terms in a geometric sequence is the constant r. The constant r is called the common ratio of the geometric sequence. Example 1 Recognizing a Geometric Sequence Are the following sequences geometric? If so, what is the common ratio? Write each sequence as a recursive function. 3, 9, 27, 81, p a. 5 6 Solution a. The sequence 5 u2 u1 3, 9, 27, 81, p 9 3 3 6 u3 u2 b 16 , e p f is geometric with a common ratio of 3. 27 9 3 u4 u3 81 27 3 Because each term is obtained by multiplying the previous term by 3, the sequence may be denoted as a recursive function. u1 3 and un 3un1 for n 2 b. The sequence 16 , e p f is geometric with a common ratio of 1 2 . u2 u1 5 4 5 2 1 2 u3 u2 5 8 5 4 1 2 u4 u3 5 16 5 8 1 2 Each term is obtained by multiplying the previous term by which , 1 2 gives the following recursive function. u1 5 2 ˛ and un 1 2 ˛un1 for n 2 is a geometric sequence with common ratio r, then for each un6 If the term preceding 5 un is un1 and un un1 r, or equivalently, un run1. ■ n 2 Section 1.6 Geometric Sequences 59 Recursive Form of a Geometric Sequence In a geometric sequence { un }, run1 and some nonzero constant r and all n 2. un for some u1 y 600 500 400 300 200 100 Example 2 Graph of a Geometric Sequence 8. Find the common ratio of the geometric sequence with List the first five terms of the sequence, write the sequence as a recursive function, and graph the function. 2 and u1 u2 Solution Because the sequence is geometric, the common ratio is Therefore, the sequence begins with sive function is given below. 5 x 2, 8, 32, 128, 512, p 2 4 6 Figure 1.6-1 un 4un1, with u1 2 The graph of the function is shown in Figure 1.6-1. u2 u1 8 2 4. and the recur- 6 ■ Notice that the graph of the geometric sequence in Example 2 does not appear to be linear. If the points were connected, the graph would be an exponential function, which is discussed in Chapter 5. Explicit Form of a Geometric Sequence Geometric sequences can also be expressed in a form where the value of the sequence can be determined by the position of the term. Example 3 Writing a Geometric Sequence in Explicit Form Confirm that the sequence defined by expressed as form. 7 n1 un 2 1 2 can also be by listing the first seven terms produced by each with un u1 2un1 7 Solution Using the recursive function, the sequence is 7 u1 u2 u3 u4 u5 u6 2 7 2 14 2 7 2 2 7 22 2 7 23 2 7 24 2 7 25 u1 u2 u3 u4 u5 u6 u7 28 3 56 4 112 5 224 6 448 60 Chapter 1 Number Patterns is u3 7 22, Notice that by the common ratio twice, and that of the sequence multiplied by the common ratio three times. product of general, which is the first term of the sequence multiplied which is the first term un is the power. In and the common ratio, r, raised to the 7 23, n 1 u1 u4 is 1 2 Figure 1.6-2a The table in Figure 1.6-2b confirms the apparent equality of the two functions. ■ 7 2n1. un Figure 1.6-2b Explicit Form of a Geometric Sequence implies that The recursive formula for u2 u3 u4 u5 n 2, 3, 4, p u1r u2r u3r u4r u1r 2 u1r 2 u1r 3 1 1 1 r u1r2 r u1r3 2 r u1r 4 2 } is a geometric sequence with common ratio r, then for If { all un n 1, un u1r n1 Example 4 Explicit Form of a Geometric Sequence Write the explicit form of a geometric sequence where the first two terms , and find the first five terms of the sequence. are 2 and 2 5 Solution The common ratio is . Using the explicit form, the geometric sequence can be written as The sequence begins un ˛u1rn1 n1 2 1 2a 1 5b 2, 2 5 , 2 52 , 2 53 , 2 54, p ■ Example 5 Explicit Form of a Geometric Sequence The fourth and ninth terms of a geometric sequence are 20 and the explicit form of the sequence. 640. Find Section 1.6 Geometric Sequences 61 Solution The fourth term can be written as The ninth term can be written as u9 u4 u1rn1, u1rn1, or 20 u1r3. 640 u1r8. or The ratio of the ninth term to the fourth term can be used to find r. u1r8 u1r3 640 20 r5 32 r 2 Substitute 2 for r into the equation defining the fourth term. u11 2 3 20 20 8 2 u1 Thus, un u1 r n1 5 2 ˛ 1 2 2 n1. Partial Sums 5 2 ■ If the common ratio r of a geometric sequence is the number 1, then 1n1u1 u1, u2, u3, p . Therefore, the sequence is just the constant sequence positive integer k, the kth partial sum of this constant sequence is for every n 1. un For any u1 u1 p u1 ku1 ⎫⎪⎪⎪⎬⎪⎪⎪⎭ k terms In other words, the kth partial sum of a constant sequence is just k times the constant. If a geometric sequence is not constant (that is, then its partial sums are given by the following formula. r 1 , 2 Partial Sums of a Geometric Sequence The kth partial sum of the geometric sequence { mon ratio r 1 is un } with com- k a n1 un u1a 1 r k 1 r b Proof If S denotes the kth partial sum, then using the formula for the nth term of a geometric sequence derives the expression for S. p uk S u1 u2 u1 S rS, u1r u1r2 p u1r k1 as shown on the next page. Use this equation to compute 62 Chapter 1 Number Patterns u1r u1r2 p u1r k1 S u1 rS u1r u1r2 p u1r k1 u1r k u1r k 1 r k S rS u1 1 r S u11 2 2 both sides of the last equation can be divided by 1 r 1, Because complete the proof. S u11 1 r k 1 r 2 u1a 1 r k 1 r b Example 6 Partial Sum Find the sum 3 2 3 4 3 8 3 16 3 32 3 64 3 128 3 256 3 512 . 1 r to ■ Solution This is the ninth partial sum of the geometric sequence e 3 1 2b n1 , f 2 a where the common ratio is r 1 2 . The formula in the box shows that n1 9 a n1 3 2 a 1 2b u1˛a 1 r9 1 r b 3 2b ˛ a 1 a 1 9 1 2b 1 2b a 3 2b a 1 9 1 2b a 3 2 3 2b a 2 ˛a 3b 1 1 ˛a 29b D 1 1 29 T 1 1 § 512 513 512 ¥ ■ Example 7 calculates the distance traveled by the ball discussed in Section 1.2 Example 4 when it hits the ground for the seventh time by using a partial sum of a geometric sequence. Example 7 Application of Partial Sum A ball is dropped from a height of 9 feet. It hits the ground and bounces to a height of 6 feet. It continues to bounce up and down. On each bounce it rises to 2 3 of the height of the previous bounce. How far has the ball traveled (both up and down) when it hits the ground for the seventh time? Section 1.6 Geometric Sequences 63 Solution First consider how far the ball travels on each bounce. On the first bounce, it rises 6 feet and falls 6 feets for a total of 12 feet. On the second bounce it rises and falls of the previous height, i.e., it travels of 12 feet. The 2 3 2 3 distance traveled is a geometric sequence with u1 12 and r 2 3 . If un denotes the distance traveled on the nth bounce, then un 12 n1 ˛ 2 3b a So un6 5 is a geometric sequence with common ratio r 2 3b a . When the ball hits the ground for the seventh time, it has completed six bounces. Therefore, the total distance it has traveled is the distance it was originally dropped, 9 feet, plus the distance traveled in six bounces. 9 u1 u2 u3 u4 u5 u6 6 9 a n1 un 9 u1˛a 9 12 6 1 r6 1 r b 2 a 3b 1 2 3 1 § ¥ 41.84 feet ■ Exercises 1.6 In Exercises 1–8, determine whether the sequence is arithmetic, geometric, or neither. 1. 2, 7, 12, 17, 22, p 2. 2, 6, 18, 54, 162, p In Exercises 9–14, the first term, and the common ratio, r, of a geometric sequence are given. Find the sixth term and the recursive and explicit formulas for the nth term. u1, 3. 13, 13 2 , 13 4 , 13 8 , p 4. 1, 1 2 , 0, , p 1 2 5. 50, 48, 46, 44, p 6. 2, 3, 9 2 , 27 4 , 81 8 , p 7. 3 16 , p 8. 6, 3.7, 1.4, 9, 3.2, p 9. u1 5, r 2 10. u1 1, r 2 11. u1 4, r 1 4 12. u1 6, r 2 3 13. u1 10, r 1 2 14. u1 p, r 1 5 In Exercises 15–18, find the kth partial sum of the geometric sequence with common ratio r. un6 5 5, r 1 2 15. k 6, u1 16. k 8, u1 9, r 1 3 64 Chapter 1 Number Patterns 17. k 7, u2 6, r 2 18. k 9, u2 6, r 1 4 In Exercises 19–22, show that the given sequence is geometric and find the common ratio. 19. n 1 2b ea 21. 5n2 5 6 f 20. 22. 23n 6 n 2 3 6 5 5 In Exercises 23–28, use the given information about and recursive the geometric sequence un6 5 . and explicit formulas for un to find u5 23. u1 256, u2 64 24. u1 1 6 , u2 1 18 25. u2 4, u5 1 16 27. u4 4 5 , r 2 5 26. u3 4, u6 32 28. u2 6, u7 192 In Exercises 29–34, find the sum. 29. 31. 33. 7 a n1 2n n 9 a n1 a 1 3b j1 6 a j1 3 2b 4˛a 30. 32. 34. 6 a k1 k 1 2b 3˛a 5 a n1 8 a t1 5 3n1 t1 0.9 6˛1 2 35. For 1987–1998, the annual revenue per share in year n of a company’s stock are approximated by un 1.191 a. Show that the sequence represents 1987. is a geometric 1.71 where n, 2 1 n 7 un6 5 sequence. b. Approximate the total revenues per share for the period 1987–1998. 36. The annual dividends per share of a company’s stock from 1989 through 1998 are approximated , where by the sequence 1989 and 1.1999 a. Show that the sequence bn6 5 0.0228 is a geometric corresponds to bn 1 n 9 n. 2 bn6 5 sequence. b. Approximate the total dividends per share for the period 1989–1998. 37. A ball is dropped from a height of 8 feet. On each bounce it rises to half its previous height. When the ball hits the ground for the seventh time, how far has it traveled? 38. A ball is dropped from a height of 10 feet. On each bounce it rises to 45% of its previous height. When it hits the ground for the tenth time, how far has it traveled? 39. If you are paid a salary of 1¢ on the second day, and your salary March, continues to double each day, how much will you earn in the month of March? on the first day of 2¢ 40. Starting with yo
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ur parents, how many ancestors do you have for the preceding ten generations? 41. A car that sold for $8000 depreciates in value 25% each year. What is it worth after five years? 42. A vacuum pump removes 60% of the air in a container at each stroke. What percentage of the original amount of air remains after six strokes? 43. Critical Thinking Suppose un6 sequence with common ratio un arithmetic sequence with common difference log r. is a geometric r 7 0 and each is an Show that the sequence log un6 7 0. 5 5 44. Critical Thinking Suppose is an arithmetic un6 5 sequence with common difference d. Let C be any positive number. Show that the sequence is a 5 Cd. geometric sequence with common ratio Cun 6 45. Critical Thinking In the geometric sequence 1, 2, 4, 8, 16, p sum of all preceding terms. show that each term is 1 plus the 46. Critical Thinking In the geometric sequence 2, 6, 18, 54, p sum of 1 and all preceding terms. show that each term is twice the of the outstanding balance. If the balance is 47. Critical Thinking The minimum monthly payment for a certain bank credit card is the larger of $5 or 1 25 less than $5, then the entire balance is due. If you make only the minimum payment each month, how long will it take to pay off a balance of $200 (excluding any interest that might accrue)? Important Concepts Section 1.1 Section 1.2 Section 1.3 Section 1.4 Section 1.5 Integers, whole, natural, and real numbers . . . . . . 3 Rational and irrational numbers . . . . . . . . . . . . . . . 4 Real number line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Coordinate plane, x-axis, y-axis, quadrants . . . . . . 5 Scatter plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Domain and range . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Function notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Definition of a sequence . . . . . . . . . . . . . . . . . . . . . 13 Sequence notation . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Recursively defined sequence . . . . . . . . . . . . . . . . 15 Recursive form of an arithmetic sequence . . . . . . 22 Explicit form of an arithmetic sequence . . . . . . . . 23 Summation notation . . . . . . . . . . . . . . . . . . . . . . . . 25 Partial sums of an arithmetic sequence . . . . . . . . . 27 Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Solution of an equation . . . . . . . . . . . . . . . . . . . . . . 30 Slope of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Properties of slope . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Slope–intercept form . . . . . . . . . . . . . . . . . . . . . . . . 33 Connection between arithmetic sequences and lines . . . . . . . . . . . . . . . . . . . . . . . 34 Point-slope form . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Vertical and horizontal lines . . . . . . . . . . . . . . . . . . 37 Parallel and perpendicular lines . . . . . . . . . . . . . . 38 Standard form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Mathematical model . . . . . . . . . . . . . . . . . . . . . . . . 43 Finite differences . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Residual . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Least–squares regression line . . . . . . . . . . . . . . . . . 47 Correlation coefficient . . . . . . . . . . . . . . . . . . . . . . . 47 Correlation and slope . . . . . . . . . . . . . . . . . . . . . . . 52 Section 1.6 Recursive form of a geometric sequence . . . . . . . . 59 Explicit form of a geometric sequence . . . . . . . . . . 60 Partial sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 65 66 Chapter Review Important Facts and Formulas A sequence is an ordered list of numbers. A sequence is defined recursively if the first term is given and there is a method of determining the nth term by using the terms that precede it. A sequence is defined explicitly if terms are determined by their position. An arithmetic sequence is a sequence in which the difference between each term and the preceding term is a constant d. Facts about an arithmetic sequence un6 5 with common difference d: • the recursive form of the sequence is un1 • the explicit form of the sequence is n 2. d for un un • the kth partial sum is k a n1 un k 2 1 u1 n 1 1 d. 2 u1 uk2 and x2, y22 1 is given by The slope of a line that passes through y2 x2 m ¢y ¢x x1, y12 1 y1 x1 . The slope-intercept form of the equation of a line is where m is the slope and b is the y-intercept. y mx b, The point-slope form of the equation of a line is where m is the slope and y y1 m x x12 1 , x1, y12 1 is a given point on the line. Ax By C, where The standard form of the equation of a line is A, B, and C are integers. The equation of a vertical line has the form x h. The equation of a horizontal line has the form y k. Parallel lines have equal slopes. The product of the slopes of perpendicular lines is 1. The difference between an actual data value and a predicted data value is called a residual. The correlation coefficient always has the same sign as the slope of the least squares regression line. A geometric sequence is a sequence in which terms are found by multiplying a preceding term by a nonzero constant r. Facts about the geometric sequence with common ratio r 0: 5 un6 run1 un rn1u1 • the recursive form is for • the explicit form is un or k the kth partial sum is a n1 un un • if r 1, n 2. u1r n1 u1 a 1 r k 1 r b Chapter Review 67 Review Exercises In Exercises 1–10, identify the smallest subset of the real numbers—natural numbers, whole numbers, integers, rational numbers, or irrational numbers— that contains the given number. Section 1.1 1. 23 2. 0.255 6. 3 7. 2121 3. e 8. 4 9 4. 11 9. 5 5. 0 10. 0.255 11. List two real numbers that are not rational numbers. In Exercises 12–15, which sets of points represent a function? Why? 12. 13. 14. 15. 51 51 51 51 , 2, 3 2 2, 3 , 2, 3 2 2, 3 3, 4 , 2 2, 4 , 1 3, 3 , 2 3, 5 1 2 4, 5 4, 5 , 5, 6 6, 7 , 2 1 2, 6 , 2 1 , 2 1 4, 3 , 5, 3 6, 3 , 2 1 5, 6 , 2 1 , 2 1 26 2, 7 26 6, 7 26 26 16. Let f be the function given by the rule f following table. 7 2x. Complete the 17. What is the domain of the function g given by g 2t 2 t 3 ? t 2 1 18. If f x 1 2 0 3 x 0 2x 3 7, then f 7 1 2 f 4 2 1 . 19. What is the domain of the function given by g r 2r 4 2r 2? 20. What is the domain of the function f x 1 2 2 1 2x 2 ? 21. The radius of an oil spill (in meters) is 50 times the square root of the time t (in hours). a. Write the rule of a function f that gives the radius of the spill at time t. b. Write the rule of a function g that gives the area of the spill at time t. c. What are the radius and area of the spill after 9 hours? d. When will the spill have an area of 100,000 square meters? 22. The function whose graph is shown below gives the amount of money (in millions of dollars) spent on tickets for major concerts in selected years. (Source: Pollstar) s n o i l l i M 1500 1200 900 600 0 1990 1991 1992 1993 1994 1995 1996 68 Chapter Review a. What is the domain of the function? b. What is the approximate range of the function? c. Over what one-year interval is the rate of change the largest? Use the graph of the function f in the figure below to answer Exercises 23–26. y 1 1 f x 23. What is the domain of f ? 24. What is the range of f ? 25. Find all numbers x such that f x 1 . 2 1 x 1 26. Find a number x such that f possible.) 1 6 f x 1 2 2 . (Many correct answers are Use the graph of the function f in the figure to answer Exercises 27–33. y 3 2 1 f −2 −1−1 −2 −3 −5 −4 −3 x 1 2 3 4 5 6 27. What is the domain of f ? 28. 3 f 1 2 31. True or false: 2f 32. True or false: 3f 2 2 2 2 1 1 29. 2 2 f 1 2 30 33. True or false: f 3 x 1 2 for exactly one number x. Section 1.2 34. The population of Gallatin is growing at the rate of 2.75% per year. The present population is 20,000. Find a recursive sequence that represents 5910ac01_1-75 9/21/05 2:26 PM Page 69 Chapter Review 69 Gallatin’s population each year. Represent the nth term of the sequence both explicitly and recursively. Find the first seven terms of the sequence. 35. Roberta had $1525 in a savings account 2 years ago. What will be the value of her account 1 year from now, assuming that no deposits or withdrawals are made and the account earns 6.9% interest compounded annually? Find the solution using both a recursive and an explicit formula. 36. Suppose that $3,000 is invested at 6.5% annual interest, compounded monthly. a. What is the balance after 6 years? b. Suppose $150 is added to the account every month. What is the balance after 6 years? 37. The “biological” specimen Geomeuricus sequencius is 5 centimeters long when born. On the second day it grows 3 centimeters. The third day it grows 1.8 centimeters, and on each following day it grows 60% of the previous day’s growth. What is its length after two weeks? What is the maximum length that it could grow? In Exercises 38–39, let 1 25 A un 38. a. Show that 1 b. Show that the first ten terms of Exercises 30–32 Section 1.2) 1. and u1 u2 1 25 n B A 2n25 n B un6 5 are Fibonacci numbers. (See 39. a. For the ninth term, compute the ratio un un1 b. As n gets large, what number does the ratio approach? This number is also referred to as the “golden ratio.” This ratio is believed to have been used in the construction of the Great Pyramid in Egypt, where the ratio equals the sum of the areas of the four face triangles divided by the total surface area. 40. For the sequence un 3un1 1 with u1 1.5 , write the first four terms. 41. For the sequence un 3un1 2 with u1 4, write the first five terms. 42. For the sequence un 3un1 with u1 1 9 , write the first four terms. Section 1.3 In Exercises 43–46, find a formula for metic.
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un ; assume that the sequence is arith- 43. u1 3 44. u2 4 and the common difference is –6. and the common difference is 3. 45. u1 5 and u3 7. 46. u3 2 and u7 1. 70 Chapter Review 47. Find the 12th partial sum of the arithmetic sequence with u12 16. u1 3 and 48. Find numbers b, c, and d such that 8, b, c, d, 23 are the first five terms of an arithmetic sequence. Section 1.4 49. The national unemployment rates for 1990–1996 were as follows. (Source: U.S. Department of Labor, Bureau of Labor Statistics) Year 1990 1991 1992 1993 1994 1995 1996 Rate (%) 5.6 6.8 7.5 6.9 6.1 5.6 5.4 Sketch a scatter plot and a line graph for the data, letting to 1990. x 0 correspond 50. The table shows the average speed (mph) of the winning car in the Indianapolis 500 race in selected years. Year 1980 1982 1984 1986 1988 1990 1992 1994 1996 Speed (mph) 143 162 164 171 145 186 134 161 148 Sketch a scatter plot and a line graph for these data, letting correspond to 1980. x 0 51. a. What is the y-intercept of the graph of the line defined by y x x 2 5 3 5 ? b. What is the slope of the line? 52. Find the equation of the line passing through (1, 3) and (2, 5). 53. Find the equation of the line passing through 2, 1 1 2 with slope 3. 54. Find the equation of the line that crosses the y-axis at perpendicular to the line 2y x 5. y 1 and is 55. a. Find the y-intercept of the line defined by 2x 3y 4 0. b. Find the equation of the line through (1, 3) that has the same y-intercept as the line in part a. 56. Sketch the graph of the line defined by 3x y 1 0. 57. Find the equation of the line through ( 4, 5 ) that is parallel to the line through (1, 3) and ( 4, 2 ). 58. As a balloon is launched from the ground, the wind blows it due east. The conditions are such that the balloon is ascending along a straight line with slope 1 5 . After 1 hour the balloon is 5000 ft directly above the ground. How far east has the balloon blown? 59. The point (u, v) lies on the line passing through (u, v) and the point y 5x 10 0, 10 1 ? 2 . What is the slope of the line Chapter Review 71 In Exercises 60–66, determine whether the statement is true or false. 60. The graph of x 5y 6 has y-intercept 6. 61. The graph of 2y 8 3x has y-intercept 4. 62. The lines 3x 4y 12 and 4x 3y 12 are perpendicular. 63. Slope is not defined for horizontal lines. 64. The line in the figure at right has positive slope. 65. The line in the figure does not pass through Quadrant III. 66. The y-intercept of the line in the figure is negative. y 1 1 −1−1 x 67. Which of the following lines rises most steeply from left to right? 68. Which of the following lines is not perpendicular to the line y x 5? a. c. e. a. c. e. y 4x 10 20x 2y 20 0 4x 1 y y 4 x 4 2x 2y 0 y x 1 5 a. c. e. y x y 2x 5 y 2x 5 b. d. y 3x 4 4x y 1 b. d. y x 5 x 1 y b. d. y 4x 7 y 4x 7 69. Which of the following lines does not pass through Quadrant III? 70. Let a and b be fixed real numbers. Where do the lines x a and y b intersect? a. Only at (b, a). c. These lines are parallel, so they don’t intersect. d. If a b, b. Only at (a, b). then these are the same line, so they have infinitely many points of intersection. e. Since these equations are not of the form y mx b, the graphs are not lines. 71. What is the y-intercept of the line 2x 3y 5 0? 72. For what values of k will the graphs of be perpendicular lines? 3y kx 2 0 2y x 3 0 and 73. The average life expectancy increased linearly from 62.9 years for a person born in 1940 to 75.4 years for a person born in 1990. a. Find an equation that gives the average life expectancy y of a person born in year x, with x 0 corresponding to 1940. b. Use the equation in part a to estimate the average life expectancy of a person born in 1980. 74. The population of San Diego grew in an approximately linear fashion from 334,413 in 1950 to 1,151,977 in 1994. 72 Chapter Review a. Find an equation that gives the population y of San Diego in year x, with x 0 corresponding to 1950. b. Use the equation in part a to estimate the population of San Diego in 1975 and 2000. In Exercises 75 –78, match the given information with the graph, and determine the slope of each line. y y 300 200 100 1000 800 600 400 200 3 6 a. 9 12 x x y y 300 200 100 600 400 200 1 2 b. 3 4 x x 2 4 6 c. 8 10 12 2 4 d. 6 8 75. A salesman is paid $300 per week plus $75 for each unit sold. 76. A person is paying $25 per week to repay a $300 loan. 77. A gold coin that was purchased for $300 appreciates $20 per year. 78. A CD player that was purchased for $300 depreciates $80 per year. Section 1.5 79. The table shows the monthly premium (in dollars) for a term life insurance policy for women who smoke. Age (years) 25 30 35 40 45 50 55 60 Premium 19.58 20.10 20.79 25.23 34.89 48.55 69.17 98.92 a. Make a scatter plot of the data, using x for age and y for amount of premium. b. Does the data appear to be linear? c. Calculate the finite differences. Do they confirm that the data is linear? Chapter Review 73 80. For which of the following scatter plots would a linear model be reasonable? Which sets of data show positive correlation, and which show negative correlation? y x c. y y a. d. y y x x x x b. e. Exercises 81–82 refer to the following table, which shows the percentage of jobs that are classified as managerial and the percentage of male and female employees who are managers. Year (since 1990) Managerial jobs (%) Female managers (%) Male managers (%) 8 5 2 0 1 3 5 12.32 12.31 12.00 11.83 11.79 11.43 11.09 6.28 6.85 7.21 7.45 7.53 7.65 7.73 16.81 16.67 16.09 15.64 15.52 14.79 14.10 81. a. Make scatter plots of each data set (managerial jobs, female managers, male managers). b. Match the following linear models with the correct data set. Explain your choices. y1 0.11x 7.34 y2 0.09x 11.74 y3 0.21x 15.48 82. a. According to the models in Exercise 81, is the percentage of female or male managers increasing at the greater rate? 74 Chapter Review b. Use the models to predict the percentage of female managers and the percentage of male managers in the year 2000. c. What year do the models indicate that the percentage of female managers will surpass the percentage of male managers? 83. The table shows the average hourly earnings of production workers in manufacturing. (Source: U.S. Bureau of Labor Statistics) Year 1991 1993 1995 1997 1999 2001 Hourly earnings ($) 11.18 11.74 12.37 13.17 13.90 14.84 a. Find a linear model for this data, with b. Use the model to estimate the average hourly wage in 1993 and in 2000. The actual average in 1993 was $11.74 and in 2000 it was $14.38. How far off is the model? corresponding to 1990. x 0 c. Estimate the average hourly earnings in 2004. 84. The table shows the total amount of charitable giving (in billions of dollars) in the United States during recent years. (Source: Statistical Abstract of the U.S.: 2001) Year Total charitable giving 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 101.4 107.2 110.4 116.5 119.2 124.0 138.6 153.8 172.1 190.8 203.5 a. Find a linear model for this data, with corresponding to 1990. b. Use your model to estimate the approximate total giving in 2002 and x 0 2005. c. Find the correlation coefficient for the model. d. Make a residual plot of the model. Is a linear model appropriate for the data? Section 1.6 un. In Exercises 85–88, find a formula for Assume that the sequence is geometric. Chapter Review 75 85. u1 2 and the common ratio is 3. 86. u1 5 and the common ratio is 1 2 . 87. u2 192 and u7 6. 88. u3 9 2 and u6 243 16 . 89. Find the 11th partial sum of the arithmetic sequence with common difference 2. u1 5 and 90. Find the fifth partial sum of the geometric sequence with u1 1 4 and common ratio 3. 91. Find the sixth partial sum of the geometric sequence with u1 5 and common ratio 1 2 . 92. Find numbers c and d such that 8, c, d, 27 are the first four terms of a geometric sequence. 93. Is it better to be paid $5 per day for 100 days or to be paid the first day, 10 the second day, 20 the third day, and have your salary increase in this fashion every day for 100 days? 5¢ ¢ ¢ 94. Tuition at a university is now $3000 per year and will increase $150 per year in subsequent years. If a student starts school now, spends four years as an undergraduate, three years in law school, and five years earning a Ph.D., how much tuition will she have paid 12 Infinite Geometric Series NOTE Although TI, Sharp, and HP calculators use the letter “u” to denote terms of a sequence, the letter “a” is traditionally used. ctions Calculus is a branch of mathematics that deals with changing quantities. It is based on the concept of quantities that can be approached more and more closely. There are two related branches of calculus: differential calculus and integral calculus. Differential calculus is used to calculate the change in one variable produced by change in a related variable, and integral calculus is used to calculate quantities like the total change of a quantity given its rate of change, area, and volume. The Can Do Calculus features found here and at the end of each chapter are short adventures into the world of calculus. This first Can Do Calculus explores infinite series, which is closely related to infinite sequences. It is an example of the limit process, the fundamental building block of both differential and integral calculus. Infinite Geometric Series n 0.6 2 2 1 1.2 2 2 2 Consider the sequence and let Sk denote its kth partial sum. S1 S2 S3 S4 2 2 2 2 0.6 0.6 0.6 0. themselves form a sequence. This The partial sums sequence is a function whose domain is the set of natural numbers. The sequence can be described by the following function: 1 1 1 S1, S2, S3, S4, . . . 2 1.92 2 2 0.6 2 2 0.6 3 2.352 3 2 0.6 4 2.6112 0.6 0.6 0.6 2 2 2 2 2 1 2 20 Figure 1.C-1 un 0.6 0. un1 By using the trace feature to find large values of n, the graph in Figure 1C-1 suggests that the terms of the sequence of partial sums are getting closer and closer to 3. Consequently, 2 0.6 2 0.6 2 2 0.6 3 2 0.6 2 wh
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ere 3 is said to be sum, or limit, of the infinite series, In the general case, an infinite series, or simply series, is defined to be an expression of the form a2 p an a5 a3 a4 p a1 in which each an is a real number. This series is also denoted by the sym- q bol a n1 an. NOTE a1, a2, a3, p If is a geometric sequence, then an expression of a3 a1 the form (sometimes written as q a2 p ) is called a geometric an a n1 series. 76 5 0 The partial sums of the series a2 a3 a4 p are a2 a2 a2 a3 a3 a4 a1 a1 a1 a1 a1 , a2 S1 S2 S3 S4 k 1 a1 and in general, for any Sk a3 a4 S1, S2, S3, S4, . . . p ak. of the sequence of partial sums If it happens that the terms get closer and closer to a particular real number S in such a way that the is arbitrarily close to S when k is large enough, then the partial sum series converges and has a limit. Additionally, S is called the sum of the 2 convergent series. The series con2 verges, and its sum is 3. This series is an infinite geometric series because it has a common ratio of 0.6. 4 p 3 2 2 2 0.6 0.6 0.6 0.6 Sk 2 1 2 1 2 1 1 2 Definition of Infinite Geometric Series If {an} is a geometric sequence with common ratio r, then the corresponding infinite series a3 a2 a4 a1 is called an infinite geometric series. By using the formula for the nth term of a geometric sequence, the corresponding geometric series can be expressed in the form ra1 p r r 3a1 2a1 a1 Under certain circumstances, an infinite geometric series is convergent and has a sum. Sum of an Infinite Geometric Series 66 1, If r 00 00 then the infinite geometric series a1 ra1 r 2a1 r 3a1 converges, and its sum is a1 1 r . NOTE A series that is Example 1 Sum of an Infinite Geometric Series the 1, not convergent is said to be divergent. If 0 r 0 series is divergent. Therefore, a geometric series is only convergent 6 1. when 0 r 0 Determine whether the infinite geometric series converges. q n1 a. a n1 6 2 1 2 q b. a n1 8 n 5 77 Solution a. The first term is 6 and the common ratio is 2. The sum of the first k terms is Sk a1 Sk The graph of as shown in Figure 1.C-2, does not approach a single value. In fact, the sums get larger for each subsequent term. So this series with a common ratio of 2 does not converge. It diverges. 1 2 b. is an infinite geometric series with a n1 partial sum of this series is the same as the kth partial sum of the The kth a1 . 8 5 and r 1 5 q 8 n 5 sequence 8 . n f 5 e 10 Figure 1.C-2 Sk a1 5b kb Sk is shown in Figure 1.C-3. If you use the trace feature and The graph of move beyond approximately the calculator will probably tell you that every partial sum is 2. Actually, the partial sums are slightly smaller than 2 but are rounded to 2 by the calculator. The graph gets very close to 2 as x gets larger, but it never reaches 2. According to the formula, the sum of the infinite series is n 10, 30 S a1 . ■ Figure 1.C-3 Example 1b is typical of the general case, as can be seen algebraically. a3 with common ratio r such Consider the geometric series 6 1. is the same as the kth partial sum of that geometric sequence a2 Sk and hence The kth partial sum an6 , p 0 r 0 a1 5 1 r k 1 r b Sk a1 a r As k gets very large, the number gets very close to 0 because 1 0 Consequently, when k is very large, 1 r k a1 1 r b is very close to 1 0 1 r b is very close to a1 a 1 r a1 a 1 r Sk . k k . 6 1. 0 r 0 so that Infinite geometric series provide another way of writing an infinite repeating decimal as a rational number. Example 2 Repeating Decimal as a Rational Number Express 6.8573573573 p as a rational number. 1000 0 0 3 0 –1 78 Solution First write 0.0573573573 the number as p as an infinite series: 6.8 0.0573573573 p . Then consider 0.0573 0.0000573 0.0000000573 0.0000000000573 p , which is the same as 0.0573 0.001 1 21 0.0573 2 0.001 1 2 1 2 This is a convergent geometric series with sum is 0.0573 0.001 1 0.0573 2 3 1 and 0.0573 p 2 r 0.001. 2 a1 a1 1 r 0.0573 1 0.001 0.0573 0.999 573 9990 . Therefore, 6.8573573573 p 6.8 0.0573 0.0000573 p 3 4 6.8 573 9990 573 9990 4567 666 68 10 68505 9990 Its ■ Exercises In Exercises 1–9, find the sum, or limit, of the infinite series, if it converges. In Exercises 10–15, express the repeating decimal as a rational number. 1. 4. 5. 6. 7. 8. q 1 2n q 2n 3 2. 3. a a n1 n1 1 0.5 0.25 0.125 0.0625 p 500 200 80 32 p q 3 4b a n1 a 9 313 3 13 1 1 13 p 2 12 1 1 12 1 2 p 4 216 6 316 9 916 2 p q 9. a n1 a 1 2n 1 nb 3 n 10. 0.22222 p 12. 5.4272727 p 11. 0.37373737 p 13. 85.131313 p 14. 2.1425425425 p 15. 3.7165165165 p 16. If 5 an6 difference infinite series convergent. is an arithmetic sequence with common ai a4 and each a3 a2 d 7 0 a1 p is not explain why the 7 0 17. Use the graphical approach illustrated in Example 1 to find the sum of the series in q a n1 a 1 2 b n . Does the graph get very close to the horizontal line through 1 3 ? Describe the behavior of the series. 79 C H A P T E R 2 Equations and Inequalities And the rockets’ red glare . . . Many Fourth of July firework displays are timed to coincide with patriotic music. To accomplish correct timing, each rocket must be detonated at precisely the correct height at the right moment. The time needed for a rocket to reach a specific height is the solution of an equation representing the height of the rocket as a function of time. See Exercise 24 of Section 2.3. 80 Solving Equations Graphically Interdependence of Sections Chapter Outline 2.1 2.2 Solving Quadratic Equations Algebraically 2.3 Applications of Equations 2.4 Other Types of Equations 2.5 Inequalities 2.5.A Excursion: Absolute-Value Inequalities Chapter Review can do calculus Maximum Area 2.1 2.2 > 2.3 > 2.4 > 2.5 > 2.5.A G raphing technology is useful for solving equations, but don’t be mis- led into thinking that technology is always the best tool. For example, graphing technology is useless if you do not know enough alge- bra to understand the information displayed on the screen. When exact answers are required, algebraic techniques are usually needed. This chap- ter and the next two chapters develop both algebraic and graphical techniques for solving equations in one variable. 2.1 Solving Equations Graphically Objectives • Solve equations using the intersect method • Solve equations using the x-intercept method A solution of an equation is a number that, when substituted for the vari3x 2 17 able, produces a true statement. For example, 5 is a solution of 2 17 is a true statement. To solve an equation means to 5 because 2 find all of its solutions. 3 1 Two equations are said to be equivalent if they have the same solutions. x 2 3 For example, are equivalent because 5 is the and only solution of each equation. 3x 2 17 Algebraic techniques provide exact solutions to linear and quadratic equations; however, there are no formulas that provide solutions to many other types of equations. For such equations, graphical approximation methods are practical alternatives. NOTE If necessary, review the material on graphing in the Technology Appendix. Knowledge of a graphing calculator is assumed throughout the remainder of this book. 81 82 Chapter 2 Equations and Inequalities Technology Tip Absolute value (ABS) is in the NUM submenu of the MATH menu of TI and in the NUM submenu of the OPTN menu of RUN mode of CASIO. The graphical intersection finder is labeled INTERSECT, in the CALC menu of TI and ISCT in the G-SOLVE menu of Casio. Complete Graphs A viewing window is said to display a complete graph if it shows all the important features of the graph—including all peaks, valleys, and points where it touches an axis—and suggests the general shape of portions of the graph that are not in the window. Many different windows may show a complete graph, but it usually is best to use a window small enough to show as much detail as possible. Later chapters develop algebraic facts that will enable you to know when graphs are complete. Until then try several different windows to see which, if any, appear to display a complete graph. The Intersection Method The following example illustrates a graphical method of approximating solutions of equations where both sides of an equation are algebraic expressions. Each side of the equation can be viewed as the output of a function, and the solutions of the equation represent inputs that produce equal outputs. Example 1 Solving an Equation Using the Intersect Method Solve 0 x2 4x 3 0 x3 x 6. Solution 0 0 y1 y2 and x2 4x 3 x3 x 6. Set Graph both equations on the same screen and find the x-coordinate of the point where the two graphs intersect. This coordinate can be approximated by zooming in and using the trace feature or by using a graphical intersection finder. As shown in is an approximate solution. Figure 2.1-1, x3 x 6 x2 4x 3 x 2.207 0 0 10 Letting x 2.207, 10 10 0 1 left side 2 4 2 2.207 3 2.207 0 1 4.870849 8.828 3 0 6.957151 2 0 right side 3 2.207 6 2.207 1 10.74996374 2.207 6 6.956963743 2 10 Figure 2.1-1 The difference between the value of the left side and the value of the right x 2.207. side is small. Therefore, the solution to the original equation is ■ The Intersection Method To solve an equation of the form intersection method, follow two steps. f(x) g(x) by using the 1. Graph 2. Find the x-coordinate of each point of intersection. on the same screen. and y1 y2 g(x) f(x) Section 2.1 Solving Equations Graphically 83 CAUTION Check several viewing windows to ensure that a complete graph is shown for each side of the equation. If the graphs do not intersect, then they have no common output value. Therefore, there are no real solutions to the equation. The x-Intercept Method A zero of a function f is an input that produces an output of 0. For exam x3 8 23 8 0. ple, 2 is a zero of the function 2 2 Note that 2 is also a solution of the equation In other words, the zeros of the function f are the solutions, or roots, of the equation f The zeros of a function also have a graphical interpretation. f because 1 x3 8 0. 0. x x f 1 2 1 2 Graphing Exploration 1. Gr
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aph y x4 2x2 3x 2 using a decimal window. (See Technology Tip.) Find the points where the graph crosses the x-axis. 2. Verify that the x-coordinates found in Step 1 are zeros of the function f. That is, the x-coordinates are solutions of x4 2x2 3x 2 0. y f x intersects the x-axis is of the form A point where the graph of 2 (a, 0) because every point on the x-axis has y-coordinate 0. The number a is called an x-intercept of the graph of f. In the preceding Exploration, the x-intercepts of and x 2. The x-intercepts of the graph are the zeros of the function f. x4 2x2 3x 2 were found to be x 1 x f 1 2 1 Let f be a function. If r is a real number that satisfies any of the following statements, then r satisfies all the statements. • r is a zero of the function f • r is an x-intercept of the graph of f • x r is a solution, or root, of the equation f(x) 0 Technology Tip A decimal window produces one-decimal- place values of the x-coordinates when using the trace feature. For a decimal window select ZDECIMAL or ZOOMDEC in the TI ZOOM menu and INIT in the Casio V-WINDOW menu. If needed, review the material on decimal windows in the Technology Appendix. Zeros, x-Intercepts, and Solutions x Because the x-intercepts of the graph of 2 zeros of f are solutions of the related equation x can be used to solve f 0. 1 y f 1 2 are the zeros of f, and the f the x-intercepts 0, x 1 2 84 Chapter 2 Equations and Inequalities The x-Intercept Method Follow three steps to solve an equation by the x-intercept method. 1. Write the equation in the equivalent form f(x) 0. 2. Graph y f(x) . 3. Find the x-intercepts of the graph. The x-intercepts of the graph are the real solutions of the equation. Technology Tip A standard viewing window displays 10 x 10 and 10 y 10. The ZOOM menu on most calculators contains a standard viewing window option. An advantage of using the x-intercept method is that solutions appear on the x-axis, and prior information about the range of the function is not needed. Example 2 Solving an Equation by Using the x-Intercept Method Solve the equation x5 x2 x3 5. Solution 10 Rewrite the equation so that one side is zero. 10 10 Graph y x5 x3 x2 5 in the standard viewing window. x5 x3 x2 5 0 10 Figure 2.1-2 Technology Tip Use the trace feature to find that the zero is between 1.3 and 1.5, then use zoom-in and trace features repeatedly, or use the graphical zero finder, to get a better approximation of 1.4242577. (See the Technology Tip for the location of the graphical zero finder.) Verify that 1.42 is an approximate solution by substituting into the original equation. x 1.42 ■ Technological Quirks The graphical zero finder is labeled ZERO in the TI CALC menu and ISCT in the Casio G-SOLVE menu. A graphical zero finder may fail to find some solutions of an equation, particularly when the graph of the equation touches, but does not cross, the x-axis. If the calculator does not show any x-intercepts on a graph or if its zero finder gives an error message, an alternative approach may be necessary, as illustrated in the next two examples. Example 3 Solving 2f(x) 0 by Solving f (x) 0 Solve 2x4 x2 2x 1 0. Solution y 2x4 x2 2x 1. The trace feature may display no y-value Graph for some points and the graphical zero finder may display an error message. See Figure 2.1-3 on the next page. Section 2.1 Solving Equations Graphically 85 2 This difficulty can be eliminated by using the fact that the only number whose square root is zero is zero itself. 3 3 2 Figure 2.1-3 NOTE Solving radical and rational equations is presented in Section 2.4, and radical and rational functions are presented in Chapter 4. That is, the solutions of tions of x4 x2 2x 1 0 . 2x4 x2 2x 1 0 are the same as the solu- As the graphs below display, the solutions of x 0.4046978 and x 1.1841347, x4 x2 2x 1 0 are which are also approximate solutions of 2x4 x2 2x 1 0. 3 3 3 3 3 3 3 Figure 2.1-4a 3 Figure 2.1-4b The solutions can be verified by substitution. ■ 5 Example 4 Solving f (x) g(x) 0 5 5 Solve 2x2 x 1 9x2 9x 2 0. 5 Figure 2.1-5 5 Solution The graph of y 2x2 x 1 9x2 9x 2 in Figure 2.1-5 is impossible to read. Using the zoom feature will display a better graph, but it may be easier to use the fact that a fraction is zero only when its numerator is zero and its denominator is nonzero. The values that make the numerator zero can easily be found by finding y 2x2 x 1. the zeros of Discard any value that makes the denominator of the original equation zero because 5 5 an input that gives an undefined output is not in the domain. 5 Figure 2.1-6 Figure 2.1-6 shows that one x-intercept of and the other is (not identified on the graph). Neither value makes the denominator zero, so they are the solutions to the given equation, which can be verified by substitution. x 1 is y 2x2 x 1 x 0.5 ■ 86 Chapter 2 Equations and Inequalities Summary of Solving Equations Graphically To solve h(x) g(x) use one of the following. • The Intersection Method y1 and 1. Graph h(x) y2 g(x). 2. Find the x-coordinate of each point of intersection. • The x-Intercept Method 1. Rewrite the equation as f(x) 0, where f(x) h(x) g (x). y f(x). 2. Graph 3. Find the x-intercepts of the graph of y f(x) x-intercepts of the graph of of the equation. f(x). The are the solutions The x-Intercept Method has the advantage of needing no information about the range of the functions. Applications Graphical solution methods can be helpful in dealing with applied problems because approximate solutions are adequate in most real-world contexts. Example 5 Equal Populations According to data from the U.S. Bureau of the Census, the approximate population y (in millions) of Chicago and Los Angeles between 1950 and 2000 are given by Chicago Los Angeles y 0.0000304x3 0.0023x2 0.02024x 3.62 y 0.0000113x3 0.000922x2 0.0538x 1.97 where 0 corresponds to 1950. In what year did the two cities have the same population? Solution 50 Graph both functions on the same screen, and find the x-value of their point(s) of intersection. As shown in Figure 2.1-7, the populations were the same when which represents September of 1978. x 28.75, ■ Figure 2.1-7 4 0 −2 Section 2.1 Solving Equations Graphically 87 Exercises 2.1 In Exercises 1–6, determine graphically the number of solutions of the equation, but don’t solve the equation. You may need a viewing window other than the standard one to find all of the x-intercepts. 1. x5 5 3x4 x 2. x3 5 3x2 24x 25. 10x5 3x2 x 6 0 26. 1 4 x4 x 4 0 27. 2x 1 2 x2 1 12 x4 0 28. 1 4 x4 1 3 x2 3x 1 0 3. x7 10x5 15x 10 0 4. x5 36x 25 13x3 5. x4 500x2 8000x 16x3 32,000 6. 6x5 80x3 45x2 30 45x4 86x In Exercises 7–34, use a graphical method to find all real solutions of the equation, approximating when necessary. 29. 31. 32. 33. 5x x2 1 x2 4 0 0 2x 3 0 30. 2x x 5 1 3x2 2x 1 x3 2 5 x x2 0 0 2x2 3 2x 2 5 34. 2x3 2 2x 5 4 7. x3 4x2 10x 15 0 8. x3 9 3x2 6x 9. x4 x 3 0 In Exercises 35–40, find an exact solution of the equation in the interval shown to the right of each equation. For example, if the graphical approximation of a solu- 10. x5 5 3x4 x 11. 2x4 x3 x 3 0 tion begins .3333, check to see if 1 3 is the exact solu- 12. 28x4 14x3 9x2 11x 1 0 13. 2 5 A x5 x2 2x 0 14. 2x4 x2 3x 1 0 15. x2 2x 5 16. 2x2 1 2x 9 0 17. 18. 19. 2x5 10x 5 x3 x2 12x 0 3x5 15x 5 x7 8x5 2x2 5 0 x3 4x 1 x2 x 6 0 20. 4 x 2 3 x 1 0 3 Use parentheses correctly. 4 21. 2x3 4x2 x 3 0 22. 6x3 5x2 3x 2 0 23. x5 6x 6 0 24. x3 3x2 x 1 0 tion. Similarly, if your approximation begins 1.414, 12 1.414 . check to see if is a solution because 12 35. 3x3 2x2 3x 2 0 36. 4x3 3x2 3x 37. 12x4 x3 12x2 25x 2 0 0 6 x 6 1 38. 8x5 7x4 x3 16x 2 0 0 6 x 6 1 39. 4x4 13x2 3 0 40. x3 x2 2x 41. According to data from the U.S. Department of Education, the average cost y of tuition and fees at public four-year institutions in year x is approximated by the equation y 0.024x4 0.87x3 9.6x2 97.2x 2196 x 0 where corresponds to 1990. If this model continues to be accurate, during what year will tuition and fees reach $4000? 42. Use the equation in Example 5 to determine the year in which the population of Los Angeles reached 2.6 million. 88 Chapter 2 Equations and Inequalities 43. According to data from the U.S. Department of Health and Human Services, the cumulative number y of AIDS cases (in thousands) as of year x is approximated by y 0.062x4 1.54x3 9.21x2 57.54x 199.36 0 x 11 2 1 x 0 corresponds to 1990. During what where year did the cumulative number of cases reach 750,000? 44. a. How many real solutions does the equation 0.2x5 2x3 1.8x k 0 have when k 0 ? b. How many real solutions does it have when k 1? c. Is there a value of k for which the equation has just one real solution? d. Is there a value of k for which the equation has no real solution? 2.2 Solving Quadratic Equations Algebraically Objectives • Solve equations by: factoring square root of both sides completing the square quadratic formula • Solve equations in quadratic form The basic strategy for solving equations is to use the basic properties of equality. • Add or subtract the same quantity from both sides of the equation. • Multiply or divide both sides of the equation by the same nonzero quantity. The properties of equality apply to all equations. They, together with other techniques that are presented in this chapter, can be used to transform a given equation into one whose solutions are easily found. This section considers quadratic equations and techniques used to find their solutions. Definition of a Quadratic Equation A quadratic, or second degree, equation is one that can be written in the form ax2 bx c 0 for real constants a, b, and c, with a 0. NOTE This chapter considers only real solutions, that is, solutions that are real numbers. Techniques Used to Solve Quadratic Equations There are four techniques normally used to algebraically find exact solutions of quadratic equations. Techniques that can be used to solve some quadratic equations include • factoring • taking the square root of both sides of an equat
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ion Techniques that can be used to solve all quadratic equations include • completing the square • using the quadratic formula Section 2.2 Solving Quadratic Equations Algebraically 89 Solving Quadratic Equations by Factoring The factoring method of solving quadratic equations is based on the Zero Product Property of real numbers. The Zero Product Property If a product of real numbers is zero, then at least one of the factors is zero. In other words, If ab 0, then a 0 or b 0 (or both). NOTE If needed, review factoring in the Algebra Appendix. Example 1 Solving a Quadratic Equation by Factoring Solve 3x2 x 10 by factoring. Solution Rearrange the terms so that one side is 0, and then factor. y 8 4 (− , 0) 5 3 3x2 x 10 0 x 2 0 3x 5 2 1 2 1 (2, 0) x Using the Zero Product Property, Subtract 10 from each side Factor the left side 3x 5 x 2 or must be 0. −8 −4 0 4 8 −4 −8 3x 5 0 or 3x 5 x 5 3 x 2 0 x 2 Figure 2.2-1 Therefore, the solutions are 5 3 and 2. See Figure 2.2-1. ■ CAUTION To guard against mistakes, always check solutions by substituting each solution into the original equation to make sure it really is a solution. Solving x 2 k x2 5 has no real solutions because the square of a numThe equation x 0. ber is never negative. The equation x2 7 because these are The equation the only numbers whose square is 7. Similar facts are true for equations of the form has only one solution, and where k is a real number. has two solutions, x2 0 27 x2 k, 27, Solutions of x 2 k For a real number k, k 66 0 k 0 k 77 0 Number of Solutions 0 1 2 Solutions 0 2k and 2k 90 Chapter 2 Equations and Inequalities CAUTION When taking the square root of both sides of an equation, ± remember to write on one side of the equation. When k is positive, the two solutions of x ± 2k, are often written as which is read “x equals plus or minus the square root of k.’’ x2 k Taking the Square Root of Both Sides of an Equation Example 2 Solving ax 2 b Solve 3x2 9. Solution 3x2 9 x2 3 x ± 23 ±1.732 Divide by 3 Take the square root Substitute both solutions into the original equation to check. ■ The method of taking the square root of both sides of an equation can be a used to solve equations of the form 2 k. x h 1 2 Example 3 Solving a(x h)2 k Solve 2 x 4 1 2 2 6 . Solution The equation is in the form the procedure outlined above can be applied. au2 k, where u represents x 4. Therefore ± 23 x 4 23 x 2.27 x 4 ± 23 or x 4 23 x 5.73 Divide by 2 Take square roots Subtract 4 Exact solutions Approximate solutions ■ Figure 2.2-2 y = 2(x + 4)2 y y = 6 −8 −4 −5.73 −2.27 8 4 0 −4 −8 Completing the Square A variation of the method of taking the square root of both sides can be used to solve any quadratic equation. It is based on the fact that an exprescan be changed into a perfect square by adding sion of the form x2 6x a suitable constant. For example, adding 9 to the expression changes it into a perfect square. x2 bx x2 6x 9 x 3 2 2 1 The number added is 9, which is of x in the original expression, completing the square, works in every case. 32, x2 6x. and 3 is one-half of 6, the coefficient This technique, which is called Section 2.2 Solving Quadratic Equations Algebraically 91 Completing the Square To complete the square of the expression square of one-half the coefficient of x, namely x2 bx, add the b 2 b The a . 2 NOTE The procedure of completing the square is used in other areas of mathematics, and knowledge of this procedure is important in later chapters. addition produces a perfect square trinomial. x2 bx Solving a Quadratic Equation by Completing the Square To solve a quadratic equation by completing the square, follow the procedure below. 1. Write the equation in the form x2 bx c . CAUTION 2. Add 2 b 2 b a to both sides so that the left side is a perfect square and the Completing the square only works when the coefficient of an equation such as 2x2 6x 1 0 is 1. In x2 right side is a constant. 3. Take the square root of both sides. 4. Simplify. first divide both sides by 2 and then complete the square. Example 4 Solving a Quadratic Equation by Completing the Square Solve 2x2 6x 1 0 by completing the square. Solution y 8 4 −8 −4 0 4 8 x −4 −8 Figure 2.2-3 2x2 6x 1 0 2x2 6x 1 x2 3x 1 2 1 2 x2 3x Subtract 1 Divide by 2 Add 2 3 2 b a 9 4 Rewrite as perfect square and simplify 7 4 A Take square root ± A 7 4 2.823 Add 3 2 or x 3 2 7 4 A 0.177 There are two real solutions. See Figure 2.2-3. ■ The technique of completing the square can be used to solve any quadratic equation. 92 Chapter 2 Equations and Inequalities Solving ax 2 bx c 0 by Completing the Square Solve ax2 bx c 0 by completing the square as follows: 1. Subtract c from both sides. ax2 bx c 2. Divide both sides by a, the leading coefficient. x2 b 3. Add the square of half of that is, 2 b 2a b a , to both sides. a x c a b a , x2 b a x 2 b 2ab 2 b 2ab c a a 4. Write the left side of the equation as a perfect square. a x b a 2ab 2 2 b 2ab a c a 5. Take the square root of both sides. x b 2a ±± 2 b 2ab c a B a 6. Subtract b 2a from both sides. x b 2a ±± 2 b 2ab c a B a The equation in step 6 can be simplified. 2 b 2ab a b2 4a2 b2 4a2 c a b2 4a2 4ac 4a2 ± ± A b2 c A 4a2 a b2 4ac 4a2 x b 2a b 2a b 2a b ± 2b2 4ac 2a ± 2b2 4ac 2a The final expression is known as the quadratic formula. The Quadratic Formula The solutions of the quadratic equation b ±± 2b2 4ac 2a x ax2 bx c 0 are Because the quadratic formula can be used to solve any quadratic equation, it should be memorized. Section 2.2 Solving Quadratic Equations Algebraically 93 Example 5 Solving a Quadratic Equation by Using the Quadratic Formula Solve x2 3 8x by using the quadratic formula. Solution Rewrite the equation as a 1, b 8, and with x2 8x 3 0, c 3. and apply the quadratic formula x 8 ± 282 4 2 1 1 1 2 1 8 252 2 3 2 21 8 ± 264 12 2 8 ± 252 2 0.4 or x 8 252 2 7.6 . Therefore, x y 4 0 x 4 −12 −8 −4 −4 −8 −12 Figure 2.2-4 The equation has two distinct real solutions, as confirmed in Figure 2.2-4. ■ The Discriminant b2 4ac in the quadratic formula, called the discriminant, The expression can be used to determine the number of real solutions of the equation ax2 bx c 0. Real Solutions of a Quadratic Equation Discriminant Value b2 4ac 77 0 b2 4ac 0 b2 4ac 66 0 Number of Real Solutions of ax2 bx c 0 2 distinct real solutions 1 distinct real solution 0 real solutions The discriminant can be used to determine if an equation has no real solutions without completing all computations. Example 6 Determining the Number of Solutions by Using the Discriminant Solve 2x2 x 3. Solution First, write the equation in general form. 2x2 x 3 0 2 3 b2 4ac 1 4 1 2 1 2 23 6 0 94 Chapter 2 Equations and Inequalities y 10 8 6 4 2 −8 −6 −4 −2 0 2 4 6 8 x Figure 2.2-5 Definition of Polynomial Equation NOTE Polynomials are discussed in Chapter 4. The discriminant of has no real solutions. 2x2 x 3 0 is negative. Therefore, 2x2 x 3 0 does not You can confirm this fact because the graph of touch the x-axis, as shown in Figure 2.2-5. Since the graph has no x-intercepts, the equation has no real solutions. y 2x2 x 3 ■ NOTE A quadratic equation with no x-intercepts has no real solution. Polynomial Equations A polynomial equation of degree n is an equation that can be written in the form where an, p anxn an1xn1 p a1x a0 , a0 are real numbers. 0 , 4x6 3x5 x4 7x3 8x2 4x 9 0 For instance, equation of degree 6. Similarly, sion of degree 3. Notice that polynomials have the following traits. is a polynomial is a polynomial expres- 4x3 3x2 4x 5 • no variables in denominators • no variables under radical signs As a general rule, polynomial equations of degree 3 and above are best solved by the graphical methods presented in Section 2.1. However, some equations are quadratic in form and can be solved algebraically. Polynomial Equations in Quadratic Form Example 7 Solving an Equation in Quadratic Form Solve 4x4 13x2 3 0. Solution To write 4x4 13x2 3 0 in quadratic form, substitute u for x2. 4x4 13x2 3 0 2 13x2 3 0 x2 2 4u2 13u 3 0 4 1 Write x4 as Substitute u for x2 x2 1 2 2 Then solve the resulting quadratic equation. u 3 4u 1 1 u 3 0 2 1 or u 3 2 0 4u 1 0 u 1 4 Factor Zero-Product Property Section 2.2 Solving Quadratic Equations Algebraically 95 Because u x2, x2 3 or x ± 23 x2 1 4 x ± 1 2 Therefore, the original equation has four solutions, x ± 1 2 as shown in Figure 2.2-6. , x ±23 and ■ y 8 4 −4 −2 0 2 4 x −4 −8 Figure 2.2-6 Exercises 2.2 In Exercises 1–12, solve each equation by factoring. 1. x2 8x 15 0 2. x2 5x 6 0 3. x2 5x 14 4. x2 x 20 5. 2y2 5y 3 0 6. 3t2 t 2 0 7. 4t2 9t 2 0 8. 9t2 2 11t 9. 3u2 u 4 10. 5x2 26x 5 11. 12x2 13x 4 12. 18x2 23x 6 In Exercises 13–24, solve the equation by taking the square root of both sides. Give exact solutions and approximate solutions, if appropriate. 13. x2 9 15. x2 40 17. 3x2 12 14. x2 12 16. x2 10 18. 1 2 v2 10 19. 5s2 30 20. 3x2 11 21. 25x2 4 0 22. 4x2 28 0 23. 3w2 8 20 24. 2t2 11 5 In Exercises 25–28, solve the equation by completing the square. 25. x2 2x 12 26. x2 4x 30 0 27. w2 w 1 0 28. t2 3t 2 0 In Exercises 29–40, use the quadratic formula to solve the equation. 29. x2 4x 1 0 30. x2 2x 1 0 31. x2 6x 7 0 32. x2 4x 3 0 33. x2 6 2x 34. x2 11 6x 35. 4x2 4x 7 36. 4x2 4x 11 37. 4x2 8x 1 0 38. 2t2 4t 1 0 39. 5u2 8u 2 40. 4x2 3x 5 In Exercises 41–46, find the number of real solutions of the equation by computing the discriminant. 41. x2 4x 1 0 42. 4x2 4x 3 0 43. 9x2 12x 1 44. 9t2 15 30t 45. 25t2 49 70t 46. 49t2 5 42t In Exercises 47–56, solve the equation by any method. 47. x2 9x 18 0 48. 3t2 11t 20 0 49. 4x 1 x 1 2 1 51. 2x2 7x 15 50. 25y2 20y 1 52. 2x2 6x 3 53. t2 4t 13 0 54. 5x2 2x 2 55. 7x2 3 2x 3 1 56. 25x 4 x 20 96 Chapter 2 Equations and Inequalities In Exercises 57–60, use a calculator to find approximate solutions of the equation. In Exercises 69–72, find a number k such that the given equation has exactly one real solution. 57. 4.42x2 10.14x 3.79 0 58. 8.06x2 25.8726x 25.047256 0 59. 3x2 82.74x 570.4923 0 60. 7.63x2 2.79x 5.32 69. x2 kx 25 0 70.
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x2 kx 49 0 71. kx2 8x 1 0 72. kx2 24x 16 0 In Exercises 61–68, find all exact real solutions of the equation. 61. y4 7y2 6 0 62. x4 2x2 1 0 63. x4 2x2 35 0 64. x4 2x2 24 0 65. 2y4 9y2 4 0 66. 6z4 7z2 2 0 67. 10x4 3x2 1 68. 6x4 7x2 3 73. Find a number k such that 4 and 1 are the solutions of x2 5x k 0. 74. Suppose a, b, and c are fixed real numbers such Let r and s be the solutions of b2 4ac 0. that ax2 bx c 0. a. Use the quadratic formula to show that a and rs c a. b. Use part a to verify that r s b ax2 bx c a c. Use part b to factor x r x s . x2 2x 1 21 1 2 and 5x2 8x 2. Section 2.3 Applications of Equations 97 2.3 Applications of Equations Objectives • Solve application problems Real-life situations are usually described verbally, but they must be interpreted and expressed as equivalent mathematical statements. The following guideline may be helpful. Applied Problems Guideline 1. Read the problem carefully, and determine what is asked for. 2. Label the unknown quantities with variables. 3. Draw a picture of the situation, if appropriate. 4. Translate the verbal statements in the problem and the relationships between the known and unknown quantities into mathematical language. 5. Consolidate the mathematical information into an equation in one variable that can be solved or an equation in two variables that can be graphed. 6. Solve for at least one of the unknown quantities. 7. Find all remaining unknown quantities by using the relationships given in the problem. 8. Check and interpret all quantities found in the original problem. Example 1 Number Relations The average of two real numbers is 41.125, and their product is 1683. Find the two numbers. Solution 1. Read: 2. Label: 3. Draw: 4. Translate: Two numbers are asked for. Let the numbers be a and b. A diagram is not appropriate in this problem. English Language two numbers Their average is 41.125. Mathematical Language a and b a b 2 41.125 [1] Their product is 1683. [2] 5. Consolidate: One technique to use when dealing with two unknowns is to express one in terms of the other, then substitute to obtain an equation in one variable. Solve equation [2] for b. ab 1683 ab 1683 b 1683 a [2] Divide both sides by a 98 Chapter 2 Equations and Inequalities Substitute the result into equation [1] and simplify. a b 2 41.125 [1] 1 0 1 50 Figure 2.3-1 a 1683 a 2 a 1683 41.125 Substitute 1683 a for b a 82.25 Multiply both sides by 2 6. Solve: Solve the equation by using the x-Intercept Method with the graph of y1 x 1683 x 82.25, where x represents a. See Figure 2.3-1. a 1683 a 82.25 The solutions of are a 44 and a 38.25. 7. Find: Find the other number, b, from the equation ab 1683. Let a 44. Similarly, let b 44. 44b 1683 b 38.25 a 38.25 and use the same equation to find 8. Check: The average of 44 and 38.25 is 44 38.25 2 The product of 44 and 38.25 is 1683. 41.125. The two numbers are 44 and 38.25. ■ Solutions in Context When solving an application problem, it is important to interpret answers in terms of the original problem. Each solution should • make sense • satisfy the given conditions • answer the original question In particular, an equation may have several solutions, some of which may not make sense in the context of the problem. For instance, distance cannot be negative, the number of people cannot be a fraction, etc. Example 2 Dimensions of a Rectangle A rectangle is twice as wide as it is high. If it has an area of 24.5 square inches, what are its dimensions? w 2h Figure 2.3-2 h Solution 1. Read: 2. Label: 3. Draw: The dimensions of the rectangle are asked for. Let w denote the width and h denote the height. See Figure 2.3-2. 99 [3] [4] Section 2.3 Applications of Equations 4. Translate: The area of a rectangle is width English Language Width is twice height. The area is height. Mathematical Language w 2h wh 24.5 24.5 in2 . 5. Consolidate: Substitute 2h for w in equation [4]. wh 24.5 h 24.5 2h 2 2h2 24.5 1 6. Solve: Solve by taking the square root of both sides. h2 12.25 h ± 212.25 ± 3.5 Divide by 2 Take square root of both sides Because height is never negative, only the positive root applies to this situation. Therefore, 7. Find: Find the width by using equation [3] and h 3.5 h 3.5 in. . 8. Check: The width is twice the height and the area is correct. w 2 3.5 2 1 7 in. Thus, the width is 7 inches and the height is 3.5 inches. ■ 3.5 7 1 2 24.5 in2 Example 3 Volume of a Rectangular Box A rectangular box with a square base and no top is to have a volume of 30,000 cm3. what are the dimensions of the box? If the surface area of the box is 6000 cm2, h s s Figure 2.3-3 Solution 1. Read: 2. Label: 3. Draw: 4. Translate: The quantities to be found are the length, width, and height of the box. Notice that the two sides of the base have the same length. Let s denote a side of the square base of the box. Let h denote the height of the box. See Figure 2.3-3. height The volume of a box is given by length width and the surface area is the sum of the area of the base and the area of the four sides of the box. English Language length, width, height surface area: Mathematical Language s, s, and h base surface area each side surface area total surface area Volume is Surface area is 30,000 cm3 6000 cm2 s2 sh s2 4sh s2h 30,000 s2 4sh 6,000 [5] [6] 100 Chapter 2 Equations and Inequalities 5. Consolidate: Solve equation [5] for h: h 30,000 s2 Substitute the expression for h into equation [6]. 10,000 s2 4s a 30,000 b s2 s2 120,000 s 6000 6000 6. Solve: To solve by using the Intersection Method, graph s2 120,000 s and y1 2.3-4, and find the points of intersection. Therefore, s 21.70 cm or s 64.29 cm. as shown in Figure y2 6000, 100 0 Figure 2.3-4 75 7. Find: 8. Check: h 30,000 Use to find h. s2 h 30,000 21.70 2 1 30,000 470.89 2 h 30,000 64.29 2 1 63.71 cm 30,000 4133.2041 2 7.26 cm Volume 63.71 21.7 2 30,000.4019 1 1 2 2 2 7.26 64.29 2 30,007.06177 2 1 1 Surface Area 2 4 21.7 6000.9180 1 2 21.7 1 21 63.71 2 2 4 64.29 6000.1857 2 1 64.29 7.26 2 2 1 1 21.7 cm 21.7 cm with a height of approxiOne base is approximately mately 63.71 cm. Another base is approximately 64.29 cm 64.29 cm with a height of approximately 7.26 cm. ■ Interest Applications Calculating interest is common in real-world applications. When an amount, P, is deposited or borrowed, P is referred to as the principal. Interest is the fee paid for the use of the money and is calculated as a percentage of the principal each year. When the duration of a loan or a bank balance is less than 1 year, simple interest is generally used. The basic rule of simple interest is I Prt P represents the principal, r represents the annual interest rate, and t represents time in years. Example 4 Stock and Savings Returns A high-risk stock pays dividends at a rate of 12% per year, and a savings account pays 6% interest per year. How much of a $9000 investment should be put in the stock and how much should be put in savings to obtain a return of 8% per year on the total investment? NOTE In Precalculus, combining basic algebraic steps to promote clarity when simplifying is encouraged as long as it does not cause confusion. Section 2.3 Applications of Equations 101 Solution Read and Label: Let s be the amount invested in stock. Then the rest of will be the amount in the sav- 9000 s, the $9000, namely ings account. Translate: Return on s dollars in stock at 12% b a Return on 9000 s dollars in savings at 6%b a 8% of $9000 > 12% of s 6% of 1 > 9000 s 2 0.08 1 > 9000 2 > > 0.12s 0.06 > 720 9000 s 0.12s 540 0.06s 720 2 1 0.12s 0.06s 720 540 0.06s 180 s 180 0.06 3000 Therefore, the investment should be as follows: • $3000 in stock 9000 3000 • 2 1 $6000 in the savings account If this is done, the total return will be 12% of $3000 plus 6% of $6000, making a total return of $360 $360 $720 —which is 8% of $9000. ■ Distance Applications The basic formula for problems involving distance and a constant rate of velocity is d rt where d represents the distance traveled at rate r for time t. The units for rate should be the distance units divided by the time units, such as miles per hour. Example 5 Distance A pilot wants to make an 840-mile round trip from Cleveland to Peoria and back in 5 hours flying time. There will be a headwind of 30 mph going to Peoria, and it is estimated that there will be a 40-mph tailwind returning to Cleveland. At what constant engine speed should the plane be flown? Solution Let r be the engine speed of the plane, and note that the headwind slows the velocity by 30 and the tailwind increases the velocity by 40. 102 Chapter 2 Equations and Inequalities Distance Actual Velocity Cleveland to Peoria Peoria to Cleveland 420 420 r 30 r 40 D rP D rC Time 420 r 30 420 r 40 The time traveling to Peoria plus the time traveling back to Cleveland is the total time traveled, or 5 hours. 1 Time to Peoria 420 r 30 2 1 Time to Cleveland 420 r 40 2 5 5 Multiply both sides by the common denominator simplify. 1 r 30 r 40 2 1 , and 2 420 r 40 1 r 30 r 40 2 21 r 30 r 40 r 30 5 1 2 1 r 40 420 r 30 1 2 21 420 2 84 2 r 30 2 5 2 2 2 1 2 1 420 84 r 40 r 40 r 40 1 r 40 r 30 1 r 30 1 1 r2 10r 1200 84r 3360 84r 2520 r2 158r 2040 0 0 r 170 r 12 2 1 r 170 0 or r 12 0 1 r 30 1 2 2 2 1 r 170 r 12 Obviously, the negative solution does not apply. Because both sides were multiplied by a quantity involving the variable, the positive solution, 170, should be checked in the original problem to make sure it is a solution. ■ Other Applications Example 6 Width of a Garden Walk x 40 x 24 A landscaper wants to put a cement walk of uniform width around a rectangular garden that measures 24 by 40 feet. She has enough cement to cover 660 square feet. How wide should the walk be in order to use all the cement? Solution Figure 2.3-5 Let x denote the width of the walk in feet and draw a picture of the situation, as shown in Figure 2.3-5. The length of the outer rectangle is the garden length plus walks on each and its width is the garden width plus walks on each end, or end, or The
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area of the walk is found by subtracting the area of the garden from the area of the outer rectangle. 40 2x, 24 2x. Section 2.3 Applications of Equations 103 Area of outer rectangle 1 1 2 Area of garden 2 Area of the walk 40 2x 24 2x > 2 1 > 660 960 128x 4x2 960 660 24 40 21 > 2 2 1 1 4x2 128x 660 0 x2 32x 165 0 165 2 1 2 Apply the quadratic formula x x 1 1 1 32 ± 2 2 4 32 2 1 2 1 32 ± 21684 2 2 x 4.5 or x 36.5 Only the positive solution makes sense, so the walk should be approximately 4.5 feet wide. Check the solution in the original problem. ■ Example 7 Box Construction A box with no top that has a volume of 1000 cubic inches is to be constructed from a -inch sheet of cardboard by cutting squares of equal size from each corner and folding up the flaps, as shown in Figure 2.3-6. What size square should be cut from each corner? 22 30 30 x x x 22 30 − 2x 30 − 2x x 22 − 2x Figure 2.3-6 Solution Let x represent the length of the side of the square to be cut from each corner. The dashed rectangle in Figure 2.3-6 is the bottom of the box. Its length is as shown in the figure. Similarly, the width of the box will be 30 2x, 22 2x, and its height will be x inches. Therefore, Length Width Height Volume of the box > 30 2x >>> 22 2x x 1000 1 2 Because the cardboard is 22 inches wide, x must be less than 11, and because x is a length, it must be positive. Therefore, the only meaningful solutions in this context are between 0 and 11. 2 1 104 1500 0 500 Chapter 2 Equations and Inequalities 30 2x 22 2x 1 x 2 1 y1 and Graph As shown in Figure 2.3-7, there are two points of intersection: one at approximately (2.23, 1000) and another at approximately (6.47, 1000), which is not identified on the graph. Because both are viable solutions, there are two boxes that meet the given conditions. y2 2 1000. Figure 2.3-7 11 Find the dimensions of the box for each possible case. Height: Length: Width: Case I 2.23 in. 30 2 22 2 1 1 25.54 in. 17.54 in. 2.23 2.23 2 2 Case II 6.47 in. 30 2 22 2 1 1 17.06 in. 9.06 in. 6.47 6.47 2 2 25.54 in 2.23 in 17.54 in Figure 2.3-8 6.47 in 9.06 in 17.06 in ■ Example 8 Mixture Problem A car radiator contains 12 quarts of fluid, 20% of which is antifreeze. Howmuch fluid should be drained and replaced with pure antifreeze so that the resulting mixture is 50% antifreeze? Solution Let x be the number of quarts of fluid to be replaced by pure antifreeze. When x quarts are drained, there are quarts of fluid left in the radiator, 20% of which is antifreeze. 12 x Amount of antifreeze in radiator after draining x quarts of fluid ± ≤ x quarts of antifreeze b a ° Amount of antifreeze in final mixture ¢ > 20% of 12 x 1 2 > x > 50% of 12 > 12 x > 0.2 2 2.4 0.2x x 6 x 0.5 1 > 12 1 2 0.8x 3.6 x 3.6 0.8 4.5 Therefore, 4.5 quarts should be drained and replaced with pure antifreeze. ■ Section 2.3 Applications of Equations 105 Exercises 2.3 In Exercises 1–4, a problem situation is given. a. Decide what is being asked for, and label the unknown quantities. b. Translate the verbal statements in the problem and the relationships between the known and unknown quantities into mathematical language, using a table like those in Examples 1–3. The table is provided in Exercises 1–2. You need not find an equation to be solved. 7. The diameter of a circle is 16 cm. By what amount must the radius be decreased in order to decrease the area by square centimeters? 48p 8. A corner lot has dimensions 25 by 40 yards. The city plans to take a strip of uniform width along the two sides bordering the streets in order to widen these roads. How wide should the strip be if the remainder of the lot is to have an area of 844 square yards? 1. The sum of two numbers is 15 and the difference of their squares is 5. What are the numbers? In Exercises 9–20, solve the problem. English Language the two numbers Their sum is 15. The difference of their squares is 5. Mathematical Language 2. The sum of the squares of two consecutive integers is 4513. What are the integers? English Language the two integers The integers are consecutive. The sum of their squares is 4513. Mathematical Language 3. A rectangle has a perimeter of 45 centimeters and an area of 112.5 square centimeters. What are its dimensions? 4. A triangle has an area of 96 square inches, and its height is two-thirds of its base. What are the base and height of the triangle? In Exercises 5–8, set up the problem by labeling the unknowns, translating the given information into mathematical language, and finding an equation that will produce the solution to the problem. You need not solve this equation. 5. A worker gets an 8% pay raise and now makes $1600 per month. What was the worker’s old salary? 6. A merchant has 5 pounds of mixed nuts that cost $30. He wants to add peanuts that cost $1.50 per pound and cashews that cost $4.50 per pound to obtain 50 pounds of a mixture that costs $2.90 per pound. How many pounds of peanuts are needed? 9. You have already invested $550 in a stock with an annual return of 11%. How much of an additional $1100 should be invested at 12% and how much at 6% so that the total return on the entire $1650 is 9%? 10. If you borrow $500 from a credit union at 12% annual interest and $250 from a bank at 18% annual interest, what is the effective annual interest rate (that is, what single rate of interest on $750 would result in the same total amount of interest)? 11. A radiator contains 8 quarts of fluid, 40% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze? 12. A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze? 13. Two cars leave a gas station at the same time, one traveling north and the other south. The northbound car travels at 50 mph. After 3 hours the cars are 345 miles apart. How fast is the southbound car traveling? 14. An airplane flew with the wind for 2.5 hours and returned the same distance against the wind in 3.5 hours. If the cruising speed of the plane was a constant 360 mph in air, how fast was the wind blowing? Hint: If the wind speed is r miles per hour, then the plane travels at the wind and at mph against the wind. mph with 360 r 360 r 2 1 1 2 15. The average of two real numbers is 41.375 and their product is 1668. What are the numbers? 106 Chapter 2 Equations and Inequalities 16. A rectangle is four times as long as it is wide. If it has an area of 36 square inches, what are its dimensions? b. It is thrown downward from the top of the same building, with an initial velocity of 52 feet per second. 17. A 13-foot-long ladder leans on a wall. The bottom of the ladder is 5 feet from the wall. If the bottom is pulled out 3 feet farther from the wall, how far does the top of the ladder move down the wall? Hint: The ladder, ground, and wall form a right triangle. Draw pictures of this triangle before and after the ladder is moved. Use the Pythagorean Theorem to set up an equation. 18. A factory that makes can openers has fixed costs for building, fixtures, machinery, etc. of $26,000. The variable cost for material and labor for making one can opener is $2.75. a. What is the total cost of making 1000 can openers? 20,000? 40,000? b. What is the average cost per can opener in each case? 19. Red Riding Hood drives the 432 miles to Grandmother’s house in 1 hour less than it takes the Wolf to drive the same route. Her average speed is 6 mph faster than the Wolf’s average speed. How fast does each drive? 20. To get to work Sam jogs 3 kilometers to the train, then rides the remaining 5 kilometers. If the train goes 40 km per hour faster than Sam’s constant rate of jogging and the entire trip takes 30 minutes, how fast does Sam jog? In Exercises 21–24, an object is thrown upward, dropped, or thrown downward and travels in a vertical line subject only to gravity with wind resistance ignored. The height h, in feet, of the object above the ground after t seconds is given by h 16t 2 v0t h0 and v0 t 0. where is the initial height of the object at starting h0 t 0, is the initial velocity (speed) of the time object at time is taken as posiv0 t 0 tive if the object starts moving upward at time and negative if the object starts moving downward at t 0. An object that is dropped (rather than thrown downward) has initial velocity The value of 0. v0 22. You are standing on a cliff 200 feet high. How long will it take a rock to reach the ground at the bottom of the cliff in each case? a. You drop it. b. You throw it downward at an initial velocity of 40 feet per second. c. How far does the rock fall in 2 seconds if you throw it downward with an initial velocity of 40 feet per second. 23. A rocket is fired straight up from ground level with an initial velocity of 800 feet per second. a. How long does it take the rocket to rise 3200 feet? b. When will the rocket hit the ground? 24. A rocket loaded with fireworks is to be shot vertically upward from ground level with an initial velocity of 200 feet per second. When the rocket reaches a height of 400 feet on its upward trip, the fireworks will be detonated. How many seconds after lift-off will this take place? 25. The dimensions of a rectangular box are consecutive integers. If the box has volume of 13,800 cubic centimeters, what are its dimensions? 26. Find a real number that exceeds its cube by 2. 27. The lateral surface area S of the right circular cone at the left in the figure below is given by S pr2r2 h2. produce a cone of height 5 inches and lateral surface area 100 square inches? What radius should be used to h h r b 21. How long does it take an object to reach the ground in each case? a. It is dropped from the top of a 640-foot-high building. 28. The lateral surface area of the right square S b2b2 4h2. pyramid at the right in the figure above is given by If the pyramid has height 10 feet and lateral surface area 100 square fee
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t, what is the length of a side b of its base? Section 2.4 Other Types of Equations 107 29. Suppose that the open-top box being made from a sheet of cardboard in Example 7 is required to have at least one of its dimensions greater than 18 inches. What size square should be cut from each corner? 30. A homemade loaf of bread turns out to be a perfect cube. Five slices of bread, each 0.6 inch thick, are cut from one end of the loaf. The remainder of the loaf now has a volume of 235 cubic inches. What were the dimensions of the original loaf? 31. A rectangular bin with an open top and volume of 38.72 cubic feet is to be built. The length of its base must be twice the width and the bin must be at least 3 feet high. Material for the base of the bin costs $12 per square foot and material for the sides costs $8 per square foot. If it costs $538.56 to build the bin, what are its dimensions? 2.4 Other Types of Equations Objectives • Solve absolute-value equations • Solve radical equations • Solve fractional equations Algebraic Definition of Absolute Value Like linear and quadratic equations, other types of equations can be solved algebraically. This section outlines procedures for solving absolute-value, radical, and fractional equations. Definition of Absolute Value The absolute value of a number c is denoted and is defined as follows. c 0 0 If c 0, then 00 If c 66 0, then c. c. c 00 c 00 00 For example, because 5 is positive, 3. 3 To determine 0 1 p 6 6 0. 3 2 0 0 Therefore, the second part of the definition applies. p 6 6 p p 6 0 0 In all cases, the absolute value of a number is nonnegative. 2 1 0 5, 5 0 p 6 0 and because , note that 3 p 3.14 is negative, and Absolute value can also be interpreted geometrically as a distance. 5 and 3 on the number line is 8 units. Observe that the distance between −10 −8 −6 −4 −2 0 2 4 6 8 10 8 units Figure 2.4-1 8 Notice that property of absolute value. 5 3 0 0 0 8. 0 This is an example of a key geometric 108 Chapter 2 Equations and Inequalities Absolute Value and Distance If c and d are real numbers, then c d 00 00 is the distance between c and d on the number line. For example, the number thus represents the distance between 5 and Figure 2.4-2. 5 22 can be written as 22 and A on the number line. See B 22 5 5 + √2 −10 −8 −6 −4 −2 – 2 0 2 4 6 8 10 5 Figure 2.4-2 In the special case when c 0 tance from c to 0 is 0 d 0, c 0 0 the distance formula shows that the diswhich is an alternative definition of , . c 0 0 0 Geometric Definition of Absolute Value If c is a real number, then is the distance from c to 0 on the number line. c 00 00 For example, line, as shown below. 0 0 3.5 denotes the distance from 3.5 to 0 on the number −3.5 −10 −8 −6 −4 −2 0 2 4 6 8 10 Figure 2.4-3 Properties of Absolute Value The following properties are helpful in simplifying absolute value expressions. Properties of Absolute Value Let c and d represent real numbers. 777 0 when c 0 c 1. c 00 d 00 c 00 cd 00 00 00 c d ` ` 00 00 0 and 00 c c 00 c d 00 00 00 00 00 00 2. 3. 4. 00 00 , where d 0 Section 2.4 Other Types of Equations 109 Illustrations of Properties 2, 3, and 4 for absolute value are shown below. Property 2: Let 3 0 0 3 and c 0 0 3 0 0 3. CAUTION When c and d have c d opposite signs, 0 0 d c is not equal to 0 0 0 0 c 3 For example, if d 5, then and But, 5 0 0 0 0 0 0 0 . Property 3: Let Property 4: Let 12 0 6 2 12 2 . 0 0 0 c 3. Then 0 c 0 3 0 0 Therefore, . 3 0 0 c 6 and d 2. 12 and d 4. 0 2 0 0 0 Therefore, cd ` ` ` Therefore and The caution shows that general case is called the Triangle Inequality. d 0 0 0 0 0 0 when c 3 and d 5. The The Triangle Inequality For any real numbers c and d, c d 00 c 00 00 00 d 00 . 00 Square Root of Squares When c is a positive number, then negative. Consider the case when 2c2 c. c 3. This is not true when c is 3 2 1 2 2 29 3, which is not 3 is equal to the absolute value of c when c is any real number. 2c 2 But That is, 2 3 1 2 2 29 3 3 0 . 0 Square Root of Squares For every real number c, 2c 2 c . 00 00 Solving Absolute-Value Equations Absolute-value equations can be solved by using the definitions. Some equations lend themselves to the geometric definition, while others are solved more easily using the algebraic definition. Graphing techniques can be used to check all solutions. 110 Chapter 2 Equations and Inequalities Technology Tip To compute absolute values on a calculator, use the ABS feature. The ABS feature is in the NUM submenu of the MATH menu of TI and in the NUM submenu of the OPTN menu of Casio. Example 1 Using Absolute Value and Distance Solve x 4 0 0 8 Solution using absolute value and distance. The equation x 4 0 0 8 can be interpreted as the distance from x to 4 is 8 units. See Figure 2.4-4. x −4 −2 −8 −6 0 2 4 6 8 10 12 14 16 x 8 units 8 units Figure 2.4-4 The two possible values of x that are solutions of the original equation are and 12, as shown. 4 ■ NOTE When dealing with long expressions inside absolute value bars, do the computations inside first, and then take the absolute value of the simplified expression. Extraneous Solutions As shown in Example 2 below, some solutions do not make the original equation true when checked by substitution. Such “fake” solutions are called extraneous solutions, or extraneous roots. Because extraneous solutions may occur when solving absolute-value equations, all solutions must be checked by substituting into the original equation or by graphing. Example 2 Using the Algebraic Definition of Absolute Value Solve x 4 0 0 5x 2 Solution by using the algebraic definition of absolute value. The absolute value of any quantity is either the quantity itself or the opposite of the quantity, depending on whether the quantity is positive or x 4 or negative. So, Therefore, the original equation can be rewritten as two equations that do not involve absolute value. is either 5x 2 4x 6 x 6 4 3 2 5x 2 0 or 5x 2 x 4 1 x 4 5x 2 2 6x 2 x 2 6 1 3 Each solution must be checked in the original equation. Section 2.4 Other Types of Equations 111 x 3 2 is a solution and checks in the original equation, as shown in Figure 2.4-5. However, do not intersect when x 1 3 x 1 3 , Therefore, the only solution of 0 is an extraneous root because the graphs which can be confirmed by substitution. x 3 2 5x 2 x 4 is . 11 3 ( , ) 2 2 y = 5x − 2 x −8 −4 0 4 8 −4 −8 Figure 2.4-5 Example 3 Solving an Absolute Value Equation Solve 0 x2 4x 3 2. 0 Solution Use the algebraic definition of absolute value to rewrite the original equation as two equations. x2 4x 3 0 x2 4x 3 2 x2 4x 5 0 2 0 2 x2 4x 3 1 x2 4x 3 2 x2 4x 1 0 2 The equation on the left can be solved by factoring, and the equation on the right by using the quadratic formula. x2 4x 5 0 x2 4x 1 0 x 5 1 21 x 1 2 0 x 5 or x 1 x x 4 ± 242 ± 220 2 1 2 4 ± 225 2 x 2 ± 25 y 8 4 y = x2 + 4x − 3 y = 2 x −8 −4 0 4 8 −4 −8 Figure 2.4-6 See Figure 2.4-6 to confirm that all four values are solutions. ■ Solving Radical Equations Radical equations are equations that contain expressions with a variable under a radical symbol. Although the approximate solutions of a radical equation can be found graphically, exact solutions can be found algebraically in many cases. The algebraic solution method depends on the following fact. NOTE Squaring both sides of an equation may introduce extraneous roots. If A and B are algebraic expressions and A B , for every positive integer n. An Bn then x 2 3, then is also a solution of For example, if x 2 3 x 2 2 9, 1 2 x 2 1 x 2 2 32. 2 2 9. Therefore, every solution of is a solution of However, 1 but 1 2 is not a solution of x 2 3. 1 112 Chapter 2 Equations and Inequalities Power Principle CAUTION Although it is always a good idea to verify solutions, solutions to radical equations must be checked in the original equation. 5 0 0 10 Figure 2.4-7 If both sides of an equation are raised to the same positive integer power, then every solution of the original equation is a solution of the new equation. However, the new equation may have solutions that are not solutions of the original one. Example 4 Solving a Radical Equation Solve 5 23x 11 x Solution Rearrange terms to get the radical expression alone on one side. 23x 11 x 5 Square both sides to remove the radical sign and solve the resulting equation. 23x 11 2 x 5 2 A B 2 1 3x 11 x2 10x 25 0 x2 13x 36 0 x 9 x 4 2 x 9 0 or If these values are solutions of the original equation, they should be x 5 x-intercepts of the graph of shows that the graph does not have an x-intercept at is an extraneous solution. But Figure 2.4-7 x 4 y 23x 11 . 2 x 4 . That is, 1 The graph suggests that can be confirmed by substitution. x 9 is a solution of the original equation, which ■ Sometimes the Power Principle must be applied more than once to eliminate all radicals. Example 5 Using the Power Principle Twice Solve 22x 3 2x 7 2. Solution Rearrange terms so that one side contains only a single radical term. 22x 3 2x 7 2 y 5 Square both sides and isolate the remaining radical. Section 2.4 Other Types of Equations 113 0 10 20 30 40 50 x 5 Figure 2.4-8 22 x 3 A 2 2x 7 2 2 B 2x 7 2x 3 2x 3 x 7 42x 7 4 x 14 42x 7 A A B B 2 2 22x 7 22 Square both sides again, and solve the resulting equation. 2 B 2 1 2 2 42x 7 B 2x 7 x 7 x 14 A x2 28x 196 42 A x2 28x 196 16 2 1 x2 28x 196 16x 112 x2 44x 84 0 x 2 0 x 42 2 1 x 2 0 x 2 2 or 1 x 42 0 x 42 Technology Tip Graphing calculators do not always show all solutions of a radical equation. See Example 3 in Section 2.1 for an illustration of a technological quirk. Verify by substitution that a solution. x 2 is an extraneous root but that x 42 is ■ Example 6 Distance Stella is standing at point A on the bank of a river that is 2.5 kilometers wide. She wants to reach point B, which is 15 kilometers downstream on the opposite bank. She plans to row downstream to point C on the opposite shore and then run to B, as shown in Figure 2.4-9. She can row downstream at a rate of 4 kilometers per hour and can run at 8 kilometers per hour. If her t
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rip is to take 3 hours, how far from B should she land? Figure 2.4-9 Solution Refer to Figure 2.4-9. The basic formula for distance can be written in different ways. d rt or t d r 114 Chapter 2 Equations and Inequalities Let x represent the distance between C and B, t represent the time required to run from C to B, and r represent the rate Stella can run (8 kilometers per hour). Therefore, t x 8 Similarly, t d 4 denotes the time needed to run from C to B. can express the time required to row distance d. NOTE To review the Pythagorean Theorem, see the Geometry Review Appendix. 15 x Since be applied to right triangle ADC. is the distance from D to C, the Pythagorean Theorem can d2 15 x 2 1 2 2.52 or equivalently d 2 15 x 1 2 2 6.25 Therefore, the total time for the trip is given by a function of x. T x 1 2 rowing time running time d 4 x 8 x 15 4 1 2 2 6.25 2 x 8 If the trip is to take 3 hours, then x 15 .25 x 8 3 Using the viewing window with function. 0 x 20 and 2 y 2, graph this x f 1 2 2 x 15 4 2 1 2 6.25 x 8 3 x 6.74 Find the zeros of f and interpret their values in the context of the problem. x 17.26. The zeros of f are These represent the distances that Stella should land from B. 6.74 represents a downstream destination from A and 17.26 represents an upstream destination from A. Therefore, Stella should land approximately 6.74 kilometers from B to make the downstream trip in 3 hours. and ■ Fractional Equations If f(x) and g(x) are algebraic expressions, the quotient is called a frac- tional expression with numerator f(x) and denominator g(x). As in all fractions, the denominator, g(x), cannot be zero. That is, if the frac- 0 is undefined. The following principle is used to solve fractional tion x x f 1 g 1 2 2 equations of the form 0. x x f 1 g 1 2 2 Section 2.4 Other Types of Equations 115 Solving f(x) g(x) 0 f(x) Let solutions of the equation and g (x) represent algebraic expressions. Then the are all values of x such that 0 f(x) g(x) f(x) 0 and g(x) 0. Example 7 Solving a Fractional Equation Solve 6x2 x 1 2x2 9x 5 0. Solution Find all solutions to 6x2 x 1 0. 3x 1 1 3x 1 0 x 1 3 21 6x2 x 1 0 0 2x 1 2 2x 1 0 or x 1 2 10 Discard any solution that makes 2x2 9x 5 0. 10 10 10 Figure 2.4-10a For x 1 3 : For x 1 2 : Therefore, x 1 3 is the solution of a solution. Figure 2.4-10a confirms that x 1 2 is extraneous 70 6x2 x 1 2x2 9x 5 x 1 3 5 0 0, and x 1 2 is not is a solution and that ■ Technology Tip 10.2 x 5 In Figure 2.4-10a, the vertical line shown at graph but is a result of the calculator evaluating the function just to the The calcu- left of lator erroneously connects these points with a near vertical segment. By 7.7 x 1.7 choosing a window such as vertical line will not be drawn. Using the trace feature in Figure 2.4-10b and just to the right of 2.2 y 10.2 is not part of the but not at x 5. x 5, , the near and x 5 identifies a hole at x 1 2 , where the function is not defined. 7.7 1.7 2.2 Figure 2.4-10b 116 Chapter 2 Equations and Inequalities Exercises 2.4 In Exercises 1–8, rewrite each statement using the geometric definition of absolute value. Represent each on a number line, and find the value(s) of the variable. 1. The distance between y and 2 is 4. 2. The distance between x and 4 is 6. 3. The distance between 3w and 2 is 8. 4. The distance between 4x and 3 is 6. 5. The distance between 2x and 4 is 5. 6. The distance between 4z and 3 is 11. 7. The distance between 3x and 2 is 5. 8. The distance between 4w and 6 is 5 2 . In Exercises 9–20, find all real solutions of each equation. 9 0 0 0 2x 3 6x 9 2x 3 0 x x 3 0 x2 4x 1 x2 5x 1 9. 11. 13. 15. 17. 19. 0 0 0 0 0 0 4x 1 4 3 0 0 10. 12. 14. 16. 18. 20. 0 0 0 0 0 0 3x 5 4x 5 3x 2 0 0 0 7 9 5x 4 2x 1 2x 1 0 x2 2x 9 6 0 12x2 5x 7 4 0 21. Explain why there are no real numbers that satisfy the equation 2x2 3x 12. 0 22. Describe in words the meaning of the inequality. 0 a b a b 0 0 Make sure to consider positive and negative values of a and b. 0 0 0 0 25. An instrument measures a wind speed of 20 feet per second. The true wind speed is within 5 feet per second of the measured wind speed. What are the possible values for the true wind speed? 26. For two real numbers s and t, the notation min(s, t) represents the smaller of the two numbers. When common value. It can be shown that min(s, t) can be expressed as shown. min(s, t) represents the s t, min s For each of the following, verify the equation. a. b. c. s 4 s 2 and s t 5 t 3 t 1 and 27. In statistical quality control, one needs to find the proportion of the product that is not acceptable. The upper and lower control limits (CL) are found by solving the following equation in which p is the mean percent defective, and n is the sample size for CL. CL Find CL when p 0.02 and n 200. In Exercises 28–63, find all real solutions of each equation. Find exact solutions when possible, approximate solutions otherwise. 28. 2x 7 4 29. 24x 9 5 30. 23x 2 7 31. 23 5 11x 3 32. 23 6x 10 2 33. 23 x2 1 2 34. 36. 23 x 1 2 4 1 2 2x2 5x 4 2 35. 2x2 x 1 1 37. 2x 7 x 5 23. Joan weighs 120 pounds and her doctor told her 38. 2x 5 x 1 that her weight is 5 percent from her ideal weight. What are the possible values, to the nearest pound, for Joan’s ideal body weight? 24. A tightrope walker is 8 feet from one end of the rope. If he takes 2 steps and each step is 10 inches long, how far is he from the same end of the rope? Give both possible answers. 39. 23x2 7x 2 x 1 40. 24x2 10x 5 x 3 41. 23 x3 x2 4x 5 x 1 42. 23 x3 6x2 2x 3 x 1 43. 25 9 x2 x2 1 44. 24 x3 x 1 x2 1 45. 23 x5 x3 x x 2 46. 2x3 2x2 1 x3 2x 1 47. 2x2 3x 6 x4 3x2 2 48. 23 x4 x2 1 x2 x 5 49. 25x 6 3 2x 3 50. 23y 1 1 2y 4 51. 22x 5 1 2x 3 52. 2x 3 2x 5 4 53. 23x 5 22x 3 1 0 54. 220 x 29 x 3 55. 26x2 x 7 23x 2 2 56. 2x3 x2 3 2x3 x 3 1 57. 2x 2 3 58. 60. 62. x2 2x 1 x 2 0 x2 x 2 x2 5x 5 0 2x2 7x 6 3x2 5x 2 0 59. 61. 63. 2x2 3x 4 x 4 0 x2 3x 2 x2 x 6 0 3x2 4x 1 3x2 5x 2 0 In Exercises 64–67, assume that all letters represent positive numbers. Solve each equation for the required letter. 64. T 2p m g A for g 65. K 1 x2 u2 A for u 66. R 2d2 k2 for d 67. A 1 a2 b2 A for b 68. A rope is to be stretched at uniform height from a tree to a fence, 20 feet from the tree, and then to the side of a building, 35 ft from the tree, at a point 30 ft from the fence, as shown in the figure. a. If 63 ft of rope is to be used, how far from the building wall should the rope meet the fence? b. How far from the building wall should the rope meet the fence if as little rope as possible is to be used? Hint: What is the x value of the lowest point on the graph? Section 2.4 Other Types of Equations 117 Tree (aerial view) 30 20 Fence 69. A spotlight is to be placed on a building wall to illuminate a bench that is 32 feet from the base of the wall. The intensity of the light I at the bench is known to be I x d3 , where x is the spotlight’s height above the ground and d is the distance from the bench to the spotlight. a. Express I as a function of x. It may help to draw a picture. b. How high should the spotlight be in order to provide maximum illumination at the bench? Hint: What is the x value of the highest point on the graph? Helicopter 1 mile Field Road Anne 70. Anne is standing on a straight road and wants to reach her helicopter, which is located 2 miles down the road from her and a mile off the road in a field. She can run 5 miles per hour on the road and 3 miles per hour in the field. She plans to run down the road, then cut diagonally across the field to reach the helicopter. a. Where should she leave the road in order to reach the helicopter in exactly 42 minutes, that is, 0.7 hour? b. Where should she leave the road in order to reach the helicopter as soon as possible? (see Exercise 68b) 118 Chapter 2 Equations and Inequalities 2.5 Inequalities Objectives • Use interval notation • Solve linear inequalities and compound linear inequalities • Find exact solutions of quadratic and factorable inequalities c 6 d, which is read “c is less than d,’’ means that c is to The statement the left of d on the real number line. Similarly, the statement which is read “c is greater than d,’’ means that c is to the right of d on the real number line. c 7 d, In the set of real numbers, any pair of numbers can be compared because the set of real numbers is ordered. That is, for any two real numbers a and b, exactly one of the following statements is true The two statements are equivalent, and both mean that c d, d c read “c is less than or equal to d,” means either c is less than d or c is equal to d. A similar statement applies to is positive. The statement c d. and The statement b 6 c 6 d, b 6 c called a compound inequality, means and simultaneously c 6 d. Interval Notation An interval of numbers is the set of all numbers lying between two fixed numbers. Such sets appear frequently enough to merit special notation. Interval Notation Let c and d be real numbers with c 66 d. [c, d] denotes the set of all real numbers x such that (c, d) denotes the set of all real numbers x such that [c, d) denotes the set of all real numbers x such that c x d. c 66 x 66 d. c x 66 d. (c, d] denotes the set of all real numbers x such that c 66 x d. All four sets above are called intervals from c to d, where c and d are the endpoints of the interval. The interval [c, d] is called the closed interval from c to d because both endpoints are included, as indicated by brackets, and (c, d) is called the open interval from c to d because neither endpoint is included, as indicated by parentheses. The last two intervals in the box above are called half-open intervals, where the bracket indicates which endpoint is included. q NOTE The symbol is read “infinity’’ but does not denote a real number. It is simply part of the notation used to denote half-lines. Section 2.5 Inequalities 119 The half-line extending to the right or left of b is also called an interval. • For the half-line to the right of b, denotes the set of all real numbers x such that
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denotes the set of all real numbers x such that x b. x 7 b. • For the half-line to the left of b, denotes the set of all real numbers x such that denotes the set of all real numbers x such that 4 2 x b. x 6 b. Similar notation is used for the entire number line. q, q denotes the set of all real numbers. 2 b, q b, q 2 2 q, b q, b 3 1 1 1 1 Basic Principles for Solving Inequalities Solving Inequalities Solutions of inequalities in one variable are all values of the variable that make the inequality true. Such solutions may be found by using algebraic, geometric, and graphical methods, each of which is discussed in this section. Whenever possible, algebra will be used to obtain exact solutions. When algebraic methods are tedious or when no algebraic method exists, approximate graphical solutions will be found. Equivalent Inequalities Like equations, two inequalities are equivalent if they have the same solutions. The basic tools for solving inequalities are as follows. Performing any of the following operations on an inequality produces an equivalent inequality. 1. Add or subtract the same quantity on both sides of the inequality. 2. Multiply or divide both sides of the inequality by the same positive quantity. 3. Multiply or divide both sides of the inequality by the same negative quantity, and reverse the direction of the inequality. Note that Principles 1 and 2 are the same as the principles used in solving linear equations, but Principle 3 states that when an inequality is multiplied or divided by a negative number, the inequality sign must be reversed. For example Multiply by 1 and reverse the inequality Solving Linear Inequalities Example 1 Solving a Compound Linear Inequality Solve 2 3x 5 6 2x 11. 120 Chapter 2 Equations and Inequalities Solution A solution of the inequality a solution of both of the following inequalities. 2 3x 5 6 2x 11 is any number that is 2 3x 5 and 3x 5 6 2x 11 Each of these inequalities can be solved by the principles listed above. Subtract 5 2 5 3x Simplify 3 3x Divide by 3 1 x Subtract 2x and Subtract 5 3x 2x 6 11 5 Simplify x 6 6 The solutions are all real numbers that satisfy both is, 1, 6 that Therefore, the solutions are the numbers in the interval as shown in Figure 2.5-1. 1 x 6 6. 1 x and x 6 6, , 3 2 −3 −2 −1 0 1 2 3 4 5 6 Figure 2.5-1 ■ Example 2 Solving a Compound Linear Inequality Solve 4 6 3 5x 6 18. Solution When a variable appears only in the middle of a compound inequality, the process can be streamlined by performing any operation on each part of the compound inequality. 4 6 35x 6 18 1 6 5x 6 15 1 5 7 x 7 3 Subtract 3 from each part 5 Divide each part by and reverse direction of the inequalities Intervals are usually written from the smaller to the larger, so the solution to the compound inequality is 3 6 x 6 1 5 . −5 −4 −3 −2 −1 0 1 2 3 54 Figure 2.5-2 The solution of the compound inequality is the interval a 3, 1 5 b , as shown in Figure 2.5-2, where open circles indicate that the endpoints are not included in the interval. ■ Section 2.5 Inequalities 121 y f Solving Other Inequalities Although the basic principles play a role in the solution of nonlinear inequalities, geometrically the key to solving such inequalities is the following fact. x x −2 3 g Figure 2.5-3a f − g y −2 3 Figure 2.5-3b The solutions of an inequality of the form intervals on the x-axis where the graph of f is below the graph of g. consist of f(x) 66 g(x) f(x) 77 g(x) The solutions of where the graph of f is above the graph of g. consist of intervals on the x-axis In Figure 2.5-3a, the blue portion of the graph of f is below the graph of correspond to the intervals on the x-axis g. The solutions of f denoted in red. In Figure 2.5-3a, and when x 7 3. 2 6 x 6 1 when The graph of f(x) g(x) 77 0 Although the above procedure can always be used, solutions of an 6 g x are often easier to find by using an inequality expressed as 1 x f equivalent inequality in the form y f(x) g(x) 2 1 lies above the x-axis when 6 0 and below the x-axis when y f Figure 2.5-3b shows the graph of the difference of the two functions shown in Figure 2.5-3a. This graph is below the x-axis in the same intervals where the graph of f is below the graph of g. Therefore, 6 0. the solution of is the same as the solution of (x) g(x) 66 0 can be rewritten in the equivalent Any inequality of the form 2 1 1 g x x by subtracting from both sides of the inequalform 2 2 1 6 g x x x ity. The procedure for solving 2 2 and find the intervals on the x-axis where the graph is below the x-axis. A similar procedure applies when the inequality sign is reversed, except that the solution is determined by x-intervals where the graph is above the x-axis. is to graph Example 3 Solving an Inequality Solve x4 10x3 21x2 7 40x 80. y Solution Rewrite the inequality as x4 10x3 21x2 x4 10x3 21x2 40x 80 1 1 2 40x 80 7 0. 2 1 f x is shown in Figure The graph of 2 1 2.5-4. The graph shows that and the other near 2. The portion of the graph above the x-axis is shown in red. has two zeros, one between and x f 1 2 2 x 2 4 6 100 80 60 40 20 0 −2 −40 −60 −80 −100 −6 −4 Figure 2.5-4 122 Chapter 2 Equations and Inequalities Graphing Exploration Use the graphical root finder of a calculator to find approximate values of the x-intercepts. The Exploration shows that the graph of f is above the x-axis approxiso the approximate mately when x 6 1.53 solutions of the original inequality are all numbers x such that or x 6 1.53 x 7 1.89, x 7 1.89. and when ■ Quadratic and Factorable Inequalities The preceding example shows that solving an inequality depends only on knowing the zeros of a function and the places where its graph is above or below the x-axis. In the case of quadratic inequalities or completely factored expressions, exact solutions can by found algebraically. Example 4 Solving a Quadratic Inequality Solve 2x2 3x 4 0. Solution y 8 4 −8 −4 0 4 8 −4 −8 Figure 2.5-5 2x2 3x 4 0 The solutions of f by using the quadratic formula. 2x2 3x 4 x 2 1 are the numbers x where the graph of lies on or below the x-axis. The zeros of f can be found x 3 ± 232 4 2 2 1 2 4 2 1 21 2 3 ± 241 4 x As shown in Figure 2.5-5, the graph lies below the x-axis between the two zeros. Therefore, the solutions of the original inequality are all numbers x such that 3 241 4 2.35 x 0.85 x 3 241 4 Exact solution Approximate solution ■ Example 5 Solving an Inequality Solve 21 1 0. Solution 1 f x 2 2 5, x 5 x are easily read from the facThe zeros of 1 2, and 8. Therefore, you need only determine where tored form to be the graph of f is on or below the x-axis. A partial graph of f that clearly shows all three x-intercepts is shown in Figure 2.5-6a. A complete graph, which does not clearly show the x-intercept at 2, is shown in Figure x 8 21 2 1 2 6 Section 2.5 Inequalities 123 2.5-6b. Using both graphs, or using the trace feature on either one, conx 5 and x 8. firms that the graph is on or below the x-axis between Therefore, the solutions of the inequality are all numbers x such that 5 x 8. 5 200,000 10 10 10 10 5 Figure 2.5-6a 1,000,000 Figure 2.5-6b ■ The procedures used in the previous examples may be summarized as follows. Solving Inequalities 1. Write the inequality in one of the following forms. f(x) 77 0 f(x) 0 f(x) 66 0 f(x) 0 2. Determine the zeros of f, exactly if possible, approximately otherwise. 3. Determine the interval, or intervals, on the x-axis where the graph of f is above (or below) the x-axis. Applications Example 6 Solving a Cost Inequality A computer store has determined the cost C of ordering and storing x laser printers. C 2x 300,000 x If the delivery truck can bring at most 450 printers per order, how many printers should be ordered at a time to keep the cost below $1600? Solution To find the values of x that make C less than 1600, solve the inequality 6 1600 or equivalently, 2x 300,000 2x 300,000 1600 6 0. x x 124 500 Chapter 2 Equations and Inequalities In this context, the only solutions that make sense are those between 0 and 450. Therefore, choose a viewing window, such as the one shown in Figure 2.5-7, and graph 0 450 2x 300,000 x f x 1 2 1600. 500 Figure 2.5-7 The graph in Figure 2.5-7 shows that the zero of f is and the graph of C is negative, i.e., below the x-axis, for values greater than 300. Therefore, to keep costs under $1600, between 300 and 450 printers should be ordered per delivery. x 300, ■ Exercises 2.5 In Exercises 1–4, express the given statement in symbols. 1. x is nonnegative. 2. t is positive. 3. c is at most 3. 4. z is at least 17. In Exercises 5–10, represent the given interval on a number line. 29. 30. 31. 32 2x 7 3x 3x 2 2 x 1 1 2 2 3x x 5 3 2x 2x 1 3 2 5. (0, 8] 8. 1 1, 1 2 6. 9. 1 0, q 2 q, 0 1 7. 10. 2, 1 2, 7 4 2 3 3 4 In Exercises 11–16, use interval notation to denote the set of all real numbers x that satisfy the given inequality. 11. 5 x 8 12. 2 x 7 13. 3 6 x 6 14 14. 7 6 x 6 135 15. x 8 16. x 12 In Exercises 17–36, solve the inequality and express your answer in interval notation. 17. 2x 4 7 18. 3x 5 7 6 19. 3 5x 6 13 20. 2 3x 6 11 21. 6x 3 x 5 22. 5x 3 2x 7 23. 5 7x 6 2x 4 24. 5 3x 7 7x 3 25. 2 6 3x 4 6 8 26. 1 6 5x 6 6 9 27. 0 6 5 2x 11 28. 4 7 3x 6 0 33. 2x 3 5x 6 6 3x 7 34. 4x 2 6 x 8 6 9x 1 35. 3 x 6 2x 1 3x 4 36. 2x 5 4 3x 6 1 4x In Exercises 37–40, the constants a, b, c, and d are positive. Solve each inequality for x. 37. ax b 6 c 38. d cx 7 a 39. 0 6 x c 6 a 40. d 6 x c 6 d In Exercises 41 – 70, solve the inequality. Find exact solutions when possible, and approximate them otherwise. 41. x2 4x 3 0 42. x2 7x 10 0 43. x2 9x 15 0 44. x2 8x 20 0 45. 8 x x2 0 46. 4 3x x2 0 47. x3 x 0 48. x3 2x2 x 7 0 49. x3 2x2 3x 6 0 50. x4 14x3 48x2 0 51. x4 5x2 4 6 0 52. x4 10x2 9 0 53. x3 2x2 5x 7 2x 1 54. x4 6x3 2x2 6 5x 2 55. 2x4 3x3 6 2x2 4x 2 56. x5 5x4 7 4x3 3x2 2 57. 59. 61. 63. 3x 1 2x 4 7 0 x2 x 2 x2 2x 65. 2 x 3 1 x 1 58. 60. 62. 64. 66. 2x 1 5x 3 0 2x2 x 1 x2 4x 4 0 x 5 2x 3 2 2x 67. 68. 69. x3 3x2 5x 29 x2 7 7 3 x4 3x3 2x2 2 x 2 7 15 2x2 6
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x 8 2x2 5x 3 6 1 Be alert for hidden behavior. 70. 1 x2 In Exercises 71–73, read the solution of the inequality from the given graph. 71. 3 2x 6 0.8x 7 y y = 3 − 2x (−1.43, 5.86) y = 0.8x + 7 8 6 4 2 x 2 4 −4 −2 0 −2 Section 2.5 Inequalities 125 72. 8 7 5x 7 − 5x| (0.4, 3) y = 3 (2.4, 3) x −8 −4 0 4 8 −4 −8 73. x2 3x 1 4 y = x2 + 3x + 1 y 10 (−3.79, 4) 8 6 4 2 (0.79, 4) y = 4 x −4 −2 0 −2 2 4 74. The graphs of the revenue and cost functions for a manufacturing firm are shown in the figure. a. What is the break-even point? b. Which region represents profit? 60,000 40,000 20,000 Cost Revenue 1000 2000 3000 4000 75. One freezer costs $623.95 and uses 90 kilowatt hours (kwh) of electricity each month. A second freezer costs $500 and uses 100 kwh of electricity each month. The expected life of each freezer is 12 years. What is the minimum electric rate in cents per kwh for which the 12-year total cost (purchase price freezer? electricity costs) will be less for the first 126 Chapter 2 Equations and Inequalities 76. A business executive leases a car for $300 per month. She decides to lease another brand for $250 per month, but has to pay a penalty of $1000 for breaking the first lease. How long must she keep the second car in order to come out ahead? 77. One salesperson is paid a salary of $1000 per month plus a commission of 2% of her total sales. A second salesperson receives no salary, but is paid a commission of 10% of her total sales. What dollar amount of sales must the second salesperson have in order to earn more per month than the first? 78. A developer subdivided 60 acres of a 100-acre tract, leaving 20% of the 60 acres as a park. Zoning laws require that at least 25% of the total tract be set aside for parks. For financial reasons the developer wants to have no more than 30% of the tract as parks. How many one-quarter-acre lots can the developer sell in the remaining 40 acres and still meet the requirements for the whole tract? 79. If $5000 is invested at 8%, how much more should be invested at 10% in order to guarantee a total annual interest income between $800 and $940? 80. How many gallons of a 12% salt solution should be added to 10 gallons of an 18% salt solution in order to produce a solution whose salt content is between 14% and 16%? 81. Find all pairs of numbers that satisfy these two conditions: Their sum is 20 and the sum of their squares is less than 362. 82. The length of a rectangle is 6 inches longer than its width. What are the possible widths if the area of the rectangle is at least 667 square inches? 83. It costs a craftsman $5 in materials to make a medallion. He has found that if he sells the 50 x dollars each, where x is medallions for the number of medallions produced each week, then he can sell all that he makes. His fixed costs are $350 per week. If he wants to sell all he makes and show a profit each week, what are the possible numbers of medallions he should make? 84. A retailer sells file cabinets for 80 x dollars each, where x is the number of cabinets she receives from the supplier each week. She pays $10 for each file cabinet and has fixed costs of $600 per week. How many file cabinets should she order from the supplier each week in order to guarantee that she makes a profit? In Exercises 85 – 88, you will need the following formula for the height h of an object above the ground at time t seconds, where denotes initial velocity and h0 v0 h 16t 2 v0t h0 denotes initial height. 85. A toy rocket is fired straight up from ground level with an initial velocity of 80 feet per second. During what time interval will it be at least 64 feet above the ground? 86. A projectile is fired straight up from ground level with an initial velocity of 72 feet per second. During what time interval is it at least 37 feet above the ground? 87. A ball is dropped from the roof of a 120-foot-high building. During what time period will it be strictly between 56 feet and 39 feet above the ground? 88. A ball is thrown straight up from a 40-foot-high tower with an initial velocity of 56 feet per second. a. During what time interval is the ball at least 8 feet above the ground? b. During what time interval is the ball between 53 feet and 80 feet above the ground? Section 2.5A Excursion: Absolute-Value Inequalities 127 2.5.A Excursion: Absolute-Value Inequalities Objectives • Solve absolute-value inequalities by the Intersection Method • Solve absolute-value inequalities by the xIntercept Method y 50 40 30 20 f Polynomial and rational inequalities involving absolute value can be solved by either of two graphing methods. Intersection Method • Graph the expressions on each side of the inequality. • Determine the intervals on the x-axis where the graph of the expression on one side of the inequality is above or below the graph of the expression on the other side of the inequality. x-Intercept Method • Rewrite the inequality in an equivalent form with 0 on one side of the inequality. • Graph the function given by the nonzero side of the inequality. • Determine the x-values where the graph is above or below the x-axis. Example 1 Solving an Absolute-Value Inequality Using the Intersection Method g Solve 0 x4 2x2 x 2 6 11x. 0 Solution x −4 −2 0 2 4 −20 Figure 2.5.A-1 x4 2x2 x 2 The solutions of the x-intervals for which the graph of the graph of 11x. g x 0 0 6 11x f x 1 2 1 2 0 can be found be determining is below x4 2x2 x 2 0 x 0.17 A graphical intersection finder shows that the points of intersection occur and the graph of f is below the graph of g when between them, as shown in Figure 2.5.A-1. Therefore, approximate solutions of the original inequality are all x such that x 1.92, and 0.17 6 x 6 1.92. ■ Example 2 Solving an Absolute-Value Inequality Using the x-Intercept Method Solve x 4 x 2 ` ` 7 3. Solution Rewrite x 4 x 2 ` ` 7 3 as x 4 x 2 ` ` 3 7 0, graph , and find the intervals on the x-axis where the graph is above the x-axis. 128 Chapter 2 Equations and Inequalities y 8 4 −8 −4 0 4 8 x −4 −8 Figure 2.5.A-2 The graph of f is above the x-axis between the two zeros, which can be found algebraically or graphically to be function is not defined at denominator. Therefore, the solutions of x 2 x 1 2 and x 5. Notice that the because a fraction cannot have a zero x 4 x 2 ` ` 3 7 0 and (2, 5). The solution can also be written out. 6 x 6 2 or 2 6 x 6 5 ■ are the x-intervals 1 2 a , 2 b 1 2 Algebraic Methods Most linear and quadratic inequalities that contain absolute values can be solved exactly by using algebra. In fact, this is often the easiest way to solve such inequalities. The key to the algebraic method is to interpret the absolute value of a number as distance on the number line. For example, the inequality 5 states that r 0 0 the distance from r to 0 is less than or equal to 5 units. A glance at the number line in Figure 2.5.A-3a shows that these are the numbers r such that 5 r 5. −8 −6 −4 −2 0 2 4 6 8 5 units 5 units Figure 2.5.A-3a Similarly, the inequality 5 states that r 0 0 the distance from r to 0 is greater than or equal to 5 units. These values are the numbers r such that Figure 2.5.A-3b. r 5 or r 5, as shown in −8 −6 −4 −2 0 2 4 6 8 5 units 5 units Figure 2.5.A-3b Similar conclusions hold in the general cases, with 5 replaced by any number k. Section 2.5A Excursion: Absolute-Value Inequalities 129 Absolute-Value Inequalities Let k be a positive real number and r any real number. k k r r 00 00 00 00 is equivalent to is equivalent to k r k. r k or r k. Example 3 Solving an Absolute-Value Inequality Solve 0 3x 7 0 11. Solution Apply the first fact in the box above, with 11 place of k, and conclude that 3x 7 3x 7 is equivalent to in place of r and 11 in 0 0 11 3x 7 11 4 3x 18 4 3 6 x Add 7 Divide by 3 Therefore, the solution to 4 3 that is, all x such that 3x is all numbers in the interval 11 x 6. ■ Example 4 Solving an Absolute-Value Inequality Solve 0 5x 2 0 7 3. Solution Apply the second fact in the box with k, and in place of 7 . 5x 2 in place of r, 3 in place of 5x 2 6 3 or 5x Therefore, the solutions of the original inequality are the numbers in either of the intervals q, 1 1 or 2 1 5 a , q b , that is, x 6 1 or x 7 1 5 . ■ Example 5 Solving an Absolute-Value Inequality Solve 0 x2 x 4 2. 0 Solution Rewrite the absolute-value inequality as two quadratic inequalities using the algebraic definition. 130 Chapter 2 Equations and Inequalities The inequality 2 x2 x 4 0 x2 x 4 2 x2 x 2 0 0 is equivalent to two inequalities. or x2 x 4 2 x2 x 6 0 The solutions are all numbers that are solutions of either one of the two inequalities shown above. The solutions are the intervals on the x-axis that are determined by the following x2 x 2 x2 x 6 is on or below the x-axis is on or above the x-axis y 8 4 f(x) = x2 − x − 2 y 8 4 g(x) = x2 − x − 6 −8 −4 0 4 8 −8 −4 0 4 8 x x −4 −8 −4 −8 Figure 2.5.A-4a Figure 2.5.A-4b As shown in Figure 2.5.A-4a, the graph of the x-axis when f x 1 2 x2 x 2 is on or below 1 x 2. As shown in Figure 2.5.A-4b, the graph of above the x-axis when g x 2 1 x2 x 6 is on or x 2 or x 3. Therefore, the solutions of the original inequality are all numbers x such that x 2 or 1 x 2 or x 3, as shown in Figure 2.5.A-4c. −2 −1 2 3 Solutions of x2 − x − 2 ≤ 0 Solutions of x2 − x − 6 ≥ 0 Solutions of either one Figure 2.5.A-4c ■ Section 2.5A Excursion: Absolute-Value Inequalities 131 Example 6 Interpreting an Absolute-Value Inequality Let a and d represent real numbers with positive. d 0 0 6 d geometrically. x a a. Interpret b. Draw the interval represented. c. Write the equivalent simplified extended inequality. d. Interpret the last inequality. Solution a. Geometrically, x a 0 0 6 d means that the distance from x to a is less than d. b. c. a − δ a a + δ δ units δ units Figure 2.5.A- Add a to each term d. The solutions of the inequality are all numbers strictly between a d and a d. Exercises 2.5.A In Exercises 1–32, solve the inequality. Find exact solutions when possible, and approximate values oth
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erwise. 1. 3. 5. 7. 9. 11. 13. 3x 2 3 2x 2x 3 5x 12 5 2x 2x . 4. 6. 8. 5x 1 4 5x 3x 1 2 3x 10. 12. 14. ` ` ` 5 6 3x 6 7 6 ` x 1 3x 15. 17. 19. 21. 23. 25. 27. 28. 29. 31. ■ 3x 1 1 2x ` ` 2 x2 4 3 0 1 x2 1 ` ` 2 x2 x 4 0 x2 3x 4 2 6 6 0 7 1 4x x3 0 1 4x 2 3x ` ` 6 1 x2 2 x2 2 6 1 7 4 0 0 x2 x 1 1 0 3x2 8x 2 x5 x3 1 0 6 2 0 6 2 16. 18. 20. 22. 24. 26. 0 0 0 0 x4 x3 x2 x 1 7 4 x3 6x2 4x 30. 2x2 2x 12 x3 x2 x 2 ` 7 2 32. 0 ` 0 ` x2 9 x2 4 0 x2 x 2 x2 x 2 ` 6 2 7 3 132 Chapter 2 Equations and Inequalities 33. Critical Thinking Let E be a fixed real number. Show that every solution of solution of 5x 4 11 0 2 0 1 x 3 0 6 E. 6 E 5 0 is also a with a 6 b. Show that the solutions of 34. Critical Thinking Let a and b be fixed real numbers are all x such that a 6 x 6 b Important Concepts Section 2.1 Complete graphs. . . . . . . . . . . . . . . . . . . . . . . . . . 82 The intersection method . . . . . . . . . . . . . . . . . . . . 82 Zeros, x-intercepts, and solutions . . . . . . . . . . . . . 83 The x-intercept method. . . . . . . . . . . . . . . . . . . . . 84 Technological quirks . . . . . . . . . . . . . . . . . . . . . . . 84 Section 2.2 Section 2.3 Section 2.4 Section 2.5 Definition of a quadratic equation . . . . . . . . . . . . 88 Techniques used to solve quadratic equations . . . 88 Solving a quadratic equation by factoring . . . . . . 89 The Zero Product Property . . . . . . . . . . . . . . . . . . 89 Solutions of . . . . . . . . . . . . . . . . . . . . . . . . 89 Completing the square . . . . . . . . . . . . . . . . . . . . . 91 Solving ax2 bx c 0 by completing x2 k the square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 The quadratic formula . . . . . . . . . . . . . . . . . . . . . 92 The discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Polynomial equations in quadratic form . . . . . . . 94 Applied problems guideline . . . . . . . . . . . . . . . . . 97 Solutions in context . . . . . . . . . . . . . . . . . . . . . . . 98 Interest applications . . . . . . . . . . . . . . . . . . . . . . 100 Distance applications . . . . . . . . . . . . . . . . . . . . . 101 Other applications. . . . . . . . . . . . . . . . . . . . . . . . 102 Algebraic definition of absolute value . . . . . . . . 107 Geometric definition of absolute value . . . . . . . . 108 Properties of absolute value . . . . . . . . . . . . . . . . 108 Square root of squares. . . . . . . . . . . . . . . . . . . . . 109 Solving absolute-value equations . . . . . . . . . . . . 109 Extraneous solutions . . . . . . . . . . . . . . . . . . . . . . 110 Solving radical equations . . . . . . . . . . . . . . . . . . 111 Power Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Fractional equations . . . . . . . . . . . . . . . . . . . . . . 114 Interval notation . . . . . . . . . . . . . . . . . . . . . . . . . 118 Equivalent inequalities . . . . . . . . . . . . . . . . . . . . 119 Solving linear inequalities . . . . . . . . . . . . . . . . . . 119 Solving other inequalities . . . . . . . . . . . . . . . . . . 121 Quadratic and factorable inequalities . . . . . . . . . 122 133 134 Chapter Review Section 2.5.A The intersection method . . . . . . . . . . . . . . . . . . . 127 The x-intercept method . . . . . . . . . . . . . . . . . . . . 127 Algebraic methods . . . . . . . . . . . . . . . . . . . . . . . 128 Important Facts and Formulas To solve an equation of the form Method, use two steps. f x 1 2 g x 1 2 with the Intersection 1. Graph 2. Find the x-coordinate of each point of intersection and y2 y1 x x 1 1 2 2 g f When f is a function and a is a real number, the following are equivalent statements: • a is a zero of the function 2 • a is an x-intercept of the graph of f • a is a solution, or root, of the equation x 1 y f 0 f x 1 2 To solve an equation by the x-Intercept Method, use three steps. 0 f x 1. Rewrite the equation in the form 2. Graph f 3. Find the x-intercepts of the graph. The x-intercepts of the graph 0. of f are the solutions of 1 The only number whose square root is zero is zero itself. x f 1 2 2 A fraction is zero only when its numerator is zero and its denominator is nonzero. Quadratic Formula If a 0, then the solutions of ax2 bx c 0 are x b ± 2b2 4ac 2 a a 0, then the number of real solutions of If 1, or 2, depending on whether the discriminant, tive, zero, or positive, respectively. ax2 bx c 0 b2 4ac is 0, , is nega- Absolute Value if x 0 x x if x 6 0 x x 0 0 0 0 represents the distance between c and d on the number line. c d 0 represents the distance between c and 0 on the number line. c 0 0 0 Review Questions In Questions 1–8, solve the equation graphically. You need only find solutions in the given interval. Chapter Review 135 0, q 2 q, 0 2 0, q 2 q, 1 2 10, q 2 0, q 2 0, q 2 5 Section 2.1 1. x3 2x2 11x 6; 2. x3 2x2 11x 6; 3. x4 x3 10x2 8x 16; 4. 2x4 x3 2x2 6x 2 0; 5. 6. x3 2x2 3x 4 x2 2x 15 0; 3x4 x3 6x2 2x x5 x3 2 0; 7. 2x3 2x2 3x 5 0; 8. 21 2x 3x2 4x3 x4 0; Section 2.2 9. Solve for x: 3x2 2x 5 0 10. Solve for y: 3y2 2y 5 11. Solve for z: 5z2 6z 7 12. Solve for x: 325x2 17x 127 0 13. Solve for x: x4 11x2 18 0 14. Solve for x: x6 4x3 4 0 15. Find the number of real solutions of the equation 20x2 12 31x. 16. For what value of k does the equation kt2 5t 2 0 have exactly one real solution for t? Section 2.3 17. A jeweler wants to make a 1-ounce ring composed of gold and silver, using $200 worth of metal. If gold costs $600 per ounce and silver $50 per ounce, how much of each metal should she use? 18. A calculator is on sale for 15% less than the list price. The sale price, plus a 5% shipping charge, totals $210. What is the list price? 19. Karen can do a job in 5 hours and Claire can do the same job in 4 hours. How long will it take them to do the job together? 20. A car leaves the city traveling at 54 mph. A half hour later, a second car leaves from the same place and travels at 63 mph along the same road. How long will it take for the second car to catch up to the first? 21. A 12-foot rectangular board is cut into two pieces so that one piece is four times as long as the other. How long is the bigger piece? 22. George owns 200 shares of stock, 40% of which are in the computer industry. How many more shares must he buy in order to have 50% of his total shares in computers? 136 Chapter Review 23. A square region is changed into a rectangular one by making it 2 feet longer and twice as wide. If the area of the rectangular region is three times larger than the area of the original square region, what was the length of a side of the square before it was changed? 24. The radius of a circle is 10 inches. By how many inches should the radius be increased so that the area increases by 5p square inches? 25. If c x 2 1 is the cost of producing x units, then c x 1 x 2 is the average cost per unit. The cost of manufacturing x caseloads of ballpoint pens is given by 600x2 600x x2 1 c x 2 1 c x where order to have an average cost of $25? 2 1 is in dollars. How many caseloads should be manufactured in 26. An open-top box with a rectangular base is to be constructed. The box is to be at least 2 inches wide, twice as long as it is wide, and must have a volume of 150 cubic inches. What should be the dimensions of the box if the surface area is 90 square inches? Section 2.4 27. Simplify: b2 2b 1 0 0 In Exercises 28–40, find all real exact solutions. 28. 30. 32 2x 1 3 x 4 0 34. 26x2 7x 5 0 36. x2 x 2 x 2 0 38. 23 1 t2 2 0 0 29. 31. 33. 35. x 3 5 2 0 3x 1 4 0 2x2 x 2 0 x2 6x 8 x 1 0 37. 2x 1 2 x 39. 2x 1 2x 1 1 40. 23 x4 2x3 6x 7 x 3 Section 2.5 41. Express in interval notation: 8 a. The set of all real numbers that are strictly greater than b. The set of all real numbers that are less than or equal to 5. 42. Express in interval notation: a. The set of all real numbers that are strictly between b. The set of all real numbers that are greater than or equal to 5, but and 9; 6 strictly less than 14. 43. Solve for x: 44. Solve for x: 3 x 4 5 x. 1 2 4 6 2x 5 6 9. 45. On which intervals is 2x 1 3x 1 6 1? Chapter Review Chapter Review 137 137 46. On which intervals is 2 x 1 6 x? 47. Solve for x: 48. Solve for x: x 1 2 x2 1 1 2 1 x2 x 7 12. x 0. 2 49. If a. c. x 3 2x 3 x 3 2x 3 3 2x x 3 7 1, then which of these statements is true? 6 1, 7 1 6 1 2x 3 x 3 2x 3 6 x 3 b. d. e. None of these 50. If a. c. e. 0 6 r s t then which of these statements is false? b. 51. Solve and express your answer in interval notation: 2x 3 5x 9 6 3x 4. In Questions 52–61, solve the inequality. 52. x2 x 20 7 0 53. x 2 x 4 3 54. 55. 57 x2 x 9 x 3 6 1 x2 x 5 x2 2 7 2 Section 2.5.A 59. y 2 3 ` 3x 2 61. 0 5 2 ` 0 56. 58. x2 x 6 x 3 7 1 x4 3x2 2x 3 x2 4 6 1 60. 1 1 x2 ` ` 12 Maximum Area ctions There are two related branches of calculus: differential calculus and integral calculus. Differential calculus is a method of calculating the changes in one variable produced by changes in a related variable. It is often used to find maximum or minimum values of a function. Integral calculus is used to calculate quantities like distance, area, and volume. This Can Do Calculus finds the maximum area of the triangle formed by folding a piece of paper using different methods. The Maximum Area of a Triangle Problem -inch piece of paper is folded over to the oppoOne corner of an site side, as shown in Figure 2.C-1. A triangle is formed, and its area 8.5 11 formula is A 1 2 (base)(height). The following Example will find the length of the base that will produce the maximum area of the triangle using numerical, graphical, and algebraic methods. Example 1 Numerical Method 8.5 11 One corner of an -inch piece of paper is folded over to the opposite side, as shown in Figure 2.C-1. The area of the darkly shaded triangle at the lower left is the focus of this problem. a. Determine the shortest and the longest base that will produce a trian- gle by folding the paper. b. Measure the height when x has the lengths given in the chart, and cal- culate the area in each case. c. Creat
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e a scatter plot of the data. d. Estimate the length of the base that produces the maximum area, and state the approximate maximum area. Solution a. The base must be greater than 0 and less than 8.5 inches, and nt in the chart indicates that no triangle can be formed with a base length of 9 inches. b. The values shown in the chart may vary from your data. Base – 1 Height 4.25 Area 2.125 – 2 4.1 4.1 – 3 3.6 5.4 – 4 3.3 6.6 – 5 2.75 6.875 – 6 2 6 – 7 1.5 – 8 0.5 5.25 2 – 9 nt - x Figure 2.C-1 138 7 0 0 7 0 0 c. The graph of the data is shown in Figure 2.C-2. d. A maximum area of 6.875 in2 appears to occur when the base length Figure 2.C-2 8.5 y y x Figure 2.C-3 is 5 in. ■ 9 In Example 1, all calculations were accomplished with measurements. where y is the height of the trianNotice that the hypotenuse is gle. (Why?) 8.5 y, Because each triangle formed was a right triangle, the Pythagorean Theorem can be used to find an expression that gives the height as a function of the length of the base. That function can then be used to write a function that gives area in terms of the length of the base. Example 2 Algebraic Method Find a function of the base to represent the area of the triangle described in Example 1, graph the function along with the scatter plot of the data found in Example 1. Find the length of the base that produces maximum area. What is the maximum area? Solution The Pythagorean Theorem yields the following equation. 8. 72.25 17y y 2 y 2 72.25 17y x y 72.25 x 2 17 Height as a function of the base Hence, the area is represented by A 1 2 1 x 2 a 2 72.25 x 17 b . Using the max- imum finder on a calculator indicates that the maximum area of occurs at approximately x 4.9 in. 6.95 in2 ■ 9 Figure 2.C-4 To get exact values of x and the area, differential calculus is needed. However, graphing technology can provide very good approximations. Exercises In each problem, find the maximum by using a numerical method like the one shown in Example 1, and then by using an analytical and graphical method like the one shown in Example 2. Answer all questions given in the two examples. 1. Ten yards of wire is to be used to create a rectangle. What is the maximum possible area of the rectangle? 2. A rectangle is bounded by the x- and y-axes and the semicircle dimensions of the rectangle with maximum area? What are the y 236 x 2. 3. A rectangle is bounded by the x- and y-axes and the line y 1 2 . What are the dimensions of 4 x 2 the rectangle with maximum area? 139 C H A P T E R 3 Functions and Graphs This is rocket science! If a rocket is fired straight up from the ground, its height is a function of time. This function can be adapted to give the height of any object that is falling or thrown along a vertical path. The shape of the graph of the function, a parabola, appears in applications involving motion, revenue, communications, and many other topics. See Exercise 50 of Section 3.3. 140 Chapter Outline 3.1 Functions 3.2 Graphs of Functions 3.3 Quadratic Functions 3.4 Graphs and Transformations 3.4.A Excursion: Symmetry 3.5 Operations on Functions 3.5.A Excursion: Iterations and Dynamical Systems 3.6 Inverse Functions 3.7 Rates of Change Chapter Review can do calculus Instantaneous Rates of Change Interdependence of Sections 3.1 > 3.2 > > > 3.3 3.4 3.5 > > 3.6 3.7 T he concept of a function and function notation are central to mod- ern mathematics and its applications. In this chapter you will review functions, operations on functions, and how to use function notation. Then you will develop skill in constructing and interpreting graphs of functions. To understand the origin of the concept of a function it may help to consider some “real-life” situations in which one numerical quantity depends on, corresponds to, or determines another. Example 1 Determining Inputs and Outputs of Functions Describe the set of inputs, the set of outputs, and the rule for the following functions: a. The amount of income tax you pay depends on your income. b. Suppose a rock is dropped straight down from a high place. Physics feet. c. The weather bureau records the temperature over a 24-hour period tells us that the distance traveled by the rock in t seconds is 16t2 in the form of a graph (Figure 3.1-1). The graph shows the temperature that corresponds to each given time. 3.1 Functions Objectives • Determine whether a relation is a function • Find the domain of functions • Evaluate piecewise-defined and greatest integer functions 70º 60º 40ºT 50º 1284 16 20 24 A.M. Noon Hours P.M. Figure 3.1-1 141 142 Chapter 3 Functions and Graphs Solution The table below summarizes the features of each function. Set of inputs Set of outputs Function rule a. b. all incomes all tax amounts tax laws number of seconds, t, after dropping the rock distance rock travels Distance 16t2 c. time temperature time/temperature graph ■ The formal definition of function incorporates the set of inputs, the function rule, and the set of outputs, with a slight change in terminology. Definition of a Function A function consists of • a set of inputs, called the domain • a rule by which each input determines one and only one output • a set of outputs, called the range The phrase “one and only one” means that for each input (element of the domain), the rule of a function determines exactly one output (element of the range). However, different inputs may produce the same output. Example 2 Determining Whether a Relation is a Function The tables below list the inputs and outputs for two relations. Determine whether each relation is a function. a. Inputs Outputs b. Inputs Outputs Section 3.1 Functions 143 Solution In table a, the input 1 has two corresponding outputs, 5 and 6; and the input 3 has two corresponding outputs, 8 and 9. So the relation in table a is not a function. In table b, each input determines exactly one output, so the relation in table b is a function. Notice that the inputs 1, 3, and 9 all produce the same output, 5, which is allowed in the definition of a function. ■ The value of a function f that corresponds to a specific input value a, is found by substituting a into the function rule and simplifying the resulting expression. See Section 1.1 for a discussion of function notation. Example 3 Evaluating a Function Find the indicated values of a. f 3 2 1 Solution b. f 1 2x2 1 . x f 2 1 5 2 c. f 0 1 2 a. To find the output of the function f for input 3, simply replace x with 3 in the function rule and simplify the result. 232 1 210 3.162 5 2 1 226 5.099 Similarly, replace x with 5 f and 0 for b and c. 2 5 3 f 2 1 b. c. 1 f 0 2 1 1 2 202 1 1 2 Technology Tip Function notation can be used directly. For example, if a function is entered as ate the function at x 3, 3 press , then ENTER. Y1, Y1 1 2 evalu- Figure 3.1-2 Example 4 Finding a Difference Quotient For f x 2 1 x 2 x 2 and h 0, find a. x h f 1 2 b Solution each output. x h 2 h c. f 1 ■ f x 1 2 When function notation is used in expressions such as the basic rule applies: Replace x in the function rule with the entire expression within parentheses and simplify the resulting expression. , 1 2 f x h a. Replace x with x h in the rule of the function x2 2xh h2 x h 2 x h 2 2 1 2 144 Chapter 3 Functions and Graphs b. By part a 2xh h2 x h 2 x 2 2xh h2 x h 2 x 2 x 2 2xh h2 h x 2 x 2 4 3 4 c. By part b 2xh h 2 h h h 1 2x h 1 h 2 2x h 1 ■ If f is a function, then the quantity , as in Example 4, is called the difference quotient of f. Difference quotients, whose significance is explained in Section 3.7, play an important role in calculus. Functions Defined by Equations Equations in two variables can be used to define functions. However, not every equation in two variables represents a function. Example 5 Determining if an Equation Defines a Function Determine whether each equation defines y as a function of x. a. b. 4x 2y 3 5 0 y2 x 1 0 Solution a. The equation 4x 2y 3 5 0 can be solved uniquely for y. 2y 3 4x 5 y 3 2x 5 2 3 2x 5 A 2 y If a number is substituted for x in this equation, then exactly one value of y is produced. So the equation defines a function whose domain is the set of all real numbers and whose rule is stated below. f x 2 y2 x 1 0 1 3 2x 5 A 2 can not be solved uniquely for y: b. The equation y2 x 1 y ± 2x 1 or y 2x 1 y 2x 1 This equation does not define y as a function of x because, for example, the input produces two outputs, 2 and x 5 2. ■ Section 3.1 Functions 145 Domains When the rule of a function is given by a formula, as in Examples 3–6, its domain (set of inputs) is determined by the following convention. Domain Convention Unless information to the contrary is given, the domain of a function f consists of every real number input for which the function rule produces a real number output. x is Thus, the domain of a polynomial function such as 2 is defined for every value of x. Howthe set of all real numbers, since 2 ever, in cases where applying the rule of a function leads to one of the following, the domain may not consist of all real numbers. x f f 1 1 x 3 4x 1 • division by zero • the square root of a negative number (or nth root, where n is even) Example 6 Finding Domains of Functions Find the domain for each function given below. a. k x 2 1 x2 6x x 1 b. f u 2 1 2u 2 Solution a. When x 1, the denominator of x2 6x x 1 is 0 and the output is not x 1, defined. When fraction is defined. Therefore, the domain of k consists of all real numbers except 1, which is written as however, the denominator is nonzero and the x 1. b. Since negative numbers do not have real square roots, is a real number only when Therefore, the domain of f consists of all real numbers greater than or equal to 2, that is, the interval that is, when u 2 0, u 2. 2, q . 2u 2 3 2 ■ Applications and the Domain Convention The domain convention does not always apply when dealing with ap 16t2. plications. Consider the distance function for falling objects, Since t represents time, only nonnegative values of t make sense here, even thoug
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h the rule of the function is defined for all values of t. Analogous comments apply to other applications. d t 1 2 A real-life situation may lead to a function whose domain does not include all the values for which the rule of the function is defined. 146 Chapter 3 Functions and Graphs Example 7 Finding the Domain of a Profit Function A glassware factory has fixed expenses (mortgage, taxes, machinery, etc.) of $12,000 per week. In addition, it costs 80 cents to make one cup (labor, materials, shipping). A cup sells for $1.95. At most, 18,000 cups can be manufactured each week. Let x represent the number of cups made per week. a. Express the weekly revenue R as a function of x. b. Express the weekly cost C as a function of x. c. Find the rule and the domain of the weekly profit function P. Solution number sold 2 a. b. c. 1 1 2 2 2 price per cup 1 1.95x cost per cup 2 2 0.80x 12,000 2 revenue cost R 1.95x 1.15x 12,000 1 1 2 0.80x 12,000 number sold 1 2 fixed expenses 2 Although this rule is defined for all real numbers x, the domain of the function P consists of the possible number of cups that can be made each week. Since only whole cups can be made and the maximum production is 18,000, the domain of P consists of all integers from 0 to 18,000. ■ Piecewise-Defined and Greatest Integer Functions A piecewise-defined function is one whose rule includes several formulas. The formula for each piece of the function is applied to certain values of the domain, as specified in the definition of the function. Example 8 Evaluating a Piecewise-Defined Function For the piecewise-defined function 2x 3 x2 1 x e f 2 1 if x 6 4 if 4 x 10 find each of the following. a. 5 f 1 2 Solution b. f 8 2 1 c. the domain of f. a. Since 5 f 5 2 2 1 2 is less than 4, the first part of the rule applies. 5 3 7 b. Since 8 is between 4 and 10, the second part of the rule applies. 1 1 f 8 2 82 1 63 c. The rule of f gives directions when x 6 4 and when the domain of f consists of all real numbers x such that is, q, 10 . 1 4 4 x 10, x 10, so that ■ Section 3.1 Functions 147 The greatest integer function is a piecewise-defined function with infinitely many pieces if 3 x 6 2 if 2 x 6 1 if 1 x 6 0 if 0 x 6 1 if 1 x 6 2 if 2 x 6 3 The rule can be written in words as follows: For any number x, round down to the nearest integer less than or equal to x. The domain of the greatest integer function is all real numbers, and the range is the set of integers. It is written as x x f . 1 2 3 4 Example 9 Evaluating the Greatest Integer Function . Evaluate the following: x 3 4 b. 3 f 1 2 c. f 0 2 1 d. f 5 4b a e. p f 1 2 5 4 4.7 3 0 b. 3 f 1 2 d. f 5 4b .25 3 4 1 x 2 1 4.7 Let f a. f 1 Solution 4.7 a. f 1 2 2 0 3 4 c. e ■ Greatest Integer Function Technology Tip The greatest integer function is denoted INT in the NUM submenu of the MATH menu of TI. It is denoted INTG in the NUM submenu of the Casio OPTN menu. 148 Chapter 3 Functions and Graphs Exercises 3.1 In Exercises 1–4, describe the set of inputs, the set of outputs, and the rule for each function. 1. The amount of your paycheck before taxes is a function of the number of hours worked. 2. Your shoe size is a function of the length of your foot. 3. In physics, the pressure P of a gas kept at a constant volume is a function of the temperature T, related by the formula constant k. P T k for some 4. The number of hours of daylight at a certain latitude is a function of the day of the year. The following graph shows the number of hours of daylight that corresponds to each day. 17. 2 f 1 2 19. h(3) 21. h 3 2b a 18. f 3 a 2b 20. h 22. h 1 1 4 2 p 1 2 23. 25. 27. 29. 31. 33 24. 26. 28. 30. 32. 34 16 12 8 4 100 200 300 x In Exercises 35–42, compute and simplify the difference quotient (shown below). Assume h 0. f(x h) f(x) h 35. 37. 39. 41 3x 7 x x 2 2x 36. 38. 40. 42 10x x 2 x 3 1 x In Exercises 5–12, determine whether the equation defines y as a function of x. In Exercises 43–56, determine the domain of the function according to the domain convention. 5. y 3x 2 12 6. y 2x 4 3x 2 2 7. y 2 4x 1 8. 5x 4y 4 64 0 9. 3x 2y 12 10. y 4x 3 14 0 11. x2 y 2 9 12. y2 3x 4 8 0 43. f x 1 2 x 2 45. 47 44. g x 2 1 1 x 2 2 46. k u 2 1 2u 1 2x 1 48. h 2 x 1 Exercises 13–34 refer to the functions below. Find the indicated value of the function. f(x) 2x 3 x 1 g(t) t 2 1 h(x) x2 1 x 2 13. 15. f f 0 1 2 22 A 14. 16. f f B 1 1 A 2 22 1 B 49. u g 1 2 0 u 0u 50 2x 1 x 2 1 51. g y 1 2 53. g 55. 56 52. f t 2 1 2t 54. f t 2 1 24 t2 29 x 9 2 2 1 2x 2 x 1 In Exercises 57–62, find the following: a. Express the total cost of the box as a function Section 3.1 Functions 149 f (0) f (2.3) a. c. e. The domain of f b. d. f (1.6) f (5 2P) 57. f x 1 2 x 4 3 58. f x 1 2 x x e if x 6 0 if x 0 59. f x 1 2 x2 2x 3x 5 e if x 6 2 if 2 x 20 60. f x 1 2 x 5 3x e if 3 6 x 0 if 0 6 x 5 61. f x 1 2 ⎧ ⎨ • ⎩ 2x 3 5 x 0 x 2 0 if x 6 1 if 1 x 2 if x 7 2 62 2x 1 if 4 x 6 2 if 2 x 1 if x 7 1 63. Find an equation that expresses the area A of a circle as a function of its a. radius r b. diameter d 64. Find an equation that expresses the area A of a square as a function of its a. side s b. diagonal d 65. A box with a square base of side x is four times higher than it is wide. Express the volume V of the box as a function of x. 66. The surface area of a cylindrical can of radius r 2pr 2 2prh. and height h is If the can is twice as high as the diameter of its top, express its surface area S as a function of r. 67. A rectangular region of 6000 square feet is to be fenced in on three sides with fencing that costs $3.75 per foot and on the fourth side with fencing that costs $2.00 per foot. a. Express the cost of the fence as a function of the length x of the fourth side. b. Find the domain of the function. t t ft 68. A box with a square base measuring is to be made of three kinds of wood. The cost of the wood for the base is $0.85 per square foot; the wood for the sides costs $0.50 per square foot, and the wood for the top $1.15 per square foot. The volume of the box must be 10 cubic feet. of the length t. b. Find the domain of the function. 69. A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at a rate of 5 mph, then sits and rests for 30 minutes, and finally walks for 90 minutes at a rate of 3 mph. a. Write a piecewise-defined function that expresses his distance traveled as a function of time. b. Find the domain of the function. 70. Average tuition and fees in private four-year colleges in recent years were as follows. (Source: The College Board) Year Tuition & fees 1995 1996 1997 1998 1999 2000 $12,432 $12,823 $13,664 $14,709 $15,380 $16,332 a. Use linear regression to find the rule of a function f that gives the approximate average tuition in year x, where corresponds to 1990. b. Find How do they x 0 and , compare with the actual data? c. Use f to estimate tuition in 2003. 10 , 71. Suppose that a state income tax law reads as follows: Annual income Amount of tax less than $2000 0 $2000–$6000 2% of income over $2000 more than $6000 $80 plus 5% of income over $6000 Write a piecewise-defined function that represents the income tax law. What is the domain of the function? 150 Chapter 3 Functions and Graphs 72. The table below shows the 2002 federal income a. Write a piecewise-defined function T such that x T x is the tax due on a taxable income of dollars. What is the domain of the function? 2 1 b. Find T 24,000 , T 1 2 35,000 2 1 , and T 1 100,000 . 2 tax rates for a single person. Taxable income Tax not over $6000 10% of income over $6000, but not over $27,950 $600 15% amount over $6000 of over $27,950, but not over $67,700 $3892.50 27% amount over $27,950 of over $67,700, but not over $141,250 $14,625 30% amount over $67,700 of over $141,250, but not over $307,050 $36,690 35% amount over $141,250 of over $307,050 $94,720 38.6% amount over $307,050 of 3.2 Graphs of Functions Objectives Functions Defined by Graphs • Determine whether a graph represents a function • Analyze graphs to determine domain and range, local maxima and minima, inflection points, and intervals where they are increasing, decreasing, concave up, and concave down • Graph parametric equations NOTE An open circle on a graph indicates that the point is not a part of the graph, and a solid circle indicates that the point is a part of the graph. A graph may be used to define a function or relation. Suppose that f is a is function defined by a graph in the coordinate plane. If the point 2 x on the graph of f, then y is the output produced by the input x, or . 1 x, y 1 y f 2 Example 1 A Function Defined by a Graph The graph in Figure 3.2.1 defines the function f. Determine the following. a. f 0 1 2 b. f 3 1 2 c. the domain of f d. the range of f y 10 5 −10 −5 0 5 10 x −5 −10 Figure 3.2-1 Section 3.2 Graphs of Functions 151 Solution a. To find f(0), notice that the point (0, 7) is on the graph. Thus, 7 is the output produced by the input 0, or f 0 b. To find f(3), notice that the point (3, 0) is on the graph. Thus, 0 is the output produced by the input 3, or f 3 7. 0. 1 1 2 2 c. To find the domain of the function, find the x-coordinates of the point farthest to the left and farthest to the right. Then, determine whether there are any gaps in the function between these values. 8, 9 ( ) is the point farthest to the left, and (7, 8) is farthest to the right. The function does not have a point with an x-coordinate of 2, so the domain of f is ) and ( d. To find the range of the function, find the y-coordinates of the 8, 2 2, 7 . 4 3 highest and lowest points, then determine if there are any gaps between these values. The highest point is (7, 8) and the lowest point ), and there are no y-values between and 8 that do not is ( correspond to at least one x-value. Thus, the range of f is 9, 8 9 9 8 , . 3 4 ■ The Vertical Line Test If a graph represents a function, then each input determines one and only one output. Thus, no two points can have the same x-coordinate and different y-coordinates. Since
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any two such points would lie on the same vertical line, this fact provides a useful test for determining whether a graph represents a function. Vertical Line Test A graph in a coordinate plane represents a function if and only if no vertical line intersects the graph more than once. Example 2 Determining Whether a Graph Defines a Function Use the Vertical Line Test to determine whether the following graphs represent functions. If not, give an example of an input value that corresponds to more than one output value. y y 8 8 4 4 −8 −8 −4 −4 0 0 −4 −4 −8 −4 −4 −8 −8 −8 −4 −4 Figure 3.2-2 152 Chapter 3 Functions and Graphs Solution y 8 4 −8 −4 0 −4 −8 y 8 4 x 8 −8 −4 0 4 8 x −4 −8 Figure 3.2-3 x 4 The vertical line intersects the graph above at (4, 2) and 4, 2 so the graph is not a func2 1 tion. The two input 4 has corresponding outputs, 2 and 2. , There is no vertical line that intersects the graph above in more than one place, so this graph defines a function. ■ Analyzing Graphs In order to discuss a graph or compare two graphs, it is important to be able to describe the features of different graphs. The most important features are the x- and y-intercepts, intervals where the graph is increasing or decreasing, local maxima and minima, intervals where the graph is concave up or concave down, and points of inflection. Increasing and Decreasing Functions A function is said to be increasing on an interval if its graph always rises as you move from left to right over the interval. It is decreasing on an interval if its graph always falls as you move from left to right over the interval. A function is said to be constant on an interval if its graph is a horizontal line over the interval. y y y x x x increasing decreasing Figure 3.2-4 constant Section 3.2 Graphs of Functions 153 7 Example 3 Where a Function is Increasing/Decreasing On what interval is the function ing? constant increasing? decreas- 5 5 Solution 2 Figure 3.2-5 peaks y valleys Figure 3.2-6 3.1 The graph of the function, shown in Figure 3.2-5, suggests that f is decreas. ing on the interval increasing on 4 Using the trace feature on a graphing calculator, you can confirm the function is constant between 0 and 2. For an algebraic proof that f is constant on , and constant on see Exercise 55. q, 0 2, q 0, 2 0 Local Maxima and Minima The graph of a function may include some peaks and valleys, as in Figure 3.2-6. A peak may not be the highest point on the graph, but it is the highest point in its neighborhood. Similarly, a valley is the lowest point in its neighborhood. x c A function f has a local maximum (plural: local maxima) at if the f x c, f c graph of f has a peak at the point for all x near c. Similarly, a function has a local minimum (plural: local minima) at . This means f f if the graph of f has a valley at This means that for all x near d. x d d d, f 22 22 Calculus is usually needed to find exact local maxima and minima. However, they can be approximated with a calculator. 4.7 4.7 Example 4 Finding Local Maxima and Minima 3.1 Figure 3.2-7 Graph f x 2 1 Solution x 3 1.8x2 x 1 and find all local maxima and minima. 1.2 0.1 1.1 Figure 3.2-8 In the decimal or standard window, the graph does not appear to have any local maxima or minima (see Figure 3.2-7). Select a viewing window such as the one in Figure 3.2-8 to see that the function actually has a local maximum and a local minimum (Figure 3.2-9). The calculator’s minimum finder and maximum finder show that the local minimum occurs when x 0.763 and the local maximum occurs when x 0.437. 1 1.2 1.2 0.1 1.1 1 0.1 1.1 Figure 3.2-9 1 ■ 154 Chapter 3 Functions and Graphs Concavity and Inflection Points Concavity is used to describe the way that a curve bends. For any two points in a given interval that lie on a curve, if the segment that connects them is above the curve, then the curve is said to be concave up over the given interval. If the segment is below the curve, then the curve is said to be concave down over the interval (see Figure 3.2-10). A straight line is neither concave up nor concave down. A point where the curve changes concavity is called an inflection point. y y y inflection point x x x concave down concave up concave up concave down Figure 3.2-10 Example 5 Analyzing a Graph Graph the function ing, using the graph and a maximum and minimum finder. 2x 3 6x 2 x 3 x f 1 2 and estimate the follow- a. all local maxima and minima of the function b. intervals where the function is increasing and where it is decreasing c. all inflection points of the function d. intervals where the function is concave up and where it is concave down Solution a.–b. The maximum and minimum finders show that the function has a local maximum at Thus, the graph shows that the function is decreasing over the intervals q, 0.0871 and increasing over the interval (0.0871, 1 1.9129). and a local minimum at x 0.0871. x 1.9129 1.9129, q and , 1 2 2 c.–d. The function is concave up on the left and concave down on the right, and the inflection point appears to be at about Thus, the funcand concave down over the tion is concave up over the interval interval q, 1 x 1. 1, q . 1 2 1 2 ■ 10 4.7 4.7 10 Figure 3.2-11 Section 3.2 Graphs of Functions 155 Graphs of Piecewise-Defined and Greatest Integer Functions The graphs of piecewise-defined functions are often discontinuous, that is, they commonly have jumps or holes. To graph a piecewise-defined function, graph each piece separately. Example 6 Graphing a Piecewise-Defined Function Graph the piecewise-defined function below. if x 1 if 1 6 x 4 x2 x 2 x e f 1 2 Solution The graph is made up of parts of two different graphs, corresponding to the different parts of the function. x 1, the graph of f For coincides with the graph of y x 2. 1 6 x 4, For the graph of f coincides with the graph of y x 2. y 8 x x 4 8 −8 0 −4 −4 −8 y 8 4 8 −8 0 −4 −4 −8 y 8 4 x 4 8 −8 0 −4 −4 −8 Combining these partial graphs produces the graph of f. Figure 3.2-12 ■ Piecewise-defined functions can be graphed on a calculator, provided that you use the correct syntax. However, the screen does not show which endpoints are included or excluded from the graph. 156 Chapter 3 Functions and Graphs Technology Tip Inequality symbols are in the TEST menu of TI-84/TI-83. TI-89 has the symbols <, >, and | on the keyboard, and other inequality symbols and logical symbols (such as “and”) are in the TEST submenu of the MATH menu. Graphing Exploration Graph the function f from Example 6 on a calculator as follows: On TI-84/TI-83 calculators, graph these two equations on the same screen: Y1 Y2 11 X 4 21 22 On a TI-89/92, graph these equations on the same screen: Y1 X 2 Y2 X 2 0 X 7 1 X 1 and X 4 0 To graph f on a Casio, graph these equations on the same screen (including commas and square brackets): Y1 Y2 6, 1 X2, 3 X 2, 4 1, 4 3 4 How does your graph compare with Figure 3.2-12? Example 7 The Absolute-Value Function 3.1 Graph f x 1 2 x . 0 0 Solution 4.7 4.7 3.1 Figure 3.2-13 The absolute-value function tion, since by definition f x 2 1 x 0 0 is also a piecewise-defined func- x 0 0 x x e if x 6 0 if x 0 Its graph can be obtained by drawing the part of the line right of the origin and the part of the line or by graphing on a calculator. ABS X y x to the to the left of the origin Y1 y x y 8 4 −8 0 −4 −4 −8 x 4 8 y 8 4 −8 0 −4 −4 −8 0 −4 −4 −8 Figure 3.2-14 x ■ Section 3.2 Graphs of Functions 157 Example 8 The Greatest Integer Function Graph the greatest integer function f x 1 2 x . 4 3 Solution The greatest integer function can easily be graphed by hand, by considering the values of the function between each two consecutive integers. x 1 For instance, between is always 2, so the graph there is a horizontal line segment, all of whose points have y-coordinate with a solid circle on the left endpoint and an open circle on the right endpoint. The rest of the graph is obtained similarly. the value of x 2 2, and 5 −4 −3 −2 0 1 2 3 4 5 −2 −3 −4 −5 Figure 3.2-15 ■ A function whose graph consists of horizontal line segments, such as Figure 3.2-15, is called a step function. Step functions can be graphed on a calculator, but some features of their graphs may not be shown. Graphing Exploration f x Graph the greatest integer function on your calculator (see the Technology Tip on page 147). Does your graph look like Figure 3.2-15? Change your calculator to “dot” rather than “connected” mode (see the Technology Tip at left) and graph again. How does this graph compare with Figure 3.2-15? Can you tell from the graph which endpoints are included and which are excluded? x 2 1 4 3 Parametric Graphing In parametric graphing, the x-coordinate and the y-coordinate of each point on a graph are each given as a function of a third variable, t, called a parameter. The functions that give the rules for the coordinates are called parametric equations. Technology Tip To change to dot mode, select DOT in the TI MODE menu. In the Casio SETUP menu, set the DRAWTYPE to PLOT. 158 Chapter 3 Functions and Graphs A parametric graph can be thought of as representing the function where f t 2 1 x, y 1 2 x x t 2 1 and y y t 2 1 are the rules for the x- and y-coordinates. Note that the graph will not necessarily pass the Vertical Line Test. Example 9 Graphing a Parametric Equation Graph the curve given by x 2t 1 y t2 3 Solution Make a table of values for t, x, and y. Then plot the points from the table and complete the graph. t 2 1 0 1 2 3 x 2t 1 y t2 Graphing Exploration (x, y) 3, 1 2 1 1, 2 1 2 1, 3 3, 2 1 1 2 2 (5, 1) (7, 6) ■ Graph the equations from Example 9 on a calculator in parametric mode in the standard viewing window. Set the range so that 10 t 10, Use the trace feature to find at least three points on the graph that are not given in the table in Example 9. with t-step 0.1. Parametric mode can be used to graph equations of the form the form x f y . 1 2 y f or x 1 2 y 10 8 6 4 2 (−3, 1) (7, 6) (5, 1) x 8 10 6 (3, −2) 2 (1, −3) −10 −8 0 −6 −4 (−1, −2) −4 −6 −8
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−10 Figure 3.2-16 Technology Tip To change to parametric mode, choose PAR in the TI MODE menu or PARM in the TYPE submenu of the Casio GRAPH menu (on the main menu). Graphing or y f(x) in Parametric Mode x f(y) NOTE The graph in part a is the same as in Example 9. To obtain the equation in part a from the parametric equations in Example 9, solve the first equation for t and substitute the result into the second equation. Section 3.2 Graphs of Functions 159 To graph y f(x) in parametric mode, let x t y f(t) To graph x f(y) in parametric mode, let x f(t) y t Example 10 Graphing in Parametric Mode Graph the following equations in parametric mode on a calculator. a. y 2 x 1 2 a b 3 Solution b. x y2 3y 1 a. Let x t and y 2 t 1 2 b a 3. b. Let x t2 3t 1 and y t. 10 10 10 10 10 10 10 10 Figure 3.2-17 Notice that the graph in part b does not pass the Vertical Line Test. The equation does not represent y as a function of x. ■ Parametric equations will be studied more thoroughly in Chapter 11. 160 Chapter 3 Functions and Graphs Exercises 3.2 In Exercises 1–4, the graph below defines a function, f. Determine the following: 9. y 8 4 −8 −4 0 4 8 x −4 −8 1. 5 f 1 2 2. f 1 1 2 3. the domain of f 4. the range of f In Exercises 5–8, the graph below defines a function, g. Determine the following: y 8 4 −8 −4 0 4 8 x −4 −8 5. g 1 1 2 6. g 5 1 2 7. the domain of g 8. the range of g In Exercises 9–14, use the Vertical Line Test to determine whether the graph defines a function. If not, give an example of an input value that corresponds to more than one output value. 10. 11. 12. y 8 4 −8 −4 0 4 8 −4 −8 8 4 y −8 −4 0 4 8 −4 −8 8 4 y −8 −4 0 4 8 −4 −8 8 4 y −8 −4 0 4 8 −4 −8 x x x x 13. 14. y 8 4 −8 −4 0 4 8 −4 −8 8 4 y −8 −4 0 4 8 −4 −8 x x In Exercises 15 and 16, the graph of a function is shown. Find the approximate intervals on which the function is increasing and on which it is decreasing. 15. 16. −6 −5 −4 −3 −2 −1 x 1 2 3 4 y y −6 −5 −4 −3 −2 −1 1 2 3 x In Exercises 17–22, graph each function. Find the approximate intervals on which the function is increasing, decreasing, and constant. 17. f 18. g 19 8x 2 8x 5 Section 3.2 Graphs of Functions 161 20. f 21. g x x 1 1 2 2 22. g x 2 1 x 4 0.7x 3 0.6x 2 1 0.2x 4x 2 x 1 In Exercises 23–28, graph each function. Estimate all local maxima and minima of the function. 23. f x 1 2 x 3 x 25. h x 2 1 x x 2 1 24. g t 2 1 26. k x 1 2 216 t 2 x 3 3x 1 27. f 28.8x 2 x 2 2x 3 x 2 1 29. a. A rectangle has a perimeter of 100 inches, and one side has length x. Express the area of the rectangle as a function of x. b. Use the function in part a to find the dimensions of the rectangle with perimeter 100 inches and the largest possible area. 30. a. A rectangle has an area of 240 in , and one side has length x. Express the perimeter of the rectangle as a function of x. 2 b. Use the function in part a to find the 2 dimensions of the rectangle with area 240 in and the smallest possible perimeter. 31. a. A box with a square base has a volume of 867 3 in . Express the surface area of the box as a function of the length x of a side of the base. (Be sure to include the top of the box.) b. Use the function in part a to find the dimensions of the box with volume 867 in and the smallest possible surface area. 3 2 32. a. A cylindrical can has a surface area of 60 in . Express the volume of the can as a function of the radius r. b. Use the function in part a to find the radius 2 and height of the can with surface area 60 in and the largest possible volume. In Exercises 33–36, graph each function. Find the approximate intervals on which the function is concave up and concave down, and estimate all inflection points. 33. f x 1 2 x3 34. f x 1 2 x3 2x 35. h 36 2x 2 x 3 3x 2 2x 1 162 Chapter 3 Functions and Graphs In Exercises 37–40, a. Graph each function. b. Find the approximate intervals on which the function is increasing, decreasing, and constant. c. Estimate all local maxima and minima. d. Find the approximate intervals on which the function is concave up and concave down. e. Estimate the coordinates of any inflection points. 37. f 39 2x 1 38. f x 3 3x 2 2 40. g x 2 4x 3 x 3 4x 2 x x 1 1 2 2 In Exercises 41–44, sketch the graph of the function. Be sure to indicate which endpoints are included and which are excluded. 41. f x 1 2 2x 3 x 2 e if x 6 1 if x 1 42. g x 1 2 x 0 3x 4 0 e if x 6 1 if x 1 43. k u 2 1 ⎧ ⎪ ⎨ ⎪ ⎩ 2u 2 u 2u2 u 3 4 if u 6 3 if 3 u 1 if u 7 1 44. f x 1 2 ⎧ ⎪ ⎨ ⎪ ⎩ if x 6 2 if 2 x 6 4 x 2 x 2x if x 4 In Exercises 45–49, a. Use the fact that the absolute-value function is piecewise-defined (see Example 7) to write the rule of the given function as a piecewise-defined function whose rule does not include any absolute value bars. b. Graph the function. 45. f x 1 2 x 0 47 49. f x 1 2 0 x 5 0 46. g 48 In Exercises 50–53, sketch the graph of the function. Be sure to indicate which endpoints are included and which are excluded. 52 53. f x 1 2 2 x 3 4 54. A common mistake is to graph the function f in Example 6 by graphing both on the same screen, with no restrictions on x. Explain why this graph could not possibly be the graph of a function. and y x2 y x 2 55. Show that the function x f 2 1 constant on the interval 0, 2 3 4 piecewise definition of absolute value in Example 0 x 2. 7 to compute x 0 0 Hint: Use the when is x f . 0 0 x 2 1 2 In Exercises 56–59, use your calculator to estimate the domain and range of the function by tracing its graph. 56. g 58 2x 2 4 57. 59. h x 2 1 f x 1 2 2x 2 4 3x 2 In Exercises 60 and 61, draw the graph of a function f that satisfies the given conditions. The function does not need to be given by an algebraic rule. 1 2 2 2 60 when x is in the interval 1, a 1 2b starts decreasing when 3 f starts increasing when 0 1 2 x 1 x 5 61. • domain • range 1 • 1 2b , 4 3 5, 6 3 3 4 1 2 4 In Exercises 62–67, graph the curve determined by the parametric equations. 2 2 2 62. 63. x 0.1t3 0.2t 2 2t 4 y 1 t 5 t 6 x t2 3t 2 y 8 t3 4 t 4 1 1 64. x t 2 6t y 1t7 5 t 9 1 65. x 1 t2 y t3 t 1 66. x t2 t 1 y 1 t t2 4 t 4 2 1 50. 51 Exercise 50.) x 1 2 3 (This is not the same function as in 67. x 3t Section 3.3 Quadratic Functions 163 In Exercises 68–71, graph the equation in parametric mode. Give the rule for x and for y in terms of t. 68. y x 3 x 2 6x 69. y x 4 3x3 x 2 70. x y 3 5y 2 4y 5 71. x y 4 3y 2 5 3.3 Quadratic Functions Objectives • Define three forms for quadratic functions • Find the vertex and intercepts of a quadratic function and sketch its graph • Convert one form of a quadratic function to another• Parabolas The rule of a quadratic function is a polynomial of degree 2. The shape of the graph of a quadratic function is a parabola. Three parabolas are shown in Figure 3.3-1, with important points labeled. y 8 −8 0 −4 −4 −8 x 8 4 vertex x-intercept y-intercept opens upward y vertex 8 4 x 0 4 −8 y-intercept −4 −4 x-intercepts −8 opens downward Figure 3.3-1 y 8 y-intercept x 4 8 −8 vertex 0 −4 −4 −8 opens upward Notice that the graph of a quadratic function • can open either upward or downward • always has a vertex which is either the maximum or minimum • always has exactly 1 y-intercept • can have 0, 1, or 2 x-intercepts A parabola is symmetric about a line through the vertex called the axis of symmetry. Quadratic Functions Quadratic functions can be written in several forms. 164 Chapter 3 Functions and Graphs Three Forms of a Quadratic Function A quadratic function can be written in any of the following forms: Transformation form: Polynomial form: x-Intercept form: f(x) a(x h)2 k f(x) ax 2 bx c f(x) a(x s)(x t) where a, b, c, h, k, s, and t are real numbers and positive, the graph opens upward; and if a is negative, the graph opens downward. a 0. If a is Transformation Form The transformation form is the most useful form for finding the vertex. For 2 k, a quadratic function written in transformation form, the vertex is the point with coordinates x h a h Since the y-intercept of a function is 0 h 0 f 2 k ah2 k f , 1 2 1 a 1 2 ah2 k. 2 shows the y-intercept is Since the x-intercepts occur when x h of the quadratic equation the x-intercepts are the solutions a 1 x h a a 1 y 0, 2 k 0 ± The x-intercepts are h k a A and h k a . A Example 1 Transformation Form f For the function y-intercepts. Then sketch the graph, find the vertex and the x- and Solution 2 x f In 1 vertex is , h 3 , and k 4 . The the y-intercept is h, k 2 1 1 3, 4 2 ah2 k 2 32 1 2 4 1 2 14 y 20 16 12 8 4 x −4 0 −4 8 4 (3, −4) Figure 3.3-2 and the x-intercepts are k a k a h h A A Section 3.3 Quadratic Functions 165 12 4.4 3 12 1.6 To graph the function, plot the vertex and intercepts, then draw the parabola. Since a is positive, the parabola opens upward. ■ Polynomial Form The polynomial form is the most useful form for finding the y-intercept. Since the y-intercept of a function is the y-intercept is c ax2 bx c The x-intercepts are the solutions of the quadratic equation 0, which can be solved either by factoring or by the quadratic formula. In general, the x-intercepts are b 2b2 4ac 2a and b 2b2 4ac 2a for values which make one x-intercept. If b2 4ac b2 4ac positive. If b2 4ac 0, is negative, then there are no x-intercepts. then there is only To find the vertex of the graph of a quadratic function in polynomial form, compare the transformation form to the polynomial form. First, multiply and distribute the terms in the transformation form, so the coefficients can be compared. x h a 2 x 2 2hx h2 a k ax2 2ahx ah2 k ax2 2ahx ah2 k 1 Match the second coefficient in each of the two equivalent forms. transformation form distribute the a multiply 2ah b 2a h Since h is the x-coordinate of the vertex in the transformation form, b 2a is the x-coordinate of the vertex in the polynomial form. The y-coordinate of the vertex can be found by substituting into the function. Thus, b 2a the vertex of the graph of a quadratic function is b 2a a , f b 2abb a 166 Chapter 3 Functions and Graphs Example 2 Polynomial Form y 9 6 3 (−1, 2) −6 −3 0
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3 x 6 For the function y-intercepts. Then sketch the graph. x 2 2x 3, x f 2 1 find the vertex and the x- and Solution In f x 1 2 x 2 2x 3, a 1, b 2, and c 3 c 3. Thus, the y-intercept is and the vertex is b 2a a b2 4ac 8, Since tions, so there are no x-intercepts. b 2abb , the quadratic equation 1 2 1, 2 2 1 1 2 2 x 2 2x 3 0 bb 2 has no solu- Figure 3.3-3 To graph the function, plot the vertex and y-intercept, then draw the parabola. Since a is positive, the parabola opens upward. ■ NOTE Not all quadratic functions can be written in x-intercept form. If the graph of a quadratic function has no x-intercepts, then there are no real values of s and t for which 21 2 x-Intercept Form The x-intercept form is the most useful form for finding the x-intercepts. For function written form, in x f the x-intercepts of the graph are s and t. Notice f that both of these values are solutions to the equation 0 s s quadratic x t x-intercept a a x s s t 0 21 2 21 s t 2 and 21 0 2 1 Since the y-intercept of a function is 0 s a 21 21 ast 2 so the y-intercept is ast. To find the vertex of a graph of a quadratic function in x-intercept form, recall that a parabola is symmetric about a line through the vertex. Thus, the vertex is exactly halfway between the x-intercepts. The x-coordinate s t 2 b of the vertex is the average of the x-intercepts, or a . To find the y-coordinate of the vertex, substitute this value into the function. The vertex is bb Example 3 x-Intercept Form For the graph of the function 21 find the vertex and the x- and y-intercepts. Then sketch the graph. Section 3.3 Quadratic Functions 167 Solution 1 2 1 x 4 x 2 21 , a 1 2 2 , s 4, and t 2 . The x-intercepts In f x 2 1 are (1, 4.5) and the y-intercept is s 4, and t 2 ast 1 2 4 1 21 2 2 4 y 8 4 −8 −4 0 4 8 x −4 −8 and the vertex is bb bb a 1, 4.5 1 2 Figure 3.3-4 To graph the function, plot the vertex and intercepts, then draw the parabola. Since a is negative, the parabola opens downward. ■ Changing from One Form to Another It is sometimes necessary to change a quadratic function from one form to another. To change either transformation or x-intercept form to polynomial form, distribute and collect like terms. To change either polynomial or transformation form to x-intercept form, factor out the leading coefficient, then factor or use the quadratic formula to find the x-intercepts. Example 4 Changing to Polynomial and x-Intercept Form Write the following functions in polynomial and x-intercept form, if possible. a. b. c.4 3x 2 3.9x 43.2 2 x 2 x 4 2 21 2 1 Solution a. To change the function f to polynomial form, distribute and collect like terms 6x 9 0.4 0.4 0.4x 2 2.4x 5.6 1 2 2 f b2 4ac 3.2 Since the function f cannot be written in x-intercept form. 2 1 is negative, there are no x-intercepts. Thus, 168 Chapter 3 Functions and Graphs b. The function g is already given in polynomial form. To find the a 3, x-intercept form, first factor out formula on the remaining expression. then use the quadratic x x g g 1 1 2 2 3x 2 3.9x 43.2 3 x 2 1.3x 14.4 1 2 x-intercepts: 1.3 ± 2 1.3 1 2 4 2 2 1 1 21 14.4 2 1.3 ± 7.7 2 4.5 and 3.2 So the x-intercept form is 3 x 4.5 g x x 3.2 1 2 1 21 2 c. The function h is already given in x-intercept form. To change the function to polynomial form, distribute and collect like terms. x 2 x 4 2 21 x 2 2x 8 2 2 2x 2 4x 16 ■ To convert a quadratic function to transformation form, it may be necessary to complete the square. Example 5 Changing to Transformation Form Write the following functions in transformation form. a. b. f g x x 1 1 2 2 3x 2 4x 1 0.3 x 1 x 2 21 2 1 Solution a. Factor a 3 shown below. out of the first two terms, then complete the square as f x 1 2 1 3 3x 2 4x 3b x b x 4 9 4 9b 3 3 aa 2 Factor 3 out of first two terms 4 9b 1 Add and subtract 1 x2 4 3 Write square 2 b 2b a x 4 9 as a perfect 3b x 2 3b a 4 3 1 3 1 Distribute the 3 over the 4 9 Combine like terms CAUTION When completing the square of an expression that is part of a function, it is not possible to divide by the leading coefficient or add a term to both sides. It is necessary to factor out the leading coefficient, then add and subtract the 2 term. b 2b a Summary of Quadratic Functions Section 3.3 Quadratic Functions 169 b. Multiply the terms in parentheses, then complete the square as shown below. 0.3 2 0.3 x 1 1 g x 1 x 2 21 x2 x 2 2 1 2 x2 x 0.25 0.25 2 0.3 1 0.3 0.3 0.3 2 g x 1 1 11 x 0.5 x 0.5 11 x 0.5 2 2 0.25 2 2 2.25 2 2 2 0.675 2 Multiply 2 Add and subtract a x2 x 0.25 Write 2 2 b 2b as a perfect square Combine like terms Distribute the 0.3 over the 2.25 ■ Summary of Quadratic Forms Below is a summary of the basic forms and the important points of a quadratic function. You should memorize the highlighted items. Name Transformation Polynomial x-Intercept Form f(x) a(x h)2 k f(x) ax2 bx c f(x) a(x s)(x t) Vertex (h, k) b 2a a , f b 2abb bb x-Intercepts h k a A and h k a A ah2 k y-Intercept Applications b 1b2 4ac 2a and b 1b2 4ac 2a s and t c ast In the graph of a quadratic function, the vertex of the parabola is either a maximum or a minimum of the function. Thus, the solution of many applications depends on finding the vertex of the parabola. Example 6 Maximum Area for a Fixed Perimeter Find the dimensions of a rectangular field that can be enclosed with 3000 feet of fence and that has the largest possible area. Solution y Let x denote the length and y the width of the field, as shown in Figure 3.3-5. Perimeter x x y y 2x 2y Area xy y x x Figure 3.3-5 170 Chapter 3 Functions and Graphs The perimeter is the length of the fence, or 3000. 2x 2y 3000 y 1500 x The area of the field is A x 1 2 xy x 1500 x 1 2 x2 1500x The graph of the function A is a parabola that opens downward, so the maximum occurs at the vertex. The function is in polynomial form, so the vertex occurs when x b 2a 1500 1 2 2 1 562,500 ft2, 750. The largest possible area is y 1500 750 750. x 750 The dimensions of the field are 750 by 750 ft. which occurs when and ■ Example 7 Maximizing Profit A vendor can sell 275 souvenirs per day at a price of $2 each. The cost to price increase decreases sales the vendor is $1.50 per souvenir. Each by 25 per day. What price should be charged to maximize profit? 10¢ Solution 10¢ price increases. Then the profit on each souand the number of souvenirs sold per day is Let x be the number of $0.50 $0.10x, venir is 275 25x. Thus, the profit per day is 0.1x 0.5 P x 25x 275 2 21 1 or, in x-intercept form, 2 1 P x 1 2 2.5 x 5 1 21 x 11 2 The graph is a parabola that opens downward, so the maximum occurs at the vertex. The function is in x-intercept form, so the vertex occurs when x s t 2 5 11 2 3. The maximum profit of $160 per day occurs for 3 price increases, so the price should be per souvenir. $2.30 $2 3 $0.10 1 2 ■ Exercises 3.3 In Exercises 1 – 4, determine the vertex of the given quadratic function and state whether its graph opens upward or downward. In Exercises 5–8, determine the y-intercept of the given quadratic function and state whether its graph opens upward or downward. 1. 3. g x 1 2 4. h x 1 2 6 x 2 1 x2 1 2 5 2 5. f 7. h x x 1 1 2 2 x2 6x 3 6. g 3x2 4x 5 8. g x2 8x 1 2x2 x 1 x x 1 1 2 2 In Exercises 9 – 12, determine the x-intercepts of the given quadratic function and state whether its graph opens upward or downward. Section 3.3 Quadratic Functions 171 35. Write a rule in transformation form for the quadratic function whose graph is the parabola with vertex (0, 1) that passes through ( 2, 7 ). 9. f 10. h x x 1 1 2 2 11. g 12 21 x 3 1 x 3 2 x 1 21 2 x 1 2b 4b a 0.4 x 2.1 x 0.7 2 21 1 In Exercises 13–21, determine the vertex and x- and y-intercepts of the given quadratic function, and sketch a graph. 13 14. g 15. 17. 19. 21 8x 2 1 2 x 1 21 x 3 x 3 2 x 4 1 21 16. 18. 20 6x 3 2x 2 4x 21 x 6 2 Write the following functions in polynomial form. 22. f 23 21 x 5 2 x 2 2 1 x 4 21 2 47 2 24. h x 1 2 3 1 25 Write the following functions in x-intercept form. 26. f 28 3x 4 27. h x 1 2 2x 2 13x 7 3 1 x 3 2 2 3 29 3b 2 3 Write the following functions in transformation form. 30. g 32. h 33 4x 5 31. f x 1 2 3x 2 6x 21 x 6 2 21 2 34. Write a rule in transformation form for the quadratic function whose graph is the parabola with vertex at the origin that passes through (2, 12). 36. Find the number c such that the vertex of x 2 8x c lies on the x-axis. f x 1 2 37. If the vertex of f b and c. x 1 2 x 2 bx c is at (2, 4), find 38. If the vertex of f x 2 bx 8 has x 1 2 y-coordinate 17 and is in the second quadrant, find b. 39. Find the number b such that the vertex of x 2 bx c lies on the y-axis. f x 1 2 40. If the vertex of f x-coordinate 7, find s. x 1 2 a x s 1 21 x 4 2 has x 2 21 18 41. If the y-intercept of f x 1 2 a 1 x 3 is 3, find a. 2 42. Find two numbers whose sum is and whose product is the maximum. 43. Find two numbers whose difference is 4 and whose product is the minimum. 44. The sum of the height h and the base b of a triangle is 30. What height and base will produce a triangle of maximum area? 45. A field bounded on one side by a river is to be fenced on three sides to form a rectangular enclosure. If the total length of fence is 200 feet, what dimensions will give an enclosure of maximum area? 46. A salesperson finds that her sales average 40 cases per store when she visits 20 stores per week. If she visits an additional store per week, her average sales per store decrease by one case. How many stores per week should she visit to maximize her sales? 47. A potter can sell 120 bowls per week at $4 per decrease in price, 20 more bowl. For each bowls are sold. What price should be charged in order to maximize revenue? 50¢ 48. When a basketball team charges $4 per ticket, 20¢ average attendance is 400 people. For each decrease in ticket price, average attendance increases by 40 people. What should the ticket price be to maximize revenue? 172 Chapter 3 Functions and Graphs 49. A ballpark concessions manager finds that each vendor sells an average of 40 boxes of p
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opcorn per game when 20 vendors are working. For every additional vendor, each averages 1 fewer box sold per game. How many vendors should be hired to maximize sales? In Exercises 50–53, use the following equation for the height (in feet) of an object moving along a vertical line after t seconds: s 16t2 v0t s0 s0 is the initial height (in ft) and is the iniwhere tial velocity (in ft/sec). The velocity is positive if it is traveling upward, and negative if it is traveling downward. v0 53. A bullet is fired upward from ground level with an initial velocity of 1500 ft/sec. When does the bullet reach its maximum height and how high is it at that time? 2 1 f x 54. Critical Thinking The discriminant of a quadratic ax 2 bx c function For each value of the discriminant listed, state which graphs below could possibly be the graph of f. a. c. b2 4ac 0 b. d. b2 4ac 72 b2 4ac 25 b2 4ac 49 is the value b2 4ac. y y 50. A rocket is fired upward from ground level with an initial velocity of 1600 ft/sec. When does the rocket reach its maximum height and how high is it at that time? 51. A ball is thrown upward from a height of 6 ft with an initial velocity of 32 ft/sec. When does the ball reach its maximum height and how high is it at that time? (i) y 52. A ball is thrown upward from the top of a 96-ft tower with an initial velocity of 80 ft/sec. When does the ball reach its maximum height and how high is it at that time? x x (ii) y x x (iii) (iv) 3.4 Graphs and Transformations Objectives • Define parent functions • Transform graphs of parent functions When the rule of a function is algebraically changed in certain ways to produce a new function, then the graph of the new function can be obtained from the graph of the original function by a simple geometric transformation. Parent Functions The functions on the next page are often called parent functions. A parent function is a function with a certain shape that has the simplest algebraic f is the simplest rule for a rule for that shape. For example, parabola. You should memorize the basic shapes of the parent functions. x 2 x 2 1 Section 3.4 Graphs and Transformations 173 NOTE The points 0, 0 and 1, 1 are labeled for reference in each 1 case. The origin is not on the graph of the constant or reciprocal function. 2 1 2 y 4 2 (1, 1) −4 0 −2 −2 −4 2 (0, 0) 4 f(x) = 1 Constant function y (1, 1) 4 2 x x −4 −2 0 2 (0, 0) 4 −4 −4 y y 4 2 −2 −2 −4 −4 (1, 1) x 2 (0, 0) 4 −4 4 2 −2 −2 −4 (1, 1) x 2 (0, 0) 4 f(x) = x Identity function f(x) = |x| Absolute-value function y y 4 2 −2 −2 −4 (1, 1) x 2 (0, 0) 4 −4 4 2 −2 −2 −4 (1, 1) x 2 (0, 0) 4 f(x) = [x] Greatest integer function f(x) = x2 Quadratic function f(x) = x3 Cubic function y 4 2 −4 0 −2 −2 −4 (1, 1) x 2 (0, 0) 4 1 f(x) = x Reciprocal function y y 4 2 −2 −2 −4 −4 (1, 1) x 2 (0, 0) 4 −4 f(x) = x Square root function Figure 3.4-1 4 2 −2 −2 −4 (1, 1) 2 (0, 0) 4 x 3 f(x) = x Cube root function The parent functions will be used to illustrate the rules for the basic transformations. Remember, however, that these transformation rules work for all functions. 174 Chapter 3 Functions and Graphs Technology Tip If the function f is entered as Y1, then the function g x 1 2 f x 2 1 c Vertical Shifts Graphing Exploration Graph these functions on the same screen and describe your results. f x 1 2 23 x g x 1 2 23 x 2 h 23 x 3 x 1 2 can be entered in Y2 as Y1 C and the function f c x x g 1 2 1 2 can be entered in Y3 as Y1 C. Vertical Shifts Notice that when a value is added to the effect is to add the value to the y-coordinate of each point. The result is to shift every point on the graph by the same amount. A similar result is true for subtracting a value from Let c be a positive number. The graph of c units. g(x) f(x) c is the graph of f shifted upward The graph of downward c units. g(x) f(x) c is the graph of f shifted Example 1 Shifting a Graph Vertically Graph g x 1 2 x 0 0 4 and h x 1 2 x 0 0 3. Solution The parent function is f f x g 0 1 h shifted upward 4 units, and the graph of shifted downward 3 units. The graph of is the graph of is the graph of 1, 5) (0, 4) x 4 8 −8 −8 0 −4 −4 −8 8 4 0 −4 −4 −8 x 8 4 (1, −2) (0, −3) Figure 3.4-2 ■ Technology Tip If the function f is entered as Y1, then the function g x 1 2 f 1 x c 2 can be entered in Y2 as Y1(X C) and the function g x 2 1 f 1 c x 2 can be entered in Y3 as Y1(X C). Section 3.4 Graphs and Transformations 175 Horizontal Shifts Graphing Exploration Graph these functions on the same screen and describe your results. x 1 x 3 x2 table of values is helpful in visualizing the direction of a horizontal shift. x x 1 x2 (x 1)2 3 4 9 16 16 9 5 4 25 16 When 1 is subtracted from the x-values, the result is that the entries in the table shift 1 position to the right. Thus, the entire graph is shifted 1 h unit to the right. Construct a table for to see that the entries shift 3 positions to the left. x 3 x 1 2 2 1 2 Horizontal Shifts Let c be a positive number. The graph of to the left. The graph of to the right. g(x) f(x c) is the graph of f shifted c units g(x) f(x c) is the graph of f shifted c units Example 2 Shifting a Graph Horizontally Graph g 1 x 3 x 2 1 and h 1 x 4 . x 1 2 Solution The parent function is f 1 x. x 2 1 The graph of shifted 3 units to the right and the graph of h 1 shifted 4 units to the left. is the graph of f 2 is the graph of 176 Chapter 3 Functions and Graphs = x3 4 P (x, y) O x Q (−x, −y) Figure 3.4-3 ■ Reflections Graphing Exploration Graph these functions on the same screen. f x 1 2 2x g 2x x 2 1 Use the TRACE feature to verify that for every point on the graph of f there is a point on the graph of g with the same x-coordinate and opposite y-coordinate. Graph these functions on the same screen. f x 1 2 2x h 2x x 1 2 Use the TRACE feature to verify that for every point on the graph of f there is a point on the graph of h with the same y-coordinate and opposite x-coordinate. The results of the first part of the Graphing Exploration show that in the if the point (a, b) is on the graph of functions is on the graph of g. These two points are reflecf, then the point tions of each other across the x-axis, as shown in Figure 3.4-4. 1x, 1x a, b and In the second part of the Graphing Exploration, the results show that in 1x, if the point (a, b) is on the graph h and the functions ) is on the graph of h. These two points are reflecof f, then the point ( tions of each other across the y-axis, as shown in Figure 3.4-4. 1x a, b x x f 2 1 2 1 Technology Tip If the function f is entered as Y1, then the function g x f x 1 2 can be entered in Y2 as Y1 and the function 1 2 g x f x 1 2 can be entered in Y3 as Y1(X). 1 2 y (−a, b) (a, b) x (a, −b) Figure 3.4-4 Section 3.4 Graphs and Transformations 177 Reflections The graph of the x-axis. The graph of the y-axis. g(x) f(x) is the graph of f reflected across g(x) f(x) is the graph of f reflected across Example 3 Reflecting a Graph Across the x- or y-Axis Graph g x 2 1 and Solution 4 1 2 . f x x 3 x The graph of x x 4 3 2 reflected across the x-axis, and the graph of x The parent function is is the graph of x f h is the graph 2 1 f reflected across the y-axis. One difference in the reflections of is whether the endpoint of each segment is included on the left or on the right. Study the functions closely, along with the parent function, to determine another difference in the two reflections (−1, 1) (0, 0) x (0, 0) x −4 −2 0 2 4 −4 −2 0 2 4 −2 −4 (1, −1) −2 −4 Figure 3.4-5 ■ Stretches and Compressions Graphing Exploration Graph these functions on the same screen. 2x g 23x h x x f 1 2 1 2 32x x 1 2 Use the TRACE feature to locate the y-value on each graph when x 1. Describe your results. Predict the graph of j 42x. x 1 2 Technology Tip If the function f is entered as Y1, then the function g x c f x 2 1 1 2 can be entered in Y2 as C Y1 and the function f 2 can be entered in Y3 as Y1(C X). c x x g 2 1 1 178 Chapter 3 Functions and Graphs In addition to reflections and shifts, graphs of functions may be stretched or compressed, either vertically or horizontally. The following example illustrates the difference between a vertical stretch and compression. Example 4 Vertically Stretching and Compressing a Graph Graph g x 2 1 2x 3 and h 1 4 x 1 2 x 3. Solution every The parent function is y-coordinate of the parent function is multiplied by 2, stretching the graph For the function x g x f 2 1 2 1 2x 3, x 3. of the function in the vertical direction, away from the x-axis. For the func- 1 4 x 2 1 x 3, every y-coordinate of the parent function is multiplied compressing the graph of the function in the vertical direction, h tion 1 4 by , toward the x-axis. 2x 1, 2) x 4 (0, 0) 8 −8 −4 0 −4 −8 −8 −4 0 −4 −8 ( 1 1, ) 4 x 4 (0, 0) 8 Figure 3.4-6 ■ The following example illustrates the difference between a vertical and a horizontal stretch. Example 5 Stretching a Function Vertically and Horizontally Graph g x 1 2 2 x 3 4 and h x 2 1 2x . 4 3 Solution x the y-values are multiThe parent function is 3 the x-values are multiplied by 2. The result plied by 2, and in is that in the graph of the first function, the “steps” get higher, and in the graph of the second function, they get narrower. x 2 2x f 1 In Section 3.4 Graphs and Transformations 179 1, 2) (0, 0) 0 −4 −2 x 2 4 h x 2 1 2x 4 3 y 4 2 (0, 0) 0 −4 −2 ( 1 2 , 1) x 2 4 Figure 3.4-7 ■ Stretches and Compressions Let c be a positive number. Vertical Stretches and Compressions (x, y) If on the graph of is a point on the graph of f(x), then (x, cy) is a point c 77 1, If vertically, away from the x-axis, by a factor of c. the graph of is the graph of f stretched g(x) c f(x) . g(x) c f(x) c 66 1, g(x) c f(x) If compressed vertically, toward the x-axis, by a factor of c. is the graph of f the graph of Horizontal Stretches and Compressions If (x, y) is a point on the graph of f(x), then point on the graph of If c 77 1, the graph of g(x) f(c x). g(x) f(c x) is the graph of f 1 c x, y b a is a compressed horizontally, toward
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the y-axis, by a factor of 1 c . If c 66 1, the graph of g(x) f(c x) horizontally, away from the y-axis, by a factor of is the graph of f stretched 1 c . NOTE Some horizontal compressions can be expressed as vertical stretches, and vice versa. The graph of the function g below can be 1 3 obtained from a horizontal compression of by a factor of or x2 x f 1 2 from a vertical stretch by a factor of 9. g x 1 2 3x 2 1 2 32x2 9x2 This is not possible for the greatest integer function (see Example 5) or trigonometric functions, which you will study in later chapters. 180 Chapter 3 Functions and Graphs Transformations can be combined to produce many different functions. There is often more than one correct order in which to perform these transformations; however, not every possible order is correct. One method is shown below. Combining Transformations For a function of the form graph f(x). a 6 0, 1. If reflect the graph across the y-axis. g(x) c f(a(x b)) d, first 2. Stretch or compress the graph horizontally by a factor of 1 a ` . ` 3. Shift the graph horizontally by b units: right if b 7 0, and left if c 6 0, 4. If b 6 0. reflect the graph across the x-axis. 5. Stretch or compress the graph vertically by a factor of 6. Shift the graph vertically by d units: up if down if d 6 0. d 7 0, and c . 0 0 Example 6 Combining Transformations Graph the following function. x g 1 2 2x 1 0 4 0 Solution x g Rewrite 2 function is f 1 1 Compress the graph horizontally by a factor of 1 2 . Shift the graph 1 2 unit to the right. 0 x 2 2x 1 4 0 as g x 1 2 x 1 2b ` 2 ` a 4. The parent x . 0 0 Reflect the graph across the x-axis. Shift the graph upward 4 units. y 4 2 1 ( , 1) 2 (1,1) x y 4 2 (1, 1) x 0 −4 −2 −2 2 (0, 0) 4 −4 0 −4 −2 −2 −4 4 2 , 0) x 2 4 (1, −1) 0 −4 −2 −2 −4 y 1 2 ( , 4) (1, 3) 4 2 0 −4 −2 −2 −4 2 4 Figure 3.4-8 x ■ Section 3.4 Graphs and Transformations 181 Graphing Exploration 1 2 1 2 x g x 2x 322x 4 1, to produce the graph of For the function list several different possible orders of the transformations performed on the graph of f on a calculator. Did any orders of transformations that you listed produce an incorrect graph? Try to determine some rules about the order in which the transformations may be performed. Compare your rules with those of your classmates. Then graph g g x x . 2 2 1 1 Example 7 Package Delivery An overnight delivery service charges $18 for a package weighing less than 1 pound, $21 for one weighing at least 1 pound, but less than 2 pounds, $24 for one weighing at least 2 pounds, but less than 3 pounds, and so on. The cost c(x) of shipping a package weighing x pounds is given by and interpret the result. 18 3 Graph Solution The parent function of The parent function is stretched 3 vertically by a factor of 3, then shifted upward 18 units. Note: although this rule is defined for all real numbers, the domain of this cost function is x 7 0. is 30 24 18 12 6 (1, 3) x −4 −2 0 2 (0, 0) 4 y (1, 21) (0, 18) 30 24 18 12 6 −4 −2 0 2 4 x Figure 3.4-9 The function is stretched vertically by a factor of 3 because the increment from one step to the next is $3, and the function is shifted 18 units upward because the lowest rate for any package is $18. ■ 182 Chapter 3 Functions and Graphs Exercises 3.4 In Exercises 1–9, identify the parent function that can be used to graph each function. Do not graph the function. 1. 3. g 4 22x 4 25. parent function: x 0 0 transformations: shift the graph 3 units to the left, reflect it across the x-axis, and shrink it vertically x f 1 2 by a factor of 1 2 26. parent function: f 1 x x 2 1 6. f x 1 2 3x 2 transformations: shift the graph 2 units to the right, stretch it horizontally by a factor of 2, and shift it upward 2 units 7. 8 223 x 5 9 In Exercises 10–21, graph each function and its parent function on the same set of axes. 10. f x 1 2 x 2 11. h x 2 1 1 x 12. 14. 16 2x 4 18. g x 2 1 3x 20 13. 15. 17 19. f x 1 2 23 21. h x 2 1 1 3x In Exercises 22–27, write a rule for the function whose graph can be obtained from the given parent function by performing the given transformations. 22. parent function: x3 transformations: shift the graph 5 units to the left and upward 4 units x f 2 1 23. parent function: 2x transformations: reflect the graph across the x-axis and shift it upward 3 units x f 2 1 24. parent function: f x 1 2 x 3 4 transformations: shift the graph 6 units to the right, stretch it vertically by a factor of 2, and shift it downward 3 units 27. parent function: f x2 x 1 2 transformations: shift the graph 3 units to the left, reflect it across the y-axis, and stretch it vertically by a factor of 1.5 In Exercises 28–33, describe a sequence of transformations that transform the graph of the parent function f into the graph of the function g. Do not graph the functions. 28. 29. 30. 31. 32 2x x x 4 3 x3 1 x 33. f x 1 2 23 2x 23 1.3x 4.2 0. In Exercises 34–41, graph each function and its parent function on the same graph. 34 35. g x 1 2 1 4 23 x 3 1 36 37 38 39. f x 1 2 3 2 x 4 40. h x 2 1 24x 3 1 41 In Exercises 42–45, use the graph of the function f in the figure to sketch the graph of the function g. Section 3.4 Graphs and Transformations 183 x 1 3 52. 54 53. 55 Exercises 56–61 refer to the parent function f (x) 21 x 2 x The graph of f is a semicircle with radius 1, as shown below. y f(x) 43. 45. g g x x 1 1 2 2 3f 1 −1 42. 44.25f x 1 2 In Exercises 46–49, use the graph of the function f in the figure to sketch the graph of the function h. Use the graph of f to sketch the graph of the function g. y f(x) x 56. 58. 60. g g g 61 57. 59. g g x x 1 1 2 2 21 x 2 4 321 x 2 21 x 2 21 21 46. 48 47. 49 2x 2 In Exercises 50–55, use the graph of the function f in the figure to sketch the graph of the function g. y f(x) x 50 51 62. In 2002, the cost of sending first-class mail was $0.37 for a letter weighing less than 1 ounce, $0.60 for a letter weighing at least one ounce, but less than 2 ounces, $0.83 for a letter weighing at least 2 ounces, but less than 3 ounces, and so on. a. Write a function c that gives the cost of x 2 mailing a letter weighing x ounces (see Example 7). 1 b. Graph c(x) and interpret the result. 63. A factory has a linear cost function ax b, where b represents fixed costs and a represents the labor and material costs of making one item, both in thousands of dollars. a. If property taxes (part of the fixed costs) are x f 1 2 increased by $28,000 per year, what effect does this have on the graph of the cost function? b. If labor and material costs for making 100,000 items increase by $12,000, what effect does this have on the graph of the cost function? 184 Chapter 3 Functions and Graphs 3.4.A Excursion: Symmetry Objectives • Determine whether a graph has y-axis, x-axis, or origin symmetry • Determine whether a function is even, odd, or neither This section presents three kinds of symmetry that a graph can have and ways to identify these symmetries both geometrically and algebraically. y-Axis Symmetry A graph of a function or relation is symmetric with respect to the y-axis if the part of the graph on the right side of the y-axis is the mirror image of the part on the left side of the y-axis, as shown in Figure 3.4.A-1. y x2 + 4y2 = 24y (−6, 3) P (−x, y) (6, 3) Q (x, y) x Figure 3.4.A-1 Each point P on the left side of the graph has a mirror image point Q on the right side of the graph, as indicated by the dashed lines. Note that • their y-coordinates are the same • their x-coordinates are opposites of each other • the y-axis is the perpendicular bisector of PQ y-Axis Symmetry A graph is symmetric with respect to the y-axis if whenever (x, y) (x, y) is on the graph, then is also on it. In algebraic terms, this means that replacing x by produces an equivalent equation. x Example 1 y-Axis Symmetry Verify that y x4 5x2 3 is symmetric with respect to the y-axis. Section 3.4.A Excursion: Symmetry 185 Solution Replace x by x in the equation. y x4 5x2 3 y x 4 5 x 2 3 1 which is equivalent to the original equation because x and Therefore, the graph is symmetric with respect to the y-axis. ■ 4 x4 Graphing Exploration Graph the equation from Example 1. Use the TRACE feature to locate at least three points (x, y) on the graph and show that for each point, is also on the graph. x, y y x 4 5x 2 3 1 2 x-Axis Symmetry A graph of a relation is symmetric with respect to the x-axis if the part of the graph above the x-axis is the mirror image of the part below the x-axis, as shown in Figure 3.4.A-2. y x = y2 − 3 (x, y) (2, 5) P Q (x, −y) (2, − 5) Figure 3.4.A-2 x Each point P on the top of the graph has a mirror image point Q on the bottom of the graph, as indicated by the dashed lines. Note that • their x-coordinates are the same • their y-coordinates are opposites of each other • the x-axis is the perpendicular bisector of PQ x-Axis Symmetry A graph is symmetric with respect to the x-axis if whenever (x,y) (x, y) is on the graph, then is also on it. In algebraic terms, this means that replacing y by produces an equivalent equation. y 186 Chapter 3 Functions and Graphs Example 2 x-Axis Symmetry Verify that y2 4x 12 is symmetric with respect to the x-axis. NOTE 0 x Except for , the graph of a 1 2 f function is never symmetric with respect to the x-axis. By the Vertical Line Test, a function’s graph cannot contain points that lie on a vertical line, such as . and x, y x, y 1 2 1 2 Solution Replacing y by y in the equation gives y 2 4x 12 1 2 which is equivalent to the original equation because 1 fore, the graph is symmetric with respect to the x-axis. Graphing Exploration y 2 2 y2. There- ■ Graph the equation mode by entering the parametric equations below. y2 4x 12 from Example 2 in parametric t2 3 1 4 t x1 y1 Use the TRACE feature to locate at least three points (x, y) on the graph and show that for each point, is also on the graph. x, y 1 2 Origin Symmetry A graph is symmetric with respect to the origin if a line through the origin and any point P on the graph also intersects
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the graph at a point Q such that the origin is the midpoint of 3.4.A-3. as shown in Figure PQ, y y = x3 4 P (x, y) O x Q (−x, −y) Figure 3.4.A-3 Using Figure 3.4.A-3, symmetry with respect to the origin can also be described in terms of coordinates and equations. Origin Symmetry Section 3.4.A Excursion: Symmetry 187 A graph is symmetric with respect to the origin if whenever (x, y) (x, y) is on the graph, then is also on it. In algebraic terms, this means that replacing x by by produces an equivalent equation. y x and y Example 3 Origin Symmetry Verify that y x3 10 x is symmetric with respect to the origin. Solution Replace x by x and y by y 1 y 2 y x 10 x 3 10 in the equation. 3 x 2 1 x Simplify 1 1 y 1 2 y x 3 10 x 3 10 a x b Multiply both sides by 1 x Simplify Therefore, the graph is symmetric with respect to the origin. ■ Graphing Exploration Graph the equation y x 3 10 x from Example 3. Use the TRACE feature to locate at least three points (x, y) on the graph and show that for each point, is also on the graph. x, y 1 2 Summary There are a number of techniques used to understand the fundamental features of a graph. Symmetry, whether line or point, is beneficial when graphing a function. Knowing a graph’s symmetry prior to graphing reduces the necessary number of coordinates needed to display a complete graph; thus, the graph can be sketched more quickly and easily. Here is a summary of the various tests for line and point symmetry: 188 Chapter 3 Functions and Graphs Symmetry Tests Symmetry with respect to y-Axis x-Axis Origin Coordinate test Algebraic test 1 x, y is on the If 2 graph, then is on the graph. 1 x, y 1 x, y is on the If 2 graph, then is on the graph. 1 x, y x Replacing x by produces an equivalent equation. y Replacing y by produces an equivalent equation. 2 2 1 x, y is on the If 2 graph, then is on the graph. 1 x, y 2 x Replacing x by y and y by produces an equivalent equation. Even Functions NOTE If the rule of a function is a polynomial in which all terms have even degree, then the function is even. (A constant term has degree 0, which is even.) Even and Odd Functions For relations that are functions, the algebraic description of symmetry can take a different form. Even Functions A function f whose graph is symmetric with respect to the y-axis is called an even function. A function f is even if f(x) f(x) for every value x in the domain of f. The graph of an even function is symmetric with respect to the y-axis. For example is even because Thus, the graph of f is symmetric with respect to the y-axis, as you can verify with your calculator. Odd Functions A function f whose graph is symmetric with respect to the origin is called an odd function. If both (x, y) and are on the graph of such a function f, then x, y 1 2 y f x 2 1 so f x 1 2 f x . 2 1 and Section 3.4.A Excursion: Symmetry 189 A function f is odd if f(x) f(x) for every value x in the domain of f. The graph of an odd function is symmetric with respect to the origin. For example, f x 2 1 x 3 2x is an odd function because x x 3 2x x x 3 2x 2x 2 1 Hence, the graph of f is symmetric with respect to the origin, as you can verify with your calculator. Odd Functions NOTE If the rule of a function is a polynomial in which all terms have odd degree, then the function is odd. Exercises 3.4.A In Exercises 1–6, graph the equation and state whether the graph has symmetry. If so, is it symmetric with respect to the x-axis, the y-axis, or the origin? In Exercises 19–24, determine whether the given graph is symmetric with respect to the y-axis, the x-axis, the origin, or any combination of the three. 1. y x 2 2 3. y x 3 2 5. y 23 x x 2 2. x 4. y 0 0 x In Exercises 7–10, determine algebraically whether or not the graph of the given equation is symmetric with respect to the y-axis. 7. x 2 y 2 1 8. y x 3 x 2 9. 4x 2 3y y 2 7 10. x 4 x 2 x y 3 1 In Exercises 11 – 14, determine algebraically whether the graph of the given equation is symmetric with respect to the x-axis. 11. x 2 6x y 2 8 0 12. x 2 8x y 2 15 13. x 2 2x y 2 2y 2 14. x 2 x y 2 y 0 In Exercises 15–18, determine algebraically whether the graph of the given equation is symmetric with respect to the origin. 15. 4x 2 3y 2 xy 6 16. x 3 y 3 x 17 18. 3x 2 4y 2x 6 19. 20. 21. 22. y y y y x x x x 190 23. Chapter 3 Functions and Graphs y 37. Odd y 24. y x x 38. Odd y x x In Exercises 25–34, determine whether the given function is even, odd, or neither. 39. a. Plot the points (0, 0), (2, 3), (3, 4), (5, 0), 6, 1 , . 4, 1 b. Suppose the points in part a lie on the graph of 1, 1 7, 3 and , , 2 1 2 1 1 2 1 4 an odd function f. Plot the points 1 7 3, f 5 5, f 22 1 (4, f(4)), and (6, f(6)). 7, f 3 22 , , 1 1 1 1 1 2 2, f , 22 2 , 22 (1, f(1)), 1 40. a. Plot the points 1,3 1 , 2 1 5, 2 , 2 1 3, 5 2, 3 , 2 1 2 , and 4, 1 . 2 1 1 b. Suppose the points in part a lie on the graph of , an even function f. Plot the points 5, f and 1, f 1 4, f 1 1 4 2, f 2 3, f , 5 3 , , 1 22 1 1 22 1 1 22 22 . 22 1 1 41. Show that any graph that has two of the three types of symmetry (x-axis, y-axis, origin) always has the third type. 42. Use the midpoint formula (see the Algebra Review Appendix) to show that (0, 0) is the midpoint of the segment joining (x, y) and x, y . 1 2 25. 27. f f 29. k 31 4x x 2 x 0 0 t 4 6t 2 5 2t 2 5 26. k t 2 1 28. h 30. f 32 5t 3u 0 x 0 x 4 x 2 1 2 27 2x 2 33 34 In Exercises 35–38, complete the graph of the given function, assuming that it satisfies the given symmetry condition. 35. Even y 36. Even y x x Section 3.5 Operations on Functions 191 3.5 Operations on Functions Objectives • Form sum, difference, product, and quotient functions and find their domains • Form composite functions and find their domains There are several ways in which two or more given functions can be used to create new functions. Sums and Differences of Functions If f and g are functions, their sum is the function h defined by this rule. For example, if 3x 2 x f x 1 2 g and 3x 2 x 1 2 x 1 g x 2 4x 2, 4x 2 2 then 3x 2 5x 2 1 2 2 1 1 2 2 Instead of using a different letter h for the sum function, we shall usually f g denote it by f is defined by the rule g Thus, the sum f g 21 2 1 2 1 2 NOTE This rule is not just a formal manipulation of symbols. If x is and g(x). The plus sign in a number, then so are addition of numbers, and the result is a number. But the plus sign in f g is addition of functions, and the result is a new function. g is Technology Tip If two functions are entered as Y1 and Y2, their sum function can be graphed by entering Y2 . Y3 Y1 Difference, product, and quotient functions can be graphed similarly. The difference function is the function defined by a similar rule. f g f g 1 x 2 21 f x 1 2 g x 1 2 The domain of the sum and difference functions is the set of all real numbers that are in both the domain of f and the domain of g. Example 1 Sum and Difference Functions For f x 1 2 29 x2 a. write the rule for b. find the domain of g and f g 1 x 2x 2 f g. 2 and f g and f g. Solution f g f g a. 1 1 x x 2 2 21 21 29 x 2 2x 2 29 x 2 2x 2 b. The domain of f consists of all x such that 9 x2 0, that is, Similarly, the domain of g consists of all x such that 3 x 3. x 2 0, all real numbers in both the domain of f and the domain of g, namely, all x such that The domain of 2 x 3. x 2. that is, f g f g and consists of ■ 192 Chapter 3 Functions and Graphs Products and Quotients of Functions The product and quotient of functions f and g are the functions defined by the following rules. fg x 2 21 1 f x 1 2 g x 1 2 and The domain of fg consists of all real numbers in both the domain of f and the domain of g. The domain of f gb a in both the domain of f and the domain of g such that consists of all real numbers x 0. x g 1 2 CAUTION The function with the 23x 3 3x rule is defined for values that are not in the domain of f and g. The domain of a sum, difference, product, or quotient function must be determined before the function is simplified. Example 2 Product and Quotient Functions For f x 1 2 23x and g 2x2 1 x 1 2 a. write the rule for fg and f g . b. find the domain of fg and f g . Solution a. fg x 2 21 1 f gb1 x 2 a 23x 2x 2 1 23x 3 3x 23x 23x 3 3x x 2 1 2x 2 1 b. The domain of f consists of all x such that 3x 0, that is, x 0. x2 1 0, Similarly, the domain of g consists of all x such that is, both the domain of f and the domain of g, that is, that The domain of fg consists of all real numbers in x 1 x 1. x 1. The or domain of f g b a consists of all these x for which g 0, so the x 2 1 value x = 1 must also be excluded. Thus, the domain of x 7 1. f g b a is ■ Products with Constant Functions If c is a real number and f is a function, then the product of f and the constant function c is g x 1 2 cf The domain of cf is the same as the domain of f. For example, if f x3 x 2 c 5, x and 5 f then 5f is the function 5 x3 x 2 5x3 5x 10 1 2 5f 1 x 2 21 x 1 2 1 2 and the domain of 5f is still all real numbers. Section 3.5 Operations on Functions 193 Composition of Functions Another way of combining functions is to use the output of one function as the input of another. This operation is called composition of functions. The idea can be expressed in function notation as shown below. input of f output of f f x 2 input of g 1 output of 22 1 Composite Functions If f and g are functions, then the composite function of f and g is (g f )(x) g( f(x)) The expression Note the order carefully; the functions are applied right to left. is read “g circle f” or “f followed by g.” g f Technology Tip To evaluate or graph composite functions, enter the equations as Y1 and Y2. Then enter Y1(Y2(X)) or Y2(Y1(X)). Example 3 Composite Functions If f x 2 1 4x2 1 and g 1 x 2 , x 1 2 find the following. a. c 21 21 Solution b. d. f g f g 1 x 2 21 21 2 1 1 a. To find g f 1 2 2 21 , first find 17 2 1 2 Next, use the result as an input in g. So g f 2 1 b. To find 2 21 f g 1 1 19 1 21 17 g 1 2 1 17 2 1 19 . 2 , first find Next, use the result as an input in f. So 1 f g 1 2
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21 5 194 Chapter 3 Functions and Graphs c. To find x 21 g f 1 g f 1 x 2 21 g 2 g f x 1 22 1 , replace x with the rule for f in g. x 1 2 x f 1 22 1 1 2 2 f x 1 1 4x 2 1 1 2 2 1 4x 2 3 , replace x with the rule for g(x) in f. d. To find 1 f g x f 2 21 f f g x 2 21 1 g x 1 1 22 g x 22 1 1 4 2 1 4 x 1 22 2 1 x 2b a 1 g 1 Notice that g f 1 x 2 21 1 f g x . 2 21 4 x 2 1 2 2 1 ■ Domains of Composite Functions g f The domain of is determined by the following convention. Domain of g f Let f and g be functions. The domain of real numbers x such that g f is the set of all • x is in the domain of f • f(x) is in the domain of g Example 4 Finding the Domain of a Composite Function If f x 2x g f and . find b. find the domain of each composite function. and x2 5, 1 x 21 a. Solution g f f g b. Domain of : number x, the domain of 22 21 x x f 2 1 2 1 A 22 x x f 1 1 2g 2 5 2x 2x 2 5 2 22 even though the rule is defined for every real is not the set of all real numbers. • The domain of f is • The domain of g is all real numbers, so all values of f x 2 1 are in the domain of g. Thus, the domain of g f is x 0. Domain of f g: • The domain of g is all real numbers. • The values of g(x) that are in the domain of f are x 2 5 0 x 25 or x 25 Thus, the domain of f g is x 25 or x 25. ■ Section 3.5 Operations on Functions 195 Expressing Functions as Composites In calculus, it is often necessary to write a function as the composition of two simpler functions. For a function with a complicated rule, this can usually be done in several ways. Example 5 Writing a Function as a Composite 33x 2 1. h x Let 1 ferent ways. 2 Solution Write h as a composition of functions in two dif- The function h can be written as the composite and 1x. g x g f, where f 3x 21 g f x 1 22 1 g 1 3x 2 1 2 Additionally, h can be written as the composite and 3x 2. x k 23x 2 1 j k, where j 2x 21 j x k 1 1 22 j 3x 2 2 1 1 23x 2 1 ■ Graphs of Composite Functions The graph of a composite function can sometimes be described in terms of transformations. Example 6 Compositions with Absolute-Value Functions x g Let in terms of transformations. and f Graph f and the composite function x , and describe the relationship between the graphs of f and x 2 1 0 1x. Solution y 10 8 6 4 2 0 −4 −2 −4 −6 −8 −10 −10 −8 −6 1x f x 2 1 (4, 2) x y 10 8 6 (−4, 2) 4 2 2 4 6 8 10 −10 −8 −6 0 −4 −2 −4 −6 −8 −10 4, 2) x 2 4 6 8 10 For f g x 0 is the graph of , the graphs of f and f x , the graph of , a reflection of the graph of f across the y-axis. are the same. For x 6 0 f g 1 2 196 Chapter 3 Functions and Graphs Applications Composition of functions arises in applications involving several functional relationships. In such cases, one quantity may have to be expressed as a function of another. Example 7 The Area of a Circular Puddle A circular puddle of liquid is evaporating and slowly shrinking in size. After t minutes, the radius r of the surface of the puddle measures inches. The area A of the surface of the puddle is given by 18 2t 3 pr 2. t r 1 A 2 r 1 2 Express the area as a function of time by finding compute the area of the surface of the puddle at A A r r t 1 21 1 t 12 minutes. 2 1 Solution Substitute r 18 2t 3 t 2 1 into the area function 21 A 18 2t 3 b a p 2 18 2t 3b a At t 12 minutes, the area of the surface of the puddle is A r 1 12 2 21 p 18 2 12 3b a 2 4p 9 1.396 square inches. , and t 22 ■ Exercises 3.5 In Exercises 1–4, find and their domains. ( f g)(x), ( f g)(x), (g f )(x), In Exercises 8–11, find the domains of fg and f g. 1. 2. 3. 4. f x 1 2 f x 1 2 3x 4x 2x 5 f x 2 1 2x g x 1 2 x 2 1 In Exercises 5–7, find ( fg)(x), f gb a (x), and g f b a (x). 5. 6. 7 3x 2 g x 1 4x 2 x 4 g 1 2x 2 1 g x x 3 2x 2 4 2 2 2x 1 x 1 2 8 10. 11. f f x x 1 1 2 2 24 x 2 g 2 3x 2 x 4 2 g x 1 23x 4 4x 3 x 1 2 In Exercises 12–14, find ( f f )(0). ( g f )(3), ( f g)(1), and 12. 13. 14 3x 2x x 1 2 In Exercises 15–18, find the indicated values, where g(t) t 2 t and f(x) 1 x. 15. 17. g f 0 1 22 1 f g 0 1 1 22 g f 2 1 2 1 3 2 16. 18. 1 f f g 3 2 21 2f 1 1 2 2g 1 1 22 1 In Exercises 19–22, find the rule of the function and its domain and the rule of g f and its domain. f g Section 3.5 Operations on Functions 197 In Exercises 37 and 38, graph both on the same screen. Use the graphs to show that g f f g. and g f f g 37. 38. f f x x 1 1 2 2 x5 23 x 1 For Exercises 39 – 42, complete the given tables by using the values of the functions f and g given below. 19. 20. 21. 22 3x 2x f x 1 2 1 2x 1 g x 2 1 x 2 1 In Exercises 23–26, find the rules of the functions ff and f f. 23. 25. f f x x 1 1 2 2 x3 1 x 24. 26 39. In Exercises 27–30, verify that (g f )(x) x for the given functions f and g. (f g)(x) x and 27. 28. 29. 30 9x 2 g x 2 9 x 1 2 23 x 1 g 1 23 41 2x 3 5 g x 2 1 In Exercises 31–36, write the given function as the composite of two functions, neither of which is the identity function, (There may be more than one possible answer.) f(x) x. 31. f 32. g 33. h 34 23 x 2 2 2x 3 23 x 3 7x 3 10x 17 7 2 1 23 7x 3 2 2 1 35. f x 1 2 1 3x 2 5x 7 36. g t 2 1 3 2t (x) 3 5 1 2 3 ( g f )(x) 40. 4 ? 5 ? ? ( f f )(x) 42(x) 5 4 4 3 2 ( f g)(x) ? 2 ? ? ? ( g g)(x) ? ? ? 4 ? g(x) In Exercises 43–46, let and the composite function graph. . x 00 00 g f Graph the function f on the same f(x) 00 00 43. 44. 45.5x 2 5 x3 4x 2 x 3 x 3 46. f x 1 2 x 0 0 2 198 Chapter 3 Functions and Graphs 47. a. Use the piecewise definition of absolute value to explain why the following statement is true: f f 1 1 b. Use part a and your knowledge of if f if 21 x e 2 1 transformations to explain why the graph of g f consists of the parts of the graph of f that lie above the x-axis together with the reflection across the x-axis of those parts of the graph of f that lie below the x-axis. 58. Express the surface area of the weather balloon in Exercise 57 as a function of time. Hint: the surface area of a sphere of radius r is 4pr2. 59. Brandon, who is 6 ft tall, walks away from a streetlight that is 15 ft high at a rate of 5 ft per second, as shown in the figure. Express the length s of Brandon’s shadow as a function of time. Hint: first use similar triangles to express s as a function of the distance d from the streetlight to Brandon. g(x) In Exercises 48–52, let and the composite function graph. . x 0 0 f g f Graph the function f on the same x A 00 00 B 48. 50.5 x 4 2 9 2 1 49. 51. f f 52. Write a piecewise definition for 2x 15 ft 6 ft d s 60. A water-filled balloon is dropped from a window 120 ft above the ground. Its height above the ground after t seconds is standing on the ground 40 ft from the point where the balloon will hit the ground, as shown in the figure. a. Express the distance d between Laura and the 120 16t 2 ft. Laura is balloon as a function of time. b. When is the balloon exactly 90 ft from Laura? d 40 ft 61. Critical Thinking Find a function f (other than the f f f identity function) such that every value of x in the domain of f. (More than one correct answer is possible.) x 21 x 1 2 for 53. If f is any function and I is the identity function f I what are I f ? x, and x I 1 2 54. In a laboratory culture, the number N(d) of bacteria (in thousands) at temperature d degrees Celsius is given by the function N d 2 1 90 d 1 20 1 4 d 32 2 and the temperature D(t) at time t hours is given 2t 4 t by the function 2 1 a. What does the composite function 0 t 14 2 N D D 1 represent? b. How many bacteria are in the culture after 4 hours? after 10 hours? 55. Suppose that a manufacturer produces n telephones. The unit cost for producing each telephone is given by the function U n 2 1 total cost number of phones 12,000 15n n The price P of each telephone is the unit cost plus P U a 30% markup. Find result. , and interpret the 21 n 1 2 56. Find the unit price in Exercise 55 if 10,000 telephones are produced. 57. As a weather balloon is inflated, its radius increases at the rate of 4 cm per second. Express the volume of the balloon as a function of time, and determine the volume of the balloon after 4 seconds. Hint: the volume of a sphere of radius r is 4 3 pr 3. Section 3.5.A Excursion: Iterations and Dynamical Systems 199 3.5.A Excursion: Iterations and Dynamical Systems Discrete dynamical systems, which deal with growth and change, occur in economics, biology, and a variety of other scientific fields. Compound interest provides a simple example of a discrete dynamical system. As discussed in Chapter 5, an investment of P dollars at 8% interest compounded annually for one year grows by a factor of 1.08 each year. For example, an investment of $1000 grows like this: 1.08 1.08 1000 1080 1166.40 1 1 1.08 1 $1000.00 $1080.00 $1166.40 $1259.71 2 2 2 Initial amount Amount after one year Amount after two years Amount after three years This process can be described in function notation as follows. Let f 1.08x. x 1 2 $1000.00 $1080 1.08 1000 f 1080 1080 1166.40 2 f f f f 1000 f f 1 1 1 $1259.71 1.08 $1166.40 1.08 2 2 2 In each step after the first one the output of the function f becomes the input for the next step, and the final result can be expressed in terms of a composite function. 2 1000 f After three years After two years 22 1000 1166.40 222 1 1 1 1 1 1 1 2 1 Initial amount After one year CAUTION Do not confuse this notation with f 2 x exponents. 2 not represent the product of two output x values, f does f x . 1 1 2 1 2 Iteration Iterations of a function are the repeated compositions of a function with itself. 1. Select a number k as the initial input. 2. Compute the output 3. Using the output from Step 2 as input, compute the output . 22 4. Continue the process repeatedly, using the output from each step as the input for the next step. Iterated Function Notation Because function notation becomes cumbersome after several steps, the following abbreviated notation is used for iterations of functions: x 2 denotes denotes denotes f f x is the initial value , 2 , 22 x 1 the first iteration the second iteration , the third iteration 222 and in general f n x 2 1 denotes th
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e nth iteration. 200 Chapter 3 Functions and Graphs NOTE Throughout this section all numerical results are displayed rounded to four decimal places, but computations are done using the full decimal expansion given by a calculator. Technology Tip If a function has been entered as Y1 in the equation memory, it can be iterated as follows: • Store the initial value as X. • Key in Y1 STO X. Pressing ENTER repeatedly produces the iterated values of the function. Example 1 Iterated Function Notation Write the first four iterations of the function using iterated function notation. f x 1 2 2x with x 0.25 Solution Orbits 0.25 0.25 0.25 0.25 20.25 0.5 320.25 0.7071 4320.25 0.8409 54320.25 0.9170 ■ For a given function, the orbit of a number, c, is the sequence of output values produced by starting with c and then iterating the function. That is, the orbit of a number c for a given function f is the sequence c Example 2 Orbits Find the orbit of x 0.25 for f 2x. x 1 2 Solution 2X Y1 Enter margin yields the orbit of x 0.25. in the equation memory. Using the Technology Tip in the Figure 3.5.A-1 ■ Converging Orbits in Example 2 has the property that as n Notice that the orbit of gets larger, the terms of the orbit get closer and closer to 1. The orbit is said to converge to 1. x 0.25 Section 3.5.A Excursion: Iterations and Dynamical Systems 201 Calculator Exploration x f Let 2 x 100. 1 2x To what number does the orbit appear to converge? and compute the first twelve terms of the orbit of Choose another positive number and compute its orbit. Does the orbit appear to converge? To what number? Example 2 and the preceding Exploration suggest that the orbit of every converges to 1. The folpositive number under the function lowing notation is sometimes used to express this fact. 2x x f 2 1 Figure 3.5.A-2 Example 3 Orbits For f 1 f n x 2 x 1 2 2x and x 7 0, S 1 as n S q Find the orbits of x 0.4 and x 2 for the function f x2. x 1 2 Solution Let Y1 X 2 and x 0.4. Figure 3.5.A-2 shows that the orbit begins 0.4, 0.16, 0.0256, 0.0006554, 0.0000004295, . . . Figure 3.5.A-3 and that the orbit appears to converge to 0. Now let x 2. Figure 3.5.A-3 shows that the orbit begins 2, 4, 16, 256, 65536, 4294967296, . . . In this case, the orbit is not converging: its terms get larger and larger without bound as n increases. The fact is expressed by saying that the orbit diverges or that the orbit approaches infinity. ■ Calculator Exploration Choose a number strictly between the function and 1 and find its orbit under Does the orbit converge? To what number? x2. x f 1 1 2 Choose another number with absolute value greater than 1 and find its orbit. How would you describe the behavior of the orbit? Example 3 and the preceding Exploration suggest that the orbit of a number x converges to 0 when but that the orbit of x approaches infinity when 1 6 x 6 1, 7 1. x 0 0 For f For f x 1 x 1 2 2 This fact is expressed as follows. x2 and 6 1, f n x2 and 7 1, f n S 0 as n S q S q as 202 Chapter 3 Functions and Graphs Fixed Points and Periodic Orbits In addition to orbits that converge and orbits that approach infinity, there are several other possibilities, some of which are illustrated in the next example. Example 4 Orbits Describe the orbit of each of the following. x 2x 2 a. b. c. d. x 1 x 1 x 4 x 22 f 1 f x2 under under the function x 2 1 x x 2 1 1 under 2 f under x x f 1 1 2 Solution a. Because b. The orbit is 1 2 f 1 the orbit is 21 1, said to be a fixed point of the function 1, 1, 1, 1, p 1 2 1. said to be an eventually fixed point of the function f 1 1 4b 4 1, 1, 1, p . f x 2 1 2x. 1 the orbit is because 1 4 4, and 4. Because 1 1 4 1 is Hence, x2. x 1 4 2 1 , 4, , 4, p . Consequently, 1 is Since it repeats the same values, the orbit is said to be periodic and 4 is said to be a periodic point. d. Direct computation shows that A 2 1 1 22 1 22 A 22 A 22 A 22 B 12 1 0 02 22, 1, 0, 1, 0 1, 0, 1, p . 2 1 0. 2 B B B 1 Therefore, the orbit is Because the orbit begins to repeat its values after a few steps, it is said to be eventually periodic and is said to be an eventually periodic point. 22 ■ Example 5 Orbit Analysis Analyze all the orbits of f x2 x 1 2 and illustrate them graphically. Solution Example 3 and the Calculator Exploration after it suggest two possibilities. When When x x 0 0 0 0 6 1, 7 1, the orbit of x converges to 0. the orbit of x approaches infinity. Section 3.5.A Excursion: Iterations and Dynamical Systems 203 The orbit of 1 is 1, 1, 1, The orbit of 0 is 0, 0, 0, 1, The orbit of 1 is p, p, 1, 1, so 1 is a fixed point. so 0 is a fixed point. p, 1 so is an eventually fixed point. Figure 3.5.A-4 illustrates the orbits of f x2. x 1 2 −5 −4 −3 −2 −1 0 Figure 3.5.A-4 1 2 3 4 5 Red: Fixed points Orange: Eventually fixed point Green: Orbits that converge to 0 Blue: Orbits that approach q ■ For a function f, if of the function. f a 1 2 a, for some value a, then a is called a fixed point Exercises 3.5.A In Exercises 1–6, find the first eight terms of the orbit of the given number under the given function. 1. x 2 and f 1.08x x 2 1 2. x 0.8 and f 3. x 0.2 and f 4. x 1.2 and f 5. x 0.5 and f 6. x 0.1 and .5x 4x 1 1 x 4x 1 x 1 x3 x 2 2 In Exercises 7–12, determine whether the orbit of the point under the function converges or approaches infinity or neither. f(x) 4x(1 x) 7. x 1 9. x 0 11. x 1 8. x 0.5 10. x 0.5 12. x 1.5 x3 x2 20. Let f In Exercises 13–18, find the real number fixed points of the function (if any) by solving the equation g(x) x. 13. 15. 17 14. 16. 18 3x 3 x 3 2x 2 2x x 2 x 2 19. Determine all the fixed points and all the x eventually fixed points of the function . Find the orbit of x for every integer value of x such that fixed, eventually fixed, periodic, or eventually periodic. . Classify each point as b. Describe a pattern in the classifications you made in part a. 21. Let x 1 0 2 x f 1 as follows. a. Show that x 0.5 b. Show that and perform an orbit analysis 0 x 0.5 is an eventually fixed point. x 0 x 1 and is a fixed point and that are periodic points. 204 Chapter 3 Functions and Graphs 3.6 Inverse Functions Objectives Inverse Relations and Functions • Define inverse relations and functions • Find inverse relations from tables, graphs, and equations • Determine whether an inverse relation is a function • Verify inverses using composition Consider the following question: What was the world population in 1997? The answer to the question can be thought of as the output of a function. The input of the function is a year, 1997, and the output is a number of people. The function can be represented by a table of values or a scatter plot, as shown below. From the table, the output corresponding to the input of 1997 was a world population of 5.85 billion people. Year 1995 1996 1997 1998 1999 2000 Population (in billions) 5.69 5.77 5.85 5.92 6.00 6.08 y 7 6.5 6 5. 1994 1996 1998 2000 Figure 3.6-1 x Now consider another question: In what year was the world population 6 billion people? In this question, the input and output are reversed. The input is a population, 6 billion, and the output is a year. To create a function to answer this question, exchange the columns in the table, or the axes of the scatter plot. From the table, the output corresponding to the input of 6 billion people was the year 1999. Population (in billions) 5.69 5.77 5.85 5.92 6.00 6.08 Year 1995 1996 1997 1998 1999 2000 y 2000 1998 r a e Y 1996 1994 5 5.5 6 6.5 7 Figure 3.6-2 x Section 3.6 Inverse Functions 205 The result of exchanging the input and output values of a function or relation is called an inverse relation. If the inverse is a function, it is called the inverse function. Graphs of Inverse Relations x, y Suppose that is a point on its inverse function or relation. This fact can be used to graph the inverse of a function. is a point on the graph of a function. Then y, x 2 1 2 1 Example 1 Graphing an Inverse Relation The graph of a function f is shown below. Graph the inverse, and describe the relationship between the function and its inverse. y (−3, 9) 8 (3, 9) (−2, 4) 4 (−1, 1) (2, 4) (1, 1) x −8 −4 0 (0, 0) 4 8 −4 −8 Figure 3.6-3 Solution Start by reversing the coordinates of each labeled point. Plot these points, then sketch the relation. y (1, 1) (4, 2) (9, 3) x 8 (9, −3) 4 (4, −2) (1, −1) 8 4 (0, 0) 0 −4 −8 −8 −4 Figure 3.6-4 ■ 206 Chapter 3 Functions and Graphs y x Exercise 55 shows that the line is the perpendicular bisector of the segment from (a, b) to (b, a), which means that the two points are reflections of each other across the line Thus, the graph of the inverse x2 x in Example 1 is a reflection of the original graph across the of 2 1 line y x. y x. f CAUTION When using a calculator to verify that an inverse is a reflection over the line it is important to use a square viewing window. y x, Graphs of Inverse Relations y 8 4 −8 −4 0 4 8 x −4 −8 Figure 3.6-5 Let f be a function. If (b, a) (a, b) is a point on the graph of its inverse. is a point on the graph of f, then The graph of the inverse of f is a reflection of the graph of f across the line y x The definition of an inverse makes it very easy to graph an inverse by using parametric graphing mode. Recall that to graph the function y f t To graph the inverse, x2 let in parametric mode, let t. x 2 1 f and and f y2 x1 y1 t t . 1 2 1 2 Example 2 Graphing an Inverse in Parametric Mode Graph the function 2 inverse in parametric mode. x f 1 0.7x 5 0.3x 4 0.2x 3 2x 0.5 and its Solution To graph f, let x1 t and y1 0.7t5 0.3t4 0.2t3 2t 0.5. To graph x2 the inverse of 0.7t5 0.3t4 0.2t3 2t 0.5 f, exchange x y2 t. and and y by letting Figure 3.6-6 shows the graph of f and its inverse on the same screen with the graph of y x. Section 3.6 Inverse Functions 207 3.1 4.7 4.7 3.1 Figure 3.6-6 ■ Algebraic Representations of Inverses Suppose that a function is represented by an equation in x and y. Then the inverse of the function is found by exchanging the input and output values, that is, by exchanging x and y. Exampl
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e 3 Finding an Inverse from an Equation Find g x 1 2 , the inverse of f 3x 2. x 1 2 Solution First, write the function in terms of x and y. y 3x 2 Exchange the x and the y. Thus, the inverse relation is that the relation can be represented in function notation. It is common to solve for y, so x 3y 2 x 3y 2 ■ Graphing Exploration Graph the functions f and g from Example 3 together on the same y x, screen with the line and describe your results. Be sure to use a square viewing window. Example 4 Finding an Inverse from an Equation Find the inverse of f x2 4x. x 2 1 208 Chapter 3 Functions and Graphs Solution Write the function in terms of x and y. Exchange x and y. y x 2 4x x y 2 4y Thus, the inverse is quadratic equation in y. x y 2 4y. To solve for y, write the relation as a Then use the quadratic formula with a 1, b 4 , and c x . y 2 4y x 0 y 4 ± 216 4x 2 y 2 ± 24 x Notice that the inverse is not a function, so it cannot be written in function notation. ■ Determining Whether an Inverse is a Function The inverse of a function is also a function if every input of the inverse corresponds to exactly one output. This means that in the original function, every output corresponds to exactly one input. A function that has this property is called a one-to-one function. One-to-One Functions A function f is one-to-one if f(a) f(b) implies that a b. NOTE By the definition implies of a function, b a that f f 1 2 1 2 a b . If a function is one-to-one, then its inverse is also a function. Determining Whether a Graph is One-to-One In Example 1, the points 1 These two points have different inputs and the same output, so f is not one-to-one. Also, the points lie on the same horizontal and line, which suggests a graphical test for whether a function is one-to-one. are both on the graph of f. 2 2, 4 2, 4 and 2, 4 2, 4 1 2 2 1 1 2 Horizontal Line Test A function f is one-to-one if and only if no horizontal line intersects the graph of f more than once. Section 3.6 Inverse Functions 209 Example 5 Using the Horizontal Line Test Graph each function below and determine whether the function is oneto-one. If so, graph its inverse function. a. b. c 7x 5 3x 4 2x 3 2x 1 x 3 3x 1 1 0.2x3 Solution Complete graphs of each function are shown below. 3.1 3.1 3.1 −4.7 4.7 −4.7 4.7 −4.7 4.7 −3.1 a. −3.1 b. Figure 3.6-7 −3.1 c. 3.1 a. The graph of f passes the Horizontal Line Test, since no horizontal line intersects the graph more than once. Hence, f is one-to-one. The inverse function is the reflection of f across the line shown in Figure 3.6–8. y x, b. The graph of g fails the Horizontal Line Test because many horizontal lines, including the x-axis, intersect the graph more than once. Therefore, g is not one-to-one, and its inverse is not a function. c. The graph of h appears to contain a horizontal line segment, so h appears to fail the Horizontal Line Test. In this case, it is helpful to use the trace feature to see that the points on the segment that appears horizontal do not have the same y-value. Thus, the graph of h passes the Horizontal Line Test, and so it is one-to-one. The function and its inverse are shown in Figure 3.6–9. ■ NOTE The function f in Example 5 is always increasing and the function h is always decreasing. Every increasing or decreasing function is one-to-one because its graph can never touch the same horizontal line twice—it would have to change from increasing to decreasing, or vice versa, to do so. 4.7 4.7 3.1 Figure 3.6-8 3.1 4.7 4.7 3.1 Figure 3.6-9 210 Chapter 3 Functions and Graphs Inverse Functions Let f be a function. The following statements are equivalent. • The inverse of f is a function. • f is one-to-one. • The graph of f passes the Horizontal Line Test. f 1, The inverse function, if it exists, is written as where if y f(x), then x f 1 (y). The notation f 1 does not mean 1 f . Restricting the Domain For a function that is not one-to-one, it is possible to produce an inverse function by considering only a part of the function that is one-to-one. This is called restricting the domain. Example 6 Restricting the Domain Find an interval on which the function f on that interval. 1 x 2 f x 2 1 is one-to-one, and find Solution The graph of f is one-to-one on the interval 1, To find a rule for 3 first write the function in terms of x and y. as shown in the graph. f , 2 0, q Exchange x and y in the equation, and solve for y. y x 2 x y 2 y ± 2x y 2x or y 2x y 8 4 0 −4 −8 −8 −4 x 4 8 Figure 3.6-10 Section 3.6 Inverse Functions 211 Because every point on the graph of the restricted function f has nonnegative coordinates, the inverse function must be y 2x . Thus, 1 f x 1 2 2x y f 8 4 y = x −1 f x −8 −4 0 4 8 −4 −8 Figure 3.6-11 ■ Composition of a Function and its Inverse The inverse of a function f is designed to send each output of f back to the input it came from, that is, f a b exactly when f 1 b 1 Consequently, if you first apply f and then apply obtain the number you started with. 1 2 a 1 to the result, you 2 f Similarly, 1 f f a 1 22 22 f a 1 2 b The results above can be generalized to all values in the domains of f and 1. f Composition of Inverse Functions A one-to-one function f and its inverse function these properties. f 1 have ( f 1 f )(x) x for every x in the domain of f ( f f 1) (x) x for every x in the domain of f 1 Also, any two functions having both properties are one-to-one and are inverses of each other. 212 Chapter 3 Functions and Graphs Example 7 Verifying the Inverse of a Function Let f x 1 2 5 2x 4 and g 4x 5 2x x 1 2 Use composition to verify that f and g are inverses of each other. Solution g f 1 x 2 21 5 4 f 2 x f 1 22 22 1 x 1 1 2x 4 20 5 1 10 5 2x 4b 4 a 5 a 5 2 2x 4b 20 10x 20 10 2 f g x 2 21 1 5 x 22 g 2 1 1 4 5 4x 5 2x 4 b 5 4x 5 4x x 2 a 5 5 x x 2 20 5 2x 4 1 2x 4 10 2x 4 10x 10 5 4x 5 x x 4 Thus, f and g are inverses of each other. ■ Exercises 3.6 In Exercises 1 and 2, write a table that represents the inverse of the function given by the table. In Exercises 3 and 4, the graph of a function f is given. Sketch the graph of the inverse function of f and give the coordinates of three points on the inverse. 1. x 1 2 3 4 5 f(x) 4 2 3 6 1 2. x 1 0 1 2 3 g(x) 3 Section 3.6 Inverse Functions 213 29. 30.1x 3 0.1x 2 0.005x 1 0.1x 3 0.005x 1 x In Exercises 31–36, each given function has an inverse function. Sketch the graph of the inverse function. 4. y 1 1 In Exercises 5–8, graph f and its inverse in parametric mode (see Example 2). 5. 7 3x 2 2 x 4 3x 2 6. 8. f f x x 1 1 2 2 23 x 2 1 2x 2 1 In Exercises 9 – 22, find the rule for the inverse of the given function. Solve your answers for y and, if possible, write in function notation (see Examples 3 and 4). 9. 11. 13. 15. 17 5x 2 4 5 2x 3 24x 7 1 x 10. 12. 14. 16. 18 3x 2 5 x5 1 1 3 2 5 23x 2 1 2x 19. f x 1 2 1 2x 2 1 20. f x 1 2 x x 2 1 21 22. f x 1 2 5 3x 1 x 2 B In Exercises 23–30, use a calculator and the Horizontal Line Test to determine whether the function f is one-to-one. 23. 24. 25. 26. 27. 28 4x 2 3 x 4 4x 3 x 3 x 5 x 3 for x 3 2x 6 for x 7 3 e x 5 2x 4 x 2 4x 5 x 3 4x 2 x 10 31. 32. 33. 34. 35. 36 2x 3 23x 2 0.3x 5 2 23 x 3 25 x 3 x 2 x2 1 for x 0 0.5x 1 for x 7 0 e In Exercises 37–44, none of the functions is one-toone. State at least one way of restricting the domain of the function so that the restricted function has an inverse that is a function. Then find the rule of the inverse function (see Example 6). 38. 40. 42 24 x 2 37. 39. 41. 43. 44 In Exercises 45–50, use composition to show that f and g are inverses of each other (see Example 7). 45. 46 2x 6 g x 2 x 2 1 3 47 48. 49. 50 2x 5 g x 1 2 3 5x 2x x 5 g 25 x x 2 1 x 3 1 g 23 x 1 x 2 1 51. Show that the inverse function of the function f 2x 1 3x 2 whose rule is is f itself. x f 1 2 y (a, b) P y = x (c, c) R (b, a) Q x 56. Let C be the temperature in degrees Celsius. Then the temperature in degrees Fahrenheit is given by C f 1 2 9 5 C 32. Let g be the function that converts degrees Fahrenheit to degrees Celsius. Find the rule of g and show that g is the inverse function of f. 214 Chapter 3 Functions and Graphs 52. List three different functions (other than the one in Exercise 51), each of which is its own inverse. Many correct answers are possible. 53. Critical Thinking Let m and b be constants with is m 0. one-to-one, and find the rule of the inverse function Show that the function mx b 1. x f f 2 1 54. Critical Thinking Prove that the function of Example 5c is one-to-one by 1 x 1 0.2x3 2 a b, h showing that it satisfies the definition: f If Hint: Use the rule of f to show that when b f 2 impossible to have 1 2 a b. f a when then then If this is the case, then it is f a b. b 1 2 1 2 55. Critical Thinking Show that the points P a, a as follows: are symmetric with respect to the and line a. Find the slope of the line through P and Q. b. Use slopes to show that the line through P and Q is perpendicular to y x. y x it has c. Let R be the point where the line y x, Since R is on PQ. intersects coordinates (c, c) for some number c, as shown in the figure. Use the distance formula to show PR that that the line of respect to the line y x. RQ. is the perpendicular bisector Therefore, P and Q are symmetric with has the same length as Conclude y x PQ. 3.7 Rates of Change Objectives • Find the average rate of change of a function over an interval • Represent average rate of change geometrically as the slope of a secant line • Use the difference quotient to find a formula for the average rate of change of a function Rates of change play a central role in calculus. They also have an important connection with the difference quotient of a function, which was introduced in Section 3.1. Average Rates of Change When a rock is dropped from a high place, the distance the rock travels (ignoring wind resistance) is given by the function 16t2 d t 2 1 with distance table shows the distance the rock has fallen at various times. measured in feet and time t in seconds. The following d t 2 1 Time t Distance d(t) 0
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0 1 16 2 64 3 3.5 4 4.5 5 144 196.5 256 324.5 400 Section 3.7 Rates of Change 215 To find the distance the rock falls from time the end of three seconds, the rock has fallen 16 had only fallen to 144 feet at the end of one second, note that at feet, whereas it 144 16 128 feet So during this time interval the rock traveled 128 feet. The distance traveled by the rock during other time intervals can be found the same way. In general, the distance traveled from time t a to time t b is d(b) d(a) feet. From the chart, the length of each time interval can be computed by takt 4 ing the difference between the two times. For example, from seconds. Similarly, the interval from is a time interval of length t 2 4 1 3 3.5 2 1.5 is of length t 3.5 t 1 to to seconds. In general, t b b a t a seconds. the length of the time interval from is Since distance traveled average speed time interval, to average speed distance traveled time interval so the average speed over the time interval from average speed distance traveled time interval d 1 t a to d b 2 b a t b is a 1 2 . Example 1 Average Speed over a Given Interval Find the average speed of the falling rock a. from b. from t 1 t 2 to to and t 4 t 4.5. Solution a. Apply the average speed formula with average speed and 256 16 4 1 240 3 80 feet per second b. Similarly, from the average speed is t 2 d 4.5 2 4.5 2 2 2 1 t 4.5 to 324 64 4.5 2 d 1 260 2.5 104 feet per second. ■ The units in which average speed is measured in Example 1 (feet per second) indicate the rate of change of distance (feet) with respect to time (seconds). The average speed, or average rate of change of distance with respect to time, as time changes from is given by t a t b average speed average rate of change d change in distance change in time to 1 d b 2 b a a 1 2 . 216 Chapter 3 Functions and Graphs Although speed is the most familiar example, rates of change play a role in many other situations as well, as illustrated in Examples 2–4 below. The average rate of change of any function is defined below. Average Rate of Change Let f be a function. The average rate of change of respect to x as x changes from a to b is the value f(x) with change in f(x) change in x f(b) f(a) b a Example 2 Rate of Change of Volume A balloon is being filled with water. Its approximate volume in gallons is V x 1 2 x 3 55 where x is the radius of the balloon in inches. Find the average rate of change of the volume of the balloon as the radius increases from 5 to 10 inches. Solution change in volume change in radius V 1 V 10 2 10 5 5 2 1 18.18 2.27 10 5 15.91 5 3.18 gallons per inch ■ NOTE Analyzing units is helpful in interpreting the meaning of a rate of change. In Example 2, the units of the answer are gallons per inch. This means that for every inch that the radius increases between 5 and 10 inches, the volume increases by an average of 3.18 gallons. Example 3 Manufacturing Costs A small manufacturing company makes specialty office desks. The cost (in thousands of dollars) of producing x desks is given by the function 2 Find the average rate of change of the cost 1 c x 0.0009x 3 0.06x 2 1.6x 5 a. from 0 to 10 desks b. from 10 to 30 desks c. from 30 to 50 desks. Section 3.7 Rates of Change 217 Solution a. As production increases from 0 to 10 desks, the average rate of change of cost is change in cost change in production c 1 c 10 1 2 10 0 0 2 15.9 5 10 10.9 10 1.09 This means that costs are rising at an average rate of 1.09 thousand dollars (that is, $1090) per desk. b. As production goes from 10 to 30 desks, the average rate of change of cost is c 1 c 30 1 30 10 2 10 2 23.3 15.9 30 10 7.4 20 0.37 so costs are rising at an average rate of only $370 per desk. c. As production goes from 30 to 50 desks 47.5 23.3 50 30 c 50 1 50 30 30 c 1 2 2 24.2 20 1.21 so the rate increases to $1210 per desk. ■ Example 4 Rate of Change of Temperature The graph of the temperature function f during a particular day is given below. The temperature at x hours after midnight is What is the average rate of change of the temperature x f . 1 2 a. from 4 a.m. to noon? b. from 3 p.m. to 8 p.m.? 68° 64° 60° 56° 52° 48° 44° 40° 4 A.M. 8 12 Noon Time Figure 3.7-1 16 20 P.M. 24 218 Chapter 3 Functions and Graphs Solution a. The graph shows that the temperature at 4 a.m. is f f 4 and the 1 Thus, the average rate of change 58°. 12 2 40° temperature at noon is of temperature is 1 2 change in temperature change in time 1 4 f f 12 1 2 12 4 2.25° per hour 2 58 40 12 4 18 8 Notice that the rate of change is positive. This is because the temperature is increasing at an average rate of per hour. x 20. The Thus, the average and 8 p.m. to 53°. b. The time 3 p.m. corresponds to x 15 f 20 68° 2.25° and 15 f graph shows that rate of change of temperature is 2 1 1 2 change in temperature change in time 2 1 15 f f 20 1 20 15 3° per hour 2 53 68 20 15 15 5 The rate of change is negative because the temperature is decreasing at an average rate of per hour. 3° ■ Geometric Interpretation of Average Rate of Change If P and Q are points on the graph of a function f, then the straight line determined by P and Q is called a secant line. Figure 3.7-2 shows the secant line joining the points (4, 40) and (12, 58) on the graph of the temperature function f of Example 3. (12, 58 68° 64° 60° 56° 52° 48° 44° 40° (4, 40) 4 A.M. 8 12 Noon Time Figure 3.7-2 16 20 P.M. 24 Using the points (4, 40) and (12, 58), the slope of 58 40 12 4 this secant line is 18 8 2.25. To say that (4, 40) and (12, 58) are on the graph of f means that slope of secant line 2.25 58. 40 Thus, and 12 4 f f 2 1 2 1 58 40 12 4 f 12 1 2 12 4 1 2 4 f average rate of change as x goes from 4 to 12. The same thing happens in the general case. Section 3.7 Rates of Change 219 Secant Lines and Average Rates of Change Let f be a function. average rate of change of f from x a to x b f(b) f (a) b a slope of secant line joining (a, f (a)) and (b, f (b)) on the graph of f The Difference Quotient Average rates of change are often computed for very small intervals, such as the rate from 4 to 4.01 or from 4 to 4.001. Since and 4.001 4 0.001, both cases are essentially the same: computing the rate for some small quantity h. Furof change over the interval from 4 to thermore, it is often possible to define a formula to determine the average rate for all possible values of h. 4.01 4 0.01 4 h Example 5 Computing Average Speed by using a Formula Find a formula for the average speed of a falling rock from time x to time x h. Use the formula to find the average speed from 3 to 3.1 seconds. Solution 1 t d d Recall that the distance the rock travels is given by the function with distance 2 average speed d measured in feet and time t in seconds. x h 2 x h 2 x 2 2xh h2 x h 2 h 16x 2 32xh 16h2 16x2 h d x 1 x 2 16x2 16x2 16 1 2 2 1 16t2 t 2 1 1 16 1 32xh 16h2 h h h 1 32x 16h h 2 32x 16h To find the average speed from 3 to 3.1 seconds, apply the formula above, with h 0.1. x 3 and average speed 32x 16h 16 32 3 1 2 96 1.6 97.6 feet per second 0.1 2 1 ■ x h In general, the average rate of change of any function f over the interval from x to can be computed as in Example 5: Apply the definition of average rate of change with average rate of change f b x h and This last quantity is just the difference quotient of f (see page 144). 220 Chapter 3 Functions and Graphs Difference Quotients and Rates of Change Let f be a function. The average rate of change of f over the interval from x to quotient. is given by the difference x h f(x h) f(x) h Example 6 Using a Rate of Change Formula Find the difference quotient of V x 3 55 x 1 2 and use it to find the average rate of change of V as x changes from 8 to 8.01. Solution Find the difference quotient of ⎧⎨⎩ 3 V(x h) x h 55 2 1 and simplify. V x 1 2 V(x) ⎧⎨⎩ x 3 55 1 55 11 55 1 3x 2 3xh h2 55 1 55 x 3 3x 2h 3xh 2 h 3 x 3 h To find the average rate of change as x changes from 8 to let x 8 h 0.01. and 3x 2 3xh h2 55 So the average rate of change is 3 82 3 8 0.01 2 2 1 0.01 1 55 2 8.01 8 0.01, 3.495 ■ Exercises 3.7 1. A car moves along a straight test track. The distance traveled by the car at various times is shown in the table below. Find the average speed of the car over each interval. a. 0 to 10 seconds b. 10 to 20 seconds c. 20 to 30 seconds d. 15 to 30 seconds 2. The yearly profit of a small manufacturing firm is shown in the tables below. What is the average rate of change of profits over the given time span? a. 1996–2000 c. 1999–2002 b. 1996–2003 d. 1998–2002 Year 1996 1997 1998 1999 Profit $5000 $6000 $6500 $6800 Time (seconds) Distance (feet) 0 0 5 10 15 20 25 30 Year 2000 2001 2002 2003 20 140 400 680 1400 1800 Profit $7200 $6700 $6500 $7000 3. Find the average rate of change of the volume of the balloon in Example 2 as the radius increases a. from 2 to 5 inches. b. from 4 to 8 inches. 4. Find the average rate of change of cost for the company in Example 3 when production increases from a. 5 to 25 desks. b. 0 to 40 desks. 5. The graph in the figure shows the monthly sales of floral pattern ties (in thousands of ties) made by a company over a 48-month period. Sales are very low when the ties are first introduced; then they increase significantly, hold steady for a while, and then drop off as the ties go out of fashion. Find the average rate of change of sales (in ties per month) over the given interval. a. 0 to 12 c. 12 to 24 e. 28 to 36 g. 36 to 40 b. 8 to 24 d. 20 to 28 f. 32 to 44 h. 40 to 48 10 16 24 32 40 48 Month 6. A certain company has found that its sales are related to the amount of advertising it does in trade magazines. The graph in the figure shows the sales (in thousands of dollars) as a function of the amount of advertising (in number of magazine ad pages). Find the average rate of change of sales when the number of ad pages increases from a. 10 to 20. c. 60 to 100. e. Is it worthwhile to buy more than 70 pages of ads if the cost of a one-page ad is $2000? if the cost is $5000? if the cost is $8000?
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b. 20 to 60. d. 0 to 100. Section 3.7 Rates of Change 221 200 s e l a 100S 0 10 20 30 40 50 60 70 80 90 100 Pages 7. When blood flows through an artery (which can be thought of as a cylindrical tube) its velocity is greatest at the center of the artery. Because of friction along the walls of the tube, the blood’s velocity decreases as the distance r from the center of the artery increases, finally becoming 0 at the wall of the artery. The velocity (in centimeters per second) is given by the function v 18,500 1 0.000065 r2 2 where r is measured in centimeters. Find the average rate of change of the velocity as the distance from the center changes from r 0.002 a. to . r 0.003 b. to . r 0.025 c. . r 0.001 r 0.002 r 0 to 8. A car is stopped at a traffic light and begins to move forward along a straight road when the light turns green. The distance (in feet) traveled by the car in t seconds is given by 0 t 30 from a. c. t 2 1 What is the average speed of the car t 10? t 10.1? t 5? t 30? t 5 t 10 t 0 t 10 to to 2t 2 b. d. for to to s In Exercises 9–14, find the average rate of change of the function f over the given interval. 9. x f 2 1 from 2 x 2 x 0 to x 2 10. x f 2 1 from 0.25x 4 x 2 2x 4 x 1 to x 4 11. x f 2 1 from x 3 3x 2 2x 6 x 1 to x 3 12. x 22x 2 x 4 f from x 0 to x 3 1 2 222 Chapter 3 Functions and Graphs 13. x f 2 1 from 2x 3 2x 2 6x 5 x 1 to x 2 14. f x 2 1 from x 2 3 2x 4 x 3 to x 6 In Exercises 15–22, compute the difference quotient of the function. 15. 17. 19 16. 18. f f x x 1 1 2 2 7x 2 x 2 3x 1 160,000 8000t t2 20. V 22 21. A r 2 1 pr 2 23. Water is draining from a large tank. After t minutes 160,000 8000t t2 gallons of water in there are the tank. a. Use the results of Exercise 19 to find the average rate at which the water runs out in the interval from 10 to 10.1 minutes. b. Do the same for the interval from 10 to 10.01 minutes. c. Estimate the rate at which the water runs out after exactly 10 minutes. 24. Use the results of Exercise 20 to find the average rate of change of the volume of a cube whose side has length x as x changes from a. 4 to 4.1. d. Estimate the rate of change of the volume at x 4. the instant when c. 4 to 4.001. b. 4 to 4.01. 25. Use the results of Exercise 21 to find the average rate of change of the area of a circle of radius r as r changes from a. 3 to 3.5. d. Estimate the rate of change at the instant when b. 3 to 3.2. c. 3 to 3.1. r 3. e. How is your answer in part d related to the circumference of a circle of radius 3? 26. Under certain conditions, the volume V of a quantity of air is related to the pressure p (which is measured in kilopascals) by the equation V 5 p. the rate at which the volume is changing at the instant when the pressure is 50 kilopascals. Use the results of Exercise 22 to estimate 27. Two cars race on a straight track, beginning from a dead stop. The distance (in feet) each car has covered at each time during the first 16 seconds is shown in the figure below. a. What is the average speed of each car during this 16-second interval? b. Find an interval beginning at t 4 during which the average speed of car D was approximately the same as the average speed of car C from t 10. t 2 to c. Use secant lines and slopes to justify the statement “car D traveled at a higher average speed than car C from t 10. t 4 to ” e c n a t s i D 1200 1000 800 600 400 200 D C Car C Car D 2 4 6 8 10 12 Time 14 16 18 28. The figure below shows the profits earned by a certain company during the last quarter of three consecutive years. a. Explain why the average rate of change of profits from October 1 to December 31 was the same in all three years. b. During what month in what year was the average rate of change of profits the greatest? 400,000 300,000 200,000 t i f o r P 100,000 1991 1992 1993 Oct. 1 Nov. 1 Dec. 1 29. The graph in the figure shows the chipmunk population in a certain wilderness area. The population increases as the chipmunks reproduce, but then decreases sharply as predators move into the area. Section 3.7 Rates of Change 223 a. During what approximate time period 30. Lucy has a viral flu. How bad she feels depends (beginning on day 0) is the average growth rate of the chipmunk population positive? b. During what approximate time period, beginning on day 0, is the average growth rate of the chipmunk population 0? c. What is the average growth rate of the chipmunk population from day 50 to day 100? What does this number mean? d. What is the average growth rate from day 45 to day 50? from day 50 to day 55? What is the approximate average growth rate from day 49 to day 51? primarily on how fast her temperature is rising at that time. The figure shows her temperature during the first day of the flu. a. At what average rate does her temperature rise during the entire day? b. During what 2-hour period during the day does she feel worst? c. Find two time intervals, one in the morning and one in the afternoon, during which she feels about the same (that is, during which her temperature is rising at the same average rate). 2000 1000 103° 102° 101° 100° 99° 98° 2 4 6 8 10 12 14 16 18 20 22 24 Hours 50 100 Days 150 200 Important Concepts Important Concepts Section 3.1 Section 3.2 Section 3.3 Section 3.4 Section 3.4.A Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 Functions defined by equations . . . . . . . . . . . . . . 144 Difference quotient . . . . . . . . . . . . . . . . . . . . . . . . 144 Domain convention . . . . . . . . . . . . . . . . . . . . . . . 145 Piecewise-defined and greatest integer functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 Functions defined by graphs . . . . . . . . . . . . . . . . 150 Vertical Line Test . . . . . . . . . . . . . . . . . . . . . . . . . 151 Increasing and decreasing functions . . . . . . . . . . 152 Local maxima and minima. . . . . . . . . . . . . . . . . . 153 Concavity and inflection points . . . . . . . . . . . . . . 154 Graphs of piecewise-defined and greatest integer functions . . . . . . . . . . . . . . . . . 155 Parametric graphing. . . . . . . . . . . . . . . . . . . . . . . 157 Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 Quadratic function . . . . . . . . . . . . . . . . . . . . . . . . 163 Transformation form . . . . . . . . . . . . . . . . . . . . . . 164 Polynomial form. . . . . . . . . . . . . . . . . . . . . . . . . . 165 x-Intercept form . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Changing from one form to another . . . . . . . . . . 167 Summary of quadratic forms . . . . . . . . . . . . . . . . 169 Quadratic applications . . . . . . . . . . . . . . . . . . . . . 169 Parent functions . . . . . . . . . . . . . . . . . . . . . . . . . . 172 Vertical shifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Horizontal shifts. . . . . . . . . . . . . . . . . . . . . . . . . . 175 Reflections across the x- and y-axis . . . . . . . . . . . 177 Vertical and horizontal stretches and compressions . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Combining transformations . . . . . . . . . . . . . . . . . 180 y-Axis symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 184 x-Axis symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 185 Origin symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 187 Symmetry tests . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 Even and odd functions . . . . . . . . . . . . . . . . . . . . 188 224 Chapter Review 225 Section 3.5 Section 3.5.A Section 3.6 Sums and differences of functions . . . . . . . . . . . . 191 Products and quotients of functions . . . . . . . . . . 192 Composition of functions . . . . . . . . . . . . . . . . . . . 193 . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Domain of g f Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 Fixed points and periodic orbits . . . . . . . . . . . . . 202 Inverse relations and functions . . . . . . . . . . . . . . 204 Graphs of inverse relations . . . . . . . . . . . . . . . . . 206 One-to-one functions . . . . . . . . . . . . . . . . . . . . . . 208 Horizontal Line Test . . . . . . . . . . . . . . . . . . . . . . . 208 Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . 210 Composition of inverse functions. . . . . . . . . . . . . 211 Section 3.7 Average rate of change. . . . . . . . . . . . . . . . . . . . . 216 Secant line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 Difference quotients and rates of change . . . . . . . 220 Important Facts and Formulas Vertical Line Test: A graph in a coordinate plane represents a function if and only if no vertical line intersects the graph more than once. A quadratic function can be written in any of the following forms: f(x) a(x h)2 k Transformation form: • • • vertex h, k 2 1 x-intercepts h ± A y-intercept ah2 k k a Polynomial form: b 2a vertex • a f(x) ax 2 bx c , f b 2abb a • • x-intercepts y-intercept c b ± 2b2 4ac 2a x-Intercept form: f(x) a(x s)(x t) s t • vertex a s t 2 , f a 2 bb • x-intercepts s and t • y-intercept The graph of g ast f 1 x 2 x 1 2 c is the graph of f shifted upward c units. 226 Chapter Review The graph of c units. The graph of the left. The graph of the right. The graph of x-axis. The graph of y-axis. x The graph of 1 pressed vertically by a factor of c. x The graph of pressed horizontally by a factor of 2 1 1 c . is the graph of f shifted downward is the graph of f shifted c units to is the graph of f shifted c units to is the graph of f reflected across the is the graph of f reflected across the is the graph of f stretched or com- is the graph of f stretched or com- Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects the graph more than once. A one-to-one function f and its inverse f 1 have these properties The average rate of change of a func
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tion f as x changes from a to b is the number The average rate of change of a function f as x changes from x to x h is given by the difference quotient of the function Review Exercises Section 3.1 1. Let f be the function given by the rule f 7 2x. Complete the table x 1 2 below. x f(x. If h 3. If f x 1 2 x 1 2 x 2 3x, then h t 2 1 2 ? 2x3 x 1, then f x 2b a ? Chapter Review Chapter Review 227 227 4. What is the domain of the function f 5. What is the domain of the function g 6. What is the domain of the function h 2x 2 ? 2t 2 t 3 ? 2r 4 2r . The cost of renting a limousine for 24 hours is given by C x 1 2 150 1.75x 150 e if 0 6 x 25 if x 7 25 where x is the number of miles driven. a. What is the cost if the limo is driven 20 miles? 30 miles? b. If the cost is $218.25, how many miles were driven? denote the greatest integer function and evaluate 8. Let 3 x 4 5 2d 18.7 a. c. c 3 ? 3 4 15.7 ? b. d. 1755 . If f x 1 2 x x 0 0 , then find 2b , and f 3 2b . a 4 3 Section 3.2 Use the graph of the function f in the figure to answer Exercises 10–13. y 3 2 1 f −2 −1−1 −2 −3 −5 −4 −3 x 1 2 3 4 5 6 10. What is the domain of f ? 11. What is the range of f ? 12. 13 ? 228 Chapter Review 14. The function whose graph is shown gives the amount of money (in millions of dollars) spent on tickets for major concerts in selected years. [Source: Pollstar] s n o i l l i M 1500 1200 900 600 0 1990 1991 1992 1993 1994 1995 1996 a. What is the domain of the function? b. What is the approximate range of the function? c. Over what one-year interval is the rate of change the largest? In Exercises 15–18, determine the local maxima and minima of the function and the intervals on which the function is increasing and decreasing. Estimate the intervals on which it is concave up and concave down, and the inflection points. 15. g 16. f 17. g 18 2x2 x 1 2x3 5x2 4x 3 x3 8x2 4x 3 0.5x4 2x3 6x2 16x 2 19. State whether the graphs below represent functions of x. Explain your reasoning. a. y b. y x x 20. Draw the graph of a function f that satisfies the given conditions. The function does not need to be given by an algebraic rule. • domain of 0 2 • • range of 7 2 • 1 3, 4 f f 2 21. Sketch the graph of the function f x 2 x 1 2x if x 0 if 0 6 x 6 4 if x 4 x 1 2 e In Exercises 22 and 23, sketch the graph of the curve given by the parametric equations. 22. x t3 3t2 1 y t2 1 3 t 2 1 2 23. x t2 4 y 2t 1 3 t 3 2 1 Section 3.3 In Exercises 24–29, find the vertex, y-intercept, and x-intercepts (if any) of the quadratic function. Sketch the graph, with these points labeled. Chapter Review 229 24 26. h 28. g x x 2 2 1 1 2x 2 4x 3 2 x 1 1 21 x 2 30. Write the function f x 2 1 2 25. g 27 2x 7 29. h x 2 1 1 x 2.4 21 x 1.7 2 x 1 2 1 2 1 in polynomial and x-intercept form. 31. Write the function f x 2 3x 4 x 1 2 in transformation and x-intercept form. 32. Write the function f polynomial form. x 2 1 2 x 3 1 21 x 1 2 in transformation and Section 3.4 In Exercises 33–38, graph each function with its parent function on the same graph. 33. f x 1 2 2x 36. h x 2 1 x 3 4 34. h x 2 1 1 x 2 37. f x 1 2 x 2 3 35. x g 1 2 1.5 x 0 0 38. g x 1 2 23 2x In Exercises 39–42, list the transformations, in the order they should be performed on the graph of to produce a graph of the function f. g( x) x2, 39. 41.25x 2 2 3 x 7 2 2 2 1 40. 42 43. The figure shows the graph of a function f. If g is the function f x 2 2 2 1 , x then which of these statements is true? g 1 a. The graph of g touches, but does not cross, the x-axis. b. The graph of g touches, but does not cross, the y-axis. c. The graph of g crosses the y-axis at d. The graph of g crosses the y-axis at the origin. e. The graph of g crosses the x-axis at x 3. y 4. y 2 2 2 x 2 230 Chapter Review 44. The graph of a function f is shown in the figure. On the same coordinate plane, carefully draw the graphs of the functions g and h whose rules are: g x 1 2 f x 1 2 and 45. U.S. Express Mail rates in 2002 are shown in the following table. Sketch the graph of the function e, whose rule is weighing x pounds by Express Mail. x e 1 2 cost of sending a package Express Mail Letter Rate—Post Office to Addressee Service Up to 8 ounces Over 8 ounces to 2 pounds $13.65 17.85 Up to 3 pounds Up to 4 pounds Up to 5 pounds Up to 6 pounds Up to 7 pounds 21.05 24.20 27.30 30.40 33.45 Section 3.4.A In Exercises 46–48, determine algebraically whether the graph of the given equation is symmetric with respect to the x-axis, the y-axis, or the origin. 46. 5y 7x 2 2x 47. x 2 y 2 2 48. x 2 y 2 6y 5 In Exercises 49–51, determine whether the given function is even, odd, or neither. 49. x g 1 2 9 x 2 50. 52. Plot the points 2, 1 , f x 2 1 1, 3 x 1 x 0 0 51. h x 2 1 3x , 1 , 3, 2 , and 4, 1 on coordinate axes. 1 2 1 2 points 2 a. Suppose the points lie on the graph of an even function f. Plot the , 1 3, f 1 b. Suppose the points lie on the graph of an odd function g. Plot the points 3 22 1 3, g 2 3 4, g 4, f , and , and 4 4 0, g 1, g 2, g 2, f 0, f 1, f 22 22 22 22 , 22 1 1 22 1 1 22 1 1 22 1 22 1 Chapter Review 231 Section 3.5 53. If a. f 1 1 x 2 f g 3x 2 1 2 21 1 x 1 and g x 1 2 and g x 1 2 x3 1, f g b. 1 21 2x2 5, 2 54. If f x 1 2 find each value. c. 2 fg 0 2 21 1 find the rule of each function and state its domain. f g x a. 1 2 21 x 2 1 55. If g x 1 2 b. 1 x 1 f g x 21 g 2 x 1 2 1 then g 1 c. fg x 2 21 1 ? 2 d. f gb 1 x 2 a In Exercises 56–59, if f ( x) 1 x 1 and g (x) x3 3 , find each value. 56. f g 1 2 21 1 57. g f 1 2 2 21 58. 2 g 1 f 1 22 59. g f 1 21 x 1 2 In Exercises 60 and 61, let f(x) x2 x 6 and g(x) |x|. Describe the relationship between the graph of f(x) and the composite function in terms of transformations. 60. f g 1 2 62. If x f 1 f g. 1 1 x 2 and g x 1 2 2x, find the domain of the composite function 61. f g 1 2 63. Find two functions f and g such that neither is the identity function and f g x 2 21 1 2 x 1 1 2 2 64. The radius of an oil spill (in meters) is 50 times the square root of the time t (in hours). a. Write the rule of a function f that gives the radius of the spill at time t. b. Write the rule of a function g that gives the area of the spill at time t. c. What are the radius and area of the spill after 9 hours? d. When will the spill have an area of 100,000 square meters? Section 3.5.A 65. Find the first eight terms of the orbit of under the function 66. Describe the set of fixed points of the function f x 1 2 x . 4 3 Section 3.6 67. The graph of a function f is shown in the figure. Sketch the graph of the inverse of f. y x f 232 Chapter Review In Exercises 68–73, find the inverse relation of each function. If the inverse is a function, write its rule in function notation. 68. 70. 72 25 x 7 69. 71. 73 2x 1 25 x3 1 In Exercises 74–76, determine whether or not the given function is one-to-one. Give reasons for your answer. If so, graph the inverse function. 74. 76.02x 3 0.04x 2 0.6x 4 75. 1 x f x 1 2 0.2x3 4x 2 6x 15 In Exercises 77–80, use composition to verify that f and g are inverses. 77. 78. 79 4x 6 g x 2 1 0.25x 1.5 x3 1 g 23 x 1 x 2 1 2x 1 x 3 g x 1 2 3x 1 x 2 80 Section 3.7 81. Find the average rate of change of the function g x3 x 1 x 2 x 1 2 as x changes from a. 1 to 1 b. 0 to 2 82. Find the average rate of change of the function f changes from a. 3 to 0 b. 3 to 3.5 c. 3 to 5 2x2 x 1 as x x 1 2 83. If f 2x 1 and g composite function x 2 1 f g 3x 2, as x changes from 3 to 5. find the average rate of change of the 84. If f x2 1 and g composite function x 2 1 f g x 2, as x changes from 1 to 1. find the average rate of change of the x 1 2 x 1 2 In Exercises 85–88, find the difference quotient of the function and simplify. 85. f 87. g x x 1 1 2 2 3x 4 x 2 1 86. g 88. f x x 1 1 2 2 4x 1 x 2 x 89. The graph of the function g in the figure consists of straight line segments. Chapter Review 233 Find an interval over which the average rate of change of g is 3 a. 0 b. d. Explain why the average rate of change of g is the same from c. 0.5 as it is from 2.5 to 0. 3 to 1 y 2 −2 −4 −2 x 2 4 6 8 90. The graph in the figure below shows the population of fruit flies during a 50-day experiment in a controlled atmosphere. a. During what 5-day period is the average rate of population growth the slowest? b. During what 10-day period is the average rate of population growth the fastest? c. Find an interval beginning at the 30th day during which the average rate of population growth is the same as the average rate from day 10 to day 20. 400 300 n o i t a l u p o 100P 200 0 10 20 30 Days 40 50 91. The profit (in hundreds of dollars) from selling x tons of an industrial P 0.2x 2 0.5x 1. chemical is given by What is the average rate of change in profit when the number of tons of the chemical sold increases from a. 4 to 8 tons? c. 4 to 4.1 tons? b. 4 to 5 tons? x 2 1 92. On the planet Mars, the distance traveled by a falling rock (ignoring atmospheric resistance) in t seconds is in order to have an average speed of 25 feet per second over that time interval? feet. How far must a rock fall 6.1t2 12 Instantaneous Rates of Change ctions Rates of change are a major theme in differential calculus—not just the average rate of change discussed in Section 3.7, but the instantaneous rate of change of a function. Instantaneous rate of change is the rate of change at a particular instant, which is like a policeman finding the speed of a car at a particular moment by using radar. Even without calculus, however, quite accurate approximations of instantaneous rates of change can be obtained by using average rates appropriately. The equation of the position (in feet) above the ground of a falling object after t seconds is given by the equation s 16t 2 v0t s0, s0 is the initial height in feet and where is the initial velocity in feet per second. The object must be moving straight upward or straight downward. v0 Example 1 Instantaneous Velocity A ball is thrown straight up from a rooftop with an initial height of 160 feet and an initial velocity of 48 feet per second. The bal
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l misses the rooftop on its way down and falls to the ground. Find the instantaneous velocity of the ball at seconds. t 2 Solution The height of the ball is given by the equation 2 48t 160. s t 2 1 16t t 2 The exact speed of the ball at can be approximated by finding the average speed over very small time intervals, say 2 to 2.01 or even shorter intervals. Over a very short time span, such as a hundredth of a second, the ball cannot change speed very much, so these average speeds should be a reasonable approximation of its speed at the instant t 2. The difference quotient of the function s, which represents the average velocity of the function s, is s 1 ¢s ¢ 16 1 2 48 1 2 t h 2 16 1 1 2 2th h t 2 48 2 1 2 160 h t h 2 h 16t 2 48t 160 1 2 160 1 2 16t 2 48t 160 2 16t 2 32th 16h 2 48t 48h 160 16t 2 48t 160 h ¢ NOTE is often used to denote change in the value of a quantity. , which is ¢s ¢t read “change in s divided by change in t,” represents the ratio of the change in position to the change in time, which is velocity. 234 32th 16h2 48h h 32t 16h 48 General expression of average velocity When t 2, the expression for average velocity is 32 2 16h 48 16 16h. 2 1 The values of the average velocity at times close to following table. t 2 are shown in the Change in Time 2 to 2 h 2 to 2.1 2 to 2.01 2 to 2.001 2 to 2.0001 2 to 2.00001 h 0.1 0.01 0.001 0.0001 0.00001 Average Velocity t 2 16 16h at 17.6 16.16 16.016 16.0016 16.00016 As the value of h gets very small, the values of the average velocity approach the value of the instantaneous velocity. Since these values feet per second. approach ■ the instantaneous velocity at t 2 16, 16 is In the table in Example 1, notice that when h is small, the term contributes very little to the average rate of change. Therefore, the instantaneous velocity is the remaining term of the difference quotient at t 2, In a similar fashion, the general expression of the difference quotient, becomes namely 16. 16h 32t 16h 48, 32t 48 Instantaneous rate of change expression when h is very small. Instantaneous rate of change can be found by using 32t 48 for different values of t. Slope of the Tangent Line In Example 1, the instantaneous rate of change was found by calculating the average rate of change over smaller and smaller intervals. This technique can also be represented on a graph. The average rate of change over an interval is the slope of the secant line that contains the points 22 If h is very small, the slope of the secant line and approaches the slope of the tangent line, a line that touches the graph at only one point. The slope of the tangent line to a curve at a point is equal to the instantaneous rate of change of the function at that point. x h, f 1 x h x, f 1 22 x . 1 1 235 1 1 , 3 , 2 4 22 22 3, f 1 and the points The graph in Figure 3.C-1 shows secant lines that contain the fixed point 2, f 1 4, f 1 for the function in 1 Example 1. As the second point approaches the fixed point, the secant lines approach the tangent line, and the slopes of the secant lines approach the slope of the tangent line. The slope of the tangent line, shown in red, is the instantaneous rate of change of the function at the point The tangent line can be used to determine when the instantaneous rate of change is 0, that is, when the object changes direction. x 2. 5, f 1 and 22 22 5 1 t 1 2 3 4 5 Example 2 Instantaneous Rate of Change Equal Zero Figure 3.C-1 When is the instantaneous rate of change of the ball in Example 1 equal to zero? What is the maximum height reached by the ball? Solution The expression for instantaneous rate of change for 160 solve for t. was found to be 32t 48. 2 48t s Set this expression equal to zero and 16t˛ t 1 2 32t 48 0 t 48 32 1.5 seconds Therefore, 1.5 seconds after the ball is thrown upward, its rate of change is zero. That is, the ball stops moving upward 1.5 seconds after being thrown, it will be at its highest at that time. Maximum height is given by s(1.5). 5 Figure 3.C-2 1.5 s 1 2 1 1 16 2 48 1.5 2 1 48 16 2.25 1 36 72 160 196 feet. 2 1.5 2 1.5 2 160 160 The graph of s and the tangent line to the curve at Figure 3.C-2. Notice that the tangent line is horizontal when are shown in t 1.5. t 1.5 ■ Writing the Equation of a Tangent Line The equation of the tangent line to a curve at a point can be found by using the instantaneous rate of change at that point and the point-slope form of a line. Example 3 Equation of a Tangent Line Write the equation of the tangent line to the function given in Example 1 when t 2. Solution t 2, s 2 the position of the ball is When 2 1 2, 192 therefore, the point on the graph is 1 16. is taneous rate of change when is 16. t 2 1 2 2 48 16 . 160 192, 2 2 From Example 1, the instan2 So, the slope of the tangent line 1 2 s 200 150 100 50 0 300 0 0 236 300 0 0 5 Figure 3.C-3 By using the point-slope form of a line with equation of the tangent line is x x12 y y1 x 2 y 192 16 1 2 y 192 16x 32 y 16x 224 m 1 m 16 and (2, 192), the Point-slope form Tangent line at 2, 192 1 2 The graph of s and the tangent line at t 2 is shown in figure 3.C-3. ■ Numerical Derivatives Most graphing calculators will give you the value of the instantaneous rate of change of functions at particular input values. The instantaneous rate of change at a particular input value is usually called a numerical derivative and is denoted as nDeriv, nDer, d/dx, dY/dX, or On TI calculators, numerical derivatives are found in the MATH or CALC menu. On a Casio, numerical derivatives are displayed when using the TRACE, Graph-to-Table, and Table & Graph features if the Derivative item is On in the SET UP screen for graphs. Check your calculator’s manual for the syntax. 0. Exercises 1. Write an equation for the height of a ball after t seconds that is thrown straight up from a bridge at an initial height of 75 ft with an initial velocity of 20 ft/sec. Find the instantaneous velocity of the ball at seconds. t 2 2. Find when the instantaneous velocity of the ball in Exercise 1 is 0 feet per second, and interpret the result. 3. Write an equation for the height of a ball after t seconds that is dropped from a tower at an initial height of 300 ft. Find the instantaneous t 3 seconds. [Hint: When velocity of the ball at an object is dropped, and not thrown upward or downward, the initial velocity is 0.] In Exercises 4–7, find the instantaneous rate of change of the function at the given value. 4. 5. t 2 4t 7 at at t 3 6. 7 at x 1 at u a 8. Find the instantaneous rate of change of 2 t 4t 2 16t 12 f 1 the equation of the tangent line. Graph the tangent line on the same graph. and use it to find f 1 and at t 4 t 2 9. The surface area of a sphere of radius r is given 4 pr by the formula instantaneous rate of change of the surface area at r 1 and interpret the result. Find the 2. S r 2 1 10. The profit, in dollars, of a manufacturer selling 2 1 x 0.05x digital phones is given by the equation 2 200x 30,000 where x is the P number of phones sold. Find the instantaneous rate of change at What are the units of the rate of change? What does the instantaneous rate of change at of the manufacturer? x 1000. x 1000 tell you about the profit 237 C H A P T E R 4 Polynomial and Rational Functions Beauty between order and chaos! Everyday life abounds in apparently random phenomena: changing weather, traffic clusters on the freeway, lightning paths, ocean turbulence, and many others. Chaos theory is an area of mathematics that analyzes such chaotic behavior. The Mandelbrot set, which is shown above, is a fascinating mathematical object derived from the complex numbers. Its beautiful boundary illustrates chaotic behavior. See Excursion 4.5.A. 238 Chapter Outline 4.1 Polynomial Functions 4.2 Real Zeros 4.3 Graphs of Polynomial Functions 4.3.A Excursion: Polynomial Models 4.4 Rational Functions 4.5 Complex Numbers 4.5.A Excursion: The Mandelbrot Set 4.6 The Fundamental Theorem of Algebra Chapter Review can do calculus Optimization Applications Interdependence of Sections 4.3 4.1 > 4.2 > > > 4.4 4.5 > 4.6 P olynomial functions arise naturally in many applications. Many com- plicated functions can be approximated by polynomial functions or their quotients, rational functions. 4.1 Polynomial Functions Objectives • Define a polynomial • Divide polynomials • Apply the Remainder Theorem, the Factor Theorem, and the connections between remainders and factors • Determine the maximum number of zeros of a polynomial A polynomial is an algebraic expression that can be written in the form anxn an1xn1 p a3x3 a2x2 a1x a0 where n is a nonnegative integer, x is a variable, and each of is a constant, called a coefficient. The coefficient term. Note the characteristics of a polynomial. a0, a1, p , an is called the constant a0 • all exponents are whole numbers • no variable is contained in a denominator • no variable is under a radical Any letter may be used as the variable in a polynomial. Examples of polynomials include the following. x3 6x2 1 2 y15 y10 7 w 6.7 12 239 240 Chapter 4 Polynomial and Rational Functions Constant and Zero Polynomials A polynomial that consists of only a constant term, such as the polynomial 12, is called a constant polynomial. The zero polynomial is the constant polynomial 0. Degree of a Polynomial The exponent of the highest power of x that appears with nonzero coefficient is the degree of the polynomial, and the nonzero coefficient of this highest power of the variable is the leading coefficient. Polynomial 6x7 4x3 5x2 7x 10 x3 12 0x9 2x6 3x7 x8 2x 4 Degree 7 3 0 8 Leading coefficient 6 1 12 1 Constant term 10 0 12 4 The degree of the zero polynomial is not defined because in that case no power of x occurs with a nonzero coefficient. A polynomial function is a function whose rule is given by a polynomial f(x) anxn an1xn1 p a1x a0 an are real numbers with 0, and n an, an1 , p , a1, a0 where is a nonnegative integer. Definition of a Polynomial Function NOTE The term polynomial may refer to a polynomial express
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ion or a polynomial function. The context should clarify the meaning. Review addition, subtraction, and multiplication of polynomials in the Algebra Review Appendix, if needed. Polynomial functions of degree less than 5 are often referred to by special names. • First-degree polynomial functions are called linear functions. • Second-degree polynomial functions are called quadratic functions. • Third-degree polynomial functions are called cubic functions. • Fourth-degree polynomial functions are called quartic functions. Polynomial Division Long division of polynomials is similar to long division of real numbers, as shown in Example 1. Example 1 Polynomial Division Divide 3x4 8x2 11x 1 by x 2. Solution Expand the dividend to accommodate the missing term, 0x3, and write Section 4.1 Polynomial Functions 241 the problem as division. Divisor S x 23x4 0x3 8x2 11x 1 d Dividend Divide the first term of the divisor, x, into the first term of the dividend, and put the result, 3x4, the entire divisor, put the result on the third line, and subtract. on the top line. Then multiply 3x3, 3x3 times 3x4 x 3x3 x 23x4 0x3 8x2 11x 1 3x4 6x3 6x3 8x2 11x 1 Next, divide the first term of the divisor, x, into the leading term of the 6x3 x 6x2, times the entire divisor, put the result on the fifth line, and on the top line. Then mul- and put the result, difference, 6x3, 6x2 tiply subtract. Divisor S Repeat the process until the remainder has a smaller degree than the divisor. Because the degree of the divisor in this case is 1, the process will stop when the subtraction results in a constant. 3x3 6x2 4x 3 x 23x4 0x3 8x2 11x 1 3x4 6x3 6x3 8x2 11x 1 6x3 12x2 4x2 11x 1 4x2 3x 1 3x 6 5 d Quotient d Dividend d Remainder 8x ■ CAUTION Synthetic division can be used only when the divisor is x c. Synthetic Division When the divisor is a first-degree polynomial such as there is a convenient shorthand method of division called synthetic division. The problem from Example 1 is reconsidered below, where it is used to illustrate the procedure used in synthetic division. Divisor x 2 or x 5, Dividend Step 1 In the first row, list the 2 from the divisor and the coefficients of the dividend in order of decreasing powers of x, inserting 0 for missing powers of x. Step 2 Skip a line, draw a line, and draw a partial box under the line beneath the last coefficient in the first row. Bring the first coefficient of the dividend below the line. > ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ 3 0 8 11 1 > 2 2 3 0 8 11 1 > 3 242 Chapter 4 Polynomial and Rational Functions Step 3 Multiply the divisor constant, 2, and the number below the line, 3. Place the product, 6, under the dividend’s next coefficient. 0 6 > 2 3 > 3 8 11 1 1 1 1 Step 4 Add the numbers in that column and write the sum below the column under the line. Step 5 Multiply the divisor and the last entry of the last row. Place the product under the dividend’s next coefficient. Step 6 Add the numbers in that column and write the sum below the column under the line. Step 7 Repeat steps 5 and 6 until a number appears in the box under the last column 11 0 6 > 6 11 8 12 > 11 8 12 > 4 8 12 4 11 8 3 1 6 5 The last number in the third row is the remainder. The other numbers in the third row are the coefficients of the quotient, arranged in order of decreasing powers of x. Because a fourth-degree polynomial was divided by a first-degree polynomial, the quotient is a third-degree polynomial. 3x3 6x2 4x 3 5. Therefore, the quotient is and the remainder is Example 2 Synthetic Division Divide x5 5x4 6x3 x2 4x 29 by x 3 and check the result. Solution write the divisor as To divide x 3 1 2 by x 3, x5 5x4 6x3 x2 4x 29 and perform synthetic division. 1 0 1 x4 2x3 x 7 29 21 The last row shows that the quotient is der is 8. and the remain- ■ Checking Polynomial Division Recall how to check a long division problem. When 4509 is divided by 31, the quotient is 145 and the remainder is 14. To check the division, multiply the quotient by the divisor and add the remainder, 4509 145 31 14 Section 4.1 Polynomial Functions 243 Checking polynomial division uses the same process. Dividend Divisor Quotient Remainder Check the division in Example 2. 8 x4 2x3 x 7 x 3 21 2 1 x5 5x4 6x3 x2 4x 21 x5 5x4 6x3 x2 4x 29 1 8 2 The Division Algorithm f(x) If a polynomial then there is a quotient polynomial polynomial is divided by a nonzero polynomial q(x) and a remainder such that r(x) Dividend Divisor Quotient Remainder h(x), > > f(x) h(x) q(x) r(x) > > where either of the divisor, h(x). r(x) 0 or r(x) has degree less than the degree The Division Algorithm can be used to determine if one polynomial is a factor of another polynomial. Remainders and Factors If the remainder is 0 when one polynomial is divided by another polynomial, the divisor and the quotient are factors of the dividend. Example 3 Factors Determined by Division Determine if 2x2 1 is a factor of 6x3 4x2 3x 2. Solution Divide 6x3 4x2 3x 2 and see if the remainder is 0. 2x2 1, by 3x 2 2x2 16x3 4x2 3x 2 6x3 3x 4x2 2 4x2 2 0 The remainder is 0, and the Division Algorithm confirms the factorization. Dividend Divisor Quotient Remainder 6x3 4x2 3x 2 2x2 1 2x2 1 3x 2 3x 2 2 2 21 21 1 1 0 244 Chapter 4 Polynomial and Rational Functions 2x2 1 Therefore, tor is the quotient, 1 is a factor of 3x 2 . 2 2 1 6x3 4x2 3x 2, and the other fac- ■ Remainders When a polynomial x 3 x 5, or f x 1 2 is divided by a first-degree polynomial, such as the remainder is a constant. For example, when x3 2x2 4x 5 is divided by x 3, 2 f x 1 x2 x 1 the quotient is x3 2x2 4x 5 f 1 2 3 Notice that 1 1 That is, the number by x 3, 3 3 0 11 21 3 1 2 f x2 x 1 and the remainder is 2. x 3 21 2 1 32 3 1 21 2 2. 2 is the same as the remainder when 2 2 2 as stated in the Remainder Theorem. f x 1 2 is divided Remainder Theorem If a polynomial is f(c). f(x) is divided by x c, then the remainder Example 4 The Remainder When Dividing by x c Find the remainder when x79 3x24 5 is divided by x 1. Solution Let f x 1 2 x79 3x24 5 and apply the Remainder Theorem with c 1. Therefore, the remainder when f 1 1 2 179 3 1 1 24 5 1 3 5 9 2 x79 3x24 5 is divided by x 1 is 9. ■■ Example 5 The Remainder When Dividing by x k Find the remainder when 3x4 8x2 11x 1 is divided by x 2. Solution x a, x 2, is not in the form The divisor, must be applied carefully. Rewrite the divisor as 3 2 11 2 4 8 2 1 2 11 8 3 4 1 2 48 32 22 1 5 2 16 so the Remainder Theorem 2 . and find Section 4.1 Polynomial Functions 245 By the Remainder Theorem, when x 2 , the remainder is 5, which can be verified by division. 3x4 8x2 11x 1 is divided by ■ Zeros and Factors 1 f x is a polynomial, then solutions of the equation 0 Recall that if are called zeros of the function. A zero that is a real number is called a real zero. The connection between zeros of a polynomial function and factors of the polynomial is given below. x f 1 2 2 Factor Theorem A polynomial function only if f(x) has a linear factor x a if and f(a) 0. The Factor Theorem states If a is a solution of f(x) 0, then x a is a factor of f(x). and If x a is a factor of f(x), then a is a solution of f(x) 0. You can see why the Factor Theorem is true by noting that the remainder when according to the Remainder is divided by Theorem. Therefore, by the Division Algorithm x a is x a f f 1 2 2 1 f Thus, when versely, when a 2 1 x a 0, f q x x a 1 x a is a factor, the remainder 1 so that is a factor. Con- x a must be 0. 1 2 Example 6 The Factor Theorem Show that orem. Find x 3 q x 1 2 Solution is a factor of such that x 3 1 x q 1 2 2 x3 4x2 2x 3 by using the Factor The- x3 4x2 2x 3. Let f x 1 2 x3 4x2 2x 3 and find 33 4 32 3 2 1 1 27 4 3 9 2 1 1 27 36 6 3 0 x 3 is a factor of 3 3 x3 4x2 2x 3 because By the Factor Theorem, f 0. 3 1 1 2 2 Dividing f x3 4x2 2x 3 x 3 can be written in factored form. by x yields a quotient of x2 x 1, and 246 Chapter 4 Polynomial and Rational Functions x3 4x2 2x 3 Dividend Divisor Quotient x 3 x2 x 1 1 21 The product can be verified by multiplication. Fundamental Polynomial Connections 2 ■ Zeros, x-Intercepts, Solutions, and Factors of Polynomials 10 10 10 y f Section 2.1 described the connection among the x-intercepts of the graph the zeros of the function f, and the solutions of This of x x connection can be extended to include linear factors of is a polynomial. 0. x 2 1 f when (x) be a polynomial. If r is a real number that satisfies Let any of the following statements, then r satisfies all the statements. • r is a zero of the function f • r is an x-intercept of the graph of the function f • • x r x r is a solution, or root, of the equation f(x) 0 is a factor of the polynomial f(x) There is a one-to-one correspondence between the linear factors of of the graph of f. that have real coefficients and the x-intercepts f(x) The box above states that a zero, an x-intercept, a solution, and the value x r of r in a linear factor of the form are all the same for a polynomial. Additionally, the x-intercepts correspond to the linear factors of of the form where r is a real number. x r, x f 1 2 Example 7 Fundamental Polynomial Connections For f x 2 1 15x3 x2 114x 72, find the following: a. the x-intercepts of the graph of f b. the zeros of f c. the solutions to d. the linear factors with real coefficients of 15x3 x2 114x 72 0 15x3 x2 114x 72 Solution f Graph 2 shown in Figure 4.1-1. x 1 15x3 x2 114x 72 in a standard viewing window, as 10 Figure 4.1-1 a. Using the zero finder, find the x-intercepts. 3, 2 3 0.6667, and 12 5 2.4 Section 4.1 Polynomial Functions 247 b. The zeros of f are 3, 2 3 0.6667, and 12 5 2.4. c. The solutions of 0.6667, x 2 3 substitution. 15x3 x2 114x 72 0 2.4, and x 12 5 are x 3, which can be verified by d. The linear factors of x 12 5 b a and . 15x3 x2 114x 72 are x 1 3 1 , 22 x 2 3b , a The product x 3 1 2a x 2 x 12 5 b is not the original polynomial. 3ba There is a constant, a, such that x 3 a 1 2a x 2 x 12 5 b 15x3 x2 114x 72. 3b a The leading coefficient of the original polynomial, 15, must be the leading
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coefficient of the product of the factors. So, a 15. 15x3 x2 114x 72 15 x 3 1 2a x 2 3b a x 2 x 12 5 b x 12 5 b x 12 5 b 3 5 x 3 1 3 3b a 2a x 2 a 3b 5x 12 5 a 3x 2 21 x 3 x 3 1 1 2 21 2 ■ Example 8 A Polynomial with Specific Zeros Find three polynomials of different degrees that have 1, 2, 3, and zeros. 5 as Solution A polynomial that has 1, 2, 3, and x 3, 5 and conditions, such as x 5 x 1 2 5 x 2, as factors. Many polynomials satisfy these as zeros must have x 1, g h k x 2 21 x 2 21 x 1 21 Notice that g has degree 4, h had degree 5, and k has degree 8. x 1 21 x 1 21 x 4 2 2 x 3 21 x2 x 1 x 5 x x x 2 21 1 21 2 2 2 1 1 1 21 1 1 1 2 ■ The Number of Zeros of a Polynomial If a polynomial argument used in Example 8 must have x c , and has four real zeros, say a, b, c, and d, then by the same 248 Chapter 4 Polynomial and Rational Functions x c as linear factors. Because its leading term is 1 2 x d must have all four factors, its degree must 1 be at least 4. In particular, this means that no polynomial of degree 3 can have four or more zeros. A similar argument holds in the general case. has degree 4. Since x b x a x4, 21 x f 1 2 2 1 2 Number of Zeros A polynomial of degree n has at most n distinct real zeros. Example 9 Maximum Number of Distinct Real Zeros State the maximum number of distinct real zeros of f. 18x4 51x3 187x2 56x 80 f x 1 2 Solution The degree of f is 4. Therefore, the maximum number of distinct real zeros of f is 4. ■ Exercises 4.1 In Exercises 1–8, determine whether the given algebraic expression is a polynomial. If it is, list its leading coefficient, constant term, and degree. 15. 16. 1 1 x4 6x3 4x2 2x 7 1 2 x 2 x6 x5 x4 x3 x2 x 1 2 x 3 1 2 2 1. 1 x3 2. 7 3. 1 x 1 x2 1 4. 7x 2x 1 6. 4x2 32x 5 5. 7. 21 x 23 2 B x 23 A B A 7 x2 5 x 15 8. x 1 k 2 1 (where k is a fixed positive integer) In Exercises 9–16, use synthetic division to find the quotient and remainder. 3x4 8x3 9x 5 9. 1 4x3 3x2 x 7 2 1 2x4 5x3 2x 8 3x3 2x2 8 1 5x4 3x2 4x 6 2 3x4 2x3 7x 4 10. 11. 12. 13. 14 In Exercises 17–22, state the quotient and remainder when the first polynomial is divided by the second. Check your division by calculating (Divisor)(Quotient) Remainder 17. 3x4 2x2 6x 1; x 1 18. x5 x3 x 5; x 2 19. x5 2x4 6x3 x2 5x 1; x3 1 20. 3x4 3x3 11x2 6x 1; x3 x2 2 21. 5x4 5x2 5; x2 x 1 22. x5 1; x 1 In Exercises 23–26, determine whether the first polynomial is a factor of the second. 2 2 23. x2 3x 1; x3 2x2 5x 6 24. x2 9; x5 x4 81x 81 25. x2 3x 1; x4 3x3 2x2 3x 1 26. x2 5x 7; x3 3x2 3x 9 In Exercises 27–30, determine which of the given numbers are zeros of the given polynomial. 27. 2, 3, 0, 1; g x 1 2 x4 6x3 x2 30x 28. 1, 1 2 , 2, 1 2 , 3; f x 1 2 6x2 x 1 29. 222, 22, 22, 1, 1; 30. 23, 23, 1, 1; k x 1 2 x3 x2 8x 8 h x 2 1 8x3 12x2 6x 9 Section 4.1 Polynomial Functions 249 47. f 48. g 49 50. f x 1 2 6x3 7x2 89x 140 x3 5x2 5x 6 4x4 4x3 35x2 36x 9 x5 5x4 5x3 25x2 6x 30 f(x) In Exercises 51– 54, each graph is of a polynomial function of degree 5 whose leading coefficient is 1, but the graph is not drawn to scale. Use the Factor Theorem to find the polynomial. Hint: What are the zeros of ? What does the Factor Theorem tell you? f(x) 51. y In Exercises 31–40, find the remainder when divided by without using division. g(x), f(x) is −3 −2 −1 1 2 3 x10 x8; g x 1 x 1 2 x6 10; g x 2 x 1 2 3x4 6x3 2x 1; g x 1 x5 3x2 2x 1; g x3 2x2 5x 4 10x75 8x65 6x45 4x32 2x15 5; 2x5 3x4 x3 2x2 x 8; g x x3 8x2 29x 44 10 1 2 x 11 52. y −3 −2 −1 1 2 3 2x5 3x4 2x3 8x 8; g x5 10x4 20x3 5x 95; g x 20 x 10 x x 1 1 2 2 53. y In Exercises 41–46, use the Factor Theorem to deteris a factor of mine whether f(x). h(x) x 1; f x5 1 x 1 2 x 1 2 ; f 2x4 x3 x 3 4 x 2 1 −3 −2 −1 1 2 3 x 2; f x 1; f x 1; f x 2 x3 3x2 4x 12 54. y x3 4x2 3x 8 14x99 65x56 51 x3 x2 4x 4 −3 −2 −1 1 2 3 In Exercises 47–50, use the Factor Theorem and a calculator to factor the polynomial, as in Example 7. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40 41. h 42. h x x 2 2 1 1 43. 44. 45. 46 250 Chapter 4 Polynomial and Rational Functions In Exercises 55–58, find a polynomial with the given degree n, the given zeros, and no other zeros. 55. n 3; zeros, 1, 7, 4 56. n 3; zeros, 1, 1 57. n 6; zeros 1, 2, p 58. n 5; zero 2 59. Find a polynomial function f of degree 3 such that 17 10 f 1 2 and the zeros of f are 0, 5, and 8. x 2 1 60. Find a polynomial function g of degree 4 such that the zeros of g are 0, g 288. 3 1 2 1 , 2, 3, and In Exercises 61 – 64, find a number k satisfying the given condition. 61. x 2 is a factor of x3 3x2 kx 2. 62. x 3 is a factor of x4 5x3 kx2 18x 18. 63. x 1 is a factor of k2 x4 2kx2 1. 64. x 2 is a factor of x3 kx2 3x 7k. 65. Use the Factor Theorem to show that for every real number c, x c is not a factor of x4 x2 1. 66. Let c be a real number and n a positive integer. a. Show that b. If n is even, show that x c is a factor of x c xn cn. [Remember: x c x xn cn. is a factor of c . 1 2 67. a. If c is a real number and n an odd positive x c integer, give an example to show that may not be a factor of xn cn. b. If c and n are as in part a, show that factor of xn cn. x c is a 68. Critical Thinking For what value of k is the kx2 2x 1 x g difference quotient of 7x 2 3.5h? to 1 2 equal 69. Critical Thinking For what value of k is the difference quotient of 2x 5 h? f x 1 2 x2 kx equal to 70. Critical Thinking When x3 cx 4 is divided by x 2, the remainder is 4. Find c. 71. Critical Thinking If 1 d2 2x3 dx2 x d is a factor of what is d? x 5, 1 2 4.2 Real Zeros Objectives • Find all rational zeros of a polynomial function • Use the Factor Theorem • Factor a polynomial completely • Find lower and upper bounds of zeros 1 1 2 2 f f x x 0. 5x 3, is the same as solving the Finding the real zeros of a polynomial The zero of a first-degree polynorelated polynomial equation, mial, such as can always be found by solving the equation 5x 3 0. Similarly, the zeros of any second-degree polynomial can be found by using the quadratic formula, as discussed in Section 2.2. Although the zeros of higher degree polynomials can always be approximated graphically as in Section 2.1, it is better to find exact values, if possible. Rational Zeros When a polynomial has integer coefficients, all of its rational zeros (zeros that are rational numbers) can be found exactly by using the following test. Section 4.2 Real Zeros 251 The Rational Zero Test If a rational number r s (written in lowest terms) is a zero of the polynomial function f(x) anxn p a1x a0 where the coefficients a0 0, then an, p , a1 are integers with an 0 and • r is a factor of the constant term • s is a factor of the leading coefficient an. and a0 The test states that every rational zero of a polynomial function with integer coefficients must meet the conditions that • the numerator is a factor of the constant term, and • the denominator is a factor of the leading coefficient. By finding all the numbers that satisfy these conditions, a list of possible rational zeros is produced. The polynomial must be evaluated at each number in the list to see if the number actually is a zero. Using a calculator can considerably shorten this testing process, as shown in the next example. Example 1 The Rational Zeros of a Polynomial Find the rational zeros of f x 2 1 2x4 x3 17x2 4x 6. Solution If f x 1 2 has a rational zero r s , then by the Rational Zero Test r must be a factor of the constant term, 6. Therefore, r must be one of the integers ± 1, ± 2, ± 3, or ± 6. 2. Therefore, s must be one of the integers ± 6 ± 1 Similarly, s must be a factor of the leading coefficient, ± 1 Consequently, the ± 1 ± 2 only possibilities for or ± . r s , are , , , , , , , . 7 7 7 7 Eliminating duplications leaves a list of the only possible rational zeros. 1, 1, 2, 2, 3, 3, 6, 6 Graph 1 the x-axis, such as necessary because only the x-intercepts are of interest. in a viewing window that includes all of these numbers on 7 y 7. A complete graph is not 7 x 7 and 2 The graph in Figure 4.2-1a shows that the only numbers in the list that could possibly be zeros are 3 , 1 2 , and 1 2 , so these are the only ones Figure 4.2-1a that need to be tested. The table feature can be used to evaluate f at x 1 2 252 Chapter 4 Polynomial and Rational Functions these three numbers, as shown in Figure 4.2-1b, where the function is entered in the editor and the independent variable is set to Ask. Y Technology Tip The table setup screen is labeled TBLSET on the TI and RANG in the Casio TABLE menu. Figure 4.2-1b The table shows that 3 and 1 2 are the rational zeros of f and 1 2 is not a zero. Therefore, the two other zeros shown in Figure 4.2-1a cannot be rational numbers, that is, the two other zeros must be irrational. ■ Zeros and the Factor Theorem Once some zeros of a polynomial have been found, the Factor Theorem can be used to factor the polynomial, which may lead to additional zeros. Example 2 Finding All Real Zeros of a Polynomial Find all the real zeros of the function given in Example 1. 2x4 x3 17x2 4x 6 f x 1 2 Solution The graph of f, shown in Figure 4.2-1a, shows that there are four x-intercepts, and therefore, four real zeros. The rational zeros, were found in Example 1. By the Factor Theorem, x 1 2 are factors of x f . 2 1 thetic division twice. First, factor x 3 out of f The other factors can be found by using syn- and 3 x 3 , 1 2 and x 3 1 2 3 2 1 6 5 17 15 2 4 6 2 2 ⎧⎪⎪⎪⎨⎪⎪⎪⎩ . x 1 2 6 6 0 Then factor x 1 2 out of 2x3 5x2 2x 2. f x 1 2 1 x 3 21 ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ 2x3 5x2 2x 2 2 Section 4.2 Real Zeros 253 2a x 1 2b1 2x2 4x 4 2 Therefore, 2x4 x3 17x2 4x 6 x 3 1 2a x 1 2b1 x 1 2b1 2x2 4x 4 x2 2x 2 2 2 2 1 x 3 2a The two remaining zeros of f are the solutions of can be found by using the quadratic formula 21 2 2 ± 212 2 x2 2x 2 0, which Therefore, 2 212 2 f x 1 2 and has two rational zeros, 2 212 2 . 3 and 1 2 , and two irrational zeros, ■ Irreducible and Completely Factored Polynomials A polynomial that cannot be written as the product of polynomials of lesser degree is said to be irreducible. When a pol
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ynomial is written as the product of irreducible factors with real coefficients, it is said to be completely factored over the set of real numbers. All linear polynomials are irreducible, and some quadratic polynomials are irreducible over the set of real numbers. NOTE Recall from 2 ± 212 2 algebra that can be simplified as follows: 2 ± 212 2 1 ± 23 2 ± 223 2 . Example 2 shows that By the Factor Theorem 2 212 2 x and 2 212 2 a 2 212 2 and b x are zeros of 2 212 2 a 2x2 4x 4. are factors. b You can verify that 2x2 4x 4 2 x c 2 212 2 a b d c x 2 212 2 a . b d Therefore, the original polynomial can be written as f x 2 1 2x4 x3 17x2 4x 6 x 1 x 3 x 2 1 2a 2b c 2 212 2 x b d c 2 212 2 a . b d a Notice that the Factor Theorem applies to irrational zeros as well as to 2 212 2 rational zeros. That is, because 2 212 2 is a zero, x is a b a factor. 254 Chapter 4 Polynomial and Rational Functions 10 Example 3 Factoring a Polynomial Completely 10 Factor f x 1 2 Solution 10 2x5 10x4 7x3 13x2 3x 9 completely. Begin by finding as many rational zeros as possible. By the Rational Zero 10 Figure 4.2-2 Test, every rational zero is of the form where s ± 1 Thus, the possible rational zeros are ± 2. or r s, r ±1, ±3, or ± 9 and ± 1, ± 3, ± 9 The graph of f shows that the only possible zeros are easily verified that both numbers are zeros of x x 3 1 and 3. It is . Consequently, by the Factor Theorem. are factors of x 1 and Division shows other factors. 2x5 10x4 7x3 13x2 3x 9 The other zeros of f are the zeros of 1 1 g 2x4 12x3 19x2 6x 9 x 3 2x3 6x2 x 3 2 2 21 2x3 6x2 x 3. x 1 21 x 1 21 x 1 2 Because every zero of zeros of f are 1 1 2 g x is also a zero of and the only rational and 3, check if either is a zero of g by using substitution 12 is not. By the Factor Theorem, 1 x 3 is a factor So 3 is a zero of g, but of x g . 1 2 Division shows that 21 21 21 21 2x3 6x2 x 3 x 3 2x2 1 2 21 2 2x2 1 Because completely factored in the last statement above. has no real zeros, it cannot be factored further. So f x 1 2 is ■ Bounds In some cases, special techniques may be needed to guarantee that all zeros of a polynomial are located. Example 4 Finding All Real Zeros of a Polynomial Show that all the real zeros of 1 and 3, and find all the real zeros of g x 1 2 g x . 2 1 x5 2x4 x3 3x 1 lie between Solution First show that g has no zero larger than 3, as follows. Use synthetic division to divide by x 3. x g 1 2 Section 4.2 Real Zeros 255 18 63 1 1 2 6 21 64 Thus, the quotient is Applying the Division Algorithm, x4 x3 2x2 6x 21 and the remainder is 64. x g 1 2 x5 2x4 x3 3x 1 x4 x3 2x2 6x 21 x 3 1 21 x 3 64 x4 x3 2x2 6x 21 2 x 7 3, the factor When , the is positive and quotient, is also positive because all its coefficients are positive. The x 7 3. x g remainder, 64, is also positive. Therefore, 1 x 7 3, g and so there are no zeros of g In particular, greater than 3. is positive whenever is never 0 when x 2 2 1 4 4 4 4 Figure 4.2-3 A similar procedure shows that g has no zero less than by x 2 by applying the Division Algorithm. g and rewrite x 1 . Divide x5 2x4 x3 3x 1 x4 3x3 2x2 2x 5 x 1 4 2 1 the factor 21 x 1 x 6 1, x 1 x 6 1 is positive when is also negative. Hence, x4 3x3 2x2 2x 5, is negative. When x is negative, all odd When powers of x are negative and all even powers are positive. Consequently, the quotient, because all odd powers are multiplied by negative coefficients. The positive quotient multiplied by the negative factor produces a negative product. 4, The remainder, is negative whenever x 6 1, 1. Therefore, all the real zeros of 2 and there are no real zeros of g less than 1 2 Finally, find all the real zeros of The only possible rational zeros are and it is easy to verify that neither is actually a zero. The graph of g in Figure 4.2-3 shows that there are exactly and 3. Because all real zeros of g lie between three real zeros between 1 and 3, g has only these three real zeros. They can be approximated by using a calculator’s zero finder. x5 2x4 x3 3x 1. lie between and 3. ± 1.3361 x 1.4268 and x 2.1012 ■ 2 is a polynomial and m and n are real numbers with Upper Bound and Lower Bound x f Suppose 1 If all the real zeros of 2 and n is called an upper bound for the real zeros of that x g m 6 n. are between m and n, m is called a lower bound Example 4 shows is a lower bound and 3 is an upper bound for the real zeros of 1 x5 2x4 x3 3x 1 If you know lower and upper bounds for the real zeros of a polynomial, you can usually determine the number of real zeros the polynomial has, as shown in Example 4. The technique used in Example 4 to test possible lower and upper bounds works in the general case. 256 Chapter 4 Polynomial and Rational Functions Bounds Test Let f(x) be a polynomial with positive leading coefficient. • If d 77 0 division of by upper bound for the real zeros of f. x d f(x) and every number in the last row in the synthetic is nonnegative, then d is an • If and the numbers in the last row of the synthetic c 66 0 division of negative, with 0 considered as either, then c is a lower bound for the real zeros of f. are alternately positive and x c f(x) by 10 Example 5 Finding All Real Zeros of a Polynomial Find all real zeros of f x 1 2 x6 x3 7x2 3x 1. Solution 10 10 10 Figure 4.2-4 By the Rational Zero Test, the only possible rational zeros are but neither is a zero of f. Hence, the real zeros are irrational. Figure 4.2-4 shows four x-intercepts between and 2, but because f has degree 6, there may be two more zeros that are not shown. Use the Bounds Test to see if all the zeros are between and 2. 2 2 ± 1, To see if 2 is an upper bound, divide f x by x 2 18 22 38 1 2 4 9 11 19 39 2 1 All nonnegative Because the divisor and every term in the last row are positive, 2 is an upper bound for the real zeros of f. Now divide f x 1 2 is a lower bound. 2 x 2 2 1 to see if by 14 14 34 1 2 4 7 7 17 35 Alternating signs Because the divisor number is negative and the signs in the last row of the synthetic division alternate, is a lower bound. 2 Therefore, all real zeros of f are between shown in Figure 4.2-4. Using a zero finder, there are four zeros. and 2, and all zeros of f are 2 x 1.5837 x 0.6180 x 0.2220 and x 1.6180 ■ The examples in this section illustrate the following guidelines for finding all the real zeros of a polynomial. Finding Real Zeros of Polynomials Section 4.2 Real Zeros 257 1. Use the Rational Zero Test to find all the rational zeros of f. [Examples 1, 3, and 6] 2. Write f(x) as the product of linear factors, one for each rational zero, and another factor g(x). [Examples 2 and 3] 3. If g(x) has degree 2, find its zeros by factoring or by using the quadratic formula. [Example 2] 4. If g(x) has degree 3 or greater, use the Bounds Test, if possible, to find lower and upper bounds for the zeros of g. Approximate the remaining zeros graphically. [Examples 4 and 5] Shortcuts and variations are always possible. For instance, if the graph of a cubic polynomial shows three x-intercepts, then it has three real zeros, which is the maximum number of zeros, and there is no point in finding bounds of the zeros. In order to find as many exact zeros as possible in Guideline 4 above, check to see if the rational zeros of f are also zeros of g. Factor accordingly, as in Example 3. g x 1 2 5 Example 6 Finding All Real Zeros of a Polynomial Find all real zeros of 5 5 f x 1 2 Solution x7 6x6 9x5 7x4 28x3 33x2 36x 20. 5 Figure 4.2-5 The graph of f shown in Figure 4.2-5 indicates 4 possible real zeros. By the Rational Zero Test, the only possible rational zeros are: ± 1, ± 2, ± 4, ± 5, ± 10, and ± 20. It can be verified that both 1 and 2 are zeros of f. The graph also suggests 2 that all the real zeros lie between and 6. The Bounds Test shows that this is indeed the case. Negative Divisor 2 1 6 9 7 28 33 36 20 2 16 50 86 116 166 260 1 8 25 43 58 83 130 240 Alternating signs Positive Divisor 6 1 6 9 7 28 33 36 20 6 0 54 366 2028 12,366 73,980 12,330 74,000 1 0 9 61 338 2061 All nonnegative Therefore, all the real zeros of f are between and 6, and the four x-intercepts shown in Figure 4.2-5 are the only real zeros of f: two are x 2.7913, rational zeros, 1 and 2, and two are irrational, as determined by using a zero finder. x 1.7913 and 2 ■ 258 Chapter 4 Polynomial and Rational Functions Exercises 4.2 When asked to find the zeros of a polynomial, find exact zeros whenever possible and approximate the other zeros. In Exercises 1 – 12, find all the rational zeros of the polynomial. 19. x3 2x2 7x 20 20. x3 15x2 16x 12 21. x5 5x4 9x3 18x2 68x 176 Hint: The Bounds Test applies only to polynomials with a x positive leading coefficient. The polynomial f has the same zeros as . Why? x f 1 2 1 2 1. x3 3x2 x 3 2. x3 x2 3x 3 22. 0.002x3 5x2 8x 3 3. x3 5x2 x 5 4. 3x3 8x2 x 20 5. Hint: The Rational Zero 2x5 5x4 11x3 4x2 Test can only be used on polynomials with nonzero constant terms. Factor of a power of x and a polynomial g(x) with nonzero constant term. Then use the Rational Zero Test on as a product . 2x6 3x5 7x4 6x3 7. 8. 9. x 1 x2 2 3 x3 1 12 Hint: The Rational Zero 1 12 Test can only be used on polynomials with integer x coefficients. Note that same zeros. (Why?) have the and 12f x f 2 1 1 2 2 3 x4 1 2 x3 5 4 x2 x 1 6 1 3 x4 x3 x2 13 3 x 2 10. 1 3 x7 1 2 x6 1 6 x5 1 6 x4 11. 0.1x3 1.9x 3 12. 0.05x3 0.45x2 0.4x 1 In Exercises 13–18, factor the polynomial as a product of linear factors and a factor is either a constant or a polynomial that has no rational zeros. such that g(x) g(x) 13. 2x3 4x2 x 2 14. 6x3 5x2 3x 1 15. x6 2x5 3x4 6x3 16. x5 2x4 2x3 3x 2 17. x5 4x4 8x3 14x2 15x 6 18. x5 4x3 x2 6x In Exercises 19–22, use the Bounds Test to find lower and upper bounds for the real zeros of the polynomial. In Exercises 23–36, find all real zeros of the polynomial. 23. 2x3 5x2 x 2 24. t4 t3 2t2 4t 8 25. 6x3 11x2 6x 1 26. z3 z2 2z 2 27. x4 x3 19x2 32x 12 28. 3x5 2x4 7x3 2x2 29. 2x5 x4 10x3 5x2 12x 6 30. x5 x3 x 31. x6 4x5 5x4 9x2 36x 45 32. x5 3x4 4x3 11x2 3x 2 33. 3x
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4 2x3 4x2 4x 1 34. x5 8x4 20x3 9x2 27x 27 35. x4 48x3 101x2 49x 50 36. 3x7 8x6 13x5 36x4 10x3 21x2 41x 10 22 37. a. Show that is an irrational number. Hint: x2 2. 22 Does this polynomial have any rational zeros? 23 b. Show that is irrational. is a zero of 38. Graph f 0.001x3 0.199x2 0.23x 6 in the x 1 2 x appear to have? standard viewing window. a. How many zeros does f 1 Without changing the viewing window, explain why Each zero corresponds to a factor of does the rest of the factorization consist of? f must have an additional zero. Hint: b. Find all the zeros of What 39. According to the FBI, the number of people murdered each year per 100,000 population can be approximated by the polynomial function x f 2 1 8.1104 1987. 0.0011x4 0.0233x3 0.1144x2 0.0126x corresponds to 0 x 10 x 0 where , 1 2 a. What was the murder rate in 1990? b. In what year was the rate 8 people per 100,000? c. In what year was the rate the highest? x x x 48 x x Section 4.2 Real Zeros 259 40. During the first 150 hours of an experiment, the growth rate of a bacteria population at time t 0.0003t3 0.04t2 0.3t 0.2 g hours is bacteria per hour. a. What is the growth rate at 50 hours? at 100 t 2 1 hours? b. What is the growth rate at 145 hours? What does this mean? c. At what time is the growth rate 0? d. At what time is the growth rate 50 bacteria per hour? e. Approximately at what time does the highest growth rate occur? 41. An open-top reinforced box is to be made from a 12-by-36-inch piece of cardboard by cutting along the marked lines, discarding the shaded pieces, and folding as shown in the figure. If the box must be less than 2.5 inches high, what size squares should be cut from the corners in order for the box to have a volume of 448 cubic inches? 36 x x x x x x 12 cut along fold along 42. A box with a lid is to be made from a 48-by-24- inch piece of cardboard by cutting and folding, as shown in the figure. If the box must be at least 6 inches high, what size squares should be cut from the two corners in order for the box to have a volume of 1000 cubic inches? 24 43. In a sealed chamber where the temperature varies, the instantaneous rate of change of temperature with respect to time over an 11-day period is given by time is measured in days and temperature in degrees Fahrenheit (so that rate of change is in degrees per day). a. At what rate is the temperature changing at the 0.0035t4 0.4t2 0.2t 6, where F t 1 2 beginning of the period the period t 11 ? 1 1 2 t 0 ? at the end of 2 b. When is the temperature increasing at a rate of 4°F per day? c. When is the temperature decreasing at a rate of 3°F per day? d. When is the temperature decreasing at the fastest rate? 44. Critical Thinking a. If c is a zero of 5x4 4x3 3x2 4x 5, f x 2 1 show that 1 c is also a zero. b. Do part a with f x 2 1 replaced by g x . 2 1 2x6 3x5 4x4 5x3 4x2 3x 2 x 2 g 1 c. Let a12x12 a11x11 p a2x2 a1x a0. f x 1 2 If c is a zero of f, what conditions must the coefficients ai satisfy so that 1 c is also a zero? 260 Chapter 4 Polynomial and Rational Functions 4.3 Graphs of Polynomial Functions Objectives • Recognize the shape of basic polynomial functions • Describe the graph of a polynomial function • • Identify properties of general polynomial functions: Continuity, End Behavior, Intercepts, Local Extrema, Points of Inflection Identify complete graphs of polynomial functions The graph of a first-degree polynomial function is a straight line, as discussed in Section 1.4. The graph of a second-degree, or quadratic, polynomial function is a parabola, as discussed in Section 3.3. The emphasis in this section is on higher degree polynomial functions. Basic Polynomial Shapes where The simplest polynomial functions are those of the form a is a constant and n is a nonnegative integer. The graphs of polynomial axn, functions of the form are of two basic types. The with different types are determined by whether n is even or odd. n 2, x x f f 2 1 2 1 axn, Polynomial Functions of Odd Degree is odd, When the degree of a polynomial function in the form its graph has the basic form shown is Figures 4.3-1a and 4.3-1b. Notice that the graph shown in Figure 4.3-1b has the same shape as the graph shown in Figure 4.3-1a, but it is the reflection of the Figure 4.3-1a across either the x-axis or the y-axis. x f 2 1 axn, f(x) axn, n odd a 7 0 y a 6 0 900,000 x 0 40,000 21 Figure 4.3-1a Figure 4.3-1b Graphing Exploration Graph each of the following functions of odd degree in the window , and compare each shape with and with those shown in Figure 4.3-1a and 4.3-1b. 30 y 30 5 x 5 • • 2x3 0.01x5 x3 2x7 Section 4.3 Graphs of Polynomial Functions 261 Polynomial Functions of Even Degree When the degree of a polynomial function in the form is even, its graph has the form shown in Figures 4.3-2a or 4.3-2b. Again, the graph x when a is negative, is the reflection of Figure 4.3-2a across of 2 1 the x-axis. axn, x f f 1 2 axn f(x) axn, n even Figure 4.3-2a Figure 4.3-2b Graphing Exploration Graph each of the following functions of even degree in the window , and compare each shape with and with those shown in Figure 4.3-2a or 4.3-2b. 30 y 30 2x4 6x6 • h • k x x 1 1 2 2 2x2 3x4 Properties of General Polynomial Functions The graphs of other polynomial functions can vary considerably in shape. Understanding the properties that follow should assist you in interpreting graphs correctly and in determining when a graph of a polynomial function is complete. Continuity Every graph of a polynomial function is continuous, that is, it is an unbroken curve, with no jumps, gaps, or holes. Furthermore, graphs of polynomial functions have no sharp corners. Thus, neither of the graphs shown in Figure 4.3-3 represents a polynomial function. Note: some calculator graphs of polynomial functions may appear to have sharp corners; however, zooming in on the area in question will show a smooth curve. 262 Chapter 4 Polynomial and Rational Functions Jump Hole Sharp corner Gap Sharp corner Figure 4.3-3 End Behavior The shape of a polynomial graph at the far left and far right of the cooris large, is called the end behavior of the x dinate plane, that is, when 0 graph. End behavior of graphs of functions of the form have common characteristics when n is odd and when n is even. The Graphing Exploration below asks you to find a generalization about the end behavior of polynomial functions of odd degree. axn x f 1 2 0 Graphing Exploration x Consider the function 2 mined by its leading term f 1 2x3 x2 6x 2x3. g x 1 2 and the function deter- • In a standard viewing window, graph f and g. Describe how the graphs look different and how they look the same. • In the viewing window 20 x 20 and 10,000 y 10,000, graph f and g. Do the graphs look almost the same? • In the viewing window 100 x 100 and 1,000,000 y 1,000,000, graph f and g. Do the graphs look virtually identical? The reason that the answer to the last question is “yes” can be understood x by observing which term contributes the most to the output value when x is large, as shown in the following chart. f 2 1 Values of Specific Terms of f(x) 2x3 x2 6x x 6x x2 100 600 10,000 50 300 2,500 70 420 4,900 100 600 10,000 g(x) 2x3 2,000,000 250,000 686,000 2,000,000 f(x) 2x3 x2 6x 1,989,400 247,200 690,480 2,009,400 Section 4.3 Graphs of Polynomial Functions 263 The chart shows that when icant compared with f the end behavior of close for large values of x. 2x3, x . 1 2 0 0 x is large, the terms x2 are insignifand they play a very minor role in determining x are relatively Hence, the values of and and x g f 6x 1 2 1 2 End Behavior of Polynomial Functions x When resembles the graph of its highest degree term. is large, the graph of a polynomial function closely 00 00 When a polynomial function has odd degree, one end of its graph shoots upward and the other end downward. When a polynomial function has even degree, both ends of its graph shoot upward or both ends shoot downward. Following are some illustrations of the facts listed in the preceding box. In Figures 4.3-4a–d, the graph of a polynomial function is shown on the left and the graph of its leading term is shown on the right. The end behavior of the graph of the polynomial is the same as the end behavior of the graph of the leading term. Note the degree of each set of graphs and whether the leading coefficient is positive or negative. 2x4 5x2 2 f x 1 2 y 4 2 2x4 g x 1 2 y 4 2 −4 −2 0 2 4 −4 −2 0 2 4 x Figure 4.3-4a 3x6 5x2 2 f x 1 2 3x6 2 −4 −2 4 2 0 −2 2 4 x 2 4 −4 −2 Figure 4.3-4b x x 264 Chapter 4 Polynomial and Rational Functions 0.4x3 x2 2x 3 f x 1 2 y 4 2 0.4x3 g x 1 2 y 4 2 −4 −2 0 2 4 −4 −2 0 2 4 x −2 −4 Figure 4.3-4c 0.6x5 4x2 2 −4 0.6x5 g x 1 2 y 4 2 −4 −2 0 2 4 −4 −2 0 2 4 x −2 −4 −2 −4 x x Figure 4.3-4d Intercepts Consider a polynomial function written in polynomial form. f x anxn an1xn1 p a1x a0 • The y-intercept of the graph of f is the constant term, a0. • The x-intercepts of the graph of f are the real zeros of f. 1 2 The graph of every polynomial function has exactly one y-intercept, and because a polynomial of degree n has at most n distinct zeros, the number of x-intercepts is limited. Intercepts The graph of a polynomial function of degree n • has one y-intercept, which is equal to the constant term. • has at most n x-intercepts. Section 4.3 Graphs of Polynomial Functions 265 That is, the number of x-intercepts can be no greater than the degree of the polynomial function. Multiplicity There is another connection between zeros and graphs. If is a factor that occurs m times in the complete factorization of a polynomial expression, then r is called a zero with multiplicity m of the related polynomial function. x r For example, and 1 are zeros of multiplicity of each zero is shown in the following chart, 1 x 1 3. The 2 21 Zero 3 1 Multiplicity 2 1 1 3 Observe in Figure 4.3-5 that the graph of f does not cross the x-axis at a zero of even multiplicity, but does cross the x-axis at odd multiplicity. 3, and 1, zero
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s of 1 Let c be a zero of multiplicity k of a polynomial f. • If k is odd, the graph of f crosses the x-axis at c. • If k is even, the graph of f touches, but does not cross, the x-axis at c. Example 1 Multiplicity of Zeros x 2 Find all zeros of State the multiplicity of each zero, and state whether the graph of f touches or crosses the x-axis at each corresponding x-intercept. x 3 x 1 3. 21 x f 1 1 2 2 1 2 2 Solution The following chart lists the zeros of f, the multiplicity of each, and whether the graph touches or crosses the x-axis at the corresponding x-intercept. Zero Multiplicity x-axis 1 2 3 2 1 3 touches crosses crosses The graph of f, shown in Figure 4.3-6, verifies that the graph touches but does not cross the x-axis at and crosses the x-axis at 2 and 3. 1 ■ y 30 20 10 −4 −2 0 2 x −10 Figure 4.3-5 Multiplicity and Graphs y 60 40 20 −2 0 2 4 x Figure 4.3-6 266 Chapter 4 Polynomial and Rational Functions Local Extrema The term local extremum (plural, extrema) refers to either a local maximum or a local minimum, that is, a point where the graph has a peak or a valley. Local extrema occur when the output values change from increasing to decreasing, or vice versa, as discussed in Section 3.2. Graphing Exploration • Graph f x 0.5x5 1.5x4 2.5x3 7.5x2 2x 5 the standard viewing window. What is the total number of local extrema on the graph? What is the degree of f ? in 2 1 • Graph x g x4 3x3 2x2 4x 5 in the standard viewing window. What is the total number of local extrema on the graph? What is the degree of x g ? 1 2 1 2 The two polynomials graphed in the Exploration are illustrations of the following fact. Number of Local Extrema A polynomial function of degree n has at most extrema. n 1 local That is, that total number of peaks and valleys on the graph is at most one less than the degree of the function. Points of Inflection Recall from Section 3.2 that an inflection point occurs where the concavity of a graph of a function changes. The number of inflection points on the graph of a polynomial is governed by the degree of the function. Number of Points of Inflection • The graph of a polynomial function of degree n, with n 2, has at most n 2 points of inflection. • The graph of a polynomial function of odd degree, with n 7 2 , has at least one point of inflection. Thus, the graph of a quadratic function, which has degree 2, has no points The graph of of inflection because it can have at most a cubic has exactly one point of inflection because it has at least 1 and at most 3 2 1. n 2 2 2 0. Technology Tip Points of inflection may be found by using INFLC in the TI-86/89 GRAPH MATH menu. Section 4.3 Graphs of Polynomial Functions 267 Complete Graphs of Polynomial Function By using the facts discussed in this section, you can often determine whether or not the graph of a polynomial function is complete, that is, shows all the important features. Example 2 A Complete Graph of a Polynomial Find a complete graph of f x 2 1 x4 10x3 21x2 40x 80. 100 Solution Because the y-intercept is 100 y 100, and as shown in Figure 4.3-7. 80, graph f in the window with 10 x 10 10 10 The three peaks and valleys shown are the only local extrema because a fourth-degree polynomial graph has at most three local extrema. 100 Figure 4.3-7 There cannot be more x-intercepts than the two shown because if the graph turned toward the x-axis farther to the right or farther to the left, there would be an additional peak, which is impossible. Finally, the end behavior of the graph resembles the graph of highest degree term. y x4, the Figure 4.3-7 includes all the important features of the graph and is therefore complete. ■ Example 3 A Complete Graph of a Polynomial Find a complete graph of f x3 1.8x2 x 2. x 1 2 Solution The graph of f, shown in Figure 4.3-8a on the next page, is similar to the graph of its leading term but it does not appear to have any local extrema. However, if you use the trace feature on the flat portion of the graph to the right of the y-axis, you should see that the y-coordinates increase, then decrease, and then increase again. y x3, Zoom in on the portion of the graph between 0 and 1, as shown in Figure 4.3-8b. Observe that the graph actually has two local extrema, one peak and one valley, which is the maximum possible number of local extrema for a cubic function. Figures 4.3-8a and 4.3-8b together provide a complete graph of f. NOTE No polynomial graph of a function of degree horizontal line segments like those shown in Figure 4.3-8a. Always investigate such segments by using trace or zoom-in to determine any hidden behavior. contains n 7 1 268 Chapter 4 Polynomial and Rational Functions 6 2.2 6 6 6 Figure 4.3-8a 0 2 Figure 4.3-8b 1 ■ Example 4 A Complete Graph of a Polynomial Determine if the graph shown in Figure 4.3-9a is a complete graph of 0.01x5 x4 x3 6x2 5x 4. f x 1 2 Solution The graph shown in Figure 4.3-9a cannot be a complete graph because, 0.01x5, when whose left end goes downward. is large, the graph of f must resemble the graph of x g x 2 1 0 0 So the graph of f must turn downward and cross the x-axis somewhere to the left of the origin. Therefore, the graph must have one more peak, where the graph turns downward, and must have another x-intercept. One additional peak and the ones shown in Figure 4.3-9a make a total of four, the maximum possible for a polynomial of degree 5. Similarly, the additional x-intercept makes a total of 5 x-intercepts. Because f has degree 5, there are no other x-intercepts. A viewing window that includes the local maximum and the x-intercept shown in Figure 4.3-9b will not display the local extrema and x-intercepts shown in Figure 4.3-9a. Consequently, a complete graph of f requires both Figure 4.3-9a and Figure 4.3-9b to illustrate the important features of the graph. 10 10,000,000 10 10 115 10 Figure 4.3-9a 5,000,000 Figure 4.3-9b 50 ■ The graphs shown in Examples 2–4 were known to be complete because they included the maximum possible number of local extrema. Many graphs, however, may have fewer than the maximum number of possible peaks and valleys. In such cases, use any available information and try several viewing windows to obtain the most complete graph. Section 4.3 Graphs of Polynomial Functions 269 Exercises 4.3 In Exercises 1–6, decide whether the given graph could possibly be the graph of a polynomial function. 5. 1. 2. 3. 4. In Exercises 7–12, determine whether the given graph could possibly be the graph of a polynomial function of degree 3, degree 4, or degree 5. 7. 8. 9. y y y x x x 270 10. 11. 12. Chapter 4 Polynomial and Rational Functions y y y x x x 16. 4 −5 −5 −5 17. 18. 5 5 5 −14 10 −5 20 −4 In Exercises 13 and 14, find a viewing window in which the graph of the given polynomial function f appears to have the same general shape as the graph of its leading term. 13. 14. f f x x 1 1 2 2 x4 6x3 9x2 3 x3 5x2 4x 2 In Exercises 15–18, the graph of a polynomial function is shown. List each zero of the polynomial and state whether its multiplicity is even or odd. 15. −5 10 −6 5 In Exercises 19–24, use your knowledge of polynomial graphs, not a calculator, to match the given function with one of graphs a–f. a. b. y y x x c. y d. y e. f. y y x x x x 19. f 21. g 23. f 24 2x 3 x3 4x 20. g 22. f x x 1 1 2 2 x2 4x 7 x4 5x2 4 x4 6x3 9x2 2 2x2 3x 1 In Exercises 25–28, graph the function in the standard viewing window and explain why that graph cannot possibly be complete. 25. f 26. g 27.01x3 0.2x2 0.4x 7 0.01x4 0.1x3 0.8x2 0.7x 9 0.005x4 x2 5 Section 4.3 Graphs of Polynomial Functions 271 28. f x 1 2 0.001x5 0.01x4 0.2x3 x2 x 5 In Exercises 29–34, find a single viewing window that shows a complete graph of the function. 29. 30. 31. 32. 33. 34 x3 8x2 5x 14 x3 3x2 4x 5 x4 3x3 24x2 80x 15 x4 10x3 35x2 50x 24 2x5 3.5x4 10x3 5x2 12x 6 x5 8x4 20x3 9x2 27x 7 In Exercises 35–40, find a complete graph of the function and list the viewing window(s) that show(s) this graph. 35. 36. 37. 38 39. g 40. f x x 1 1 2 2 0.1x5 3x4 4x3 11x2 3x 2 x4 48x3 101x2 49x 50 0.03x3 1.5x2 200x 5 0.25x6 0.25x5 35x4 7x3 823x2 25x 2750 2x3 0.33x2 0.006x 5 0.3x5 2x4 7x3 2x2 41. a. Explain why the graph of a cubic polynomial function has either two local extrema or none at all. Hint: If it had only one, what would the x graph look like when is very large? 0 b. Explain why the general shape of the graph of a cubic polynomial function must be one of the following. 0 a. b. c. d. 42. The figure shows an incomplete graph of an even polynomial function f of fourth degree. (Even functions were defined in Excursion 3.4.A.) a. Find the zeros of f. b. Explain why f x x c where a, b, c, d are the zeros of f. x b x a k 21 21 1 2 1 x d 2 21 272 Chapter 4 Polynomial and Rational Functions c. Experiment with your calculator to find the value of k that produces the graph in the figure. d. Find all local extrema of f. e. List the approximate intervals on which f is increasing and those on which it is decreasing. 45. The figure below is a partial view of the graph of a cubic polynomial whose leading coefficient is negative. Which of the patterns shown in Exercise 41 does this graph have? −10 20 −10 10 43. A complete graph of a polynomial function g is 46. The figure below is a partial view of the graph of a fourth-degree polynomial. Sketch the general shape of the graph and state whether the leading coefficient is positive or negative. shown below. a. Is the degree of b. Is the leading coefficient of even or odd? g x x g 2 1 positive or 1 2 negative? c. What are the real zeros of d. What is the smallest possible degree of 15 10 5 −5 −4 −2 −10 −15 x 2 4 6 44. Do Exercise 43 for the polynomial function g whose complete graph is shown here. y 8 4 −4 −8 −3 −2 −1 −12 x 1 2 3 In Exercises 47–56, sketch a complete graph of the function. Label each x-intercept and the coordinates of each local extremum; find intercepts and coordinates exactly when possible, otherwise approximate them. 47. f 49. h 50 51. g x 1 2 52. h 53. f x x 1 1 2
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2 54. g x 1 2 55. h 56. f x x 2 2 1 1 x3 3x2 4 48. g x 2 1 4x 4x3 3 0.25x4 2x3 4x2 0.25x4 2x3 3 3x3 18.5x2 4.5x 45 2x3 x2 4x 2 x5 3x3 x 1 0.25x4 x2 0.5 8x4 22.8x3 50.6x2 94.8x 138.6 32x6 48x4 18x2 1 g 57. Critical Thinking a. Graph x 2 the viewing window with 0 y 6 coincide with the horizontal line 1 0.01x3 0.06x2 0.12x 3.92 3 x 3 and and verify that the graph appears to between y 4 in x 3. and x 1 every x with equation In other words, it appears that is a solution of the 1 x 3 0.01x3 0.06x2 0.12x 3.92 4. Explain why this is impossible. Conclude that the actual graph is not horizontal between and x 3. b. Use the trace feature to verify that the graph is x 1 Find a viewing window that shows actually rising from left to right between and this. x 3. x 1 c. Show that it is not possible for the graph of a 1 2 f x to contain a horizontal polynomial segment. Hint: A horizontal line segment is part of the horizontal line for some constant k. Adapt the argument in part a, k 4. which is the case y k 58. Critical Thinking 2 1 f x x be a polynomial of odd degree. a. Let f Explain why Hint: Why must the graph of f cross the x-axis, and what does this mean? b. Let be a polynomial of even degree, with a must have at least one real zero. g x 1 2 negative leading coefficient and a positive 1 2 Section 4.3.A Excursion: Polynomial Models 273 constant term. Explain why least one positive and at least one negative zero. must have at g x 2 1 59. Critical Thinking The graph of x2 20 x 18 x 2 1 2 1 f 21 x 2 x 10 2 has x-intercepts at each of its zeros, that is, at x 18, 2, and 10. It is also true x 2. f x that a. Draw the x-axis and mark the zeros of ± 120 ± 4.472, has a relative minimum at 21 x f . 1 1 2 2 1 2 Then use the fact that to sketch the general shape of the graph, as was done for cubics in Exercise 41. has degree 6 (Why?) x f 2 1 x 1 2 b. Now graph f in the standard viewing window. Does the graph resemble your sketch? Does it even show all the x-intercepts between 10 c. Graph and 10? x f 19 x 11 window include all the x-intercepts, as it should? in the viewing window with 10 y 10. Does this and 2 1 d. List viewing windows that give a complete graph of f x . 2 1 4.3.A Excursion: Polynomial Models Objectives • Fit a polynomial model to data Linear regression was used in Section 1.5 to construct a linear function that modeled a set of data points. When the scatter plot of the data points looks more like a higher degree polynomial graph than a straight line, similar least squares regression procedures are available on most calculators for constructing quadratic, cubic, and quartic polynomial functions to model the data. Example 1 A Polynomial Model The following data, which is based on statistics from the Department of Health and Human Services, gives the cumulative number of reported cases of AIDS in the United States from 1982 through 2000. Find a quadratic, a cubic, and a quartic regression equation and determine which equation best models the data. 274 Chapter 4 Polynomial and Rational Functions Technology Tip Quadratic, cubic, and quartic regression are denoted by QuadReg, CubicReg, QuartReg in the CALC submenus of the TI STAT menu and by x 3, submenu of the Casio STAT menu. x 2, in the CALC REG x 4 900,000 0 40,000 21 Figure 4.3.A-1b Year 1982 1984 1986 1988 1990 Cases 1563 10,845 41,662 105,489 188,872 Year 1991 1992 1993 1994 1995 Cases 232,383 278,189 380,601 457,789 528,421 Year 1996 1997 1998 1999 2000 Cases 595,559 652,439 698,527 743,418 784,518 Solution x 0 Let correspond to 1980 and plot the data points (2, 1563), (4, 10845), etc., to obtain the scatter plot shown in Figure 4.3.A-1a. The points are not in a straight line, but could be part of a polynomial graph of degree 2 or more. 900,000 0 40,000 Figure 4.3.A-1a 21 Using the same procedure as for linear regression, find a quadratic, a cubic, and a quartic regression equation for the data. See the Technology Tip in the margin on this page for the specific calculator procedure needed. The polynomial functions shown below have rounded coefficients, but the graph in Figure 4.3.A-1b shows the data points along with the quadratic regression equation and was produced using full coefficients. 1758.0x2 9893.3x 59,024.3 219.18x3 9111.66x2 59,991.75x 103,255.32 20.29x4 681.94x3 4318.57x2 15,550.81x 17,877.25 Quadratic Cubic Quartic The graphs of f, g, and h are virtually identical in the viewing window shown. Although any one of f, g, or h provides a reasonable model for the given data, knowledge of polynomial graphs suggests that the cubic and quartic models, should not be used for predicting future results. As x gets y 219.2x3 larger, the graphs of g and h will resemble respectively those of and , which turn downward. However, the cumulative number of cases cannot decrease because even when there are no new cases, the cumulative total stays the same. y 20.3x4 Graphing Exploration Graph the functions f, g, and h in the window with and 0 y 1,800,000. In this window, can you distinguish the graphs of f and g? Assuming no medical breakthroughs or changes in the current social situation, does the graph of f seem to be a plausible model for the next few years? What about the graph of g? 0 x 27 ■ Section 4.3.A Excursion: Polynomial Models 275 Example 2 Estimating Data Values The population of San Francisco in selected years is given in the table. Year 1950 1960 1970 1980 1990 2000 Population 775,357 740,316 715,674 678,974 723,959 776,733 [Source: U.S. Census Bureau] x 0 Let correspond to 1950. Find a polynomial regression model that is a reasonably good fit and estimate the population of San Francisco in 1995 and in 2004. 900,000 Solution The scatter plot of the data shown in Figure 4.3.A-2a suggests a parabola. However, the data points do not climb quite so steeply in the later years, so a higher-degree polynomial graph might fit the data better. The quadratic, cubic, and quartic regression models are shown below. 5 0 Figure 4.3.A-2a 60 128.14x2 6632.39x 783,517.18 2.48x3 58.10x2 3230.49x 776,067.76 0.11x4 13.20x3 387.24x2 168.65x 774,230.65 Quadratic Cubic Quartic 900,000 900,000 900,000 5 0 60 5 0 60 5 0 60 Quadratic Cubic Quartic Figure 4.3.A-2b The quartic appears to be the best fitting function. Use h to estimate the population in 1995 and 2004 by finding h(45) and h(54). 45 h 1 2 745,843.98 and h 803,155.18 54 2 1 That is, the estimated population of San Francisco in 1995 was approximately 745,844 and the estimated population of San Francisco in 2004 was approximately 803,155. ■ In Example 2, a model may not be accurate when applied outside the 5,817,115, range of points used to construct it. For instance, suggesting that the population of San Francisco in 1776 was about 5,817,115. 174 f 1 2 276 Chapter 4 Polynomial and Rational Functions NOTE The following table lists the minimum number of data points required for polynomial regression. In each case, the minimum number of data points required is the number of coefficients in the polynomial function modeling the data. If you have exactly the required minimum number of data points, no two of them can have the same first coordinate. Model Minimum number of data points Quadratic Regression Cubic Regression Quartic Regression 3 4 5 When using the minimum number of data points, the polynomial regression function will pass through all the data points and will fit exactly. When using more than the minimum number of data points required, the fit will generally be approximate rather than exact. Exercises 4.3.A In Exercises 1–4, a scatter plot of data is shown. State the type of polynomial model that seems most appropriate for the data (linear, quadratic, cubic, or quartic). If none of them is likely to provide a reasonable model, say so. 4. y x 1. y 2. y 3. y x x x 5. The table, which is based on the United States FBI Uniform Crime Report, shows the rate of property crimes per 100,000 population. Year 1982 1984 1986 1988 1990 1992 Crimes 5032.5 4492.1 4862.6 5027.1 5088.5 4902.7 Year 1994 1996 1997 1998 1999 2000 Crimes 4660.0 4450.1 4318.7 4051.8 3743.6 3617.9 a. Use cubic regression to find a polynomial x 0 function that models this data, with corresponding to 1980. b. According to this model, what was the property crime rate in 1987 and 1995? c. The actual crime rate was about 3698 in 2001. 8. The table, which is based on data from the Section 4.3.A Excursion: Polynomial Models 277 What does the model predict? d. Is this model a reasonable one? 6. The table, which is based on the U.S. National Center for Educational Statistics, shows actual and projected enrollment (in millions) in public high schools in selected years. Year 1975 1980 1985 1990 Enrollment 14.3 13.2 12.4 11.3 Year 1995 2000 2005 2010 Enrollment 12.5 13.5 14.4 14.1 a. Use quartic regression to find a polynomial function that models this data, with corresponding to 1975. x 0 b. According to this model, what was the enrollment in 1998 and 1999? c. According to the model, in what year between 1975 and 2000 was enrollment at its lowest level? d. Does this estimate appear to be accurate? 7. The table shows the air temperature at various times during a spring day in Gainesville, Florida. Association of Departments of Foreign Languages, shows the fall enrollment (in thousands) in college level Spanish classes. Year 1970 1974 1977 1980 1983 Enrollment 389.2 362.2 376.7 379.4 386.2 Year 1986 1990 1995 1998 Enrollment 411.3 533.9 606.3 656.6 a. Sketch a scatter plot of the data, with x 0 corresponding to 1970. b. Find a cubic polynomial model for this data. Use the following table for Exercises 9–10. It shows the median income of U.S. households in 1999 dollars. Year 1985 1987 1989 1991 Median Income $36,568 38,220 38,836 36,850 Year 1993 1995 1997 1999 Median Income $36,019 037,251 038,411 040,816 Time 6 a.m. 7 a.m. 8 a.m. 9 a.m. 10 a.m. 11 a.m. noon (F) Temp 52 56 61 67 72 77 80 Time 1 p.m. 2 p.m. 3 p.m. 4 p.m. 5 p.m. 6 p.m. Temp 82 86 85 83 78 72 (F°) [Source: U.S.
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Census Bureau] 9. a. Sketch a scatter plot of the data from 1985 to 1999, with x 0 corresponding to 1985. b. Decide whether a quadratic or quartic model seems more appropriate. c. Find an appropriate polynomial model. a. Sketch a scatter plot of the data, with x 0 corresponding to midnight. b. Find a quadratic polynomial model for the data. c. What is the predicted temperature for noon? for 9 a.m.? for 2 p.m.? 278 Chapter 4 Polynomial and Rational Functions d. Use the model to predict the median income in a. Sketch a scatter plot of the data with x 0 2002. e. Does this model seem reasonable after 2002? 10. a. Sketch a scatter plot of the data from 1989 to 1999, with x 0 corresponding to 1989. b. Find both a cubic and a quartic model for this data. c. Is there any significant difference between the models from 1989 to 1999? What about from 1999 to 2005? d. According to these models, when will the median income reach $45,000? 11. The table shows the U.S. public debt per person, in dollars, in selected years. Year 1981 1983 1985 1987 1989 1991 Debt $4,338 5,870 7,598 9,615 11,545 14,436 Year 1993 1995 1997 1999 2001 Debt $17,105 18,930 20,026 20,746 20,353 [Source: U.S. Department of Treasury, Bureau of Public Debt] corresponding to 1980. b. Find a quartic model for the data. c. Use the model to estimate the public debt per person in 1996. How does your estimate compare with the actual figure of $19,805? 12. The table shows the total advertising expenditures, in billions of dollars, in selected years. Year 1990 1992 1994 Expenditures $129.59 132.65 151.68 Year 1996 1998 2000 Expenditures $175.23 201.59 236.33 [Source: Statistical abstract of the United States 2001] a. Sketch a scatter plot of the data with x 0 corresponding to 1990. b. Find a quadratic model for the data. c. Use the model to estimate expenditures in 1995 and 2002. d. If this model remains accurate, when will expenditures reach $350 billion? 4.4 Rational Functions Objectives • Find the domain of a rational function • Find intercepts, vertical Recall that a polynomial is an algebraic expression that can be written as anxn an1xn1 p a2x2 a1x a0 where n is a nonnegative integer. asymptotes, and horizontal asymptotes A rational function is a function whose rule is the quotient of two polynomials, such as • Identify holes • Describe end behavior • Sketch complete graphs 1 x f x 2 1 4x 3 2x 1 x t 1 2 2x3 5x 2 x2 7x 6 k x 1 2 Section 4.4 Rational Functions 279 Although a polynomial function is defined for every real number x, a rational function is defined only when its denominator is nonzero. Domain of Rational Functions The domain of a rational function is the set of all real numbers that are not zeros of its denominator. Example 1 The Domain of a Rational Function Find the domain of each rational function. a. f x 1 2 1 x2 Solution b. g x 1 2 x2 3x 1 x2 x 6 a. The domain of f x is the set of all real numbers except x 0, 1 x2 2 1 because the denominator is 0 when undefined. x 0, making the fraction x2 3x 1 x2 x 6 x2 x 6 0. b. The domain of g x 2 1 is the set of all real numbers except the solutions of into 1 x 3. except x 2 x 2 21 Therefore, the domain of g is the set of all real numbers x 2 Because x2 x 6 0 the solutions to x2 x 6 are x 3. x 3 and , 2 factors and ■ Properties of Rational Graphs Because calculators often do a poor job of graphing rational functions, the emphasis in this section is on the algebraic analysis of rational functions. Such analysis should enable you to interpret misleading screen images. Intercepts As with any function, the y-intercept of the graph of a rational function f The x-intercepts of the occurs at graph of a rational function occur when its numerator is 0 and its denominator is nonzero. provided that f is defined at x 0. 0 f , 2 1 Intercepts of Rational Functions If f has a y-intercept, it occurs at f(0). The x-intercepts of the graph of a rational function occur at the numbers that • are zeros of the numerator • are not zeros of the denominator 280 Chapter 4 Polynomial and Rational Functions Locating the intercepts can help you determine if you correctly entered the parentheses when graphing a rational function on a graphing calculator. Example 2 Intercepts of a Rational Graph Find the intercepts of f x2 x 2 x 1 . x 1 2 y 16 8 (0, 2) (−1, 0) 4 (2, 0) −8 −4 0 −8 −16 Figure 4.4-1 x 8 Solution The y-intercept is f 0 1 2 02 0 2 0 1 2. 2 1 x2 x 2 0 The x-intercepts are solutions of x 1 0. Solutions of x2 x 2 0 that are not solutions of can be found by factoring. x 1 x2 x 2 0 x 2 0 1 x 1 or x 2. x 1 0, 21 2 1 Neither graph of f, as shown in Figure 4.4-1. nor 2 is a solution of so both are x-intercepts of the ■ Continuity There are breaks in the graph of a rational function wherever the function is not defined, that is, at the zeros of the denominator. Except for breaks, the graph is a continuous unbroken curve. Additionally, the graph has no sharp corners. Vertical Asymptotes Unlike polynomial functions, a rational function has breaks in its graph at all points where the function is not defined. Vertical asymptotes occur at every number that is a zero of the denominator but not of the numerator. The key to understanding the behavior of a rational function near these asymptotes is a fact from arithmetic. The Big-Little Concept If c is a number far from 0, then 1 c is a number close to 0. If c is close to 0, then 1 c is far from 0. In less precise, but more suggestive terms 1 big little and 1 little big Section 4.4 Rational Functions 281 For example, 5000 is big and 1 5000 is little. Similarly, 1 1000 is little and 1000 is big. Note that even though 1000 is negative, it is far 1 1 1000 from zero and therefore is large in absolute value. The role played by the Big-Little Concept when graphing rational functions is illustrated in Example 3. Example 3 A Rational Function Near a Vertical Asymptote Without using a calculator, describe the graph of x 2. x 2. Then sketch the graph for values near x 1 2x 4 x f 1 2 near Solution because the denominator is 0 there. The function is not defined at When x is greater than 2 but very close to 2, x 1, • The numerator, is very close to x 2 • The denominator, 4 0. 2 2 1 2 2x 4, 2 1 3. is a positive number very close to Figure 4.4-2a y 10 5 0 −5 −10 x 1 2 3 4 Figure 4.4-2b By the Big-Little Concept, x 1 2x 4 3 little f x 2 1 3 1 little 3 big 1 2 very big when This fact can be confirmed by a table of values for x 2.01, 2.001, 2.0001, etc., as shown in Figure 4.4-2a. In graphical terms, the points with x-coordinates slightly greater than 2 have very large x 2. y-coordinates, so the graph shoots upward just to the right of That is, near x f 2 1 x 2 f increases without bound as x approaches 2 from the right. x 1, A similar analysis when x is less than 2 but very close to 2 shows that the numerator, is very close to 3 and the denominator is negative and very close to 0. Using the Big-Little Concept, the quotient is a negative number far from 0. As x approaches 2 from values less than 2, the quotient becomes a larger and larger negative number. Therefore, the graph of f shoots downward just to left of x 2. That is, f decreases without bound as x approaches 2 from the left. x 2 is shown in Figure 4.4–2b. The portion of the graph of f near ■ The dashed vertical line in Figure 4.4-2b is included for easier visualization, but it is not part of the graph. Such a line is called a vertical asymptote of the graph. The graph approaches a vertical asymptote very closely, but never touches or crosses it because the function is not defined at that value of x. 282 Chapter 4 Polynomial and Rational Functions All rational functions have vertical asymptotes at values that are zeros of their denominators but not zeros of their numerators. Vertical Asymptotes A rational function has a vertical asymptote at x c , provided • c is a zero of the denominator • c is not a zero of the numerator Near a vertical asymptote, the graph of a rational fraction may look like the graph in Figure 4.4-2b, or like one of the graphs in Figure 4.4-3. c x c x c x vertical asymptotes at x c Figure 4.4-3 Holes When a number c is a zero of both the numerator and denominator of a x c, rational function, the function might have a vertical asymptote at or it might behave differently. You have often cancelled factors to reduce fractions. x2 4 x 2 1 But the functions x 2 x 2 21 x 2 2 x 2 x2 4 x 2 x p 1 2 and q x 2 x 1 2 are not the same, because when x 2 22 4 0 2 2 0 2 2 4. 2 x p q 1 1 2 2 , which is not defined, but For any number other than 2, the two functions have the same values, is a straight line and hence, the same graphs. The graph of that includes the point (2, 4), as shown in Figure 4.4-4a. The graph of p(x) is the same straight line, but with the point (2, 4) omitted. That is, there x 2 is a hole in the graph of p at because p is not defined there. The graph of p is shown in Figure 4.4-4b2 −1 1 2 3 −2 Figure 4.4-4a y 4 2 −2 1 2 3 −2 Figure 4.4-4b Section 4.4 Rational Functions 283 y The graph of g x 1 2 of f x 1 2 1 x. At x 0 shown in Figure 4.4-5, is the same as the graph x2 x3, neither function is defined. There is a vertical asymp- x 0 4 8 tote rather than a hole at at multiplicity 3 in the denominator. x 0, Note that the vertical asymptote occurs which is a zero of multiplicity 2 in the numerator, but of larger x 0. 9 6 3 −3 −6 −9 −8 −4 Figure 4.4-5 Holes Technology Tip To avoid erroneous vertical lines, use a window with a vertical asymptote in the center of the screen. In Figure 4.4-6b, the asymptote at x 2 8 nology Appendix for further information. is halfway between and 12. See the Tech- Let f(x) g(x) h(x) of both g and h. be a rational function and let d denote a zero • If the multiplicity of d as a zero of g is greater than or equal to its multiplicity as a zero of h, then the graph of f has a hole at x d. • Otherwise, the graph has a vertical asymptote at x d. Accurate Rational Function Graphs Getting an accurate graph of
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a rational function on a calculator often depends on choosing an appropriate viewing window. For example, the following are graphs of f x 1 2 x 1 2x 4 in different viewing windows. 10 6 10 10 8 12 10 Figure 4.4-6a 6 Figure 4.4-6b x 2, but not at The vertical segment shown in Figure 4.4-6a is not a vertical asymptote. It is a result of the calculator evaluating f just to the left of and just x 2, to the right of and then erroneously connecting these points with a near vertical segment that looks like an asymptote. In the accurate graph shown in Figure 4.4-6b, the calculator attempted to plot a point with was not defined, skipped a pixel and did not join the points on either side of the one skipped. and when it found that x 2 x 2 2 f 1 2 284 Chapter 4 Polynomial and Rational Functions A calculator graph may also fail to show holes in graphs that should have them. Even if a window is chosen so that the graph skips a pixel where the hole should be, the hole may be difficult to see. End Behavior As with polynomials, the behavior of a rational function when is large is called its end behavior. Known facts about the end behavior of polynomial functions make it easy to determine the end behavior of rational functions in which the degree of the numerator is less than or equal to the degree of the denominator. x 0 0 Example 4 End Behavior of Rational Functions List the vertical asymptotes and describe the end behavior of the following functions. Then sketch each graph. a. f x 1 2 3x 6 5 2x b. g x 1 2 x x2 4 c. h x 1 2 2x3 x x3 1 Solution a. The zero of the denominator of f 3x 6 5 2x is 5 2 x 1 2 and it is not a x zero of the numerator. So the vertical asymptote occurs at x 5 2 is large, a polynomial function behaves like its highest When degree term, as shown in Section 4.3. The highest degree term of the numerator of f is 3x and the highest degree term of the denominator is is large, the function reduces to 2x. x . 0 0 Therefore, when 3 2 . 0 0 approximately 3x 6 5 2x 3x 6 2x 5 3x 2x 3 2 f x 1 2 Thus, when 0 horizontal line x 0 is large, the graph of f gets very close to the y 3 2 which is called a horizontal asymptote of , the graph. The dashed lines in Figure 4.4-7 indicate the vertical and horizontal asymptotes of the graph. b. The zeros of the denominator of g x 2 1 x x2 4 are ± 2 and neither is a zero of the numerator. So the graph has vertical asymptotes at x 2 x 2. and at When x 0 0 is large, x g 1 2 x x2 4 x x2 1 x and 1 x is very close to 0 by the Big-Little Concept. Therefore, the (the x-axis) when is large and this line is a horizontal asymptote of the graph, as graph of g approaches the horizontal line x 0 shown in Figure 4.4-8. 0 y 0 y 8 4 −8 −4 0 4 8 x −4 −8 Figure 4.4-7 Technology Tip When the vertical asymptotes of a rational function occur at numbers such as 2.1, 2, 1.9, p ,2.9, 3, etc., a decimal window normally produces an accurate graph because the calculator actually evaluates the function at the asymptotes, finds that it is undefined, and skips a pixel. CAUTION Unlike a vertical asymptote that is never crossed by a graph, a graph may cross a horizontal or oblique asymptote at some values of x . 0 0 y 4 2 −8 −4 0 4 Figure 4.4-9 Section 4.4 Rational Functions 285 y 4 2 −4 −2 0 2 4 x −2 −4 Figure 4.4-8 c. The only real zero of the denominator of h 2x3 x x3 1 x 1 2 is x 1, which is not a zero of the numerator. So, the graph has a vertical asymptote at x 1. When x 0 0 is large, 2x3 x x3 1 2x3 x3 2 1 x h 1 2 2 Therefore, the graph of h has a horizontal asymptote at shown in Figure 4.4-9. y 2 , as x ■ The function in Example 4b illustrates a useful fact. When the degree of the numerator is less than the degree of the denominator of a rational function, the x-axis is the horizontal asymptote of the graph. When the numerator and denominator have the same degree, as in Examples 4a and 4c, the horizontal asymptote is determined by the leading coefficients of the numerator and denominator: Function 3x 6 2x 5 2x3 x x3 1 f x 1 2 h x 1 2 Horizontal asymptote y 3 2 y 2 1 2 Other Asymptotes When the degree of the numerator of a rational function is greater than the degree of its denominator, the graph will not have a horizontal asymptote. To determine the end behavior in this case, the Division Algorithm must be used. Example 5 A Slant Asymptote Describe the end behavior of the graph of f x2 x 2 x 5 . x 1 2 286 Chapter 4 Polynomial and Rational Functions y 40 20 x 15 30 45 −45 0 −30 −15 −20 −40 Figure 4.4-10 y 90 60 30 x 6 12 18 −18 −12 0 −6 −30 Figure 4.4-11 Solution Use synthetic or long division to divide the denominator into the numerator, and rewrite the rational expression by using the Division Algorithm. Dividend Divisor Quotient Remainder x2 x 2 1 21 x 5 x2 18 2 1 1 x 5 x 4 21 x 5 x 4 2 2 18 18 x 5 x 5 21 x 5 1 x 4 18 x 5 2 0 0 x When x 5 is large, is also large, and by the Big-Little Concept 18 x 5 and the graph of f approaches is very close to 0. Therefore, y x 4 y x 4 the line is called a slant or oblique asymptote of the graph. Note that is the quotient without the remainder in the division of the numerator by the denominator. gets large (see Figure 4.4-10). The line x 4, x 4 as x x f 1 2 0 0 ■ Example 6 A Parabolic Asymptote Describe the end behavior of the graph of f x3 3x2 x 1 x 1 . x 1 2 Solution Divide the denominator into the numerator and rewrite the function. x3 3x2 x 1 x 1 1 f x 1 2 x2 4x 5 6 x 1 2 Quotient Remainder Divisor When x 0 0 is large, so is x 1 f x2 4x 5 close to 0. Therefore, of f approaches the parabola The curve x2 4x 5 y x2 4x 5 is the quotient in the division. x 2 1 and by the Big-Little Concept 6 x 1 x . The graph 0 y x2 4x 5, as shown in Figure 4.4-11. is called a parabolic asymptote. Note that for large values of is very 0 ■ Section 4.4 Rational Functions 287 End Behavior of Rational Functions Let f(x) axn % cxk % be a rational function whose numerator has degree n and whose denominator has degree k. • If n 66 k, then the x-axis is a horizontal asymptote. • If n k, then the line y a c is a horizontal asymptote. • If n 77 k, then the quotient polynomial when the numerator is divided by the denominator is the asymptote that describes the end behavior of the graph. Graphing Rational Functions Notice that when the degree of the numerator and the denominator are the same, the horizontal asymptote is the horizontal line determined by the quotient of the leading coefficients of the numerator and denominator. Graphs of Rational Functions The facts presented in this section can be used in conjunction with a calculator to find accurate, complete graphs of rational functions. 1. Analyze the function algebraically to determine its vertical asymptotes, holes, and intercepts. 2. Determine the end behavior of the graph. If the degree of the numerator is less than or equal to the degree of the denominator, find the horizontal asymptote by using the facts in the box above. Otherwise, divide the numerator by the denominator. The quotient is the nonvertical asymptote of the graph. 3. Use the preceding information to select an appropriate viewing window, or windows, to interpret the calculator’s version of the graph, and display a complete graph of the function. 10 Example 7 A Complete Graph of a Rational Function −10 10 Find a complete graph of f x 1 x2 x 6 . x 2 1 Solution −10 Figure 4.4-12a The graph of f is shown in Figure 4.4-12a. It is hard to determine whether or not the graph is complete, so analyze the function algebraically. 288 Chapter 4 Polynomial and Rational Functions Begin by writing the function in factored form. Then read off the relevant information. x 1 x2 21 2 1 x 3 and Vertical Asymptotes: x 2 Intercepts: y-intercept: 0 1 02 0 6 1 6 f 0 1 2 x-intercept: x 1 Horizontal Asymptote: y 0 zeros of the denominator but not the numerator zero of numerator but not of denominator degree of numerator is less than degree of denominator Interpreting the above information suggests that a complete graph of f looks similar to Figure 4.4-12b. y 6 4 2 −2 −1 1 2 3 4 x −2 −4 Figure 4.4-12b ■ Example 8 A Complete Graph of a Rational Function Find a complete graph of f x3 2x2 5x 6 x2 3x 2 . x 1 2 Solution The denominator is easily factored. To factor the numerator, note that the ± 1, ± 2, ± 3, ± 6 only possible rational zeros of and 3 actually are zeros and by the Rational Zeros Test. Verify that use the Factor Theorem to write the numerator in factored form. Then reduce the fraction. x3 2x2 5x 6 2, 1 and are Section 4.4 Rational Functions 289 Holes: x3 2x2 5x 6 x2 3x 21 x 1 2 , where x 2 x 1 x 3 21 x 2 21 x 1 2 2 21 1 x 2. Therefore, the graph of f is the same as the graph of 1 g x 1 2 x 1 x 3 21 x 1 2 x2 4x 3 x 1 except there is a hole when x 2. Because g 1 2 2 1 2 1 2 3 21 2 1 2 1 3 5 21 21 15, the hole occurs at 2, 15 . 2 1 Intercepts: y-intercept: x-intercepts 21 0 1 2 1 1 3 21 1 2 3 The x-intercepts of f are the same as the 0 x-intercepts of g. Solving x 1 or x 3. yields x 1 x 3 21 2 1 Vertical Asymptote: The vertical asymptote is x 1. End Behavior: Dividing the numerator by the denominator prox 5. duces a quotient of Therefore, the slant asymptote that describes the end behavior of the y x 5. function is the line The graph of f is shown in Figure 4.4-13. y 16 8 0 −8 −8 −4 x 4 8 Figure 4.4-13 ■ 290 Chapter 4 Polynomial and Rational Functions Exercises 4.4 In Exercises 1–6, find the domain of the function. 1. f x 1 2 3x 2x 5 2. g x 2 1 x3 x 1 2x2 5x 3 In Exercises 23–50, analyze the function algebraically: list its vertical asymptotes, holes, and horizontal asymptote. Then sketch a complete graph of the function. 3. h x 1 2 6x 5 x2 6x 4 4. g x 1 2 x3 x2 x 1 x5 36x 5. f x 1 2 x5 2x3 7 x3 x2 2x 2 6. h x 1 2 x5 5 x4 12x3 60x2 50x 125 In Exercises 7–12, use algebra to determine the location of the vertical asymptotes and holes in the graph of the function. 7. f x 1 2 x2 4 x2 5x 6 8. g x 1 2 x 5 x3 7x2 2x 9. f x 1 2 x x3 2x2 x 10. g x 1 2 x x3 5x 11. f x 1 2 x2 4x 4 x 2 x 2 21 1 3 2 12. h x 1 2
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x 3 x2 x 6 In Exercises 13–22, find the horizontal or other asympis large, and tote of the graph of the function when find a viewing window in which the ends of the graph are within 0.1 of this asymptote. x 00 00 13. f x 1 2 3x 2 x 3 15. h x 2 1 5 x x 2 14. g x 1 2 3x2 x 2x2 2x 4 16. f x 1 2 4x2 5 2x3 3x2 x 17. g x 1 2 5x3 8x2 4 2x3 2x 18. h x 1 2 8x5 6x3 2x 1 0.5x5 x4 3x2 x 19. f x 1 2 x3 1 x2 4 20. g x 1 2 x3 4x2 6x 5 x 2 21. h x 1 2 x3 3x2 4x 1 x 4 22. f x 1 2 x3 3x2 4x 1 x2 x 23. f x 1 2 1 x 5 25. k x 1 2 3 2x 5 27. f x 1 2 3x x 1 29. f x 1 2 2 x x 3 24. q x 1 2 26 28. p x 1 2 x 2 x 30. g x 1 2 3x 2 x 3 1 x 1 2 32. g x 1 2 x 2x2 5x 3 x 31. f x 1 2 33. f x 1 2 35. h x 1 2 36. f x 1 2 2 1 x 3 x2 x 2 1 3 x2 6x 5 x 5 2 1 x2 1 x3 2x2 x 1 38. k x 1 2 x2 1 x2 1 40. F x 2 1 x2 x x2 2x 4 34. g x 1 2 x 2 x2 1 x 5 2 21 x 1 2 37. f x 1 2 4x2 1 x2 39. q x 1 2 x2 2x x2 4x 5 x 3 x 3 1 x 5 21 x 4 2 x 3 21 2 41. p x 1 2 42. p x 1 2 1 21 x3 3x2 x4 4x2 43. f x 1 2 x2 x 6 x 2 44. k x 1 2 x2 x 2 x 45. Q x 2 1 4x2 4x 3 2x 5 46. K x 1 2 3x2 12x 15 3x 6 47. f x 1 2 x3 2 x 1 48. p x 1 2 x3 8 x 1 50. f x 1 2 x4 1 x2 49. q x 1 2 x3 1 x 2 In Exercises 51–60, find a viewing window or windows that show(s) a complete graph of the function using asymptotes, intercepts, end behavior, and holes. Be alert for hidden behavior. 51. f x 1 2 x3 4x2 5x x2 9 x2 4 21 2 1 52. g x 1 2 x2 x 6 x3 19x 30 53. h x 1 2 2x2 x 6 x3 x2 6x 54. f x 1 2 x3 x 1 x4 2x3 2x2 x 1 55. f x 1 2 2x4 3x2 1 3x4 x2 x 1 56. g x 1 2 x4 2x3 x5 25x3 57. h x 1 2 3x2 x 4 2x2 5x 58. f x 1 2 2x2 1 3x3 2x 1 59. g x 1 2 x 4 2x3 5x2 4x 12 60. h x 1 2 x2 9 x3 2x2 23x 60 In Exercises 61–66, find a viewing window or windows that show(s) a complete graph of the function—if possible, with no erroneous vertical line segments. Be alert for hidden behavior. 61. f x 1 2 2x2 5x 2 2x 7 62. g x 1 2 2x3 1 x2 1 63. h x 1 2 x3 2x2 x 2 x2 1 64. f x 1 2 3x3 11x 1 x2 4 65. g x 1 2 2x4 7x3 7x2 2x x3 x 50 66. h x 1 2 2x3 7x2 4 x2 2x 3 67. a. Graph f x 2 6 x 6 1 1 x and in the viewing window with 6 y 6. Section 4.4 Rational Functions 291 . Without using a calculator, describe how the graph of p 2 x 3 x 1 2 4 can be obtained from the graph of 2 e. Show that the function x f 1 1 x. x p 1 2 of part d is a rational function by rewriting its rule as the quotient of two first-degree polynomials. If r, s, and t are constants, describe how the f. t can be obtained from r 1 2 q x graph of x s 1 x. g. Show that the function the graph of x f 2 1 rational function by rewriting its rule as the quotient of two first-degree polynomials. q x 1 2 of part f is a 68. The graph of f 2x3 2x2 x 1 3x3 3x2 2x 1 x 1 2 has a vertical asymptote. Find a viewing window that demonstrates this fact. 69. a. Find the difference quotient of f 1 x x 2 1 and express it as a single fraction in lowest terms. b. Use the difference quotient in part a to x determine the average rate of change of 1 x changes from 2 to 2.1, from 2 to 2.01, and from 2 to 2.001. Estimate the instantaneous rate x 2. of change of at as x f f 2 2 c. Use the different quotient in part a to 1 x determine the average rate of change of 1 x changes from 3 to 3.1, from 3 to 3.01, and from 3 to 3.001. Estimate the instantaneous rate x 3. of change of at as x f f 2 1 2 change of d. How are the estimated instantaneous rates of x 2 1 x2 related to the x 3? and x 2 values of x 3 and 1 x at at x g f 2 2 1 70. Do Exercise 69 for the functions f x g 1 2 2 x3 . x 1 2 1 x2 and b. Without using a calculator, describe how the 71. a. When x 0, what rational function has the 2 x Hint: x 1 x 2 . graph of g can be obtained from the 1 2 f graph of g 1 c. Without using a calculator, describe how the graphs of each of the following functions can be obtained from the graph of same graph as f x 1 2 ? Hint: Use the definition of absolute value. b. When x 6 0, what rational function has the same graph as f x 2 1 part a. ? See the hint for 292 Chapter 4 Polynomial and Rational Functions c. Use parts a and b to explain why the graph of has two vertical asymptotes What are they? Confirm your answer by graphing the function. 72. The percentage c of a drug in a person’s bloodstream t hours after its injection is approximated by c 5t 4t2 5 . t 2 1 a. Approximately what percentage of the drug is in the person’s bloodstream after four and a half hours? b. Graph the function c in an appropriate window for this situation. c. What is the horizontal asymptote of the graph? What does it tell you about the amount of the drug in the bloodstream? d. At what time is the percentage the highest? What is the percentage at that time? 73. A box with a square base and a volume of 1000 cubic inches is to be constructed. The material for the top and bottom of the box costs $3 per 100 square inches and the material for the sides costs $1.25 per 100 square inches. a. If x is the length of a side of the base, express the cost of constructing the box as a function of x. b. If the side of the base must be at least 6 inches long, for what value of x will the cost of the box be $20? 74. A truck traveling at a constant speed on a reasonably straight, level road burns fuel at the rate of of the truck in miles per hour and gallons per mile, where x is the speed is given by x g x g 1 2 2 1 800 x2 200x . g x 1 2 a. If fuel costs $1.40 per gallon, find the rule of x the cost function 2 fuel for a 500-mile trip as a function of the x speed. Hint: 500 needed to go 500 miles. (Why?) gallons of fuel are that expresses the cost of 1 g c 2 1 b. What driving speed will make the cost of fuel for the trip $250? c. What driving speed will minimize the cost of fuel for the trip? 75. Pure alcohol is being added to 50 gallons of a coolant mixture that is 40% alcohol. a. Find the rule of the concentration function x that expresses the percentage of alcohol in the resulting mixture as a function of the number x of gallons of pure alcohol that are added. Hint: The final mixture contains 50 x gallons. c 2 1 2 1 c x is the amount of alcohol in the (Why?) So final mixture divided by the total amount 50 x. How much alcohol is in the original 50-gallon mixture? How much is in the final mixture? b. How many gallons of pure alcohol should be added to produce a mixture that is at least 60% alcohol and no more than 80% alcohol? c. Determine algebraically the exact amount of pure alcohol that must be added to produce a mixture that is 70% alcohol. 76. A rectangular garden with an area of 250 square meters is to be located next to a building and fenced on three sides, with the building acting as a fence on the fourth side. a. If the side of the garden parallel to the building has length x meters, express the amount of fencing needed as a function of x. b. For what values of x will less than 60 meters of fencing be needed? 77. A certain company has fixed costs of $40,000 and variable costs of $2.60 per unit. a. Let x be the number of units produced. Find the rule of the average cost function. (The average cost is the cost of the units divided by the number of units.) b. Graph the average cost function in a window with 0 x 100,000 and 0 y 20. c. Find the horizontal asymptote of the average cost function. Explain what the asymptote means in this situation, that is, how low can the average cost possibly be? 78. Radioactive waste is stored in a cylindrical tank, whose exterior has radius r and height h as shown in the figure. The sides, top, and bottom of the tank are one foot thick and the tank has a volume of 150 cubic feet including top, bottom, and walls. r h a. Express the interior height h1 (that is, the height of the storage area) as a function of h. b. Express the interior height as a function of r. c. Express the volume of the interior as a function of r. d. Explain why r must be greater than 1. 79. The relationship between the fixed focal length F of a camera, the distance u from the object being photographed to the lens, and the distance v from the lens to the film is given by 1 F 1 u 1 v. u F v a. If the focal length is 50 mm, express v as a function of u. b. What is the horizontal asymptote of the graph of the function in part a? c. Graph the function in part a when 50 mm 6 u 6 35,000 mm. Section 4.5 Complex Numbers 293 d. When you focus the camera on an object, the distance between the lens and the film is changed. If the distance from the lens to the camera changes by less than 0.1 millimeter, the object will remain in focus. Explain why you have more latitude in focusing on distant objects than on very close ones. 80. The formula for the gravitational acceleration in units of meters per second squared of an object relative to the earth is g r 2 1 3.987 1014 6.378 106 r 1 2 2 where r is the distance in meters above the earth’s surface. a. What is the gravitational acceleration at the earth’s surface? b. Graph the function g(r) for c. Can you ever escape the pull of gravity? Does r 0. the graph have any r-intercepts? 4.5 Complex Numbers Objectives • Write complex numbers in standard form • Perform arithmetic operations on complex numbers • Find the conjugate of a complex number • Simplify square roots of negative numbers • Find all solutions of polynomial equations If restricted to nonnegative numbers, you cannot solve the equation x 5 0. Enlarging the number system to include negative integers makes it possible to find the solution to this equation. By enlarging the number system to include rational numbers, it is possible to solve equations that have no integer solution, such as Similarly, the equation x2 2 x 12 x 12 are real numhas no rational solution, but ber solutions. The idea of enlarging a number system to include solutions to equations that cannot be solved in a particular number system is a natural one. 3x 7. and Complex Numbers x2 1 21 have no solutions in the real numEquations such as ber system because are not real numbers. In order to solve such equations, that is, to find the square roots of negative numbers, the number system must b
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e enlarged again. There is a number system, called the complex number system, with the desired properties. and and x2 4 24 294 Chapter 4 Polynomial and Rational Functions Properties of the Complex Number System 1. The complex number system contains all real numbers. 2. Addition, subtraction, multiplication, and division of complex numbers obey the same rules of arithmetic that hold in the real number system, with one exception: the exponent laws hold for integer exponents, but not necessarily for fractional ones (see p. 297). 3. The complex number system contains a number, denoted i, such that i 2 1. 4. Every complex number can be written in the standard form a bi, where a and b are real numbers. a bi c di if and only if a c 5. and b d. Numbers of the form bi, where b is a real number, are called imaginary numbers. Sums of real and imaginary numbers, numbers of the form a bi, are called complex numbers. For example, 5 2i 7 4i 18 3 2 i 3 12i are all complex numbers. Just as every integer is a rational number because it can be written as a fraction with denominator of 1, every real number a is a complex numSimilarly, every imaginary number ber because it can be written as 0 bi. bi is a complex number because it can be written as a 0i. Example 1 Equating Two Complex Numbers Find x and y if 2x 3i 6 4yi. Solution Property 5 of the Complex Number System states that two complex numbers b d. c di a bi and and So, are equal exactly when 2x 6 and a c 3 4y y 3 4 x 3 Direct substitution verifies the solution. 2 1 2 3 3i 6 4 a 6 3i 6 3i 3 4b i ■ NOTE The mathematicians who invented the complex numbers in the seventeenth century were very uneasy about a number i such that i2 1. Consequently, they called numbers of the form bi, where b is a real number i 21, and numbers. imaginary The existence of imaginary numbers is as real as any of the familiar numbers that are called real numbers, such as 3, 2 3 or 22. NOTE Hereafter, in the c di, a bi complex numbers it is assumed and that a, b, c, and d are real numbers. Section 4.5 Complex Numbers 295 Arithmetic of Complex Numbers Because the usual laws of arithmetic hold, it is easy to add, subtract, and multiply complex numbers. As the following examples demonstrate, all symbols can be treated as if they were real numbers, provided that i2 is replaced by 1. Unless directed otherwise, express answers in the standard form a bi. Example 2 Adding, Subtracting, and Multiplying Complex Numbers Perform the indicated operation and write the result in the form a bi. 3 7i 2 1 a. 4i a b. d. 1 1 4 3i 8 6i 1 2 3 4i 2 2 i 21 2 Solution 1 i 2 4 3i b. a. 1 1 3 7i 2 8 6i 1 i 3 7i 4 3i 8 6i 4i 1 2 2 1 i b 2 3 4i a 2 i 1 1 21 c. d. 2 4i 2 4i 1 a 2 3 4i 2 1 2 6 5i 8i 2i2 8i 2 2 1 3 4i 1 6 8i 3i 4i2 6 4 5i 10 5i 2 2 i 4 6i 2 i 4 9i 2 2 8i ■ The familiar multiplication patterns and exponent laws for integer exponents hold in the complex number system. Example 3 Products and Powers of Complex Numbers Perform the indicated operation and write the result in the form a bi. a. 1 3 2i 3 2i 2 21 b. 4 i 2 2 1 Solution 3 2i 4 i b. a. 1 1 2 3 2i 21 2 42 2 32 i 4 2 9 4i2 9 4 9 4 13 i2 16 8i i2 16 8i 1 15 8i 1 2i 1 2 2 2 1 1 21 2 ■ Powers of i Observe that i1 i i2 1 i3 i2 i 1 i i i4 i2 i2 1 i5 i4 i 1 i i 1 21 2 1 1 Definition of i 296 Chapter 4 Polynomial and Rational Functions The powers of i form a cycle. Any power of i must be one of four values: i, 1, i, in, divide n by 4 and match the remainder to one of the powers listed above. or 1. To find higher powers of i, such as Example 4 Powers of i Find i54. Solution The remainder when 54 is divided by 4 is 2, so i54 i2 1. ■ Complex Conjugates The conjugate of the complex number a bi the conjugate of 3 4i, and the conjugate of 3 4i and each real number a, every real number is its own conjugate. 3i 0 3i are called conjugate pairs. Because a bi. 3 4i a bi 0 3i 3i. is is is the number For example, the conjugate of a bi, and 3 4i is The numbers for a 0i a 0i The product of conjugate pairs is a real number, as shown below. NOTE The result of Example 4 can also be seen i54 by rewriting exponent rules. using i54 i52 i2 i413 i2 i4 13i2 2 1 1 113 1 1. 2 be a complex number. Then the product of a bi and its con- a bi Let jugate a bi a bi is a bi 1 21 and b2 2 a2 1 1 a2 b2. are nonnegative real numbers, so is 2 a2 b2i2 a2 b2 bi 2 2 1 a2 b2 Because a2 Quotients of Complex Numbers The procedure used to find the quotient of two complex numbers uses the fact that the product of conjugate pairs is a real number. Example 5 Quotients of Two Complex Numbers Express the quotient Solution To find the quotient 3 4i 1 2i 3 4i 1 2i in standard form. , multiply both the numerator and denomi- 1 2i. nator by the conjugate of the denominator, 3 4i 3 4i 1 2i 1 2i 1 2i 1 2i 1 2i 3 4i 2 2 1 2i 1 2i 2 2 3 6i 4i 8 1 1 4 2 1 1 2 12 1 2i 3 1 4i 2i 1 1 1 3 8 6i 4i 1 4 1 2i 1 2 2 1 1 1 2 2 3 6i 4i 8i2 1 4i 2 11 2i 5 11 5 2 5 i ■ Section 4.5 Complex Numbers 297 Square Roots of Negative Numbers Because i2 1, 21 is defined to be i. Similarly, because 1 2 52i2 25 25, 5i 225 1 is defined to be 5i. In general, 2 1 2 Square Roots of Negative Numbers Let b be a positive real number. 2b is defined to be i2b because i2b A B 2 i2 2b B A 2 1 b b Example 6 Square Roots of Negative Numbers Write each of the following as a complex number. b. 1 27 3 c. A 7 24 5 29 B BA a. 23 Solution a. b. c. Technology Tip Most calculators that do complex number arithmetic will return a complex number when asked for the square root of a negative number. Make sure the MODE is set to “rectangular” or a bi. “ ” by definition 23 i23 1 27 3 7 24 A 1 i27 3 5 29 BA 1 3 B i 27 3 7 i24 A 7 2i 5 i29 B BA 5 3i 2 1 21 35 21i 10i 6i2 35 11i 6 41 11i 1 2 1 ■ CAUTION 2cd 2c 2d —or equivalently in exponential The property 1 notation —which is valid for positive real numbers, does not hold when both c and d are negative. To avoid difficulty, 2 c 2 d cd 1 2 2 1 1 NOTE When i is multiplied by a radical, it is always write square roots of negative numbers in terms of i before doing any simplification. i2b to make customary to write 2b i instead of clear that i is not under the radical. For example, 220 25 i220 i25 i2 220 5 2100 10. Therefore, 220 25 220 5. 298 Chapter 4 Polynomial and Rational Functions Because every negative real number has two square roots in the complex number system, complex solutions can be found for equations that have no real solutions. For example, the solutions of x2 25 are x ± 225 ± 5i Every quadratic equation with real coefficients has solutions in the complex number system. Example 7 Complex Solutions to a Quadratic Equation Find all solutions to 2x2 x 3 0. Solution Apply the quadratic formula. x 1 ± 212 4 2 3 2 2 1 ± 223 4 223 Because solutions. However, Thus, the equation has solutions in the complex number system. is not a real number, this equation has no real number 223 i223. is an imaginary number, namely, 223 x 1 ± 223 4 Note that the two solutions, 1 4 conjugates. 1 ± i223 4 223 4 i and ± 223 1 4 4 223 4 1 4 i, i are complex ■ NOTE See the Algebra Review Appendix to review factoring the difference of two cubes. Example 8 Zeros of Unity Find all solutions of x3 1. Solution Rewrite the equation as tern to factor the left side. x3 1 0 and use the difference of cubes pat- x3 1 x3 1 0 x 1 0 1 x 1 0 or x2 x 1 0 x2 x 1 21 2 x 1 or x 1 ± 212 4 1 1 2 1 Quadratic formula 1 ± 23 2 1 ± i23 2 1 2 ± 23 2 i Section 4.5 Complex Numbers 299 real complex solutions, Therefore, the equation has one real solution, 23 x 1 2 2 solutions is said to be a cube root of one or a cube root of unity. Observe that the two nonreal cube roots of unity are complex conjugates. and two non- Each of the x3 1 x 1 2 x 1, 23 2 and i. i ■ Examples 7 and 8 illustrated the following useful fact. Conjugate Solutions a bi If coefficients, then its conjugate, the equation. is a solution of a polynomial equation with real a bi, is also a solution of Calculator Exploration The following exploration demonstrates how matrices can be used for complex number arithmetic. a bi is expressed in matrix notation as the 1. The complex number . For example, matrix a a. Write b a b ab 3 4i, 1 2i, in your calculator as [A], [B], [C]. 1 i and 3 6i is written as 6 3b in matrix form and enter them a . 3 6 b. We know B 4 3 is A 4 3 4 6i. that 4 6 a 3 4i 1 6 4b , 1 2 1 2i 2 4 6i. Verify that which represents the complex number c. Use matrix addition, subtraction, and multiplication to find the following. Interpret the answers as complex numbers. C , 4 3 3 4 i 1 2i 3 4 d. In Example 5 we saw that B A 4 3 11 .2 0.4i. 5 1 A B 4 3 4 . Use the Do this problem in matrix form by computing x key for the exponent. 1 3 e. Do each of the following calculations and interpret the answer in terms of complex numbers. A B 1, C 3 4 3 4 3 1. 3. 5. 7. 8. 9. 11. 13. 15. 29. 31. 33. 35. 300 Chapter 4 Polynomial and Rational Functions Exercises 4.5 In Exercises 1–54, perform the indicated operation and write the result in the form a bi. 2 3i 2 8i 6 i 2 4 2i 2i b 2. 4. 6. 5 7i 14 3i 1 1 3 5i 2 2 2 1 3 7i 1 2 23 i A B A 25 2i B 47. 49. 50. 51. 53. 216 236 225 2 A BA 5 23 A 2 25 A 1 1 22 249 3 BA 1 29 B 1 210 B BA 48. 264 24 B 52. 54. 3 227 23 A 1 24 3 29 B 22 2 a i b 23 2 a i b 1 2 a 23 2 i b 3 4 a 523 2 i b 2 i 21 1 3 2i 3 5i 2 4 i 2 21 1 2 5i 2 2 23 i 23 i B BA 1 A 10. 12. 14. 16. 2 i 21 4 3i 5 2i 2 4 3i 1 1 21 2 i i 2 1 i 1 21 1 2 a i ba 1 4 2i b In Exercises 55–58, find x and y. 55. 3x 4i 6 2yi 56. 8 2yi 4x 12i 57. 3 4xi 2y 3i 58. 8 xi 1 2 y 2i 2 In Exercises 59–70, solve the equation and express each solution in the form a bi. 17. i15 18. i26 19. i33 20. 53 i 2 1 21. 107 i 2 1 22. 213 i 2 1 23. 1 5 2i 24. 1 i 25. 1 3i 26. i 2 i 27. 3 4 5i 28. 2 3i i 69. x4 1 0 59. 3x2 2x 5 0 60. 5x2 2x 1 0 61. x2 x 2 0 62. 5x2 6x 2 0 63. 2x2 x 4 65. 2x2 3 6x 67. x3 8 0 64. x2 1 4x 66. 3x2 4 5x 68. x3 125 0 x4 81 0 70. i i2 i3 p i15 i i2 i3 i4 i5 p i15 71. Simplify: 72. Simplify: 73. Critical Thinking If is a complex z
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a bi number, then its conjugate is usually denoted Prove that for any complex that is, is a real number exactly number when z a bi. z a bi, z z z. z, 74. Critical Thinking The real part of the complex a bi number a bi The imaginary part of real number b (not bi). See Exercise 73 for notation. is defined to be the real number a. is defined to be the a. Show that the real part of z a bi b. Show that the imaginary part of z is . is z z 2 z z 2i . 75. Critical Thinking If z a bi numbers, not both 0), express (with a and b real 1 z in standard form. 1 4 5i i 1 2 3i 30. 32. 1 1 1 1 2i 34. 1 2 i 2 i 1 21 2 i 2 2 2 3i 4 i 2 21 3 i 2 3i i 3 i 3 i 4 i 37. 236 39. 214 41. 216 36. 6 2i 3 i 38. 281 40. 250 42. 212 43. 216 249 44. 225 29 45. 215 218 46. 212 23 Section 4.5.A Excursion: The Mandelbrot Set 301 4.5.A Excursion: The Mandelbrot Set Chapter 1 introduced the recursive process of beginning with a value and adding a specific number repeatedly, and Section 3.5.A discussed iterating real-valued functions. This section will extend the iterative process to the set of complex numbers and complex-valued functions and will illustrate the Mandelbrot set, which is used to produce many fractal images. Before reading this section, review Section 3.5.A for terminology and processes, if needed. The Complex Plane a bi, Every complex number where a and b are real numbers, corresponds to the point (a, b) in the coordinate plane. Therefore, complex numbers can be plotted in a coordinate plane, where the horizontal axis represents the real axis and the vertical axis represents the imaginary axis. For example, several complex numbers are plotted in Figure 4.5.A-1. i 6i 4i 2 + 3i −1 + 2i −6 −4 0 −2 −2i 4 6 2 − i 5 Real −3 − 4i −4i −4i −6i Figure 4.5.A-1 Orbits of Complex Numbers The concepts and processes described in Section 3.5.A apply equally well to functions with complex number inputs. For instance, the process of finding the orbit of a complex number under a complex-valued function is the same as finding the orbit of a real number under a real-valued function. For example, the orbit of i under was illustrated in Section 4.5, where it was shown that the powers of i form an orbit of period 4. iz i2 1 i i 1 2 1 i2 i i 2 1 i 1 i iz NOTE The complex plane is discussed in detail in Section 10.3. When dealing with functions that have the complex numbers as their domains, the input variable is usually denoted as z, not x. i 2i i −2 −1 0 1 2 Real −i −2i Figure 4.5.A-2 The orbit of i under is shown graphically in Figure 4.5.A-2. 302 Chapter 4 Polynomial and Rational Functions Technology Tip Calculators that handle complex numbers allow z z2 c you to compute orbits of f 1. Store the complex number c in memory C. easily. 2 1 2. Press 0 STO X ENTER 3. Press to store 0 in X. X 2 C ENTER to produce f 0 . STO X 1 2 4. Pressing ENTER repeatedly produces the iterated values. Orbits of complex numbers of many functions are very interesting, but the discussion that follows will be limited to the orbit of 0 under f for different values of c. z2 c z 1 2 Example 1 The Orbit of 0 for f(z) z 2 c Describe the orbit of 0 under ber c. f z 2 1 z2 c for the given complex num- a. c 0.25 0.25i b. c 1.2 0.05i Solution is as follows. a. For z f 1 2 c 0.25 0.25i, z2 0.25 0.25i The orbit of 0 under z2 0.25 0.25i. 1 f z 2 z2 0.25 0.25i 02 0.25 0.25i 0.25 0.25i 0.25 0.25i .25 0.25i 2 0.25 0.125i 0.203125 0.187500i 0.243896 0.173828i 0.220731 0.165208i .2439 0.1738i 0.2207 0.1652i 0.2286 0.1771i 2 2 2 0.25 0.125i 1 0.203125 0.187500i 2 Use iteration on your calculator to find orbit of 0 is approaching a number near c 0.25 0.25i, can be shown that for . 0 It suggests that the f 15 0.2277 0.1718i. In fact, it 2 1 0 f n 1 c 1.2 0.05i, 2 b. For The orbit of 0 under z2 1.2 0.05i. 2 S 0.2276733451 0.1717803749i as n S q f z 1.2 0.05i z2 1 2 z2 1.2 0.05i .2 0.05i 1.1485 0.0168i 1.1860 0.0527i 1.1641 0.0194i f 8 1.1761 0.0515i f 10 1.1725 0.0243i f 12 1.1697 0.0476i f 14 1.1762 0.0296i f 16 is as follows. 0.2375 0.0700i 0.1188 0.0115i 0.2039 0.0751i 0.1547 0.0049i 0.1805 0.0712i 0.1741 0.0070i 0.1660 0.0613i 0.1825 0.0197i NOTE All decimals are shown rounded to four decimal places, but calculations are done using the decimal capacity of the calculator 11 f 13 f 15 Viewing additional iterations, the orbit of 0 when appears to oscillate between two values that are close to and as illustrated in Figure 4.5.A-3. 0.17 0.04i c 1.2 0.05i 1.17 0.04i Section 4.5.A Excursion: The Mandelbrot Set 303 i 1 0.5 −1.5 −1 −0.5 0 0.5 Real 0.5 Figure 4.5.A-3 ■ Example 2 The Orbit of 0 for f(z) z 2 c Describe the orbit of 0 under ber c. f z 1 2 z2 c for the given complex num- a. c 1 i b. c 0.25 0.625i Solution a. The first seven iterations of f z 2 1 z2 1 i 1 2 are shown in Figure 4.5.A-4. If each of these numbers is plotted in the complex plane, successive iterations get farther and farther from the origin at a very fast rate. For instance 9407 193i 88,454,401 3,631,103i 7.81 1015 6.42 1014 1 2 1 i 2 The distance formula shows that after only five iterations, the distance to the origin is about 9509 and after seven iterations, it is gigantic. In this case, the orbit of 0 is said to approach infinity, which is sometimes expressed as follows. z2 S q as n S q. , then f n 1 i If 0 z f 0 through b. The iterations through are shown in Figure 4.5.A-5a and f 15 are shown in Figure 4.5.A-5b. As you can see, successive iterations stay fairly close to the origin through the 16th iteration and then quickly move farther and farther away 21 f 20 1 f 21 f 22 20 0 0 1 1 2 2 2 167.4 522.8i 254,270 175,058i 2.95 1010 8.59 1010 i 2 2 1 Therefore, the orbit of 0 approaches infinity. In other words, 0.25 0.625i S q as n S q. , then f n z2 if Figure 4.5.A-4 Figure 4.5.A-5a Figure 4.5.A-5b 304 Chapter 4 Polynomial and Rational Functions The Mandelbrot Set The Mandelbrot set is defined by whether or not the orbit of 0 under the approaches infinity for each complex number c. function z2 c z f 1 2 The Mandelbrot Set The Mandelbrot set is the set of complex numbers c such that the orbit of 0 under the function approach infinity. f(z) z2 c does not 1 2 To avoid awkward repetition in the following discussion, the orbit of 0 under the function will be referred to as “the orbit of c.” z2 c z f Example 1 shows that the orbit of converges and the c 1.2 0.05i oscillates. Neither orbit approaches infinity, so orbit of both numbers are in the Mandelbrot set. Example 2 shows that the orbits of approach infinity, so these numbers are not in the Mandelbrot set. c 0.25 0.625i c 1 i and c 0.25 0.25i Diagram of the Mandelbrot Set Although the Mandelbrot set is defined analytically, it is usually viewed geometrically by plotting the numbers in the Mandelbrot set as points in the complex plane. The Mandelbrot set is the white region in Figure 4.5.A-6, in which the tick marks on each axis are unit apart. Note that 1 2 the Mandelbrot set is symmetric with respect to the real axis, but not with respect to the imaginary axis. Figure 4.5.A-6 Section 4.5.A Excursion: The Mandelbrot Set 305 Figure 4.5.A-6 illustrates the following fact in which the complex numbers are considered as points in the complex plane. If c lies outside the circle of radius 2 with center at the origin, then c is not in the Mandelbrot set. Furthermore, it can be proved that If c lies inside the circle of radius 2 with center at the origin, but some number in its orbit lies outside the circle, then c is not in the Mandelbrot set. Determining whether a particular point c is in the Mandelbrot set can be quite difficult, particularly if the numbers in its orbit move away from the origin very slowly. Even after hundreds of iterations, it may not be clear whether or not the orbit approaches infinity. Because of round-off errors, a calculator is inadequate for such calculations, and even computers have their limitations. Border of the Mandelbrot Set The border of the Mandelbrot set, which consists of the points just outside of the set, is shown in various colors in Figure 4.5.A-6. The colors indicate how quickly the orbit of the point approaches infinity. The colors are determined by the number of iterations n needed for a number in the orbit of the point to be more than 3 units from the origin, as indicated in the following table. Medium Red Light Red Red Light Yellow Yellow Blue Black n 5 6 n 7 8 n 9 10 n 12 13 n 19 20 n 49 n 50 is in the red region because the second iterExample 2 shows that ation produces a point more than 3 units from the origin and that 0.25 0.625i is in the yellow region. 1 i The border of the Mandelbrot set is very jagged and chaotic. The varying rates at which the orbits of these border points approach infinity produce some interesting patterns of great complexity. When specific areas are magnified, you can see shapes that resemble islands, seahorse tails, and elephant trunks. The most fascinating aspect of the set is that some of these islands have the same shape as the entire set. Consequently, the Mandelbrot set is said to be self-similar under magnification. The following figures show the region near the point increasing magnification. 0.75 0.3i under The purple points in Figure 4.5.A-7a indicate the Mandelbrot set. Each subsequent image is the magnification of the region denoted by the white square in the previous image. The image shown in Figure 4.5.A-7d is a copy of the set that is contained within the set, and even though the graphs shown do not indicate that all of the purple regions are connected, they are. 306 Chapter 4 Polynomial and Rational Functions Figure 4.5.A-7a Figure 4.5.A-7b Figure 4.5.A-7c Figure 4.5.A-7d Exercises 4.5.A f 1(0), f 2(0), Then determine the distance from In Exercises 1–6, compute where f(z) z2 c. f 3(0) to the origin in the complex plane, where the distance from a point to the origin is given by a bi f 3(0) and 2a2 b2. 1. c 0.3 2. c 0.5i 3. c 0.5 0.5i 4. c 1 0.5i 5. c 1.2 0.5i 6. c 0.75 0.25i In Exercises 7–12, show
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that c is not in the Mandelbrot set by finding a number in its orbit that is more than 2 units from the origin. How many iterations are needed to find the first such number? 7. c 0.4 9. c 0.7i 8. c 1.1 0.4i 10. c 0.2 0.8i 11. c 0.5 0.7i 12. c 0.4 0.6i In Exercises 13–18, determine whether or not c is in the Mandelbrot set. 13. c i 15. c 1 14. c i 16. c 0.1 0.3i 17. c 0.2 0.6i 18. 0.1 0.8i Section 4.6 The Fundamental Theorem of Algebra 307 4.6 The Fundamental Theorem of Algebra Objectives • Use the Fundamental Theorem of Algebra • Find complex conjugate zeros • Find the number of zeros of a polynomial • Give the complete factorization of polynomial expressions NOTE The graph of a polynomial with complex coefficients cannot be drawn on a coordinate plane. The complex numbers were constructed in order to obtain a solution for Every the equation quadratic equation with real coefficients has solutions in the complex number system, as discussed in Section 4.5. A natural question arises: that is, a zero of the polynomial x2 1. x2 1, x f 2 1 Does the complex number system need to be enlarged, perhaps many times, to find zeros of higher degree polynomial functions? This section explains why the answer is no. In order to give the full answer, the discussion will not be limited to polynomials with real coefficients but will consider polynomials with complex coefficients, such as x3 ix2 4 3i 2 1 x 1 or 3 2i 2 1 x6 3x 5 4i . 2 1 The discussion of polynomial division in Section 4.1 can easily be extended to include polynomials with complex coefficients. In fact, All of the results in Section 4.1 are valid for polynomials with complex coefficients. 1 f x 2 i For example, for 2 x f i is a zero of f and that . 1 checked using the same procedures as before. 1 i 1 i 1 2 is a factor of x2 x i i2 i2 i i2 2 i 1 i 1 2 i 0 2 , it is easy to verify that Both statements can be Therefore, i is a zero of f. x x2 2 x2 ix 1 2i 1 2i 1 2i Therefore, x i is a factor of 1 i x f 1 x and 2 2 i x2 2i . 2 4 Because every real number is also a complex number, polynomials with real coefficients are just special cases of polynomials with complex coefficients. In the rest of this section, “polynomial” means “polynomial with complex, possibly real, coefficients” unless specified otherwise. Fundamental Theorem of Algebra Every nonconstant polynomial has a zero in the complex number system. 308 Chapter 4 Polynomial and Rational Functions Although this is a powerful result, neither the Fundamental Theorem nor its proof provides a practical method for finding a zero of a given polynomial. You may think it strange that you can prove a zero exists without actually finding one, but such “existence” proofs are quite common in mathematics. Factorization over the Complex Numbers Number of Zeros f(x) be a polynomial of degree Let with leading coefficient a. Then there are n, not necessarily distinct, complex numbers such that c1, c2, p , cn n 77 0 f(x) a(x c1)(x c2) p (x cn) Furthermore, c1, c2 p , cn are the only zeros of f. That is, every polynomial of degree can be written as the product of n linear factors. The statement follows from the Fundamental Theorem and the Factor Theorem. By the Fundamental Theorem, has a complex zero x 2 is a factor, so x c1 c1, f 1 n 7 0 and by the Factor Theorem x c12 x f 2 1 1 is nonconstant, then it has a complex zero 1 2 g x . x g If Theorem and a factor 2 1 x c2 x f 1 2 so that x c121 1 x c221 h x 1 22 . c2 by the Fundamental This process can be continued until the final factor is a constant a, at which point you have the factorization shown in the preceding box. Because the n zeros of distinct zeros may be less than n. c1, c2, p , cn of f may not all be distinct, the number Every polynomial of degree zeros in the complex number system. n 77 0 has at most n different Suppose f has repeated zeros, meaning that some of the are Recall that a zero c is said to have the same in the factorization of x c f multiplicity k if 2 is a factor. Consequently, if every zero is counted as many times as its multiplicity, then the statement in the preceding box implies that . 2 is a factor of but no higher power of x c c1, c2 , p , cn polynomial of degree n has exactly n complex zeros. Example 1 Finding a Polynomial Given Its Zeros Find a polynomial a zero of multiplicity 3, and f x f 1 2 24. 2 1 2 of degree 5 such that 1, 2, and 5 are zeros, 1 is Section 4.6 The Fundamental Theorem of Algebra 309 Solution must be a factor of . There f x 1 2 and 5. so x 1 2 Because 1 is a zero of multiplicity are two other factors corresponding to the zeros x x 2 2 x 5 and 2 3, 1 2 x 1 3 1 2 f The product of these factors has degree 5, as does 21 2 where a is the leading coefficient. 24, Because 21 a 1 2 24 2 5 12a 24 a 2 Therefore, f x 1 2 3 x 1 2 21 2x5 12x4 4x3 40x2 54x 20 x 2 x 5 2 1 1 2 y 100 (2, −24) x 2 4 6 −4 0 −2 −100 −200 −300 Figure 4.6-1 The graph of f is shown in Figure 4.6-1. ■ NOTE Complex zeros are not shown on the graph of a polynomial. Polynomials with Real Coefficients Recall that the conjugate of the complex number is the number a bi. We usually write a complex number as a single letter, say z, and z, sometimes read “z bar.” For instance, if indicate its conjugate by z 3 7i, z 3 7i. Conjugates play a role whenever a quadratic then polynomial with real coefficients has complex zeros. a bi y 8 4 −8 −4 0 4 8 −4 −8 Figure 4.6-2 Conjugate Zero Theorem Example 2 Conjugate Zeros Find the zeros of f x 2 1 x2 6x 13. x Solution 2 1 6 ± 2 The quadratic formula shows that f has two complex zeros. 6 ± 4i 2 z 3 2i. 6 ± 216 2 The complex roots are f has no real zeros, as shown in Figure 4.6-2. 2 4 1 13 and its conjugate 6 1 2 2 1 z 3 2i 3 ± 2i Notice that ■ The preceding example is a special case of a more general theorem. f(x) Let number z is a zero of f, then its conjugate be a polynomial with real coefficients. If the complex is also a zero of f. z 310 Chapter 4 Polynomial and Rational Functions Example 3 A Polynomial with Specific Zeros Find a polynomial with real coefficients whose zeros include the numbers 2 and 3 i. Solution 3 i Because Factor Theorem, x 2 2 is a zero, is a zero, its conjugate, 3 i x 1 is a factor. So, consider the polynomial x and 1 2 2 3 i, 3 i must also be a zero. By the are factors. Similarly, because −8 −4 0 4 8 x f 1 x y 16 8 −8 −16 Figure 4.6-3 Obviously, form shows that are zeros of f. Multiplying out the factored , 3 i, and has real coefficients. 3 i 1 x 3 i 3 i x2 1 x2 3x ix 3x ix 9 i2 x2 6x 10 x3 8x2 22x 20 2 3 21 21 21 2 4 can be factored The next-to-last line of the calculation also shows that as a product of a linear and a quadratic polynomial, each with real coefficients. The graph of f is shown in Figure 4.6-3. x f 1 2 ■ The technique used in Example 3 works because the product, x x2 6x 10 has real coefficients. The proof of the following result shows why this must always be the case. Factorization over the Real Numbers Every nonconstant polynomial with real coefficients can be factored as a product of linear and irreducible quadratic polynomials with real coefficients in such a way that the quadratic factors, if any, have no real zeros. That is, every nonconstant polynomial with real coefficients can be written ax2 bx c, where each as the product of factors in the form quadratic factor is irreducible over the set of real numbers. x k or 1 2 f x a Proof The box on page 308 shows that for any polynomial x c121 c1, c2, p , cn x cr 1 are the zeros of f. If some cr is a real number, then the where is a linear polynomial with real coefficients. If some cj is a factor nonreal complex zero, then its conjugate must also be a zero. Thus, some ck is the conjugate of with a and b real numbers. Thus, x cn2 x c22 a bi, a bi f x , 2 1 and say, cj, p ck cj 1 Section 4.6 The Fundamental Theorem of Algebra 311 x cj21 x ck2 1 1 1 x 2 4 x a bi 1 a bi a bi a bi x 2 4 3 3 x2 x 2 x2 ax bix ax bix a2 x2 2ax a2 b2 1 x ck2 x cj21 1 2 1 2 1 a bi bi 1 a bi 2 21 2 2 is a quadratic expression with real Therefore, the factor coefficients because a and b are real numbers. Its zeros, and , are noncj real. By taking the real zeros of f one at a time and the nonreal ones in is obtained. conjugate pairs in this fashion, the desired factorization of ck x f 1 2 y 12 6 −4 −2 0 2 4 x −6 −12 Figure 4.6-4 Example 4 Completely Factoring a Polynomial over the Real Numbers Completely factor numbers given that Solution x f 2 1 1 i x4 2x3 x2 6x 6 over the set of real is a zero of f. 1 i, is also a zero of f, and f x 2 1 1 i is a zero of f, its conjugate, Because has the following quadratic factor. x f Dividing factors as A 3 x by 1 x 23 2 1 i x 2 4 3 1 x2 2x 2 x 23 BA x B x 23 A 1 2 f 1 i 2 4 x2 2x 2 1 shows that the other factor is . Therefore, x 23 x2 2x 2 B 1 BA x4 2x3 x2 6x6 2 is the complete factorization of bers. The graph of f is shown in Figure 4.6-4. x2 3, which over the real num- ■ Complete Factorization of Polynomials The techniques illustrated in this chapter can be used to completely factor some polynomials into linear factors. Example 5 Completely Factoring a Polynomial over the Complex Numbers Completely factor numbers and then over the set of complex numbers. x4 5x3 4x2 2x 8 x f 1 2 over the set of real Solution Because the degree of the polynomial is 4, there are exactly 4 complex zeros. Find all rational zeros: The possible rational zeros are factors of 8. ± 1, ± 2, ± 4, ± 8 312 Chapter 4 Polynomial and Rational Functions Graph sible zeros are the rational zeros. x4 5x3 4x2 2x 8 x f 2 1 and determine which of the pos- y 20 10 −8 −4 0 4 8 x −10 −20 Figure 4.6-5 The graph suggests that they are. The graph shows that there are no other real zeros. and 4 are zeros and you can easily verify that 1 Find all rational factors: Two linear factors of f are x 1 and x 4. x 1 2 Find remaining factors: Use synthetic division twice to find another factor of ⎧⎪⎪⎨⎪⎪⎩ ⎧⎪⎨⎪⎩ ⎧⎪⎪⎨⎪⎪⎩ x3 x2 2 2 1 2 2 So, x4 5x3 4x2 2x 8 x 1 1 21 Use the quadratic formula to find
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the two zeros of x 2 ± 2 21 21 1 x2 2x 2 21 . 21 2 ⎧⎪⎪⎨⎪⎪⎩ x2 2x 2 x2 2x 2 ± 2i 2 2 2 2 2 ± 24 2 1 i x 1 and The complex factors are Section 4.6 The Fundamental Theorem of Algebra 313 The complete factorizations of the set of complex numbers are shown below. x f 1 2 over the set of real numbers and over x4 5x3 4x2 2x 8 x 1 x 1 1 1 21 21 x 4 x 4 21 2 3 x2 2x 2 x 1 i 2 2 4 3 1 Both products can be verified by multiplication. x 1 i 1 2 4 ■ An expression that has even degree of 4 or greater may have only complex roots. Zeros and factors of such functions and expressions may be difficult to find, and the techniques illustrated in this chapter are of little use. However, functions of odd degree must have at least one real zero, and the corresponding expression must have at least one real linear factor. Therefore, a cubic expression can easily be approximately factored by estimating one zero, which yields the real linear factor, using synthetic division to determine the quadratic factor, and then using the quadratic formula to estimate the remaining two zeros. Exercises 4.6 In Exercises 1–6, determine if without using synthetic or long division. g(x) is a factor of 1. 2. 3. 4. 5. 6 x10 x8 g x 1 x 2 1 x6 10 g x 2 x 2 1 3x4 6x3 2x 1 g x5 3x2 2x 1 g x3 2x2 5x 10x75 8x65 6x45 4x32 2x15 5 x 1 f(x) 11. 13. 15. 16. 18. 19. 21 x2 2x 5 3x2 2x 7 12. 14. f f x x 1 1 2 2 x2 4x 13 3x2 5x 2 x3 27 Hint: Factor first. x3 125 17. f x 1 2 x3 8 x f Hint: Let x6 64 u x3 1 2 and factor u2 64 first. x4 1 x4 3x2 10 f f x x 1 1 2 2 20. 22. f f x x 1 1 2 2 x4 x2 6 2x4 7x2 4 In Exercises 7–10, list the zeros of the polynomial and state the multiplicity of each zero. with real In Exercises 23–44, find a polynomial coefficients that satisfies the given conditions. Some of the problems have many correct answers. f(x) 7. f x 1 2 x54 x 4 a 5b 8. g x 2 1 3 x 1 a 6b a x 1 5b a x 1 4b 9. 10. h x 2 1 k x 1 2 2x15 x p 14 x p 1 13 1 x 27 A 7 B A 3 1 2 x 25 5 B 1 2 4 2x 1 23. degree 3; only zeros are 1, 7, 4 24. degree 3; only zeros are 1 and 1 25. degree 6; only zeros are 1, 2, p 26. degree 5; only zero is 2 2 27. degree 3; zeros 3, 0, 4; f 80 5 1 2 In Exercises 11 –22, find all the zeros of f in the comas a product of plex number system; then write linear factors. f(x) 28. degree 3; zeros 2 29. zeros include 2 i and 2 i , 2; f 0 1 1, 1 2 2 314 Chapter 4 Polynomial and Rational Functions 30. zeros include 1 3i and 1 3i 54. x4 6x3 29x2 76x 68; zero 2 of multiplicity 2 31. zeros include 2 and 2 i 32. zeros include 3 and 4i 1 33. zeros include 3, 1 i, 1 2i 34. zeros include 1, 2 i, 3i 1 35. degree 2; zeros 1 2i and 1 2i 36. degree 4; zeros 3i and 3i, each of multiplicity 2 37. degree 4; only zeros are 4, 3 i, and 3 i 38. degree 5; zeros 2 of multiplicity 3, i, and 39. degree 6; zeros 0 of multiplicity 3 and 3, 1 i, each of multiplicity 1 i 1 i, 40. degree 6; zeros include i of multiplicity 2 and 3 41. degree 2; zeros include 42. degree 2; zeros include 1 i; f 6 0 1 2 3 i; f 3 2 1 2 43. degree 3; zeros include i and 1; f 1 2 1 8 44. degree 3; zeros include 2 3i and 2; f 3 2 1 2 In Exercises 45–48, find a polynomial with complex coefficients that satisfies the given conditions. 45. degree 2; zeros i and 1 2i 46. degree 2; zeros 2i and 1 i 47. degree 3; zeros 3, i, and 2 i 48. degree 4; zeros 22, 22, 1 i, and 1 i In Exercises 49–56, one zero of the polynomial is given; find all the zeros. 49. x3 2x2 2x 3; zero 3 50. x3 x2 x 1; zero i 51. x4 3x3 3x2 3x 2; zero i 52. x4 x3 5x2 x 6; zero i 53. x4 2x3 5x2 8x 4; zero 1 of multiplicity 2 55. x4 4x3 6x2 4x 5; zero 2 i 56. x4 5x3 10x2 20x 24; zero 2i 57. Let and be complex z a bi w c di numbers (a, b, c, d are real numbers). Prove the given equality by computing each side and comparing the results. z w z w a. (The left side says: “First find z w and then take the conjugate.” The right side says: “First take the conjugates of z and w and then add.”) z w z w b. 58. Let 1 2 h x x g and assume that there are cn1 1 2 be polynomials of degree n and numbers c1, c2, ..., cn, n 1 such that ci2 g 1 x g 2 1 g x 2 ci2 1 x Prove that . 2 1 h f a zero of is nonzero, 2 what is its largest possible degree? To avoid a contradiction, conclude that h h x 2 0. for every i. Hint: Show that each ci is If 59. Suppose f ax3 bx2 cx d has real x 1 2 coefficients and z is a complex zero of f. a. Use Exercise 57 and the fact that is a real number, to show that az3 bz2 cz d f az 3 bz2 cz d f z 0 0. b. Conclude that f f is also a zero of Note: r r, 1 2 1 2 when r 60. Let 2 1 f x be a polynomial with real coefficients and z a complex zero of f. Prove that the conjugate is also a zero of f. Hint: Exercise 59 is the case when x f has degree 3; the proof in the general case is 2 similar. z 1 61. Use the Factorization over the Real Numbers statement to show that every polynomial with real coefficients and odd degree must have at least one real zero. f x 62. Give an example of a polynomial with complex, nonreal coefficients and a complex number z such that z is a zero of f but its conjugate is not. Therefore, the conclusion of the Conjugate Roots Theorem may be false if doesn’t have real coefficients Important Concepts Section 4.1 Section 4.2 Section 4.3 Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 Constant polynomial . . . . . . . . . . . . . . . . . . . . . . 240 Zero polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . 240 Degree of a polynomial . . . . . . . . . . . . . . . . . . . . 240 Leading coefficient . . . . . . . . . . . . . . . . . . . . . . . . 240 Polynomial function . . . . . . . . . . . . . . . . . . . . . . . 240 Polynomial division . . . . . . . . . . . . . . . . . . . . . . . 240 Synthetic division . . . . . . . . . . . . . . . . . . . . . . . . . 241 Division Algorithm. . . . . . . . . . . . . . . . . . . . . . . . 243 Remainders and factors . . . . . . . . . . . . . . . . . . . . 243 Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . 244 Zero of a polynomial . . . . . . . . . . . . . . . . . . . . . . 245 Real zero. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 Factor Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . 245 Fundamental polynomial connections . . . . . . . . . 246 Number of zeros of a polynomial . . . . . . . . . . . . 248 The Rational Zero Test . . . . . . . . . . . . . . . . . . . . . 251 Factoring polynomials . . . . . . . . . . . . . . . . . . . . . 252 Irreducible polynomials . . . . . . . . . . . . . . . . . . . . 253 Complete factorization . . . . . . . . . . . . . . . . . . . . . 253 Bounds Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 Real zeros of polynomials . . . . . . . . . . . . . . . . . . 257 axn 2 1 f x Graph of . . . . . . . . . . . . . . . . . . . . . . . 260 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 End behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 Multiplicity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 Local extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 Points of inflection . . . . . . . . . . . . . . . . . . . . . . . . 266 Complete polynomial graphs . . . . . . . . . . . . . . . . 267 Section 4.3.A Polynomial models. . . . . . . . . . . . . . . . . . . . . . . . 273 Section 4.4 Rational function . . . . . . . . . . . . . . . . . . . . . . . . . 278 Domain of rational functions . . . . . . . . . . . . . . . . 279 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 315 316 Chapter Review Section 4.5 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 Vertical asymptotes. . . . . . . . . . . . . . . . . . . . . . . . 280 Big-Little Concept. . . . . . . . . . . . . . . . . . . . . . . . . 280 Holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 End behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 Horizontal asymptotes . . . . . . . . . . . . . . . . . . . . . 284 Other asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . 285 Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . 294 Imaginary numbers . . . . . . . . . . . . . . . . . . . . . . . 294 Arithmetic of complex numbers. . . . . . . . . . . . . . 295 Powers of i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Complex conjugates . . . . . . . . . . . . . . . . . . . . . . . 296 Square roots of negative numbers . . . . . . . . . . . . 297 Complex solutions to quadratic equations . . . . . . 298 Section 4.5.A The Mandelbrot set . . . . . . . . . . . . . . . . . . . . . 301 Section 4.6 The Fundamental Theorem of Algebra . . . . . . . . 307 Factorization over the complex numbers . . . . . . . 308 Number of zeros. . . . . . . . . . . . . . . . . . . . . . . . . . 308 Conjugate Zero Theorem . . . . . . . . . . . . . . . . . . . 309 Factorization over the real numbers. . . . . . . . . . . 310 Complete factorization of polynomials. . . . . . . . . 311 Important Facts and Formulas When f is a polynomial and r is a real number that satisfies any of the following statements, then r satisfies all the statements. • r is a zero of the polynomial function • r is an x-intercept of the graph of f • r is a solution, or root, of the equation • 0 f is a factor of the polynomial expression • There is a one-to-one correspondence between the linear factors of of the graph of f. x f 1 2 that have real coefficients and the x-intercepts A polynomial of degree n has at most n distinct real zeros. All rational zeros of a polynomial have the form where r is a fac- r s , tor of the constant term and s is a factor of the leading coefficient. The end behavior of the graph of a polynomial function is similar to the end behavior of the graph of the highest degree term of the polynomial. Zeros of even
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multiplicity touch but do not cross the x-axis. Zeros of odd multiplicity cross the x-axis. The number of local extrema of the graph of a polynomial function is at most one less than the degree of a polynomial. Chapter Review 317 The number of points of inflection of the graph of a polynomial function is at most two less than the degree of the polynomial. The graph of a rational function has a vertical asymptote at every number that is a zero of the denominator and not a zero of the numerator. The x-intercepts of the graph of a rational function occur at the numbers that are zeros of the numerator but are not zeros of the denominator. Every complex number can be written in the standard form i2 1 and i 21 a bi. a bi If jugate a bi is also a zero. is a zero of a polynomial with real coefficients, then its con- A polynomial of degree n has exactly n complex zeros counting multiplicities. Every polynomial expression with real coefficients can be factored into linear and irreducible quadratic factors with real coefficients. Every polynomial expression can be factored into linear factors with complex coefficients. Review Exercises Section 4.1 1. Which of the following are polynomials? a. 23 x2 d. g. 23 x4 2x 2x2 b. e. x 1 x p3 x c. f. x3 1 22 22 2x2 0 2. What is the remainder when h. 3. What is the remainder when x 1? x 0 x4 3x3 1 is divided by x2 1? x112 2x8 9x5 4x4 x 5 is divided by 4. Is x 1 a factor of f x 1 2 14x87 65x56 51? Justify your answer. 5. Use synthetic division to show that x6 5x5 8x4 x3 17x2 16x 4, x 2 is a factor of and find the other factor. 6. Find a polynomial f of degree 3 such that f 1 2 1 0, f 1 1 2 0, and f 5. 0 1 2 Section 4.2 7. Find the zero(s) of 2 x 5 a 7 b 3x x 2 5 4. 8. Find the zeros of 3x2 2x 5. 9. Factor the polynomial x3 8x2 9x 6. Hint: 2 is a zero. 10. Find all real zeros of x6 4x3 4. 318 Chapter Review 11. Find all real zeros of 9x3 6x2 35x 26. Hint: Try x 2. 12. Find all real zeros of 3y3 1 13. Find the rational zeros of y4 y2 5 . 2 x4 2x3 4x2 1. 14. Consider the polynomial 2x3 8x2 5x 3. a. List the only possible rational zeros. b. Find one rational zero. c. Find all the zeros of the polynomial. 15. a. Find all rational zeros of x3 2x2 2x 2. b. Find two consecutive integers such that an irrational zero of lies between them. x3 2x2 2x 2 16. How many distinct real zeros does x3 4x have? 17. How many distinct real zeros does x3 6x2 11x 6 have? 18. Find the zeros of x4 11x2 18. has x3 2x 1 19. The polynomial a. no real zeros. b. only one real zero. c. three rational zeros. d. only one rational zero. e. none of the above. 20. Show that 5 is an upper bound for the real zeros of x4 4x3 16x 16. 21. Show that 1 is a lower bound for the real zeros of x4 4x3 15. In Exercises 22 and 23, find the real zeros of the polynomial. 22. x6 2x5 x4 3x3 x2 x 1 23. x5 3x4 2x3 x2 23x 20 Section 4.3 24. List the zeros of the polynomial and the multiplicity of each zero. 3 3 5 x 4 x 2 x 17 x2 4 1 2 1 21 25. List the zeros of the polynomial and the multiplicity of each zero. 2 x 3 x 4 x2 9 21 2 2 2 1 1 26. Draw the graph of a function that could not possibly be the graph of a polynomial function, and explain why. 27. Draw a graph that could be the graph of a polynomial function of degree 5. You need not list a specific polynomial nor do any computation. 28. Which of the statements is not true about the polynomial function f whose graph is shown in the figure on the next page? Chapter Review 319 y 3 2 1 x −3 −2 −1 1 2 3 −2 f 2 could possibly be a fifth-degree polynomial a. f has three zeros between b. c. d. e. 0 21 f is positive for all x in the interval and , 0 3 4 1 1 2 2 29. Which of the statements i–v about the polynomial function f whose graph is shown in the figure below are false? y 4 −4 2 −6 −4 −2 i ii iii f has 2 zeros in the interval , 3 1 2 x iv v 2 2 0 f 1 f has degree 2 4 In Exercises 30–33, find a viewing window (or windows) that shows a complete graph of the function. Be alert for hidden behavior. 30. f x 1 2 31. g 32. h x x 2 2 1 1 33. f x 1 2 0.5x3 4x2 x 1 0.3x5 4x4 x3 4x2 5x 1 4x3 100x2 600x 32x3 99x2 100x 2 In Exercises 34–37, sketch a complete graph of the function. 34. f 36. h x x 1 1 2 2 x3 9x x4 x3 4x2 4x 2 35. g 37. f x x 2 2 1 1 x3 2x2 3 x4 3x 2 Section 4.3.A 38. HomeArt makes plastic replicas of famous statues. Their total cost to produce copies of a particular statue is shown in the table on the next page. a. Sketch a scatter plot of the data. b. Use cubic regression to find a function C(x) that models the data—that 320 Chapter Review is, the cost of making x statues. Assume C is reasonably accurate when x 100. c. Use C to estimate the cost of making the seventy-first statue. d. Use C to approximate the average cost per statue when 35 are made and when 75 are made. Recall that the average cost of x statues is C x 1 2 x . Number of statues Total cost 0 10 20 30 40 50 60 70 $2,000 2,519 2,745 2,938 3,021 3,117 3,269 3,425 39. The following table gives the estimated cost of a college education at a public institution. Costs include tuition, fees, books, and room and board for four years. a. Sketch a scatter plot of the data (with b. Use quartic regression to find a function C that models the data. Estimate the cost of a college education in 2007 and in 2015. corresponding to 1990). x 0 Enrollment Year Costs Enrollment Year Costs 1998 2000 2002 2004 2006 $46,691 52,462 58,946 66,232 74,418 2008 2010 2012 2014 $ 83,616 93,951 105,564 118,611 Source: Teachers Insurance and Annuity Association College Retirement Equities Fund Section 4.4 In Exercises 40–43, sketch a complete graph of the function. Label the x-intercepts, all local extrema, holes, and asymptotes. 40. g x 2 1 2 x 4 42. k x 1 2 4x 10 3x 9 41. h x 1 2 3 x x 2 43. f x 1 2 x 1 x2 1 Chapter Review 321 In Exercises 44 and 45, list all asymptotes of the graph of the function. 44. f x 1 2 x2 1 x3 2x2 5x 6 45. g x 2 1 x4 6x3 2x2 6x 2 x2 3 In Exercises 46–49, find a viewing window (or windows) that shows a complete graph of the function. Be alert for hidden behavior. 46. f x 1 2 x 3 x2 x 2 48. h x 1 2 x4 4 x4 99x2 100 47. g x 1 2 x2 x 6 x3 3x2 3x 1 49. k x 1 2 x3 2x2 4x 8 x 10 50. Which of these statements is true about the graph of x 1 x2 1 x 3 2 x2 1 ? 2 1 2 f x 1 1 21 21 a. The graph has two vertical asymptotes. x 3. b. The graph touches the x-axis at c. The graph lies above the x-axis when d. The graph has a hole at e. The graph has no horizontal asymptotes. x 1. x 6 1. Section 4.5 In Exercises 51–58, solve the equation in the complex number system. 51. x2 3x 10 0 52. x2 2x 5 0 53. 5x2 2 3x 55. 3x4 x2 2 0 57. x3 8 0 54. 3x2 4x 5 0 56. 8x4 10x2 3 0 58. x3 27 0 59. One zero of x4 x3 x2 x 2 is i. Find all zeros. 60. One zero of x4 x3 5x2 x 6 is i. Find all zeros. 61. Give an example of a fourth-degree polynomial with real coefficients whose zeros include 0 and 1 i. 62. Find a fourth-degree polynomial f whose only zeros are such that f 1 1 50. 2 2 i and 2 i Section 4.5.A 63. Find the orbit of 1 for . i b 64. Find the orbit of 0 for z2 c whether c is in the Mandelbrot set. z f 2 1 using the following values of c. State a. c 1 b. c 0.5 0.6i c. c 0.3 0.5i Section 4.6 Factor each of the following over the set of real numbers and over the set of complex numbers. 65. x3 6x2 11x 6 66. x3 3x2 3x 2 67. x4 x3 x2 x 2 68. 2x3 3x2 9x 4 69. x4 2x2 1 70. 9x5 30x4 43x3 114x2 28x 24 12 Optimization Applications ctions Many real-world situations require you to find the largest or smallest quantity satisfying certain conditions. For instance, automotive engineers want to design engines with maximum fuel efficiency. Similarly, a cereal manufacturer who needs a box of volume 300 cubic inches might want to know the dimensions of the box that requires the least amount of cardboard, which is the cheapest to make. The exact solutions of such minimum/ maximum problems require calculus. However, graphing technology can provide very accurate approximate solutions. Example 1 Maximum of a Rational Function Find two negative numbers whose product is 50 and whose sum is as large as possible. Solution Let x and z be the two negative numbers, and let y be their sum. Then xz 50 xz 50 y x z. for z yields Solving so that and z 50 x y x z x 50 x . The desired quantity is the value of x that makes y as large as possible. Since x must be negative, graph 20 x 0. y x 50 x in a window with Each point x, y 1 2 on the graph represents the following: • x represents one of the two negative number, and 50 x represents the other negative number • y is the sum of the two negative numbers The largest y possible is the point on the graph with largest y-coordinate, that is, the highest point on this part of the graph. Either zoom-in or use the maximum finder to approximate the highest point. As shown in Figure 4.C-1, the largest value of y is approximately which occurs when Therefore, the numbers are approximately 7.071067 The exact solution, as found by 7.071069. 14.14214 and using calculus, is close to the graphical solution. —which is approximately 7.071068 and is very ■ x 7.071067. 50 7.071067 150 –20 2 0 Figure 4.C-1 –26 322 Example 2 Largest Volume of a Box A box with no top is to be made from a inch sheet of cardboard by cutting squares of equal size from each corner and bending up the flaps, as shown in Figure 4.C-2. To the nearest hundredth of an inch, what size square should be cut from each corner in order to obtain a box with the largest possible volume? What is the volume of this box? 22 30 30 x x x 22 30 − 2x x 22 − 2x Figure 4.C-2 Solution Let x denote the length of the side of the square to be cut from each corner. Then, Volume of box Length Width Height > > 1300 0 0 11 Figure 4.C-3 ⎧⎪⎨⎪⎩ ⎧⎪⎨⎪⎩ 30 2x > 22 2x x 2 2 660x 1 4x˛ 3 104x˛ x x 2 1 3 104x˛ 2 660x gives the volume y of the box Thus, the equation square from each corner. Because the that results from cutting an shortest side of the cardboard is 22 inches, the length x of the side of the cut-out square must be l
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ess than 11. (Why?) 2 660x, On the graph of 3 104x˛ y 4x˛ 4x˛ x f 1 2 • the x-coordinate of each point is the size of the square to be cut from each corner. • the y-coordinate of each point is the volume of the resulting box. The box with the largest volume corresponds to the point with the largest y-coordinate, that is, the highest point in the viewing window. A maximum find shows that the highest point is approximately (4.182, 1233.809), as shown in Figure 4.C-3. Therefore, a square measuring approximately 4.18 4.18 inches should be cut from each corner, producing a box of approximately 1233.81 cubic inches. ■ 323 Example 3 Minimum Surface Area of a Cylinder A cylindrical can of volume 58 cubic inches (approximately 1 quart) is to be designed. For convenient handling, it must be at least 1 inch high and 2 inches in diameter. What dimensions will use the least amount of material? h Solution The cylinder can be constructed by rolling a rectangular sheet of metal into a tube and then attaching the top and bottom, as shown in Figure 4.C-4. The surface area of the can, which determines the amount of material needed, has the following formula: Surface Area Area of rectangular sheet Area of top Area of bottom Ch 2pr > Ch 2 > pr 2 > pr 2 When the sheet is rolled into a tube, the width c of the sheet is the circumference of the ends of the can, so C 2pr . Surface Area Ch 2pr 2 2prh 2pr 2 C h r Figure 4.C-4 The volume of the cylinder of radius r and height h is is to have volume 58 cubic inches, pr 2 h. Since the can pr 2 h 58, or equivalently, h 58 pr 2 . Therefore, surface area 2prh 2pr 2 2pr 58 pr 2b a 2pr 2 116 r 2pr 2. Note that r must be 1 or greater because the diameter 2r must be at least 2. Furthermore, r cannot be more than 5 because if then would be at least the volume which is greater than 58. h 1, r 7 5 and pr 25 p h 1 , 2 1 21 2 The situation can be represented by the graph of the equation 2px2. The x-coordinate of each point represents a possible radius, and the y-coordinate represents the surface area of the corresponding can. A graphical minimum finder shows that the coordinates of the lowest point are approximately (2.098, 82.947), as shown in Figure 4.C-5. y 116 x If the radius is 2.098, then the height is 58 2.0982 p 4.19. The dimensions of can that uses the least amount of materials are approximately a radius of 2.1 inches and a height of 4.2 inches. ■ 5 Figure 4.C-5 220 1 –55 324 Exercises 1. Find the highest point on the part of the graph of 3 3x 2 y x that is shown in the given window. The answers are not all the same. a. c. 2. Find the lowest point on the part of the graph of y x3 3x 2 window. a. c. 0 x 2 3 x 2 that is shown in the given b. 2 x 2 3. An open-top box with a square base is to be constructed from 120 square centimeters of material. What dimensions will produce a box a. of volume 100 cm3? b. with largest possible volume? 4. A 20-inch square piece of metal is to be used to make an open-top box by cutting equal-sized squares from each corner and folding up the sides (as in Example 2). The length, width, and height of the box are each to be less than 12 inches. What size squares should be cut out to produce a box with a. volume 550 in3? b. largest possible volume? 5. A cylindrical waste container with no top, a diameter of at least 2 feet, and a volume of 25 cubic feet is to be constructed. What should its radius be under the given conditions? a. 65 square feet of material will be used to construct it b. the smallest possible amount of material will be used to construct it (how much material is needed?) 6. If x c 1 2 is the cost of producing x units, then c x 1 x 2 is x the average cost per unit. Suppose the cost of producing x units is given by 2 10,000x c 1 than 300 units can be produced per week. a. If the average cost is 0.13x3 70x $1100 units are being produced? 2 and that no more per unit, how many b. What production level should be used in order to minimize the average cost per unit? What is the minimum average cost? 7. If the cost of material to make the can in Example 3 is 5 cents per square inch for the top and bottom and 3 cents per square inch for the sides, what dimensions should be used to minimize the cost of making the can? [The answer is not the same as in Example 3.] 8. A certain type of fencing comes in rigid 10-foot segments. Four uncut segments are used to fence in a garden on the side of a building, as shown in the figure. What value of x will result in a garden of the largest possible area? What is that area? x 9. A rectangle is to be inscribed in a semicircle of radius 2, as shown in the figure. What is the largest possible area of such a rectangle? Hint: The width of the rectangle is the second coordinate of the point P (Why?), and P is on the top half of the circle x 2 4. 2 y y x2 + y2 = 4 P 2 −2 −x 0 x 2 x 10. Find the point on the graph of y 5 x closest to the point (0, 1) and has positive coordinates. Hint: The distance from the point x, y 1 express y in terms of x. on the graph to (0, 1) is x 0 2 2 1 2 2 1 2 that is y 1 2; 2 325 C H A P T E R 5 Exponential and Logarithmic Functions Doorway to the past The image above is of Pueblo Benito in Chaco Canyon, New Mexico. It was the home of the Anasazi people of the desert southwest for several centuries, and it includes timbers (shown above the doorway) that were used to date the buildings by using carbon-14 dating, which involves an exponential equation. See Exercise 55 Section 5.6. 326 Chapter Outline 5.1 Radicals and Rational Exponents 5.2 Exponential Functions 5.3 Applications of Exponential Functions 5.4 Common and Natural Logarithmic Functions 5.5 Properties and Laws of Logarithms 5.5.A Excursion: Logarithmic Functions to Other Bases 5.6 5.7 Solving Exponential and Logarithmic Equations Exponential, Logarithmic, and Other Models Chapter Review can do calculus Tangents to Exponential Functions Interdependence of Sections 5.1 > 5.2 > 5.3 > 5.4 > 5.5 > 5.6 > 5.7 Section 5.1 contains prerequisite review material for this chapter. If students are familiar enough with the objectives of this section, it may be skipped. Exponential and logarithmic functions are essential for the mathe- matical description of a variety of phenomena in the physical sciences, engineering, and economics. Although a calculator is necessary to evaluate these functions for most values, you will not be able to use your calculator efficiently or interpret its answers unless you understand the properties of these functions. When calculations can readily be done by hand, you will be expected to do them without a calculator. 5.1 Radicals and Rational Exponents Objectives n th Roots • Define and apply rational and irrational exponents • Simplify expressions containing radicals or rational exponents NOTE All constants, variables, and solutions in this chapter are real numbers. c 0, Recall that when the equation in a similar fashion as solutions of the equation the square root of c is the nonnegative solution of Cube roots, fourth roots, and higher roots are defined x n c. x 2 c. This equation can be solved graphically by finding the x-coordinate of the . (Review finding intersection points of the graphs of y ax n solutions graphically in Section 2.1 and the shape of the graph of in Section 4.3, if needed.) y x n y c and Depending on whether n is even or odd and whether c is positive or negative, may have two, one, or no solutions, as shown in the following figures. x n c 327 328 Chapter 5 Exponential and Logarithmic Functions Solutions of xn c n odd Exactly one solution for any c y y = xn c y = c x c 77 0 One positive and one negative solution y y = xn c y = c x n even c 0 One solution x 0 y y = xn x y = 0 c 66 0 No solution y y = xn c x y = c Figure 5.1-1 Figure 5.1-2 Figure 5.1-3 Figure 5.1-4 The figures illustrate the following definition of nth roots. nth Roots Let c be a real number and n a positive integer. The nth root of c is denoted by either of the symbols 1 n2c or c n and is defined to be • the solution of • the nonnegative solution of when n is odd; or x n c x n c when n is even and c 0. Examples of nth roots are shown below. 1 23 8 81 81 3 2 because 2 is the solution of x 3 8. 8 1 1 4 3 because 3 is the nonnegative solution of x 4 81. 2 24 1 2 Expressions involving nth roots can often be simplified or written in a variety of ways by using a basic fact of exponents. d or equivalently, cd 2n n c c˛2n 1 n ˛d 2n cd 1 n 1 1 2 Example 1 Operations on nth Roots Simplify each expression. a. c. 28˛ 212 23 8x6 ˛y4 b. d. 212 275 5 2c 5 2c A B ˛A , where c 7 0 B Section 5.1 Radicals and Rational Exponents 329 Solution a. b. c. d. 28 212 28 12 296 216 6 216 26 426 212 275 24 3 225 3 24˛23 225˛23 223 523 323 23 8x 6y4 23 8 23 x3 x3 23 y3 y 2x 2 y˛23 y 5 2c 52 2 25 c 5 2c 2c A B ˛A B When using a calculator, exponent notation for nth roots is usually preferred over radical notation. B A ■ CAUTION Example 2 Evaluating nth Roots When using exponent notation to evaluate nth roots with a calculator, be sure to use parentheses when raising to the fractional power. Example: To enter press 9^(1/3). 23 9 , Use a calculator to approximate each expression to the nearest tenthousandth. 1 5 1 11 a. 40˛ b. 225 ˛ Solution a. Because 1 5 0.2, the expressions 1 40˛ 5 and 0.2 40˛ are equivalent, as shown at right. 5 2.0913 1 40˛ b. The fraction 1 11 repeating decimal 1 11 fraction is equivalent to the 0.090909 p . The Figure 5.1-5 is not equivalent to this decimal if it is rounded off, as Figure 5.1-6 shown at left. Therefore, it is better to leave the exponent in fractional form. 225 1 11 ˛ 1.6362 ■ Rational Exponents Rational exponents of the form 1 n are called nth roots. Rational exponents Rational exponents of the form r rs m n s, 2 ˛ c˛ 1 it is reasonable to can also be of the form m n , such as 3 2. 4˛ c˛ can be defined in such a way that the laws of exponents, such as are still valid. For example, because 1 2 2 ˛ say that 3 2 3 4˛ 4 2R˛3, Q 1 2 ˛ These
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expressions are equivalent. 2 264 8 3 8 2 64 2 24 3 4˛ 1 4. 4 2 1 3 2 4˛ 3 2 4˛ This illustrates the definition of rational exponents. 330 Chapter 5 Exponential and Logarithmic Functions Definition of Rational Exponents Let c be a positive real number and let be a rational number t k with positive denominator. t c k is defined to be the number (c t) ct k 2k t. c t 2k c BB AA In radical notation, 1 k (c 1 k)t 3.78 13 can be expressed as Every terminating decimal is a rational number; therefore, expressions 378 100. such as Although the definition of rational exponents requires c to be positive, it remains valid when c is negative, provided that the exponent is in lowest terms with an odd denominator, such as In Exercise 89, you will explore why these restrictions are necessary when c is negative. 8 13 2 3. 2 1 CAUTION 1 2 1 2 3 23 8 8 1 8 2 23 64 4, 2 3 on some calculators may produce either an error Although entering message or a complex number. If this occurs, you can get the correct answer by entering one of the equivalent expressions below. 3 or and 4 is a real number Laws of Exponents c You have seen that the law of exponents is valid for rational exponents. In fact, all of the laws of exponents are valid for rational exponents. 2 1 rs s cr Laws of Exponents Let c and d be nonnegative real numbers and let r and s be rational numbers. Then 1. 2. crcs crs cr cs crs (c 0) 3. (cr)s c rs 4. 5. 6. (cd)r cr dr r cr d r c db a (d 0) cr 1 c r (c 0) and d 1 , If c 1 cr cs cr dr • • r s. if and only if if and only if c d. Example 3 Simplifying Expressions with Rational Exponents Write the expression 3 3 4 s 8r 2 3 2 1 using only positive exponents. Section 5.1 Radicals and Rational Exponents 331 Solution 3 4 s 8r 12 23 82 1 23 64 1 4r s2 1 2 2 2 cd r crdr 1 definition and 2 simplify s crs cr 1 2 simplify and c r 1 cr The expression 1 2 4r s 2 can also be written as 42r s 2 if it is more convenient. ■ Example 4 Simplifying Expressions with Rational Exponents Simplify the expression Solution ab ac b c a 1 c r c s c rs 2 Example 5 Simplifying Expressions with Rational Exponents Simplify the expression 5 2y4 x 7 4 xy 2. 2 2 1 1 Solution 14 4 2 7 4 xy 2 2 1 y 21 y 21 7 2 5 2 2 5 2 2y 4 x x 21 1 2 y 4 x x 21 cd c r 2 1 commutative c r c s c rs simplify ■ ■ Example 6 Simplifying Expressions with Rational Exponents Let k be a positive rational number. Write the expression without radicals, using only positive exponents. 210 c5k 2 k 1 2 2 c 1 Solution 102c5k 3 k c 1 2 1 10 c5k 5k 10 ˛ definition s c rs c r 1 simplify and c r c s c rs simplify s c rs c r 1 2 ■ 332 Chapter 5 Exponential and Logarithmic Functions Rationalizing Denominators and Numerators Transforming fractions with radicals in the denominator to equivalent fractions with no radicals in the denominator is called rationalizing the denominator. Before the common use of calculators, fractions with rational denominators were preferred because they were easier to calculate or estimate. With calculators today there is no computational advantage to rationalizing denominators. However, the skill of rationalizing numerators or denominators is useful in calculus. Example 7 Rationalizing the Denominator Rationalize the denominator of each fraction. a. 7 25 Solution b. 2 3 26 a. Multiply the fraction by 1 using a suitable radical fraction. 7 25 7 25 1 7 25 25 25 a b 725 5 a2 b2 to determine b. Use the multiplication pattern a suitable radical fraction equivalent to 1. a b 1 21 2 2 3 26 3 26 3 26 3 26 B 3 26 1 2 3 26 2 3 26 2 A 3 26 B 1 6 226 9 6 6 226 3 A 2 ■ NOTE When rationalizing a denominator or numerator which contains a radical expression, use a suitable radical fraction, equal to one, that contains the conjugate of the expression. Example 8 Rationalizing the Numerator Assume h 0. Rationalize the numerator of 2x h 2x h . Section 5.1 Radicals and Rational Exponents 333 Solution Multiply the fraction by 1 using a suitable radical fraction. 2x h 2x h 2x h 2x 2x h 2x 1 A 2x h 2x h 2x h 2x h 2x h 2 2x B 2x h 2x h x h x 2x h 2x h 2x h 2x h h A A B A A 1 2x h 2x B 2 B B Irrational Exponents ■ at is defined when t is an The example (not proof) below illustrates how irrational number. 1022, the exponent could be replaced with the equivalent nonEach of the decimal approximations given below is a more accurate approximation than the preceding 1.414213562 p . To compute terminating decimal of one. 22 1.4, 1.41, 1.414, 1.4142, 1.41421, p . We can raise 10 to each of these rational numbers. 10 10 10 10 10 10 1.4 25.1189 1.41 25.7040 1.414 25.9418 1.4142 25.9537 1.41421 25.9543 1.414213 25.9545 22, gets closer and closer to a real number whose decimal expansion The pattern suggests that as the exponent r gets closer and closer to 10r begins is defined to be this number. 25.954 p . Similarly, for any 1022 So a 7 0, at is a well-defined positive number for each real exponent t. The fact below shall be assumed. The laws of exponents are valid for all real exponents. 334 Chapter 5 Exponential and Logarithmic Functions 5. 20.0081 6. 20.000169 47. 2x7 x 5 2 x 3 2 48. x A 1 2 y3 2 x0 y7 A B Exercises 5.1 Note: Unless directed otherwise, assume all letters represent positive real numbers. In Exercises 1–15, evaluate each expression without using a calculator. 2. 23 64 3. 24 16 1. 2144 4. 23 27 7. 23 0.008 8. 23 0.125 9. 20.56 10. 2 3 4 2 1 13. 64 1 2 3 2 11. 4 3 27 14. 2 3 1 64b a 12. 1 4 81 15. 3 2 16 In Exercises 16–40, simplify each expression without using a calculator. 16. 2315 18. 20.0812 20. 22. 24 23 0.05 1 2 26 212 17. 23 1216 19. 21. 28 2 2 11 1 23 0.418 23. 28 296 24. 23 18 23 12 25. 23 32 23 16 26. 28. 210 28 25 23 324 23 6 23 2 27. 29. 26 214 263 23 54 23 32 23 4 30. 227 223 31. 425 220 32. 33. 34. 35. 36. 37. 2 23 A 1 23 3 22 A 4 23 B A B A B A 225 4 B A 322 426 A A 3 22 B 5 223 B 325 2 B 2 A 5220 245 2280 B 38. 23 40 223 135 523 320 39. 11 2 2 2 23 2 5 7 2 10 2 2 1 40. 1 32 1 2 1 2 27 94 1 3 2 In Exercises 41–56, simplify each expression. 41. 216a8 b 2 42. 224x6 y 4 43. 2c2 d6 4 24c3 d 45. 4x 2y 29 1 18 2 46. 23 a b 23 44. 10 b 2a 2a14 d 12 4 a b 1 2 2 23 a b 2 5 d c 2 3 c6d3 BA 4 3 B 50 15 9 5b 7b 52. 1 6a 1 2 2ab 2 a2b 3 2 1 2 1 3 2 3b 2 3b 1 2 2 2 4a 2a 1 1 3 5 1 5 2 2 54. A ax 2 1 x B 56 ab 4 bc 1 2 x1 2 3 4 b a ab 1 bx 2 x b 1 49. 51. 53. 55. A 1 1 1 1 A 7a 5a 2a 4a 2 2 2 2 In Exercises 57–66, write each expression without radicals, using only positive exponents. 57. 23 a 2 b2 59. 34 24 a3 61. 25 t 216t5 63. 65. 23 xy 2 3 5 B c c51 1 2 2 3 42 5 6 c 2 A 1 58. 24 a 3 b3 60. 323 a 3b4 62. 2x 23 x 2 24 x 3 64. Q 34 r 14 s 3 7 21 5 R 66 In Exercises 67–72, simplify each expression. 68. 70. 1 2 x x A 3x A 1 3 y 3 2 2x 2x 69. 71. 72 CA BA 2 3 B x y BD In Exercises 73–78, rationalize the denominator and simplify your answer. 73. 76. 3 28 1 23 5 210 74. 77. 2 26 2 2x 2 75. 78. 3 2 212 2x 2x 2c B 67. 1 2 x Section 5.1 Radicals and Rational Exponents 335 In Exercises 79 – 84, factor the given expression. For example, 2 2 (x 2 2)(x 2 1). x x 1 1 1 79. 2 3 x 1 3 6 x 81. x 4x 1 2 3 83. 4 5 81 x 80. 2 5 11x 1 5 30 x 82. 1 3 7x 1 6 10 x 84. 2 3 6x 1 3 9 x In Exercises 85–88, rationalize the numerator and simplify your answer. Assume h 0. 85. 86. 87. 88. 2x h 1 2x 1 h 22x h 3 22x 2x2 1 h 2 1 x h h 2 2x2 x 89. Some restrictions are necessary when defining fractional powers of a negative number. x2 4, a. Explain why the equations x4 4, etc., have no real solutions. Conclude 1 4, c cannot be defined when c 4. is the same as it should be true that 1 6 x6 4, 1 c 2, c that 1 b. Since 3 2 1 3 c c 6, is false when that is, that c 8. 2 , 6 23 c 26 c 2. Show that this 90. a. Suppose r is a solution of the equation xn c Verify that rs is a xn d. and s is a solution of solution of xn cd. b. Explain why part a shows that c 2n cd 2n 2n d. 91. Write laws 3, 4, and 5 of exponents in radical notation in the case when r 1 m and s 1 n. 92. a. Graph f x5 and explain why this x 2 1 function has an inverse function. b. Show algebraically that the inverse function is x 1 5. x g 1 2 93. If n is an odd positive integer, show that xn 1 2 x f has an inverse function and find the rule of the inverse function. Hint: Exercise 92 is the case when n 5. 94. A long pendulum swings more slowly than a short pendulum. The time it takes for a pendulum to complete one full swing, or cycle, is called its period. The relationship between the period T (in seconds) of the pendulum and its length x (in meters) is given by the function T x 1 2 2p A x 9.8 . Find the period for pendulums whose lengths are 0.5 m and 1.0 m. 95. In meteorology, the wind chill C can be 3.712V 5.81 0.25V A calculated by using the formula C 0.0817 where V is the wind speed in miles per hour and t is the air temperature in degrees Fahrenheit. Find the wind chill when the wind speed is 12 miles per hour and the temperature is t 91.4 35°F. B 2 1 91.4, 96. The elevation E in meters above sea level and the boiling point of water, T, in degrees Celsius at that elevation are related by the equation 580 E 1000 Find the approximate boiling point of water at an elevation of 1600 meters. 100 T 100 T 2. 1 2 1 2 97. Accident investigators can usually estimate a motorist’s speed s in miles per hour by examining the length d in feet of the skid marks on the road. The estimate of the speed also depends on the road surface and weather conditions. If f represents the coefficient of friction between rubber and the road surface, then motorist’s speed. The coefficient of friction f between rubber and concrete under wet conditions is 0.4. Estimate, to the nearest mile per hour, a motorist’s speed under these conditions if the skid marks are 200 feet long. gives an estimate of the s 230fd 98. Using a viewing window with 0 x 4 and graph the following functions on the 0 y 2, same screen, x , 4 in order of increasing size and justify In each of the following cases, arrange and your answer by using the graphs. a. x 99. Using a viewing window with 3 x 3 and graph the following functi
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ons on 1.5 y 1.5, the same screen, in order of increasing size and justify your In each of the following cases, arrange x answer by using the graphs. a. c. d, x x and 336 Chapter 5 Exponential and Logarithmic Functions 100. Graph f 2x in the standard viewing x 2 1 window. Then, without doing any more graphing, describe the graphs of these functions. a. 2x 3 Hint: g x 3 f see x ; 1 2 1 2 x g Section 3.4. 2 1 b. c. h k x 1 x 1 2 2 2x 2 2x 3 2 101. Do Exercise 100 with 23 in place of 2 . 5.2 Exponential Functions Objectives • Graph and identify transformations of exponential functions • Use exponential functions to solve application problems Graphs of Exponential Functions For each positive real number a, with base a whose domain is all real numbers and whose rule is Some examples are shown below. there is an exponential function ax. x f 1 2 a 1, f x 1 2 10x g x 1 2 2x h x 1 2 x 1 2b a k f The shape of the graph of an exponential function on the size of a, as shown in the following figures. x 1 2 x 3 2b a x 2 1 ax depends only Graph of f(x) a x a > 1 y 0 << a << 1 y 1 x 1 x • graph is above x-axis • y-intercept is 1 • is increasing f(x) • graph is above x-axis • y-intercept is 1 • is decreasing f(x) • approaches the negative f(x) x-axis as x approaches • approaches the positive f(x) x-axis as x approaches a 0 a 1, or For is a constant function, not exponential. Even roots of negative numbers are not defined in the set of real is not defined for any rational exponent that numbers, so when the function a 6 0 ax , x f 2 1 ax Section 5.2 Exponential Functions 337 has an even number as its denominator. Because within any interval there are infinitely many rational numbers that have an even denominator, a 6 0 f . Therefore, the function is not well-behaved for , so it is not defined for those values. has an infinite number of holes in every interval when a 6 0 ax x 2 1 The following two Graphing Explorations illustrate the effect that the value of a has on the shape of the graph of an exponential function for a 7 1 0 6 a 6 1 and for . Graphing Exploration a. Using a viewing window with 3 x 7 and 2 y 18, graph each function below on the same screen, and observe the behavior of each to the right of the y-axis. 1.3x 2 As the graphs continue to the right, which graph rises least steeply? most steeply? 2x 10x How does the steepness of the graph of of the y-axis seem to be related to the size of the base a? x f 1 2 to the right ax b. Using the graphs of the same three functions in the viewing and 0.5 y 2, 4 x 2 observe the window with behavior to the left of the y-axis. As the graph continues to the left, how does the size of the base a seem to be related to how quickly the graph of falls toward the x-axis? x f 2 1 ax Graphing Exploration Using a viewing window with graph each function below on the same screen, and observe the behavior of each. and 1 y 4, 4 x 4 f x 1 2 0.2x g x 1 2 0.4x h x 2 1 0.6x k 0.8x x 1 2 Notice that the bases of the exponential functions are increasing in size: 0 6 0.2 6 0.4 6 0.6 6 0.8 6 1 As the graphs continue to the right, which graph falls least steeply? most steeply? How does the steepness of the graph of be related to the size of the base a? f x 2 1 ax seem to The graphing explorations above show that the graph of or falls less steeply as the base a gets closer to 1. f x 1 2 ax rises 338 Chapter 5 Exponential and Logarithmic Functions 10 Example 1 Translations f The graph of is shown in Figure 5.2-1. Without graphing, describe the transformation from the graph of f to the graph of each function below. Verify by graphing. x 2 1 2x 6 a. g x 1 2 2x3 b. h x 1 2 2x3 4 Solution a. If f x 2x, then g x 2x3 f x 3 . So the graph of g is the 1 2 2 1 1 graph of f shifted horizontally 3 units to the left, as shown in Figure 5.2-2. x f 2x, x h 2 1 h(x) is the graph of f 1 and vertically 4 units downward, as shown in Figure 5.2-3. shifted horizontally 3 units to the right 2x3 4 x So the graph of x 3 2x 4. then f 2 2 2 1 2 1 b. If ■ 6 9 ax x The graphs of exponential functions of the form increase at an 2 1 2x x f in Figure 5.2-3. explosive rate. To see this, consider the graph of 2 1 x 50 If the x-axis were extended to the right, then would be at the right 2x f x edge of the page. At this point, the graph of units high. The scale of the y-axis in Figure 5.2-3 is about 12 units per inch, or 144 units per foot, or 760,320 units per mile. Therefore, the height of the graph at x 50 250 is is f 1 2 250 760,320 1,480,823,741 miles, which would put that part of the graph well beyond the planet Saturn! Since most quantities that grow exponentially do not change as dramatically as the graph of exponential functions that model real-life growth or decay are usually modified by the insertion of appropriate constants. These functions are generally of the form 2x, x f 1 2 such as the functions shown below. Pakx.20.45x g 2 x 1 2 3.5 10 1 0.03x h x 1 2 2 Their graphs have the same shape as the graph of or fall more or less steeply, depending on the constants P, k, and a. x f 1 2 but may rise 1.0762x 2 6 21 1 ax, 6 6 3 2 Figure 5.2-1 10 2 Figure 5.2-2 7 5 Figure 5.2-3 5.1 4.7 4.6 1.1 Figure 5.2-4 Example 2 Horizontal Stretches f The graph of is shown in Figure 5.2-4. Without graphing, describe the transformation from the graph of f to the graph of each function below. Verify by graphing. x 1 2 3x g x 2 1 30.2x h x 1 2 30.8x k x 1 2 3 x p 0.4x 3 x 1 2 Section 5.2 Exponential Functions 339 30.8x Solution The graphs of g x 1 2 izontally by a factor of 30.2x 1 0.2 h and 5 x 2 1 and 30.8x 1 0.8 are the graph of f stretched hor- 1.25 , respectively. The graph of x 3 3 0.4x k 1 p x 2 x 1 2 is the graph of f reflected across the y-axis. The graph of 2.5 is the graph of f stretched horizontally by a factor of 1 0.4 and reflected across the y-axis. The graphs are identified in Figure 5.2.5. ■ −0.4x 3 −x 3 3x 5.1 30.2x 4.7 −1.1 Figure 5.2-5 10 Example 3 Vertical Stretches 5 5 p The graph of is shown in Figure 5.2-6. Without graphing, describe the transformation from the graph of p to the graph of each function below. Verify by graphing. x 2 1 0.4x 3 4 3 0.4x .4x Solution 4 3 0.4x 1 q x The graph of cally by a factor of 4. The graph of p x-axis. The graphs are identified in Figure 5.2-7. stretched vertix is the graph of 2 stretched vertically by a factor of 2 and reflected across the p is the graph of x 2 2 3 0.4x 0.4x .4x 3 3 ■ Exponential Growth and Decay In this section, you will see that exponential functions are useful for modeling situations in which a quantity increases or decreases by a fixed factor. In Section 5.3 you will learn how to construct these types of functions. −4.7 5 10 Figure 5.2-6 −0.4x 4 . 3 3 −0.4x 10 −5 (−2)3 −0.4x −10 Figure 5.2-7 Example 4 Finance If you invest $5000 in a stock that is increasing in value at the rate of 3% per year, then the value of your stock is given by the function f where x is measured in years. 5000 1.03 x, x 1 2 1 2 a. Assuming that the value of your stock continues growing at this rate, how much will your investment be worth in 4 years? b. When will your investment be worth $8000? Solution a. Letting x 4, 5000 In 4 years your stock is worth about $5627.54. 4 5627.54 . 1.03 4 f 2 1 2 1 b. Find the value of x for which equation 5000 1.03 1 2 x 8000. 8000. f x 1 2 In other words, solve the 340 20,000 0 7,000 30 0 5 1.25 0 0.25 Chapter 5 Exponential and Logarithmic Functions Figure 5.2-8 20 120 The point of intersection of the graphs of y 8000 is approximately (15.901, 8000). f x 1 2 5000 1.03 1 x 2 and Therefore, the stock will be worth $8000 in about 16 years. ■ Example 5 Population Growth Based on data from the past 50 years, the world population, in billions, g can be approximated by the function corresponds to 1950. x 0 2.5 where 1.0185 x, x 2 2 1 1 a. Estimate the world population in 2015. b. In what year will the population be double what it is in 2015? Solution a. Since x 0 2015 corresponds to x 65. x 1 Find g(65). 2.5 1.0185 1 65 g 1 2 65 8.23 2 corresponds to 1950, to 1951, and so on, the year The world population in 2015 will be about 8.23 billion people. b. Twice the population in 2015 is 2 16.46; number x such that g x 1 2 16.46 8.23 1 that is, solve 2 billion. Find the 2.5 1.0185 x 16.46. 1 2 x 2.5 A graphical intersection finder shows that the approximate coordinates of the point of intersection of the graphs of y 16.46 g 102.81, or 103 when rounded to the nearest year, corresponds to the year 2053. Notice that it takes only 38 years for the world population to double. are (102.81, 16.46). The x-coordinate 1.0185 and 2 1 1 2 x ■ Figure 5.2-9 Example 6 Radioactive Decay The amount from one kilogram of plutonium M years can be approximated by the function amount of plutonium remaining after 10,000 years. 239Pu 1 x 2 1 that remains after x Estimate the 0.99997x. 2 Solution Because M is an exponential function with a base smaller than 1 but very close to 1, its graph falls very slowly from left to right. The fact that the graph falls so slowly as x gets large means that even after an extremely long time, a substantial amount of plutonium will remain. 12,000 Figure 5.2-10 x 10,000, M Therefore, almost three-fourths of the origWhen inal plutonium remains after 10,000 years! This is the reason that nuclear waste disposal is such a serious concern. x 1 2 0.74. ■ 3x ex 14 4 2 Figure 5.2-11 2x 4 800 0 −200 100 Figure 5.2-12 Section 5.2 Exponential Functions 341 The Number e and the Natural Exponential Function There is an irrational number, denoted e, that arises naturally in a variety of phenomena and plays a central role in the mathematical description of the physical universe. Its decimal expansion begins as shown below. e 2.718281828459045p Most calculators have an x exponential function the display will show the first part of the decimal expansion of e. key that can be used to evaluate the natural e1 When you evaluate using a calculator, f 1 2 ex ex. Figure 5.2-11 shows that the
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graph of and graphs of y 2x y 3x, and less steeply than the graph of y 2x x f y 3x. has the same shape as the but it climbs more steeply than the graph of 2 1 ex Example 7 Population Growth If the population of the United States continues to grow as it has since 1980, then the approximate population, in millions, of the United States in year t, where corresponds to the year 1980, will be given by the function t 0 227e0.0093t. P t 1 2 a. Estimate the population in 2015. b. When will the population reach half a billion? Solution a. The year 2015 corresponds to t 35. 227e0.0093 Find P(35). 2 314.3 35 1 35 P 1 2 Therefore, the population in 2015 will be approximately 314.3 million people. b. Half a billion is 500 million. Find the value of t for which P A graphical intersection finder shows that the approximate coordinates of the point of intersection of the graphs of are approximately (85, 500). A t-value of P 85 corresponds to the year 2065. Therefore, the population will reach half a billion approximately by the year 2065. and y 500 227e0.0093t t t 1 2 2 1 500. ■ Other Exponential Functions In most real-world applications, populations cannot grow infinitely large. The population growth models shown previously do not take into account factors that may limit population growth in the future. Example 8 illustrates a function, called a logistic model, which is designed to model situations that have limited future growth due to a fixed area, food supply, or other factors. 342 25,000 0 0 Chapter 5 Exponential and Logarithmic Functions Example 8 Logistic Model The population of fish in a certain lake at time t months is given by the t 0. There is an upper limit on the fish p function , where t 20,000 t 1 24e 4 1 2 100 population due to the oxygen supply, available food, etc. Graph the function, and find the upper limit on the fish population. Figure 5.2-13 Solution p y 20,000 The graph of is a horizontal asymptote of the graph. If so, the upper limit on the fish population is 20,000. at the left suggests that the horizontal line t 1 2 You can verify this by rewriting the rule of p as shown below. p t 2 1 20,000 t 1 24e 4 20,000 1 24 e t 4 As t increases, t 4 increases and grows very large. As grows very large, t e 4 t e 4 24 t e 4 gets very close to 0. As 24 t e 4 gets closer and closer to 0, p gets closer t 2 1 and closer to 20,000 1 0 , or 20,000. Because e t 4 is positive and 24 t e 4 never quite is always slightly larger than 1 and reaches 0, the denominator of is always less than 20,000. p t 2 1 p 1 t 2 ■ When a cable, such as a power line, is suspended between towers of equal height, it forms a curve called a catenary, which is the graph of a function of the form shown below for suitable constants A and k. f x A ekx e kx 1 The Gateway Arch in St. Louis, shown in Figure 5.2-14, has the shape of an inverted catenary, which was chosen because it evenly distributes the internal structural forces. 1 2 2 Graphing Exploration 10 y 80, Using the viewing window with graph each function below on the same screen, and observe their behavior. 10 5 x 5 e0.4x e e2x e e3x e 10 10 and Y3 Y2 0.4x Y1 2x 3x 1 2 1 2 2 1 How does the coefficient of x affect the shape of the graph? Predict the shape of the graph of answer by graphing. y Y1 80. Confirm your Figure 5.2-14 Section 5.2 Exponential Functions 343 In Exercises 22–29, find a viewing window (or windows) that shows a complete graph of the function. 22. x k 1 2 x e 24. f x 1 2 x ex e 2 26. g x 2 1 2x x 23. f x 1 2 x 2 e 25. h x 2 1 x ex e 2 27. x k 1 2 2 ex e x 28. x f 1 2 5 1 e x 29. x g 1 2 10 1 9e x 2 In Exercises 30–34, determine whether the function is even, odd, or neither. (See Excursion 3.4A.) 30. x f 1 2 10x x ex e 2 x 2 e 32. 34. f f x x 1 1 2 2 31. g x 1 2 2x x 33. f x 1 2 x ex e 2 35. Use the Big-Little concept (see Section 4.4) to x is approximately equal to ex ex e explain why when x is large. In Exercises 36–39, find the average rate of change of the function. (See Section 3.7). 36. f x 1 2 37. g 38. h x x 2 2 1 1 39. f x 1 2 x 2x 2 1 3x 2x as x goes from 1 to 3 as x goes from 1 to 1 x 2 5 as x goes from 1 to 0 ex e x as x goes from 3 to 1 Exercises 5.2 In Exercises 1–6, list the transformations needed to transform the graph of into the graph of the given function. (Section 3.4 may be helpful.) h(x) 2x 1. 3. 5 2x 5 3 2x 1 2 2x2 5 2. 4. 6 2x 1 2x1 2 5 2x1 1 2 7 In Exercises 7–13, list the transformations needed to h(x) 3x transform the graph of into the graph of the given function. (Section 3.4 may be helpful.) 7. x f 1 2 3x 4 9. x k 1 2 1 4 1 3x 2 11. f 13. g x x 1 1 2 2 32x 4 1 0.15x 3 2 8. x g 1 2 x 3 10. g x 1 2 30.4x 12. x f 1 2 8 5 3x 1 2 In Exercises 14–19, sketch a complete graph of the function. 14. 16. 18. x 4 23x 25x 15. 17. 19. x f x 1 2 5 2b a 3 x 2 2x5 g g x x 1 1 2 2 In Exercises 20–21, match the functions to the graphs. Assume c 77 1. a 77 1 and 20. 21. ax ax 3 ax5 cx 3cx cx5 3cx In Exercises 40–43, find the difference quotient of the function. (See Section 3.7.) A B C D 40. 42. f f x x 1 1 2 2 10x 2x 2 x 41. g 43. f x x 1 1 2 2 5x 2 ex e x In Exercises 44–49, list all asymptotes of the graph of the function and the approximate coordinates of each local extremum. (See Section 4.3.) 44. 46. 48 2x 1 x 2 2 e x 2 e 2 45. 47. g x 1 2 k x 1 2 49 2x 2 6x2 xe x 2 20 344 Chapter 5 Exponential and Logarithmic Functions 50. If you deposit at $750 2.2% interest, compounded annually and paid from the day of deposit to the day of withdrawal, your balance at time t is given t. by after 2 years? after 3 years and 9 months? How much will you have 750 1.022 B t 2 1 1 2 51. The population of a colony of fruit flies t days 100 3 from now is given by the function a. What will the population be in 15 days? in 25 p t 1 2 t 10. days? b. When will the population reach 2500? 52. A certain type of bacteria grows according to the 2 x 5000e0.4055x, f function 1 measured in hours. a. What will the population be in 8 hours? b. When will the population reach 1 million? where the time x is 53. According to data from the National Center for Health Statistics, the life expectancy at birth for a person born in year x is approximated by the function below. D x 1 2 79.257 1 9.7135 1024 e 1900 x 2050 1 2 0.0304x a. What is the life expectancy of someone born in 1980? in 2000? b. In what year was life expectancy at birth 60 years? 54. The number of subscribers, in millions, to basic cable TV can be approximated by the function g x 1 2 76.7 1 16 0.8444x x 0 where corresponds to 1970. (Source: The Cable TV Financial Datebook and The Pay TV Newsletter) a. Estimate the number of subscribers in 1995 and in 2005. b. When does the number of subscribers reach 70 million? c. According to this model, will the number of subscribers ever reach 90 million? 55. The estimated number of units that will be sold t N 100,000e by a certain company t months from now is given by a. What are the current sales ? sales be in 2 months? in 6 months? What will t 0 0.09t. 1 2 1 2 b. From examining the graph, do you think that sales will ever start to increase again? Explain. 56. a. The function 2 1 t g 0.0479t 1 e percentage of the population (expressed as a decimal) that has seen a new TV show t weeks after it goes on the air. What percentage of people have seen the show after 24 weeks? gives the b. Approximately when will 90% of the people have seen it? 57. a. The beaver population near a certain lake in year t is approximated by the function p t 2 1 now 2000 1 199e when t 0 1 2 0.5544t. What is the population and what will it be in 5 years? b. Approximately when will there be 1000 beavers? f 58. Critical Thinking Look back at Section 4.3, where the basic properties of graphs of polynomial functions were discussed. Then review the basic discussed in properties of the graph of this section. Using these various properties, give an argument to show that for any fixed positive number a, where polynomial function such that words, no exponential function is a polynomial function. a 1, it is not possible to find a cnxn p c1x c0 g x for all numbers x. In other ax g ax x x 2 1 1 1 2 2 59. Critical Thinking For each positive integer n, let fn be the polynomial function below. 1 x x2 2! x3 3! x4 4! fn1 x 2 p xn n! 5 y 55, a. Using the viewing window with ex 4 x 4 f41 x and and graph the same screen. Do the graphs appear to coincide? g x 1 2 2 on 2 2 x f41 , then by that of b. Replace the graph of x f51 2 x , f7 1 x f6 1 , and so on until you find a by 2 fn1 polynomial whose graph appears to x coincide with the graph of viewing window. Use the trace feature to move from graph to graph at the same value of x to see how accurate this approximation is. in this ex x g 2 2 1 c. Change the viewing window so that and 10 y 400. 6 x 6 Is the polynomial you found in part b a good x g approximation for in this viewing 2 window? If not, what polynomial is a good approximation? 1 Section 5.3 Applications of Exponential Functions 345 5.3 Applications of Exponential Functions Objective • Create and use exponential models for a variety of exponential growth or decay application problems In Section 5.2, you used several exponential functions that modeled exponential growth and decay. In this section you will learn how to construct such exponential models in a variety of real-life situations. Compound Interest When interest is paid on a balance that includes interest accumulated from the previous time periods it is called compound interest. Example 1 Compounding Annually If you invest in the account at the end of 10 years? $6000 at 8% interest, compounded annually, how much is Solution After one year the account balance is Principal Interest 6000 0.08 6000 1 6000 1 2 1 0.08 2 6000 1.08 . 2 1 The account balance has changed by a factor of 1.08. If this amount is left in the account, the balance will again change by a factor of 1.08 after the second year. 6000 1.08 1.08 2 2 4 1 1 3 , or 6000 1.08 1 2 2 Because the balance will change by a factor
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of 1.08 every year, the balance in the account at the end of year x is given by B x 1 2 6000 1.08 1 x. 2 Therefore, the balance (to the nearest penny) in the account after 10 years is 10 B 1 2 6000 1.08 2 1 10 $12,953.55. ■ The pattern illustrated in Example 1 can be generalized as shown below. Compound Interest If P dollars is invested at interest rate r (expressed as a decimal) per time period t, then A is the amount after t periods. A P(1 r)t Notice that in Example 1, or years, is t 10. P 6000, r 0.08, and the number of periods, 346 Chapter 5 Exponential and Logarithmic Functions Example 2 Different Compounding Periods Determine the amount that a annual interest rate of 6.4% $4000 investment over three years at an is worth for each compounding period. a. annually b. quarterly c. monthly d. daily Solution a. Use P 4000, r 0.064, and t 3 in the compound interest formula. 1 b. Quarterly compounding means that interest is compounded every 2 A 4000 1.064 3 $4818.20 one-fourth of a year or 4 times a year. Therefore, • the interest rate per period is r 0.064 , and • the number of periods in 3 years is 4839.32 A 4000 1 0.064 4 b a c. Monthly compounding means that interest is compounded every 1 12 of a year or 12 times a year. Therefore, • the interest rate per period is • the number of periods in 3 years is A 4000 1 0.064 12 b a and , r 0.064 12 t 12 12 3 3 . 2 1 2 $4844.21 1 d. Daily compounding means that interest is compounded every 1 365 of a year, or 365 times a year. Therefore, • the interest rate per period is • the number of periods in 3 years is A 4000 1 0.064 365 b a and , r 0.064 365 t 365 365 3 3 . 2 1 2 $4846.60 1 ■ Notice in Example 2 that the more often interest is compounded, the larger the final amount will be. Example 3 shows you how to write and solve an exponential equation to determine how long it will take for an investment to be worth a given amount. Section 5.3 Applications of Exponential Functions 347 Example 3 Solving for the Time Period $5000 7% If the investment be worth is invested at $6800? annual interest, compounded daily, when will Solution Use the compound interest formula with the final amount P 5000. Because the interest is compounded every 1 365 A 6800 and of a year, the interest rate per period is . r 0.07 365 A P 1 6800 5000 t 1 r 2 1 0.07 365 b a t 8,000 The point of intersection of the graphs of 5000 y1 t 1 0.07 365 b a and 6800 y2 be worth is approximately (1603.5, 6800). Therefore, the investment will $6800 after about 1603 days, or about 4.4 years. ■ 0 0 3,000 Figure 5.3-1 Continuous Compounding and the Number e As you have seen in previous examples, the more often interest is compounded, the larger the final amount will be. However, there is a limit that is reached, as you will see in Example 4. Example 4 The Number e Suppose you invest for one year at 100% annual interest, compounded n times per year. Find the maximum value of the investment in one year. $1 Solution Use the compound interest formula. The annual interest rate is 1.00, so the interest rate per period is 1 n, and the number of periods is n nb a n 1 1 nb a 1 Observe what happens to the final amount as n grows larger and larger. 348 Chapter 5 Exponential and Logarithmic Functions Compounding period n Annually Semiannually Quarterly Monthly Daily Hourly 1 2 4 12 365 8760 n 1 1 nb a 1 1 1 a 1b 2 2 1 1 a 2b 2.25 4 1 1 a 4b 2.4414 12 1 1 12b a 2.6130 365 1 1 a 365b 2.71457 8760 1 1 8760b a 2.718127 Every minute 525,600 Every second 31,536,000 1 a 1 525,600b 525,600 2.7182792 1 a 1 31,536,000b 31,536,000 2.7182825 The maximum amount of the mately no matter how large n is. $2.72, $1 investment after one year is approxi- ■ When the number of compounding periods increases without bound, the process is called continuous compounding. Note that the last entry in the preceding table is the same as the number e to five decimal places. Example 4 is the case when . A similar result occurs , in the general case and leads to the following formula. r 100% P 1 t 1 , and Continuous Compounding If P dollars is invested at an annual interest rate of r, compounded continuously, then A is the amount after t years. A Pert Example 5 Continuous Compounding 5% If you invest much is in the account at the end of 3 years? $4000 at annual interest compounded continuously, how Section 5.3 Applications of Exponential Functions 349 Solution Use the continuous compounding formula with t 3. P 4000, r 0.05, and A Pert 4000e0.05 4647.34 3 1 2 After 3 years the investment will be worth $4647.34. ■ Exponential Growth Compound interest is one type of exponential growth; other exponential growth functions are very similar to the compound interest formula, as you will see in Example 6. Example 6 Population Growth The world population in 1950 was about 2.5 billion people and has been increasing at approximately 1.85% per year. Write the function that gives the world population in year x, where corresponds to 1950. x 0 Solution If the population increases each year by then it increases each year by a factor of 1.0185. Notice that this pattern of population growth is the same as that of compound interest. 1.85%, Year 1950 1951 1952 1953 Population (in billions) 2.5 2.5(1.0185) 1.0185 2.5 1 2 2 1.0185 2.5 1 3 2 … . .. 1950 x 1.0185 2.5 1 x 2 So, the function that gives the world population, in billions, in year x, f where corresponds to 1950 is x 0 2.5 1.0185 x. x 1 2 1 2 ■ Exponential Growth Exponential growth can be described by a function of the form f(x) Pax, f(x) x 0 is the quantity at time x, P is the initial quantity and where when changes when x increases by 1. If the quantity at rate r per time period, then is the factor by which the quantity a 1 r a 77 1 f(x) is growing and f(x) Pax P(1 r)x. 350 Chapter 5 Exponential and Logarithmic Functions Example 7 Bacteria Growth At the beginning of an experiment, a culture contains 1000 bacteria. Five hours later, there are 7600 bacteria. Assuming that the bacteria grow exponentially, how many bacteria will there be after 24 hours? Solution Use the exponential growth formula with 1000ax x f 1 2 P 1000. Because there are 7600 bacteria after 5 hours, 1000a5 7600. lation grows. and Solve for a to find the factor by which the bacteria popu- 5 1 2 f 7600 1000a5 7600 a5 7.6 a 25 7.6 a 7.6 1 5 7.60.2 7.60.2. 1000 7.60.2x f x Therefore, the function’s growth factor is Find f 24 , 2 1 1 2 the bacteria population after 24 hours. 1000 7.60.2 24 2 16,900,721 24 f 1 1 2 After 24 hours, the bacteria population will be approximately 16,900,721. ■ Exponential Decay Sometimes a quantity decreases by a fixed factor as time goes on, as shown in Example 8. Example 8 Filtering 30% When tap water is filtered through a layer of charcoal and other purifyof the chemical impurities in the water are removed. If ing agents, the water is filtered through a second purifying layer, of the remaining impurities in the water are removed. How many layers are needed to ensure that of the impurities are removed from the water? 95% 30% Solution of the With the first layer, of the impuimpurities remain. With the second layer, rities remain. The table below shows this pattern of exponential decay. of the impurities are removed and of the 70% 70% 30% 70% 0.5 0 0.1 12 Figure 5.3-2 Exponential Decay Section 5.3 Applications of Exponential Functions 351 Layer Impurities remaining 1 2 3 o x 0.7 70% 0.72 0.49, or 49% 0.73 0.343, or 34.3% o 0.7x Therefore, the percentage of impurities remaining in the water after it passes through x layers of purifying material is given by the function 0.7x. f x 1 2 of the impurities are removed, 5% remain. So, find the value 95% When of x for which 0.05. f x 1 2 The point of intersection of the graphs of is approximately (8.4, 0.05). Because you cannot have a fractional part of a filter, 9 layers are needed to ensure that 95% of the impurities are removed from the water. and y2 y1 0.7x 0.05 ■ Example 8 illustrates exponential decay. Notice that the impurities were and that the amount of impurities remainremoved at a rate of 1 0.3 0.7. ing in the water was changing by a factor of This pattern is true in general for exponential decay. 30% 0.3 Exponential decay can be described by a function of the form f(x) Pax, x 0 where f(x) is the quantity at time x, P is the initial quantity when changes when x increases by 1. If the quantity decaying at rate r per time period then is the factor by which the quantity 0 66 a 66 1 a 1 r is and and f(x) f(x) Pax P(1 r)x. The half-life of a radioactive substance is the time it takes a given quantity of the substance to decay to one-half of its original mass. The half-life depends only on the substance, not on the size of the sample. Because radioactive substances decay exponentially, their decay can be described where x is measured in the same time by a function of the form units as the half-life. The constant a can be determined from the half-life of the substance. Pax, x f 2 1 352 Chapter 5 Exponential and Logarithmic Functions For example, suppose that the half-life of a substance is 25 years. Then after 25 years, the initial amount P decays to 0.5P, or 0.5P. 25 f 1 2 f 2 0.5P 25 1 Pa25 0.5P a25 0.5 a 0.5 1 25 Radioactive Decay The function for this radioactive decay is x f 1 2 Pax P 1 1 25 0.5 x P 1 2 0.5 x 25. 2 The amount of a radioactive substance that remains is given by the function f(x) P(0.5) x h, where P is the initial amount of the substance, corresponds to the time when the radioactive decay began, and h is the half-life of the substance. x 0 Example 9 Radioactive Decay When a living organism dies, its carbon-14 decays exponentially. An archeologist determines that the skeleton of a mastodon has lost of its carbon-14. The half-life of carbon-14 is 5730 years. Estimate how long ago the mastodon died. 64% Solution Use the exponential decay formula for radioactive decay, with h 5730. 1 Because the mastodon has lost 14, or 0.36P, remains. So, find the value of x for which 1
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of its carbon-14, 2 64% 2 f x P 0.5 x 5730 0.36P P 1 0.36 0.5 0.5 x 5730 x 5730 x 5730 and 0.5 The point of intersection of the graphs of y1 is approximately (8445.6, 0.36) as shown in Figure 5.3-3. Therefore, the mastodon died about 8445.6 years ago. 0.36 y2 ■ 36% f x 1 2 of its carbon 0.36P. 2 1.5 0 0.5 15,000 Figure 5.3-3 Section 5.3 Applications of Exponential Functions 353 Exercises 5.3 1. If $1,000 is invested at 8% interest, find the value of the investment after 5 years for each compounding period. a. annually c. monthly b. quarterly d. weekly 17. A $8900 years 18. A $9500 years at 11.3% compounded monthly for 3 at 9.4% compounded monthly for 6 2. If $2500 is invested at 11.5% interest, what is the value of the investment after 10 years for each compounding period? a. annually b. monthly c. daily In Exercises 3–12, determine how much money will be in a savings account with an initial deposit of $500 and the interest rate indicated below. 3. 2% compounded annually for 8 years 4. 2% compounded annually for 10 years 5. 2% compounded quarterly for 10 years 6. 2.3% compounded monthly for 9 years 7. 2.9% compounded daily for 8.5 years 8. compounded weekly for 7 years and 7 3.5% months 9. 3% compounded continuously for 4 years 10. 3.5% compounded continuously for 10 years 11. 2.45% compounded continuously for 6.2 years 12. 3.25% compounded continuously for 11.6 years A sum of money P that can be deposited today to yield some larger amount A in the future is called the present value of A. In Exercises 13–18, find the present value of the given amount A. Hint: Substitute the given amount A, the interest rate r per period, and the number of periods t into the compound interest formula, and solve for P. 13. A $5000 years 14. A $3500 years 15. A $4800 years 16. A $7400 years at 6% compounded annually for 7 at 5.5% compounded annually for 4 at 7.2% compounded quarterly for 5 at 5.9% compounded quarterly for 8 In Exercises 19–26, use the compound interest formula. Given three of the quantities, A, P, r, and t, find the remaining one. 19. A typical credit card company charges 18% annual interest, compounded monthly, on the unpaid balance. If your current balance is and you do not make any payments for 6 months, how much will you owe? $520 20. When his first child was born, a father put $3000 in a savings account that pays compounded quarterly. How much will be in the account on the child’s 18th birthday? annual interest, 4% 13.2% $10,000 to invest for 2 years. Fund A 21. You have pays B pays and Fund C pays monthly. Which fund will return the most money? interest, compounded annually. Fund interest, compounded quarterly, interest, compounded 12.7% 12.6% 22. If you invest $7400 more money: an interest rate of quarterly or an interest rate of continuously? for 5 years, which will return 5% compounded 4.8% compounded 23. If you borrow 14% $1200 at monthly, and pay off the loan (principle and interest) at the end of 2 years, how much interest will you have paid? interest, compounded 24. A developer borrows $150,000 at 6.5% interest, compounded quarterly, and agrees to pay off the loan in 4 years. How much interest will she owe? 25. A manufacturer has settled a lawsuit out of court by agreeing to pay 1.5 million dollars 4 years from now. How much should the company put in an account paying $1.5 compounded monthly, in order to have million in 4 years? Hint: See Exercises 13–18. annual interest, 6.4% 26. Ellen wants to have $30,000 available in 5 years for a down payment on a house. She has inherited $25,000. invested at in order for the investment to reach a value of $30,000? How much of the inheritance should be interest, compounded quarterly, 5.7% 354 Chapter 5 Exponential and Logarithmic Functions 27. Suppose you win a contest and have a choice of $3000 prizes. You can take in 4 years. If money can be invested at $4000 interest, compounded annually, which prize is more valuable in the long run? now or you can receive 6% 28. If money can be invested at 7% interest, compounded quarterly, which is worth more: $9000 in 5 years? now or $12,500 29. If an investment of $1000 grows to $1407.10 in seven years with interest compounded annually, what is the interest rate? 30. If an investment of $2000 $2700 years, with an annual interest rate that is compounded quarterly, what is the annual interest rate? grows to in 31 2 31. If you put $3000 in a savings account today, what interest rate (compounded annually) must you receive in order to have after 5 years? $4000 32. If interest is compounded continuously, what annual rate must you receive if your investment of $2100 is to grow to in 6 years? $1500 33. a. At an interest rate of 8%, compounded annually, how long will it take to double an of investment of $1200? $500? $100? of b. What conclusion does part a suggest about doubling time? 34. At an interest rate of 6%, compounded annually, how long will it take to double an investment of P dollars? 35. How long will it take to double an investment of interest, compounded continuously? $500 7% at 36. How long will it take to triple an investment of $5000 at 8% interest, compounded continuously? 37. a. Suppose P dollars is invested for 1 year at 12% interest, compounded quarterly. What interest rate r would yield the same amount in 1 year with annual compounding? r is called the effective rate of interest. Hint: Solve the equation P 1 0.12 a 4 b 4 P 1 1 r 2 for r. The left side of the equation is the yield after 1 year at 12% interest, compounded quarterly, and the right side is the yield after 1 year at interest, compounded annually. r% b. Complete the following table: Effective rate 12% 12% interest compounding period annually quarterly monthly daily 38. This exercise investigates the continuous compounding formula, using a realistic interest rate. Consider the value of years at year, for increasing values of n. In this case, the $4000 interest, compounded n times per deposited for 3 5% interest rate per period is 0.05 n , and the number of periods in 3 years is 3n. So, the value at the end of 3 years is given by: A 4000 3n 1 0.05 n b a 4000 1 0.05 n b c a n 3 d a. Complete the following table: n 1 0.05 n b a n 1000 10,000 500,000 1,000,000 5,000,000 10,000,000 b. Compare the entries in the second column of e0.05, and the table in part a to the number complete the following sentence: As n gets larger and larger, the value of 1 0.05 n b a number gets closer and closer to the ? . n c. Use your answer to part b to complete the following sentence: As n gets larger and larger, the value of A 4000 to the number 1 0.05 n b ? c a d . n 3 gets closer and closer d. Compare your answer to part c with the value given by the continuous compounding formula. Section 5.3 Applications of Exponential Functions 355 39. A weekly census of the tree-frog population in a state park is given below. corresponds to the 1989–1990 school year. b. According to this model, what are the expenditures per pupil in 1999–2000? c. In what year did expenditures first exceed Week 1 2 3 4 5 6 $7000 per pupil? Population 18 54 162 486 1458 4374 44. There are now 3.2 million people who play bridge a. Find a function of the form f Pax that x 2 1 describes the frog population at time x weeks. b. What is the growth factor in this situation (that is, by what number must this week’s population be multiplied to obtain next week’s population)? c. Each tree frog requires 10 square feet of space and the park has an area of 6.2 square miles. Will the space required by the frog population exceed the size of the park in 12 weeks? in 14 weeks? ( 1 square mile 52802 square feet) 40. The fruit fly population in a certain laboratory triples every day. Today there are 200 fruit flies. a. Make a table showing the number of fruit flies present for the first 4 days (today is day 0, tomorrow is day 1, etc.). b. Find a function of the form Pax that x f describes the fruit fly population at time x days. c. What is the growth factor here (that is, by what 1 2 number must each day’s population be multiplied to obtain the next day’s population)? d. How many fruit flies will there be a week from now? 41. The population of Mexico was 100.4 million in 2000 and is expected to grow by approximately 1.4% each year. a. If x g is the population, in millions, of Mexico corresponds to the year x 0 in year x, where 2000, find the rule of the function g. (See Example 6.) 2 1 b. Estimate the population of Mexico in the year 2010. 42. The number of dandelions in your lawn increases a week, and there are 75 dandelions now. x is the number of dandelions in week x, by a. If 5% f 1 2 find the rule of the function f. and the number increases by a. Write the rule of a function that gives the a year. 3.5% number of bridge players x years from now. b. How many people will be playing bridge 15 years from now? c. When will there be 10 million bridge players? 45. At the beginning of an experiment a culture contains 200 h-pylori bacteria. An hour later there are 205 bacteria. Assuming that the h-pylori bacteria grow exponentially, how many will there be after 10 hours? after 2 days? (See Example 7.) 46. The population of India was approximately 1030 million in 2001 and was 865 million a decade earlier. What will the population be in 2006 if it continues to grow exponentially at the same rate? 47. Use graphical methods to estimate the following values. 313 a. b. 413 c. 513 48. Kerosene is passed through a pipe filled with clay in order to remove various pollutants. Each foot of pipe removes a. Write the rule of a function that gives the percentage of pollutants remaining in the kerosene after it has passed through x feet of pipe. (See Example 8.) of the pollutants. 25% b. How many feet of pipe are needed to ensure 90% that from the kerosene? of the pollutants have been removed 49. If inflation runs at a steady per year, then the amount that a dollar is worth today decreases by 3% a. Write the function rule that gives the val
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ue of each year. 3% a dollar x years from today. b. How much will the dollar be worth in 5 years? in 10 years? c. How many years will it take before today’s b. How many dandelions will there be in 16 weeks? dollar is worth only a dime? 43. Average annual expenditure per pupil in 50. a. The half-life of radium is 1620 years. Find the $5550 elementary and secondary schools was 1989–1990 and has been increasing at about each year. a. Write the rule of a function that gives the in 3.68% expenditure per pupil in year x, where x 0 rule of the function that gives the amount remaining from an initial quantity of 100 milligrams of radium after x years. b. How much radium is left after 800 years? after 1600 years? after 3200 years? 356 Chapter 5 Exponential and Logarithmic Functions 51. a. The half-life of polonium-210 is 140 days. Find the rule of the function that gives the amount of polonium-210 remaining from an initial 20 milligrams after t days. b. How much polonium-210 is left after 15 weeks? after 52 weeks? c. How long will it take for the 20 milligrams to decay to 4 milligrams? 52. How old is a piece of ivory that has lost 58% of its carbon-14? (See Example 9.) 53. How old is a mummy that has lost 49% of its carbon-14? 5.4 Common and Natural Logarithmic Functions Objectives • Evaluate common and natural logarithms with and without a calculator • Solve common and natural exponential and logarithmic equations by using an equivalent equation • Graph and identify transformations of common and natural logarithmic functions Technology Tip 10x The graph of can be obtained in para- x f 1 2 metric mode by letting x t and y 10t, where t is any real number. The graph of the inverse function g can then be obtained by letting x 10t and y t, where t is any real number. From their invention in the seventeenth century until the development of computers and calculators, logarithms were the only effective tools for numerical computation in astronomy, chemistry, physics, and engineering. Although they are no longer needed for computation, logarithmic functions still play an important role in the sciences and engineering. In this section you will examine the two most important types of logarithms, those to base 10 and those to base e. Logarithms to other bases are considered in Excursion 5.5A. Common Logarithms The graph of the exponential function is shown in Figure 5.4-1. Because it is an increasing function, it is a one-to-one function, as explained in Section 3.6. Recall that the graphs of inverse functions are reflections of one another across the line The exponential function f and its inverse function are graphed in Figure 5.4-2. y x. 10x x x f 1 2 10x 1 2 y 1 f(x) = 10x x y y = x f(x) 1 x 1 g(x) Figure 5.4-1 Figure 5.4-2 The inverse function of the exponential function is called the common logarithmic function. The value of this function at the number x is denoted as log x and called the common logarithm of the number x. x f 1 2 10x The functions f 1 10x and x 2 log v u if and only if 10u v x g 1 2 are inverse functions. log x Section 5.4 Common and Natural Logarithmic Functions 357 Because logarithms are a special kind of exponent, every statement about logarithms is equivalent to a statement about exponents. Logarithmic statement log v u Equivalent exponential statement 10u v log 29 1.4624 log 378 2.5775 101.4624 29 102.5775 378 Example 1 Evaluating Common Logarithms Without using a calculator, find each value. a. log 1000 b. log 1 c. log210 d. log 3 1 2 Solution a. If b. If log 1000 x, log 1 x, then 10x 1000. Because then 10x 1. Because 103 1000, log 1000 3. . 100 1, log 1 0 c. If log210 x, then 10x 210. Because 10 1 2 210, log210 1 2 . d. If 1 log 3 x, 10x 3. exponent of 10 that produces numbers. then 2 Because there is no real number is not defined for real 3, log 3 1 2 ■ Every scientific and graphing calculator has a LOG key for evaluating logarithms. For example, log 0.6 0.2218 and log 327 2.5145 A calculator is necessary to evaluate most logarithms, but you can get a rough estimate mentally. For example, because log 795 is greater than log 100 2 you can estimate that log 795 is between 2 and 3 and closer to 3. and less than log 1000 3, Example 2 Using Equivalent Statements Solve each equation by using an equivalent statement. a. log x 2 b. 10x 29 Solution a. If log b. If x 2, 10x 29, Figure 5.4-3. then 102 x. Therefore, x 100. then log 29 x. Therefore, x 1.4624, as shown in ■ NOTE Logarithms are rounded to four decimal places and an equal sign is used rather than the “approximately equal” sign. The word “common” will be omitted except when it is necessary to distinguish the common logarithm from another type of logarithm. Figure 5.4-3 358 Chapter 5 Exponential and Logarithmic Functions Natural Logarithms The exponential function is very useful in science and engineering. Consequently, another type of logarithm exists, based on the number e instead of 10. x f 1 2 ex The graph of the exponential function Because it is an increasing function, it is one-to-one. The function and its inverse function are graphed in Figure 5.4-5. is shown in Figure 5.4-4. ex x x f f 1 1 2 2 ex y 1 f(x) = ex x y y = x f(x) 1 x 1 g(x) Figure 5.4-4 Figure 5.4-5 This inverse function of the exponential function is called the natural logarithmic function. The value of this function at the number x is denoted as ln x and called the natural logarithm of the number x. x f 1 2 e x The functions f x 1 and ex ln x 2 ln v u if and only if eu v are inverse functions. x g 2 1 Again, as with common logarithms, every statement about natural logarithms is equivalent to a statement about exponents. Logarithmic statement v u ln Equivalent exponential statement eu v ln 14 2.6391 ln 0.2 1.6094 e2.6391 14 1.6094 0.2 e Example 3 Evaluating Natural Logarithms Use a calculator to find each value. c. ln a. ln 0.15 b. ln 186 1 5 2 Solution Figure 5.4-6 ln 0.15 1.8971, 186 5.2257, 5 a. b. ln c. ln e that produces 2 1 5. which means that 1.8971 0.15. e which means that e5.2257 186. is undefined for real numbers because there is no exponent of ■ Section 5.4 Common and Natural Logarithmic Functions 359 In a few cases you can evaluate ln x without a calculator. ln e 1 because e1 e ln 1 0 because e0 1 Example 4 Solving by Using an Equivalent Statement Solve each equation by using an equivalent statement. a. ln x 4 b. e x 5 Solution x 4, then a. If ln b. If e x 5, e4 x 5 x then ln . Therefore, . Therefore, x 54.5982. x 1.6094. ■ Graphs of Logarithmic Functions Because the graphs of exponential functions have the same basic shape and each logarithmic function is the inverse of an exponential function, the graphs of logarithmic functions have common characteristics. The following table compares the graphs of exponential and logarithmic functions. Exponential functions Logarithmic functions Examples x f 1 2 10x; f ex x 1 2 Domain all real numbers g x 2 1 log x; g ln x x 2 1 all positive real numbers Range all positive real numbers all real numbers x f 1 2 increases as x increases g x 2 1 increases as x increases approaches the x-axis x f 1 as x decreases 2 approaches the y-axis as x g x approaches 0 1 2 Reference points f(x) 10x g(x) log x 1, a 1 10b , 1 0, 1 , 1 2 1, 10 2 1 10 a , 1 1, 0 , 1 b , 1 2 10, 1 2 f(x) ex 1, a 1 e b , 1 0, 1 1, e , 2 1 2 g(x) ln x 1 e , 1 a 1, 0 , 1 b e, 1 , 2 1 2 Example 5 Transforming Logarithmic Functions Figure 5.4-7 y exponential y = x (0, 1) (1, 0) x logarithmic Figure 5.4-8 Describe the transformation from the graph of of Give the domain and range of h. 2 log log x to the graph 360 Chapter 5 Exponential and Logarithmic Functions Solution The graph of after a horizontal translation of 3 units right and a vertical stretch by a factor of 2. is the graph of log x Domain of h: The domain of log x The horizontal translation of 3 units to the right changes the domain to all real numbers greater than 3. is all positive real numbers. g x 2 1 Range of h: log x The range of vertical stretch has no effect on the range. is all real numbers, so the x g 2 1 The graphs of g and h are shown in Figure 5.4-9. The points a 9 1, 0 1 , and 2 1 , and 2 2 log 10, 1 2 13, 2 2 1 x 3 2 4, 0 1 h x continues to approach the asymptote at 2 1 1 on the graph of g are translated to the points b on the graph of h. Although the graph of you know that it appears to stop abruptly at x 3, 3 a , x 3. 1 10 1 10 , 1 , b , 2 ■ 3 –1 –3 Figure 5.4-9 3 5 5 5 Figure 5.4-10 Example 6 Transforming Logarithmic Functions Describe the transformation from Give the domain and range of h. g x 1 2 ln x to h x 2 1 ln 1 2 x 2 3. Solution g x 2 3, x h Because after a horizontal reflection across the y-axis followed by a horizontal translation of 2 units to the right and a vertical translation of 3 units downward. its graph is that of 22 g x 1 1 1 2 2 1 ln x Domain of h: The domain of g x ln x is all positive real numbers. 1 2 The reflection across the y-axis first changes the domain to all negative real numbers. Then the translation of 2 units to the right changes the domain from all negative real numbers to all real numbers less than 2. ln x The range of vertical translation does not affect the range. is all real numbers, so the g x 2 1 Range of h: , The graphs of g and h are shown in Figure 5.4-10. The points on the graph of g are translated to points on the graph of h. 2 e, 2 1, 3 and and 1, 0 e ■ Example 7 Solving Logarithmic Equations Graphically If you invest money at an interest rate r, compounded annually, then gives the time in years that it would take to double. D r 2 1 D r 1 2 ln 2 ln 1 1 r 2 Section 5.4 Common and Natural Logarithmic Functions 361 a. How long will it take to double an investment of $2500 at 6.5% annual interest? b. What annual interest rate is needed in order for the investment in part a to double in 6 years? Solution a. The annual interest rate r is 0.065. Find D(0.065). 0.065 D 1 2 ln 2 1 0.065 2 ln 1 11.0067 Therefore, it will take approximately 11 years to double
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an investment of $2500 6.5% at b. If the investment doubles in 6 years, then 6 annual interest rate r, solve 6. To find the r 2 1 by graphing. The point of intersection of the graphs of and Y2 6 is approximately (0.1225, 6). Therefore, an annual interest rate of 12.25% 5.4-11. is needed for the investment to double in 6 years. See Figure ■ annual interest. D ln 1 Y1 ln 2 1 r 2 ln 2 ln 1 1 r 2 7 0 1 1 Figure 5.4-11 Exercises 5.4 Unless stated otherwise, all letters represent positive numbers. In Exercises 1–4, find the value of each logarithm. 1. log 10,000 3. log 210 1000 2. log 0.001 4. log23 0.01 In Exercises 5–14, translate the given logarithmic statement into an equivalent exponential statement. 5. log 1000 3 6. log 0.001 3 7. log 750 2.8751 8. log 0.8 0.0969 9. ln 3 1.0986 10. ln 10 2.3026 11. ln 0.01 4.6052 12. ln s r 13. ln 1 x2 2y 2 z w 14. log a c 2 1 d In Exercises 15 – 24, translate the given exponential statement into an equivalent logarithmic statement. 15. 10 2 0.01 16. 103 1000 17. 100.4771 3 18. 107k r 19. e3.25 25.79 21. 12 7 5.5527 e 23. 2 r w e 20. 4 0.0183 e 22. ek t 24. e4uv m In Exercises 25–36, evaluate the given expression without using a calculator. 25. log 10243 26. log 102x2y2 27. ln e15 28. ln e3.78 31. eln 931 34. ln ex22y 29. ln 1e 32. eln 34.17 35. eln x2 30. ln 25 e 33. ln exy 36. eln 2x3 In Exercises 37–40, find the domain of the given function. 37. f 39. h x x 1 1 2 2 ln 1 log x 1 x 1 2 2 38. g x 1 2 40. k x 1 2 ln 1 log x 2 2 2 x 1 2 362 Chapter 5 Exponential and Logarithmic Functions 41. Compare the graphs of 2 log x. log x2 How are they alike? How are they and x x f 2 1 g different? 1 2 42. Compare the graphs of 3 log x. log x3 How are they alike? How are they and h x x 2 1 k different? 2 1 In Exercises 43–48, describe the transformation from g(x) ln x to the given function. Give the domain and range of the given function. 43. f 45. 47 ln x ln ln 1 1 x 4 x 3 2 2 4 44. 46. 48 ln x 7 ln ln 1 1 x 2 x 2 2 2 2 In Exercises 49–52, sketch the graph of the function. 49. 51. x f 1 2 h x 2 1 log x 3 1 2 log x 2 50. g 52. f x x 1 1 2 2 2 ln x 3 ln x 1 2 3 In Exercises 53–58, find a viewing window (or windows) that shows a complete graph of the function. 53. f x 1 2 55. h 57. f x x 1 1 2 2 x ln x ln x 2 x 10 log x x 54. g x 2 1 ln x x 56. k x 1 2 58. f x 1 2 2 ln x e log x x In Exercises 59–62, find the average rate of change of the function. (See Section 3.7.) 59. 60. 61. 62 ln x 2 2 1 x ln x, , as x goes from 3 to 5 as x goes from 0.5 to 1 log 1 x2 x 1 2 , as x goes from 5 to 3 x log x 0 0 , as x goes from 1 to 4 f 63. a. What is the average rate of change of 3 h? as x goes from 3 to b. What is the value of h when the average rate of 3 h, as x goes from 3 to ln x, ln x, x x f 1 2 change of is 0.25? 1 2 64. a. Find the average rate of change of f as x goes from 0.5 to 2. ln x2, x 1 2 b. Find the average rate of change of 2, as x goes from 3.5 to 5. x 3 ln g x 1 2 1 2 c. What is the relationship between your answers in parts a and b? Explain why this is so. 65. a. Use the doubling function D from Example 7 to find the time it takes to double your money at each of these interest rates: 4%, 6%, 8%, 12%, 18%, 24%, and 36%. b. Round the answers in part a to the nearest year . , , , , , , and 72 8 72 6 72 36 72 12 72 18 Use this and compare them with these numbers: 72 72 24 4 evidence to state a “rule of thumb” for determining approximate doubling time, without using the function D. This rule of thumb, which has long been used by bankers, is called the Rule of 72. 66. The height h above sea level (in meters) is related to air temperature t (in degrees Celsius), the atmospheric pressure p (in centimeters of mercury at height h), and the atmospheric pressure c at sea level by: h 1 30t 8000 c pb a ln 2 If the pressure at the top of Mount Rainier is 44 centimeters on a day when sea level pressure is 75.126 centimeters and the temperature is what is the height of Mount Rainier? 7°, 67. A class is tested at the end of the semester and weekly thereafter on the same material. The average score on the exam taken after t weeks is given by the following “forgetting function”. g t 2 1 77 10 ln t 1 1 2 a. What was the average score on the original exam? b. What was the average score after 2 weeks? after 5 weeks? 68. Students in a precalculus class were given a final exam. Each month thereafter, they took an equivalent exam. The class average on the exam taken after t months is given by the following function. F t 2 1 82 8 ln t 1 2 1 a. What was the average score on the original exam? b. What was the average score after 6 months? after 10 months? 69. One person with a flu virus visited the campus. The number T of days it took for the virus to infect x people is given by T. T 0.93 ln 7000 x 6999x b a Section 5.5 Properties and Laws of Logarithms 363 a. How many days did it take for 6000 people to become infected? b. After 2 weeks, how many people were infected? 70. Critical Thinking For each positive integer n, let fn be the polynomial function whose rule is x fn1 2 x x 2 2 x 3 3 x4 4 x 5 5 p – x n n if n is odd if n is even. In the viewing window with where the sign of the last term is and 1 x 1 and 1 x g 2 For what values of x does approximation of g? 4 y 1, f4 1 and ln x x 1 2 2 1 graph on the same screen. f4 appear to be a good 71. Critical Thinking Using the viewing window in Exercise 70, find a value of n for which the graph of the function (as defined in Exercise 70) fn x ln 1 x appears to coincide with the graph of g from graph to graph to see how good this approximation actually is. Use the trace feature to move . 1 2 2 1 72. A bicycle store finds that N the number of bikes sold, is related to d, the number of dollars spent on advertising. N 51 100 ln d 100 a 2 b a. How many bikes will be sold if nothing is is spent? if $1000 spent on advertising? if $10,000 is spent? b. If the average profit is worthwhile to spend $10,000? What about $25 $1000 per bike, is it on advertising? c. What are the answers in part b if the average profit per bike is $35? 5.5 Properties and Laws of Logarithms Objectives • Use properties and laws of logarithms to simplify and evaluate expressions The definitions of common and natural logarithms differ only in their bases. Therefore, common and natural logarithms share the same basic properties and laws. Basic Properties of Logarithms Logarithms are only defined for positive real numbers. That is, NOTE Any number raised to the zero power, except zero, is 1. x0 1, where x 0 log v and ln v are defined only when v 77 0. y log x y ln x and both contain the point (1, 0) because The graphs of 100 1 and e0 1. The values of log nential statements. 104 log 1 0 and ln 1 0 and ln e9 can be found by writing equivalent expo- In general, If log 104 x, then 10x 104. So x 4. If ln e9 x, then ex e9. So x 9. log 10k k, for every real number k. ln ek k, for every real number k. 364 Chapter 5 Exponential and Logarithmic Functions By definition, log 678 is the exponent to which 10 must be raised to produce 678. 10log 678 678 Similarly, ln 54 is the exponent to which e must be raised to produce 54. eln 54 54 In general, 10logv v and eln v v, for every v 77 0. The facts presented above are summarized in the table below. Common logarithms Natural logarithms 1. log v is defined only when and 2. 3. log log 10 1 for every real log 1 0 10k k v 77 0 1. ln v is defined only when ln e 1 and 2. for every real 3. ln v 77 0 number k 10log v v 4. for every v 77 0 4. for every v 77 0 ln 1 0 ek k number k eln v v Properties 3 and 4 are restatements of the fact that the composition of inverse functions produces the identity function. Basic Properties of Logarithms That is, if f x 1 2 10x f g g f 21 21 1 1 g and log x 10x 2 log x, then 10log x x for all x 7 0 2 log 10x x for all x ex and x x g f ln x. Analogous statements are true for 1 The properties of logarithms can be used to simplify expressions and solve k 2x2 7x 9 equations. For example, applying Property 3 with allows you to rewrite the expression 2x2 7x 9. ln e2x 27x9 as 1 2 2 Example 1 Solving Equations by Using Properties of Logarithms Use the basic properties of logarithms to solve the equation ln x 1 1 2 2. 2 3 2 Solution Because f x 1 2 8 ex x1 is a function, if 2 e2 e ln x 1 e2 1 x e2 1 x 6.3891 ln 1 x 1 2 2, then eln 1 x1 2 e2. Apply Property 4 with v x 1 Figure 5.5-1 The intersection of the graphs of Figure 5.5-1, confirms the solution. Y1 ln 1 x 1 2 and Y2 2, shown in ■ Section 5.5 Properties and Laws of Logarithms 365 Laws of Logarithms bm The Product Law of Exponents states that rithms are exponents, the following law holds. ˛bn bmn. Because loga- Product Law of Logarithms For all v, w 77 0, log (vw) log v log w ln (vw) ln v ln w. Proof According to Property 4 of logarithms, 10log w w. Then, by the Product Law of Exponents: vw 10log v 10log w 10log vlog w 10log v v and Again by Property 4 of logarithms: 10log vw vw Therefore, one-to-one, natural logarithms. 10log vw 10log vlog w log vw log v log w ; and because exponential functions are . A similar argument can be made for Example 2 Using the Product Law of Logarithms Use the Product Law of Logarithms to evaluate each logarithm. log 3 0.4771 ln 7 1.9459 and log 11 1.0414, ln 9 2.1972, and find ln 63. find log 33. a. Given that b. Given that Solution a. b. log 33 log ln 63 ln 1 7 9 1 2 2 3 11 log 3 log 11 0.4771 1.0414 1.5185 ln 7 ln 9 1.9459 2.1972 4.1431 ■ CAUTION Graphing Exploration A common error in applying the Product Law of Logarithms is to write the false statement ln 7 ln 9 7 9 ln 16 ln 1 2 instead of the correct statement ln 7 ln 9 7 9 ln 63. ln 1 2 10 x 10 Using the viewing window with graph both functions below on the same screen. and 8 y 8, f x 1 2 ln x ln 9 g x 1 2 ln 1 x 9 2 Explain how the graph illustrates the caution in the margin. The Quotient Law of Exponents states that bm bn bmn. When the expo- nents are logarithms, the Quotient Law is still valid. 366 Chapter 5 Expone
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ntial and Logarithmic Functions Quotient Law of Logarithms For all v, w 77 0, log a ln a v wb v wb log v log w ln v ln w. The proof of the Quotient Law of Logarithms is similar to the proof of the Product Law of Logarithms. Example 3 Using the Quotient Law of Logarithms Use the Quotient Law of Logarithms to evaluate each logarithm. Given that log 28 1.4472 and log 7 0.8451, find log 4. Given that ln 18 2.8904 and ln 6 1.7918, find ln 3. b. a. Solution a. log 4 log 28 7 b a log 28 log 7 1.4472 0.8451 0.6021 b. ln 3 ln 18 6 b a ln 18 ln 6 2.8904 1.7918 1.0986 CAUTION Do not confuse ln 7 9b a 0.2513 with the quotient ln 7 0.8856 . ln 9 They are different numbers. Graphing Exploration Using the viewing window with both functions below on the same screen. 0 x 8 and 4 y 2, graph f x 2 1 ln x 9b a g ln x ln 9 x 2 1 Explain how the graph illustrates the caution in the margin. ■ The Power Law of Exponents, which states that translated into a logarithmic statement. 1 bm 2 k bmk, can also be Power Law of Logarithms For all k and v 77 0, logvk klog v, ln vk k ln v. Section 5.5 Properties and Laws of Logarithms 367 Proof According to Property 4 of logarithms, Law of Exponents: 10log v v. Then, by the Power v k 10log v 1 2 k 10k log v Again by Property 4 of logarithms: 10log v k v k 10log v k 10k log v So, and therefore, can be made for natural logarithms. log v k k log v . A similar argument Example 4 Using the Power Law of Logarithms Use the Power Law of Logarithms to evaluate each logarithm. a. b. Given that log 6 0.7782, find log 26. Given that ln 50 3.9120, find ln 23 50. Solution a. log26 log 6 b. ln23 50 ln 50 1 2 1 2 3 1 3 1 log 6 1 2 1 ln 50 1 3 1 0.7782 3.9120 2 2 0.3891 1.3040 The laws of logarithms can be used to simplify various expressions. Example 5 Simplifying Expressions Write ln 3x 4 ln x ln 3xy as a single logarithm. Solution ln 3x 4 ln x ln 3xy ln 3x ln x4 ln 3xy ln 3xy 2 ln 1 ln ln 3x x4 3x5 3xyb x4 y b a a Power Law Product Law Quotient Law Example 6 Simplifying Expressions Simplify ln 2x x b a ln A 24 ex2 . B ■ ■ 368 Chapter 5 Exponential and Logarithmic Functions Solution 2x x b ln a ln A 24 ex2 B 1 2 1 4 2 1 4 ex2 ln ln 1 ln 2 ex2 1 ex2 x x b a 1 ln ln x 2 2 1 ln x 1 1 2 4 ln x 1 1 4 1 2 ln x 1 1 4 1 2 ln e 1 ln x 1 1 2 2 4 1 2 ln e ln x2 ln e 2 ln x ln x ln e 1 4 1 4 Power Law Product Law 2 Power Law 2 ln e 1 ■ NOTE The zero earthquake has ground motion amplitude of less than 1 micron on a standard seismograph 100 kilometers from the epicenter. Applications A logarithmic scale is a scale that is determined by a logarithmic function. Because logarithmic growth is slow, measurements on a logarithmic scale can sometimes be deceptive. The Richter scale is an illustration of this. of an earthquake on the Richter scale is given by The magnitude log R i R i 1 2 2 1 , where i is the amplitude of the ground motion of the earth- i i0b a i0 is the amplitude of the ground motion of the zero earthquake. quake and A moderate earthquake might have 1000 times the ground motion of the zero earthquake, or Its magnitude would be i 1000i0. log 1000i0 i0 a b log 1000 3 An earthquake with 10 times this ground motion, or have a magnitude of i 10,000i0, would log 10,000i0 i0 a b log 10,000 4 So a tenfold increase in ground motion produces only a 1-point change on the Richter scale. In general, increasing the ground motion by a factor of the Richter magnitude by k units. 10k increases Example 7 Richter Scale The 1989 World Series earthquake in San Francisco measured 7.0 on the Richter scale, and the great earthquake of 1906 measured 8.3. How much more intense was the ground motion of the 1906 earthquake than that of the 1989 earthquake? Section 5.5 Properties and Laws of Logarithms 369 Solution The difference in Richter magnitude is 101.3 20 earthquake was in terms of ground motion. Therefore, the 1906 times more intense than the 1989 earthquake 8.3 7.0 1.3. ■ Exercises 5.5 In Exercises 1 – 4, solve each equation by using the basic properties of logarithms. 1. log 3. ln . 4. 2x log 2 1 5 ln 1 3 x 1 2 8 In Exercises 5–10, use laws of logarithms and the values given below logarithmic expression. to evaluate each log 7 0.8451 log 5 0.6990 log 3 0.4771 log 2 0.3010 5. log 8 7. log 5 7b a 9. log 0.6 6. log 12 8. log 3 14b a 10. log 1.5 In Exercises 11–20, write the given expression as a single logarithm. 11. ln x2 3 ln y 12. ln 2x 2 ln x ln 3y 25. ln A 23 x2 1y B 26. ln a 2x 2y 23 y b 27. a. Graph y x y e ln x and in separate viewing windows. For what values of x are the graphs identical? b. Use the properties of logarithms to explain your answer in part a. 28. a. Graph and y x y ln e x in separate viewing windows. For what values of x are the graphs identical? b. Use the properties of logarithms to explain your answer in part a. In Exercises 29–34, use graphical or algebraic means to determine whether the statement is true or false. 29. ln x 0 0 ln x 0 0 31. log x5 5 log x 33. ln x3 ln x 1 3 2 30. ln 1 xb a 1 ln x 32. 34. x 7 0 ex ln x xx 1 2 log 2x 2log x In Exercises 35 and 36, find values of a and b for which the statement is false. 13. 14. log x2 9 1 log 3x 2 15. 2 ln x 3 1 2 2 y log x 3 2 1 log x log 3 ln x2 ln x 1 2 2 4 35. 36. 16. ln a e 2x b ln A 2ex B 17. 3 ln e2 e 2 1 3 37. If log a log b log a bb a log a b 2 1 ln b7 7, log a log b what is b? 18. 2 2 log 20 19. log 10x log 20y 1 20. ln e2 x 1 2 ln ey 1 2 3 u ln x In Exercises 21–26, let given expression in terms of u and v. For example, ln x3y ln x3 ln y 3 ln x ln y 3u v. v ln y. and Write the 21. ln x2y5 1 2 23. ln A 2x y2 B 22. ln 1 x3y2 2 2x y b 24. ln a 38. Suppose f x 2 constants. If and B? 1 A ln x B, 10 f and 1 1 2 where A and B are , what are A f 1 e 1 2 39. If f x A ln x B 1 2 find A and B. and f e 1 2 5 and f 8, e2 2 1 40. Show that g x 1 2 ln function of f x 1 2 a x 1 xb 1 1 e x . is the inverse (See Section 3.6.) 370 Chapter 5 Exponential and Logarithmic Functions In Exercises 41–44, state the magnitude on the Richter scale of an earthquake that satisfies the given condition. 41. 100 times stronger than the zero quake 42. 104.7 times stronger than the zero quake 43. 350 times stronger than the zero quake 44. 2500 times stronger than the zero quake Exercises 45–48 deal with the energy intensity i of a sound, which is related to the loudness of the sound by the function L(i) 10 log i i0b , a where i0 is the minimum intensity detectable by the human ear and L(i) is measured in decibels. Find the decibel measure of the sound. 45. ticking watch (intensity is 100 times ) i0 46. soft music (intensity is 10,000 times ) i0 47. loud conversation (intensity is 4 million times ) i0 48. Victoria Falls in Africa (intensity is 10 billion times )i0 49. How much louder is the sound in Exercise 46 than the sound in Exercise 45? 50. The perceived loudness L of a sound of intensity I L k ln I, where k is a certain is given by constant. By how much must the intensity be increased to double the loudness? (That is, what must be done to I to produce 2L?) 51. Compute each of the following pairs of numbers: a. log 18 and ln 18 ln 10 b. log 8950 and ln 8950 ln 10 c. What do the results in parts a and b suggest? 52. Find each of the following logarithms. c. b. e. log 8.753 log 8753 log 87.53 log 87,530 a. d. f. How are the numbers 8.753, 87.53, 875.3, 8753, and 87,530 related to one another? How are their logarithms related? State a general conclusion that this evidence suggests. log 875.3 5.5.A Excursion: Logarithmic Functions to Other Bases Objectives • Evaluate logarithms to any base with and without a calculator • Solve exponential and logarithmic equations to any base by using an equivalent equation • Identify transformations of logarithmic functions to any base • Use properties and laws of logarithms to simplify and evaluate logarithmic expressions to any base Common and natural logarithms were defined by considering the inverse functions of the exponential functions In this section, you will see that a similar procedure can be carried out with any positive number b in place of 10 and e. 10x ex. and x x f f 2 1 1 2 NOTE b 7 1. valid for In the discussion below, b is a fixed positive number with The discussion on exponents and logarithms to base b is also 0 6 b 6 1, but in that case the graphs have a different shape. Defining Logarithmic Functions to Other Bases bx 2 1 f x Because is an increasing function, it is a one-to-one function and therefore has an inverse function. (See Section 3.6) Recall that the graphs y x. of inverse functions are reflections of one another across the line and its inverse function are graphed in An exponential function Figure 5.5.A-1. bx x f 2 1 y y = x f(x) 1 x 1 g(x) Figure 5.5.A-1 Section 5.5.A Excursion: Logarithmic Functions to Other Bases 371 This inverse function g is called the logarithmic function to the base b. and is called the logThe value of g(x) at the number x is denoted arithm to the base b of the number x. logb x Because the functions f x 2 logb v u 1 bx logb x and g x 1 if and only if 2 bu v. are inverse functions, Because all logarithms are exponents, every statement about logarithms is equivalent to a statement about exponents. Logarithmic statement logb v u log3 81 4 log4 64 3 log125 5 1 3 log8a 1 4b 2 3 Equivalent exponential statement bu v 34 81 43 64 125 1 3 5 8 3 1 2 4 Example 1 Evaluating Logarithms to Other Bases Without using a calculator, find each value. 25 log2 16 b. a. c. 9 log1 3 log51 2 Solution a. If b. If c. If log2 16 x, 9 x, log 1 3 then then log51 25 x, 2 Because Because 2x 16. x 1 a 3b then 9. 5x 25. 24 16, log2 16 4. 9, log 2 1 3b a 1 3 9 2 exponent of 5 that produces a negative number, defined. log51 2 Because there is no real number is not 25 ■ Example 2 Solving Logarithmic Equations Solve each equation for x. a. log5 x 3 b. log6 1 x c. log1 6 1 3 2 x d. log6 6 x 372 Chapter 5 Exponential and Logarithmic Functions Solution a. If b. If c. If log5 x 3, log6 1 x, 3 log1 6 2 1 then then x, then 53 x. 6x 1. 1 6
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b 3 log1 1 6 6x 6. a Therefore, Therefore, x x 125. x 0. 3. x 2 Therefore, x 1. d. If log6 6 x, then negative number, has no real solution. Because no real power of 1 6 is a ■ Basic Properties of Logarithms to Other Bases Logarithms are only defined for positive real numbers. That is, logb v is defined only when v 77 0. The graph of b 7 0. That is, y logb x contains the point (1, 0) because b0 1 for any logb 1 0 The value of statement. log5 54 can be found by writing an equivalent exponential In general, If log5 54 x, then 5x 54. So x 4. logb bk k for every real number k . By definition, duce 104. Therefore, log3 104 is the exponent to which 3 must be raised to pro- In general, 3log3 104 104. blogb v v for every v 77 0 . The facts presented above are summarized in the table below. Basic Properties of Logarithms For b 77 0 and b 1 , 1. 2. 3. 4. logb v logb 1 0 logb bk k blogb v v is defined only when v 77 0 and logb b 1 for every real number k for every v 77 0 Properties 3 and 4 are restatements of the fact that the composition of inverse functions produces the identity function. Section 5.5.A Excursion: Logarithmic Functions to Other Bases 373 If f x 2 1 bx and 21 2 1 logb x, f 1 g 1 x x 2 2 logb x bx 2 2 then blogb x x for all x 7 0 logb bx x for all x Equations that involve both logarithmic and constant terms may be solved by using basic properties of logarithms. blogb v v for b 7 0 and b 1 Example 3 Solving Logarithmic Equations Solve the equation log31 x 1 2 4. Solution log31 3log3 4 x 1 2 x1 2 34 1 x 1 34 x 82 exponentiate both sides b log b v v ■ Laws of Logarithms to Other Bases Because all logarithms are a form of exponents, the laws of exponents translate to the corresponding laws of logarithms to any base. Laws of Logarithms For all b, v, w, and k, with b, v, and w positive and b 1: Product Law: logb(vw) logb v logb w Quotient Law: logba v wb logb v logb w Power Law: logb(vk) k logb v Example 4 Using the Laws of Logarithms Use the Laws of Logarithms to evaluate each expression, given that log7 2 0.3562, log7 3 0.5646, a. log7 5 0.8271. and b. c. log7 2.5 log7 48 log7 10 Solution a. Use the Product Law. 2 5 log7 10 log71 2 log7 2 log7 5 0.3562 0.8271 1.1833 374 Chapter 5 Exponential and Logarithmic Functions b. Use the Quotient Law. log7 2.5 log7 a 5 2b log7 5 log7 2 0.8271 0.3562 0.4709 c. Use the Product and Power Laws. log7 48 log71 2 3 24 log7 3 log7 24 log7 3 4 log7 2 0.5646 4 1.9894 1 0.3562 2 Example 5 Using the Laws of Logarithms Simplify and write each expression as a single logarithm. 2 125x log3 y log31 2 a. b. x 2 log31 3 log51 Solution x2 4 a. log31 x 2 log3 y log31 2 x2 4 2 2 log31 x2 4 2 4 log33 1 log3S 1 log3S log3 x 2 y 2 x 2 y 2 x2 21 1 y 2 x 2 2 T NOTE log5 x be expressed as log 5˛a 1 xb . can also 1 or log5 x b. 3 log51 125x 2 log5 125 log5 x 3 1 3 3 log5 x log5 x 2 Change-of-Base Formula Scientific and graphing calculators have a LOG key and a LN key for calculating logarithms. No calculators have a key for logarithms to other bases. One way to evaluate logarithms to other bases is to use the formula below. Change-of-Base Formula For any positive number v, logb v log v log b and logb v ln v ln b ■ ■ Section 5.5.A Excursion: Logarithmic Functions to Other Bases 375 Proof By Property 4 of the Basic Properties of Logarithms blog b v v . ln 1 logb v blog b v v ln v blog b v 2 ln v ln b 2 1 logb v ln v ln b take logarithms of both sides apply the Power Law A similar argument can be made by taking common logarithms of both sides. Example 6 Evaluating Logarithms to Other Bases Evaluate log8 9. Solution Use the change-of-base formula and a calculator. log8 9 log 9 log 8 1.0566 or log8 9 ln 9 ln 8 1.0566 ■ Figure 5.5A-2 Graphing Logarithmic Functions to Other Bases The graph of a logarithmic function to any base b shares characteristics with the graphs of natural logarithms and common logarithms. The following table compares the graphs of exponential and logarithmic functions for base b, where b is any real number, b 7 0 b 1 , and . y y = x (1, b) Exponential function f(x) bx Logarithmic function g(x) logb x (0, 1) y = bx (b, 1) x (1, 0) Domain all real numbers all positive real numbers Range all positive real numbers all real numbers y = logb x Figure 5.5A-3 x f 1 2 increases as x increses g x 1 2 increases as x increases approaches the x-axis x f 1 as x decreases 2 approaches the y-axis as x g x approaches 0 1 2 Reference points 1, a 1 b b , 0, 1 1, 0 , b 1 b, 1 , 1 2 2 Example 7 Transforming Logarithmic Functions Describe the transformation from Give the domain and range of h. g x 1 2 log2 x to h x 2 1 log21 x 1 2 3. 376 Chapter 5 Exponential and Logarithmic Functions Solution g x 1 3, h x Because after a horizontal translation of 1 unit to the left and a vertical translation of 3 units down. its graph is the graph of g x 2 1 2 1 2 1 log2 x Domain of h: The domain of g x log2 x is all positive real 1 2 numbers. The horizontal translation of 1 unit to the left changes the domain to all real numbers greater than 1. Range of h: The range of log2 x vertical translation has no effect on the range. is all real numbers, so the x g 1 2 The points 12 the points , 0 , 2 0, 3 , 1 b and (2, 1) on the graph of g are translated to , and 1 2 1, 2 2 on the graph of h. To graph these Y1 ln x ln 2 for g x 2 1 log2 x and functions with a calculator, graph ln Y2 1 x 1 ln 2 2 3 for h x 1 2 log21 x 1 2 3. The graphs of g and h are shown in Figure 5.5A-4. ■ 2 4 8 Figure 5.5A-4 Exercises 5.5.A Note: Unless stated otherwise, all letters represent positive numbers and b 1. In Exercises 1–10, translate the given exponential statement into an equivalent logarithmic statement. 1. 10 2 0.01 2. 103 1000 3. 23 10 10 1 3 4. 100.4771 3 5. 107k r 6. 101 ab 2 c 7. 78 5,764,801 8. 3 1 2 8 9. 2 1 3 9 10. b14 3379 In Exercises 11 – 20, translate the given logarithmic statement into an equivalent exponential statement. 11. log 10,000 4 12. log 0.001 3 13. log 750 2.8751 14. log 0.8 0.0969 15. log5 125 3 16. log8 a 1 4b 2 3 17. log2 a 1 4b 2 18. log2 22 1 2 x2 2y z w 19. log 20. log 1 1 a c 2 2 d In Exercises 21–28, evaluate the given expression without using a calculator. 21. log 10243 24. log3.51 3.51 x21 2 2 27. log23 1 27 2 22. 25. 1717 log17 1 log16 4 2 28. log23 a 1 9b 23. log 102x2y2 26. log2 64 In Exercises 29–36, find the missing entries in each table. 29. x log4 ? Section 5.5.A Excursion: Logarithmic Functions to Other Bases 377 30. x log5 x g x 2 1 31. x log6 x h x 1 2 32. x x k 1 2 log3˛ 1 x 3 2 33. x 2 log7 x f x 1 2 34. x 3 log x g x 1 2 35. x 1 25 ? ? 2 10 .75 1 h x 2 1 3 log2 1 x 3 2 ? ? 36. x 2 ln x x k 1 2 1 e ? 1 ? 25 25 ? 1 ? 6 ? 27 ? ? 216 ? 12 ? 49 ? 100 1000 ? 1 ? e ? ? 29 ? e2 ? In Exercises 37–40, a graph or a table of values is given for the function f (x) logb x . Find b. 37. y 3 2 1 −1 −2 38. y 5 4 3 2 1 −5 −2 0.05 1.05 1 25 4 x 5 10 15 20 25 1 0 1 0 400 2 5 2 225 1 2 125 6 39. x f x 2 1 40. x f x 1 2 In Exercises 41–46, solve each equation for x. 41. log3 243 x 42. log81 27 x 43. log27 x 1 3 45. logx 64 3 44. log5 x 4 46. logx ˛a 1 9b 2 3 In Exercises 47–60, write the given expression as the logarithm of a single quantity. (See Example 5.) 47. 2 log x 3 log y 6 log z 48. 49. 50. 51. x 3 5 log8 x 3 log8 y 2 log8 z log x log˛1 y 2 log3 ˛1 1 2 ˛ log2˛ 1 2 log3 ˛1 log˛1 y 3 25c2 52. 2 2 2 x2 9 2 log3 y 1 3 ˛ log2 ˛1 27b6 2 x 53. 2 log4 7c 1 2 54. 1 3 ˛ log5 10 11 55. 2 ln x 1 1 z 3 2 ln 2 2 ln x 2 1 z 3 1 2 2 56. ln 1 378 Chapter 5 Exponential and Logarithmic Functions 1 57. log2 59. 2 ln 1 2x 2 1 e2 e 2 2 58. 60. 25z 2 2 log5 1 4 4 log5 20 1 2 80. If logb 9.21 7.4 and logb 359.62 19.61, then what is logb 359.62 logb 9.21 ? In Exercises 61–68, use a calculator and the change-ofbase formula to evaluate the logarithm. In Exercises 81–84, assume that a and b are positive, with b 1. a 1 and 61. log2 10 62. log2 22 63. log7 5 81. Express logb ˛u in terms of logarithms to the base a. 64. log5 7 65. log500 1000 66. log500 250 67. log12 56 68. log12 725 In Exercises 69–72, describe the transformation from f to g, and give the domain and range of g. 82. Show that logb a 1 loga b . 83. How are log u and log100 u related? 84. Show that alog b blog a. 69. 70. 71. 72 log5 x and g log7 x and g log2 x and g log4 x and 3x 4 log5 ˛1 2 log7˛ 1 1 3 log 2˛1 x 1 7 2 3 log4 ˛1 2x 2 85. If logb x 1 2 logb v 3, show that x b3 2 ˛2v. 1 86. Graph the functions log ˛1 x 7 2 f x log x log 7 on the same screen. For what and 1 2 1 2 x g g values of x is it true that x What do 2 1 log 6 log 7 you conclude about the statement log In Exercises 73–78, answer true or false. Explain your answer. 87. Graph the functions f x 1 2 log ˛a x 4b and 73. logb ˛a r 5b logb r logb 5 74. logb a logb c logb ˛a a c b 75. logb r t logb ˛1 1 t r 2 76. logb 77. log5 78. logb 1 1 1 logb c logb d cd 2 log5 x 2 5x ab 5 1 2 t t˛1 2 t logb a 2 397398 log 398 2.5999 or 79. Which is larger: 2.5988 and increasing function. 398397? and Hint: x f log 397 10x is an 1 2 log x log 4 . g x 1 2 say about a statement such as Are they the same? What does this log 48 log 4 48 4 b log ˛a ? In Exercises 88 – 90, sketch a complete graph of the function, labeling any holes, asymptotes, or local extrema. 88. f x 1 2 89. h 90. g x x 2 2 1 1 log5 x 2 x log x2 log20 x2 Section 5.6 Solving Exponential and Logarithmic Equations 379 5.6 Solving Exponential and Logarithmic Equations Objectives • Solve exponential and logarithmic equations • Solve a variety of application problems by using exponential and logarithmic equations Exponential and logarithmic equations have been solved in this chapter so far by using the graphing method or by writing equivalent statements that can be easily solved. Most of them could also have been solved algebraically by using the techniques presented in this section, which depend primarily on the properties and laws of logarithms. By definition of a function, if This results in two statements. u v and f is a function, then u f 1 2 f v . 2 1 If u v, the
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n bu bv for all real numbers b 77 0. If u v, then log b u log bv for all real numbers b 77 0. Because exponential and logarithmic functions are one-to-one functions, the converse is also true. If b u b v, then u v. If log b u log b v, then u v. Exponential Equations The easiest exponential equations to solve are those in which both sides are powers of the same base. Example 1 Powers of the Same Base Solve the equation 8x 2x1. Confirm your solution with a graph. Solution Write the equation so that each side is a power of the same base. 1 8x 2x1 x 2x1 23 2 23x 2x1 3x x 1 2x 1 x 1 2 If bu bv , then u v . Y1 8x Y2 2x1, To find a window for the graphs of consider the basic shapes of the graphs and any transformations. Because both bases of these exponential functions are greater than 1, the graphs are increasing. Because there is no vertical shift on either function, both graphs are asymptotic to the x-axis. The intersection of the graphs of and Y2 shown at the left, confirms the solution. 2x1, 8x and Y1 ■ 4 4 4 2 Figure 5.6-1 380 Chapter 5 Exponential and Logarithmic Functions CAUTION Example 2 Powers of Different Bases ln 2 ln 5 ln 2 ln 5 and ln 2 5b a ln 2 ln 5 4 4 4 2 Figure 5.6-2 4 4 4 2 Figure 5.6-3 Solve the equation 5x 2. Confirm your solution with a graph. Solution 5x 2 ln 5x ln 2 x ln 5 ln 2 x ln 2 ln 5 x 0.6931 1.6094 0.4307 take logarithms on each side use the Power Law The intersection of the graphs of confirms the solution. Y1 5x and Y2 2, shown at the left, ■ Example 3 Powers of Different Bases Solve the equation 24x1 31x. Confirm your solution with a graph. Solution 24x1 31x 24x1 ln 2 ln 2 2 1 31x 1 1 x ln 1 4x 1 1 2 ln 3 2 4x ln 2 ln 2 ln 3 x ln 3 4 ln 2 ln 3 4x ln 2 x ln 3 ln 3 ln 2 ln 3 ln 2 x 1 2 x ln 3 ln 2 4 ln 2 ln 3 x 0.4628 Take logarithms on each side Power Law Distributive Property Rearrange terms and isolate x The intersection of the graphs of left, confirms the solution. Y1 24x1 and Y2 31x, shown at the ■ When you multiply each side of an equation by the same expression, extraneous solutions may be introduced, as shown in Example 4. Example 4 Using Substitution Solve the equation ex e x 4. Confirm your solution with a graph. Section 5.6 Solving Exponential and Logarithmic Equations 381 Solution First multiply each side by ex to eliminate negative exponents. x 4 ex e x e x e e x ex 4 1 2 x 4ex ex ex ex e e2x 1 4ex Product Law. 2 1 e2x 4ex 1 0 Let u ex and substitute. u2 4u 1 0 4 – 2 2 4 1 1 21 – 220 u 4 – 225 u 2 – 25 ex 2 25 is negative, ex 2 25 2 6 ex Replace u with to get 2 25 be positive and ex 2 25. or ex 2 25 Because has no solution. ex can only 7 7 ln ex ln x ln e ln 2 25 2 25 A A x 1.4436 B B ln e 1 6 Figure 5.6-4 and The intersection of the graphs of ure 5.6-4, confirms that there is exactly one solution Y1 ex e Y2 x 4, shown in Fig- ■ Applications of Exponential Equations When a living organism dies, its carbon-14 decays. The half-life of carbon-14 is 5730 years, so the amount of carbon-14 remaining at time t is given by where P is the mass of carbon-14 that was present initially. The function M can be used to determine the age of fossils and some relics. P t 5730, 0.5 M t 2 2 1 1 Example 5 Radiocarbon Dating The skeleton of a mastodon has lost 58% of its original carbon-14. When did the mastodon die? Solution If the mastodon has lost 58% of its original carbon-14, then 42% of the iniM tial amount, or 0.42P, remains and To determine when the 0.5 mastodon died, solve 0.42P. for t. 0.42P P 2 1 t 5730 t 1 2 382 Chapter 5 Exponential and Logarithmic Functions 10,000 Figure 5.6-5 1 0 0 14,000 0.5 t 5730 2 t 5730 2 0.5 1 0.5 0.42P P 0.42 1 ln 0.42 ln 1 ln 0.42 t t 5730 2 ln 0.5 2 ln 0.42 5730 1 t 5730 1 ln 0.5 t 7171.3171 2 Therefore, the mastodon died approximately 7200 years ago. The interY1 shown in Figure section of the graphs of 5.6-5, confirms the solution. 0.42 x 5730, and 0.5 Y2 1 2 ■ Example 6 Compound Interest If $3000 is to be invested at 8% per year, compounded quarterly, in how many years will the investment be worth $10,680? Solution The interest rate per quarter r is 0.08 4 will take the investment to be worth $10,680 use the compound interest formula or 0.02. To find the time t that it .02 t 1.02 t 2 10,680 3000 1 10,680 3000 1 1.02 2 1.02 2 1 ln 1.02 t ln 3.56 ln 1.02 3.56 1 ln 3.56 ln ln 3.56 t 1 t 2 64.1208 quarters 0 0 Figure 5.6-6 Therefore, it will take 64.12 quarters, or 100 10,680 section of the graphs of Figure 5.6-6, confirms the solution. Y1 and 16.03 64.12 4 3000 Y2 years. The inter- 1 0.02 x, 2 1 shown in ■ Example 7 Population Growth A biologist knows that if there are no inhibiting or stimulating factors, the population of a certain type of bacteria will increase exponentially. The population at time t is given by the function Pert, S t 2 1 Section 5.6 Solving Exponential and Logarithmic Equations 383 where P is the initial population and r is the continuous growth rate. The biologist has a culture that contains 1000 bacteria, and 7 hours later there are 5000 bacteria. a. Write the function for this population. b. When will the population reach 1 billion? Solution a. The initial population P is 1000. To find the growth rate r, use the fact that S 7 1 2 5000. 1 2 t 1000ert S 5000 1000er 1 5 e7r ln 5 ln e7r ln 5 7r ln e ln 5 7r 7 2 ln e 1 r ln 5 7 0.2299 Therefore, the function for this population is S t 2 1 1000e0.2299t 5000 Y1 The intersection of the graphs of in Figure 5.6-7, confirms the value of r. b. Find the value of t when S(t) is 1 billion. 1000e0.2299t 1,000,000,000 e0.2299t 1,000,000 ln e0.2299t ln 1,000,000 0.2299t ln e ln 1,000,000 0.2299t ln 1,000,000 t ln 1,000,000 0.2299 60.0936 hours and Y2 1000e7x, shown ln e 1 The bacteria population will reach 1 billion after about 60 hours. The 1000e0.2299x 1,000,000,000, intersection of the graphs of shown in Figure 5.6-8, confirms the solution. and Y1 Y2 ■ Example 8 Inhibited Population Growth A population of fish in a lake at time t months is given by the function F. F t 2 1 20,000 t 1 24e 4 How long will it take for the fish population to reach 15,000? 6,000 0 0 Figure 5.6-7 0.5 1.5109 0 108 Figure 5.6-8 100 384 Chapter 5 Exponential and Logarithmic Functions Solution Find the value of t when F(t) is 15,000. 15,000 20,000 1 24e t 4 15,000 1 1 24e 1 24e t 4 20,000 2 4 20,000 t 15,000 24e 4 4 t 3 4 1 t e 3 ln e t 4 ln 1 1 24 1 72 t 4 ln e ln 1 ln 72 0 ln 72 t 4 t 4 ln 72 17.1067 ln e 1 and ln 1 0 Therefore, it will take a little more than 17 months for the population to and reach 15,000. The intersection of the graphs of 15,000 Y1 30 Y2 20,000 x 1 24e 4 Figure 5.6-9 shown in Figure 5.6-9 confirms the solution. ■ 25,000 0 5,000 Logarithmic Equations Properties of one-to-one functions are useful when solving logarithmic equations, as shown in Example 9. Example 9 Equations with Only Logarithmic Terms Solve the equation tion with a graph. ln 1 x 3 ln 1 2 2x 1 2 ln x . 2 1 2 Confirm your solu- Solution First use the Product and Power Laws to rewrite the equation. ln 1 x 3 ln 2 x 3 3 1 ln 21 ln 2 2x 1 ln x 1 2 1 ln x2 2x 1 2 4 2 ˛ ln x2 2x2 5x 3 2x2 5x 3 x2 x2 5x 3 0 1 2 y ln x is a one-to-one function 6 0 6 6 0 6 Figure 5.6-10 Figure 5.6-11 10 30 Section 5.6 Solving Exponential and Logarithmic Equations 385 Use the Quadratic Formula to solve for x 237 2 5 1 1 2 2 1 2 2 21 5.5414 or x 5 237 1 0.5414 2 x 3, x 5 237 2 0.5414 can- Because ln x 3 1 2 is undefined for not be a solution. Therefore, the only solution of the original equation is x 5 237 2 2x 1 ln and The intersection of the graphs of 5.5414. 2 shown in Figure 5.6-10, confirms the solution. ■ x 3 ln ln x Y1 Y2 , 1 2 1 1 2 2 Equations that involve both logarithmic and constant terms may be solved by using the basic property of logarithms. 10log v v and eln v v Example 10 Equations with Logarithmic and Constant Terms Solve the equation a graph. ln 1 x 3 2 5 ln x 3 1 2 . Confirm your solution with Solution First get all the logarithmic terms on one side of the equal sign and the constants on the other. Then rewrite the side that contains the logarithms as a single logarithm. ln 1 x 3 2 ln ln 2 ln ln ln 5 5 5 x 3 2 2 5 x3 2 e eln 2 5 x 3 e 2 5 x e ln 1 2 3 15.1825 x 3 The intersection of the graphs of shown in Figure 5.6-11, confirms the solution. and Y2 Y1 1 2 5 ln 1 x 3 , 2 ■ Example 11 Equations with Logarithmic and Constant Terms Solve the equation with a graph. log 1 x 16 2 2 log x 1 . 2 1 Confirm your solution 386 Chapter 5 Exponential and Logarithmic Functions Solution log 1 x 1 2 1 x 16 log 3 1 log 1 log 1 log 2 log x 16 16 2 4 21 2 x2 17x 16 2 x217x16 2 102 10log 1 x2 17x 16 100 x2 17x 84 0 x 4 0 x 21 x 4 0 or x 21 0 x 4 x 21 x 1 log 21 1 2 x 16 1 log Because and 1 be a solution. Therefore, the only solution is x 16 Y1 the graphs of 5.6-12, confirms the solution. 2 log and Y2 2 2 1 2 log 1 x 21. x 1 2 are not defined for x 4, it cannot The intersection of shown in Figure , ■ 3 0 3 30 Figure 5.6-12 Exercises 5.6 In Exercises 1–8, solve the equation without using logarithms. 1. 3x 81 2. 3x 3 30 3. 3x1 95x 4. 45x 162x1 5. 35x9x 2 27 6. 2x 25x 1 16 24. 4x 6 2x 8 25. e2x 5e x 6 0 Hint: Let u ex. 26. 2e 2x 9ex 4 0 27. 6e 2x 16e x 6 28. 8e 2x 8e x 6 29. 4x 6 4 x 5 7. 9x 2 3 5x2 8. 4x 21 8x In Exercises 9 – 29, solve the equation. Give exact answers (in terms of natural logarithms). Then use a calculator to find an approximate answer. 9. 3x 5 11. 2x 3x1 13. 312x 5x5 15. 213x 3x1 17. e2x 5 19. 1.4x 21 6e 21. 2.1e x 2 ln 3 5 10. 5x 4 12. 4x2 2x1 14. 43x1 3x2 16. 3z3 2z 18. 3x 2 e 20. 3.4e x 3 5.6 22. 7.8e x 3 ln 5 14 In Exercises 30–32, solve the equation for x. 30. ex e ex e x x t 31. x ex e 2 t 32. ex e ex e x x t 33. Prove that if ln u ln v, basic property of inverses u v. then e ln v v. Hint: Use the 34. a. Solve 7 x 3 using natural logarithms. Give an exact answer, not an approximation. b. Solve 7 x 3 using common logarithms. Give an exact answer, not an approximation. c. Use the change-of-base formula in Excursion 5.5.A to show that your answer
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s in parts a and b are the same. In Exercises 35–44, solve the equation. (See Example 9.) 23. 9x 4 3x 3 0 u 3x. Hint: Note that 9x 2; let 3x 1 2 35. ln 1 3x 5 2 4x 1 ln 11 ln 2 log x 1 2 1 2 log 2 36. log 1 Section 5.6 Solving Exponential and Logarithmic Equations 387 2 59. Krypton-85 loses 6.44% of its mass each year. What is its half-life? 37. log 3x 1 log 2 log 4 log x 2 1 x 6 2 ln 10 ln 38. ln 1 2 x 1 1 2 1 ln 2 39. 2 ln x ln 36 40. 2 log x 3 log 4 41. ln x ln x 1 1 2 ln 3 ln 4 42. ln 1 6x 1 ln x 1 2 2 ln 4 43. ln x ln 3 ln x 5 44. ln 1 2x 3 2 2 1 ln x ln e In Exercises 45–52, solve the equation. ln x 1 45. ln 46. ln x 9 2 2x 1 1 1 2 47. log x log 48. log x 1 1 2 1 ln log 1 1 2 60. Strontium-90 loses 2.5% of its mass each year. What is its half-life? 61. The half-life of a certain substance is 3.6 days. How long will it take for 20 grams to decay to 3 grams? 62. The half-life of cobalt-60 is 4.945 years. How long will it take for 25 grams to decay to 15 grams? Exercises 63–68 deal with the compound interest formula which was discussed in Section 5.3 and used in Example 6 of this section. A P(1 r)t, 63. At what annual rate of interest should $1000 be invested so that it will double in 10 years, if interest is compounded quarterly? 64. Find how long it takes $500 to triple if it is invested at 6% in each compounding period. a. annually b. quarterly c. daily 49. log 2x2 1 2 50. log23 x2 21x 2 3 51. ln x2 1 ln 1 2 x 1 2 1 1 ln x 1 1 2 52. ln ln 1 1 x 1 x 1 2 2 2 Exercises 53 – 62 deal with the half-life function M(x) c(0.5) which was discussed in Section 5.3 and used in Example 5 of this section. x h, 53. How old is a piece of ivory that has lost 36% of its carbon-14? 54. How old is a mummy that has lost 49% of its carbon-14? 55. Find when part of the Pueblo Benito ruins was built if the doorway timbers have 89.14% of their original carbon-14. (See the image on the first page of this chapter.) 56. How old is a wooden statue that has only one- third of its original carbon-14? 57. A quantity of uranium decays to two-thirds of its original mass in 0.26 billion years. Find the halflife of uranium. 58. A certain radioactive substance loses one-third of its original mass in 5 days. Find its half-life. 65. a. How long will it take to triple your money if you invest $500 at a rate of 5% per year compounded annually? b. How long will it take at 5% compounded quarterly? 66. At what rate of interest compounded annually should you invest $500 if you want to have $1500 in 12 years? 67. How much money should be invested at 5% interest compounded quarterly so that 9 years later the investment will be worth $5000? This answer is called the present value of $5000 at 5% interest. 68. Find a formula that gives the time needed for an investment of P dollars to double, if the interest rate is r% compounded annually. Hint: Solve the A 2P. compound interest formula for t, when Exercises 69 – 76 deal with functions of the form f(x) Pekx, where k is the continuous exponential growth rate. See Example 7. 69. The present concentration of carbon dioxide in the atmosphere is 364 parts per million (ppm) and is increasing exponentially at a continuous yearly k 0.004 rate of 0.4% (that is, will it take for the concentration to reach 500 ppm? ). How many years 70. The amount P of ozone in the atmosphere is currently decaying exponentially each year at a 388 Chapter 5 Exponential and Logarithmic Functions continuous rate of % that is, k 0.0025 . How 1 long will it take for half the ozone to disappear 2 1 4 that is, when will the amount be a answer is the half-life of ozone. P 2 b ? Your 71. The population of Brazil increased exponentially from 151 million in 1990 to 173 million in 2000. a. At what continuous rate was the population growing during this period? b. Assuming that Brazil’s population continues to increase at this rate, when will it reach 250 million? 72. Outstanding consumer debt increased exponentially from $781.5 billion in 1990 to $1765.5 billion in 2002. (Source: Federal Reserve Bulletin) a. At what continuous rate is consumer debt growing? b. Assuming this rate continues, when will consumer debt reach $2500 billion? 73. The probability P percent of having an accident while driving a car is related to the alcohol level of the driver’s blood by the formula where k is a constant. Accident statistics show that the probability of an accident is 25% when the blood alcohol level is P 25, a. Find k. Use b. At what blood alcohol level is the probability t 0.15. not 0.25. P e kt, of having an accident 50%? 74. Under normal conditions, the atmospheric pressure (in millibars) at height h feet above sea P level is given by positive constant. a. If the pressure at 18,000 feet is half the pressure where k is a 1015e kh, h 1 2 at sea level, find k. b. Using the information from part a, find the atmospheric pressure at 1000 feet, 5000 feet, and 15,000 feet. 75. One hour after an experiment begins, the number of bacteria in a culture is 100. An hour later there are 500. a. Find the number of bacteria at the beginning of the experiment and the number 3 hours later. b. How long does it take the number of bacteria at any given time to double? 76. If the population at time t is given by ce kt, find a formula that gives the time it takes for the population to double. S t 1 2 77. The spread of a flu virus in a community of 45,000 people is given by the function f t 2 1 45,000 1 224e 0.889t , 1 t is the number of people infected in f where week t. a. How many people had the flu at the outbreak 2 of the epidemic? after 3 weeks? b. When will half the town be infected? 78. The beaver population near a certain lake in year t is approximately p t . 0.5544t 2000 1 199e 1 2 a. When will the beaver population reach 1000? b. Will the population ever reach 2000? Why? 79. Critical Thinking According to one theory of N c learning, the number of words per minute N that a person can type after t weeks of practice is kt where c is an upper limit given by , that N cannot exceed and k is a constant that must be determined experimentally for each person. a. If a person can type 50 wpm (words per 1 e 2 1 minute) after 4 weeks of practice and 70 wpm after 8 weeks, find the values of k and c for this person. According to the theory, this person will never type faster than c wpm. b. Another person can type 50 wpm after 4 weeks of practice and 90 wpm after 8 weeks. How many weeks must this person practice to be able to type 125 wpm? 80. Critical Thinking Wendy has been offered two jobs, each with the same starting salary of $24,000 and identical benefits. Assuming satisfactory performance, she will receive a $1200 raise each year at the company A, whereas the company B will give her a 4% raise each year. a. In what year (after the first year) would her salary be the same at either company? Until then, which company pays better? After that, which company pays better? b. Answer the questions in part a assuming that the annual raise at company A is $1800. Section 5.7 Exponential, Logarithmic, and Other Models 389 5.7 Exponential, Logarithmic, and Other Models Objectives • Model real data sets with power, exponential, logarithmic, and logistic functions Many data sets can be modeled by suitable exponential, logarithmic, and related functions. Most calculators have regression procedures for constructing the models described in the table below. Model Power Equation y axr Exponential y ab x or y ae kx Examples y 5x2.7 y 2 1.64 1 x 2 y 3.5x 0.45 y 2e0.4947x Logarithmic y a b ln x y 5 4.2 ln x y 2 3 ln x Logistic y a 1 be kx y 20,000 1 24e 0.25x y 650 1 6e 0.3x Exponential Models In the table of values for the exponential model examine the patterns in the ratios of successive y-values. y 3 2x that follows, x y 0 3 4 48 8 768 12 16 12,288 196,608 48 3 16 768 48 16 12,288 768 16 196,608 12,288 16 At each step, x changes from x to y changes from and the ratio of successive y-values is always the same. x 4, 3 2x to 3 2x4, 3 2x4 3 2x 3 2x 24 3 2x 24 16 A similar argument applies to any exponential model and shows that if x changes by a fixed amount k, then the ratio of the corresponding y-values is the constant above, b is 2 and k is 4. This fact identifies the model that would best represent the data. In the exponential model y 3 2x bk. y ab x When the ratio of successive entries in a table of data is approximately constant, an exponential model is appropriate. Example 1 U.S. Population Before the Civil War In the years before the Civil War, the population of the United States grew rapidly, as shown in the following table. Find a model for this growth. 390 Chapter 5 Exponential and Logarithmic Functions Year 1790 1800 1810 1820 Population in millions 3.93 5.31 7.24 9.64 Year 1830 1840 1850 1860 Population in millions 12.86 17.07 23.19 31.44 [Source: U.S. Bureau of the Census] 50 Solution x 0 The data points, with corresponding to 1790, are shown in Figure 5.7-1. Their shape suggests either a polynomial graph of even degree or an exponential graph. Since populations generally grow exponentially, an exponential model is likely to be a good choice. This can be confirmed by looking at the successive entries in the table. Year 1790 Population 3.93 Year 1830 Population 12.86 5.31 3.93 1.351 1800 5.31 1840 17.07 1810 7.24 7.24 5.31 9.64 7.24 1.363 1.331 1850 23.19 1820 9.64 1860 31.44 12.86 9.64 1.334 1830 12.86 17.07 12.86 1.327 23.19 17.07 1.359 31.44 23.19 1.356 Because the ratios are almost constant, as they would be in an exponential model, use regression to find an exponential model. The procedure is the same as for linear and polynomial regression. An exponential regression produces this model. y 3.9572 1.0299x 1 2 The graph of the exponential model in Figure 5.7-2 appears to fit the data well. In fact, you can readily verify that the model has an error of less than 1% for each of the data points. Furthermore, as discussed before this example, when x changes by 10, the value of y chang
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es by approximately 1.029910 1.343, which is very close to the successive ratios of the data. ■ −5 0 Figure 5.7-1 100 NOTE Throughout this section, coefficients are rounded for convenient reading, but the full expansion is used for calculations and graphs. 50 −5 0 100 Figure 5.7-2 Section 5.7 Exponential, Logarithmic, and Other Models 391 Logistic Models A logistic model represents growth that has a limiting factor, such as food supplies, war, new diseases, etc. Logistic models are often used to model population growth, as shown in Example 2. Example 2 U.S. Population After the Civil War After the Civil War, the population of the United States continued to increase, as shown in the following table. Find a model for this growth. Year 1870 1880 1890 1900 1910 Population in millions 38.56 50.19 62.98 76.21 92.23 Solution Year 1920 1930 1940 1950 Population in millions 106.02 123.20 132.16 151.33 Year 1960 1970 1980 1990 2000 Population in millions 179.32 202.30 226.54 248.72 281.42 The model from Example 1 does not remain valid, as can be seen in Figure 5.7-3, which shows its graph together with all the data points from 1790 through 2000, where corresponds to 1790. x 0 500 −5 5 220 Figure 5.7-3 The rate of growth has steadily decreased since the Civil War. For instance, 50.19 38.56 the ratio of the first two entries is and the ratio of the last 1.302 two is 281.42 248.72 1.131. So an exponential model may not be the best choice. Other possibilities are polynomial models, which grow at a slower rate, or logistic models, in which the growth rate decreases with time. Figure 5.7-4 on the next page shows these models compared to an exponential model. 392 Chapter 5 Exponential and Logarithmic Functions Exponential Model y 6.0662 1.02039x y Quartic Model 2.76 10 x4 7.94 10 1 0.0093x2 0.1621x 5.462 8 2 1 5 Logistic Model x3 2 y 442.10 1 56.329e 0.022x 500 500 500 −5 5 250 −5 5 250 −5 5 250 Figure 5.7-4 The quartic and logistic models fit the data better than does the exponential model. The quartic model indicates unlimited future growth, but the logistic model has the population growing more slowly in the future. ■ Exponential versus Power Models In Example 1, the ratios of successive entries of the data table were used to determine that an exponential model was appropriate. Another way to determine if an exponential model might be appropriate is to consider the exponential function as shown below. y abx, y ab x ln y ln 1 ln y ln a ln b x ln y ln a x ln b ab x 2 Take the natural logarithm of each side Product Law Power Law Because ln a and ln b are constants, let k ln a and m ln b. ln y ln a x ln b ln y x ln a ln b 1 ln y mx k 2 The points (x, line with slope m and y-intercept k. Consequently, a guideline for determining if an exponential model is appropriate is as follows. lie on a straight ln y) (x, y) are data points and if the points If approximately linear, then an exponential model may be appropriate for the data. (x, ln y) are Similarly, consider the power function y axr y axr ln y ln 1 ln y ln a ln xr ln y ln a r ln x axr 2 Take the natural logarithm of each side Product Law Power Law Section 5.7 Exponential, Logarithmic, and Other Models 393 Because r and ln a are constants, let Then: k ln a. ln y ln a r ln x ln y r ln x ln a ln y r ln x k lie on the straight line with slope r and Thus, the points y-intercept k. Consequently, a guideline for determining if a power model is appropriate is as follows ln x, ln y 1 2 (x, y) (ln x, ln y) If are data points and if the points approximately linear, then a power model may be appropriate for the data. are Example 3 Different Planet Years The length of time that a planet takes to make one complete rotation around the sun is that planet’s “year.’’ The table below shows the length of each planet’s year, relative to an Earth year, and the average distance of that planet from the Sun in millions of miles. Find a model for this data in which x is the length of the year and y is the distance from the Sun. Planet Year Distance Planet Year Distance Mercury Venus Earth Mars 0.24 0.62 1.00 1.88 Jupiter 11.86 Solution 36.0 67.2 92.9 141.6 483.6 Saturn Uranus 29.46 84.01 886.7 1783.0 Neptune 164.79 2794.0 Pluto 247.69 3674.5 Figure 5.7-5 shows the data points for the five planets with the shortest years. Figure 5.7-6 shows all of the data points, but on this scale, the first four points look like a single large point near the origin. 600 5,000 −1 0 15 −25 300 Figure 5.7-5 0 Figure 5.7-6 Technology Tip Suppose the x- and ycoordinates of the data points are stored in lists L1 L2 and Keying in , respectively. LN L2 STO S L4 L4 produces the list , whose entries are the natural logarithms of the numbers in list , and stores it in the statistics editor. You can L4 then use lists plot the points (x, ln y). and L1 L2 to 394 Chapter 5 Exponential and Logarithmic Functions 1 x, ln y Plotting the point produces the graph for each data point shown in Figure 5.7-7. Its points do not form a linear pattern, so an exponential model is not appropriate. The points shown in Figure 5.7-8 do form a linear pattern, which suggests that a power model will work. ln x, ln y x, y 1 2 2 1 2 10 −5 0 10 0 250 −3 7 Figure 5.7-7 Figure 5.7-8 A power regression produces this model: y 92.8932x0.6669 Its graph in Figures 5.7-9 and 5.7-10 show that it fits the original data points well. 600 5,000 −1 0 15 −25 0 Figure 5.7-9 Figure 5.7-10 300 ■ Logarithmic Models Consider the logarithmic function Because a and b are constants, let y m and k ln x 1 2 y b ln x a m b k a. Then: ln x, y line with slope m and The points y-intercept k. Consequently, a guideline for determining if a logarithmic model is appropriate is as follows. lie on the straight 2 1 (x, y) If are approximately linear, then a logarithmic model may be appropriate for the data. are data points and if the points (ln x, y) Section 5.7 Exponential, Logarithmic, and Other Models 395 Example 4 Logarithmic Population Growth Find a model for population growth in El Paso, Texas, given the information in the following table. Year 1950 1970 1980 1990 2000 Population 130,485 322,261 425,259 515,342 563,662 [Source: U.S. Bureau of the Census.] Solution The scatter plot of the data, where Figure 5.7-11, suggests a logarithmic curve with a very slight bend. corresponds to 1950, shown in x 50 700,000 700,000 40 0 Figure 5.7-11 120 3 0 Figure 5.7-12 6 To determine whether a logarithmic model is appropriate for this data, ln 50, 130,485 plot points (ln 100, 563,662). Because these points, shown in Figure 5.7-12, appear to be approximately linear, a logarithmic model seems appropriate. that is ln x, y p ; 1 2 2 1 Using logarithmic regression, the model is: y 2,382,368.345 640,666.815 ln x 700,000 CAUTION When using logarithmic models, you must have data points with positive first coordinates because logarithms of negative numbers and 0 are not defined. 40 0 Figure 5.7-13 120 The graph of this model, shown in Figure 5.7-13, shows that it is a good fit for the data. ■ 396 Chapter 5 Exponential and Logarithmic Functions Exercises 5.7 4. y In Exercises 1–10, state which of the following models might be appropriate for the given scatter plot of data. More than one model may be appropriate. Model A. Linear Corresponding function y ax b B. Quadratic y ax2 bx c 5. y C. Power D. Cubic y axr y ax 3 bx 2 cx d E. Exponential y ab x F. Logarithmic y a b ln x G. Logistic y a 1 be kx 6. y 7. y 8. y 1. y 2. y 3 Section 5.7 Exponential, Logarithmic, and Other Models 397 9. y 10. y x x In Exercises 11 and 12, compute the ratios of successive entries in the table to determine whether an exponential model is appropriate for the data. 11. 12. x y 0 3 2 4 6 8 10 15.2 76.9 389.2 1975.5 9975.8 x y 1 3 3 21 5 55 7 9 11 105 171 253 13. a. Show algebraically that in the logistic model for the U.S. population in Example 2, the population can never exceed 442.10 million people. b. Confirm your answer in part a by graphing the logistic model in a window that includes the next three centuries. 14. According to estimates by the U.S. Bureau of the Census, the U.S. population was 287.7 million in 2002. Based on this information, which of the models in Example 2 appears to be the most accurate predictor? 15. Graph each of the following power functions in a window with 1.5 a. x x f 0 x 20. b. x g 1 2 x0.75 c. h x 2 1 x2.4 1 2 16. Based on your graphs in Exercise 15, describe the a 7 0 y axr, when general shape of the graph of and r is as described below. 0 6 r 6 1 a. r 6 0 b. c. r 7 1 In Exercises 17–20, determine whether an exponential, power, or logarithmic model (or none or several of these) is appropriate for the data by determining which (if any) of the following sets of points are approximately linear, where the given data set consists of the points (x, y) . 5 6 17. 18. 19. 20. (x, ln y) 5 6 5 (ln x, ln y) 6 5 (ln x, y 25 5 81 7 9 11 175 310 497 3 385 5 17 6 74 10 27 9 14 12 2.75 15 35 20 40 15 0.5 25 43 18 0.1 30 48 x y 5 2 10 15 20 25 30 110 460 1200 2500 4525 21. The table shows the number of babies born as twins, triplets, quadruplets, etc., in recent years. Year Multiple births 1989 1990 1991 1992 1993 1994 1995 92,916 96,893 98,125 99,255 100,613 101,658 101,709 398 Chapter 5 Exponential and Logarithmic Functions a. Sketch a scatter plot of the data, with x 1 corresponding to 1989. b. Plot both of the following models on the same screen with the scatter plot: 93,201.973 4,545.977 ln x x f 1 2 and g x 2 1 102,519.98 1 0.1536e 0.4263x c. Use the table feature to estimate the number of multiple births in 2000 and 2005. d. Over the long run, which model do you think is the better predictor? 22. The graph shows the Census Bureau’s estimates of future U.S. population. Infant Year mortality rate* Year Infant mortality rate* 1920 1930 1940 1950 1960 1970 76.7 60.4 47.0 29.2 26.0 20.0 1980 1985 1990 1995 2000 12.6 10.6 9.2 7.6 6.9 *Rates are infant (under 1 year) deaths per 1000 live births. a. Sketch a scatte
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r plot of the data, with 425 400 375 350 325 300 275 250 U.S. Population Projections: 2000–2050 403.687 377.350 351.070 324.927 299.862 281.422 2000 2010 2020 2030 2040 2050 Year a. How well do the projections in the graph compare with those given by the logistic model in Example 2? b. Find a logistic model of the U.S. population, using the data given in Example 2 for the years from 1900 to 2000, with corresponding to 1900. x 0 c. How well do the projections in the graph compare with those given by the model in part b? 23. Infant mortality rates in the United States are shown in the following table. corresponding to 1900. b. Verify that the set of points x, ln y , 2 are the original data points, is 1 x, y 1 approximately linear. 2 where c. Based on part b, what type of model would be appropriate for this data? Find such a model. 24. The average number of students per computer in the U.S. public schools (elementary through high school) is shown in the table below. Fall of school year Students per computer 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 32 25 22 20 18 16 14 10.5 10 7.8 6.1 5.7 5.4 Section 5.7 Exponential, Logarithmic, and Other Models 399 a. Sketch a scatter plot of the data, with x 1 corresponding to 1987. b. Find an exponential model for the data. c. Use the model to estimate the number of students per computer in 2003. d. In what year, according to this model, will each student have his or her own computer in school? e. What are the limitations of this model? 25. The number of children who were home-schooled in the United States in selected years is shown in the table below. Fall of school year Number of children (in 1000s) 1985 1988 1990 1992 1993 1994 1995 1996 1997 1999 2000 183 225 301 470 588 735 800 920 1100 1400 1700 [Source: National Home Education Research Institute] a. Sketch a scatter plot of the data, with x 0 corresponding to 1980. b. Find a quadratic model for the data. c. Find a logistic model for the data. d. What is the number of home-schooled children predicted by each model for the year 2003? e. What are the limitations of each model? 26. a. Find an exponential model for the federal debt, based on the data in the following table. Let x 0 correspond to 1960. Accumulated gross federal debt Amount (in billions of dollars) 284.1 313.8 370.1 533.2 907.7 1823.1 3233.3 4974.0 5674.2 Year 1960 1965 1970 1975 1980 1985 1990 1995 2000 b. Use the model to estimate the federal debt in 2003. 27. The table gives the life expectancy of a woman born in each given year. Life Expectancy (in years) 51.8 54.6 61.6 65.2 71.1 Year 1910 1920 1930 1940 1950 Life Expectancy (in years) 73.1 74.7 77.5 78.8 79.4 Year 1960 1970 1980 1990 2000 [Source: National Center for Health Statistics] a. Find a logarithmic model for the data, with x 10 corresponding to 1910. b. Use the model to find the life expectancy of a woman born in 1986. For comparison, the actual expectancy is 78.3 years. c. Assume the model remains accurate. In what year will the life expectancy of a woman born in that year be at least 81 years? 400 Chapter 5 Exponential and Logarithmic Functions 28. The table gives the death rate in motor vehicle accidents, per 100,000 population, in selected years. Year 1970 1980 1985 1990 1995 2000 Death Rate 26.8 23.4 19.3 18.8 16.5 15.6 a. Find an exponential model for the data, with x 0 corresponding to 1970. b. Use the model to predict the death rate in 1998 and in 2002. c. Assuming the model remains accurate, when will the death rate drop to 13 per 100,000? 29. Worldwide production of computers has grown dramatically, as shown in the first two columns of the following table. a. Sketch a scatter plot of the data, with x 1 corresponding to 1985. b. Find an exponential model for the data. c. Use the model to complete column 3 of the table. d. Fill in column 4 of the table by dividing each entry in column 2 by the preceding one. e. What does column 4 tell you about the appropriateness of the model? Worldwide shipments (in thousands) Predicted number of shipment (in thousands) Ratio 14.7 15.1 16.7 18.1 21.3 23.7 27 32.4 38.9 47.9 60.2 70.9 84.3 Year 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 [Source: Dataquest Section 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 Section 5.5.A Section 5.6 Section 5.7 nth root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 Rational exponents . . . . . . . . . . . . . . . . . . . . . . . 330 Laws of exponents . . . . . . . . . . . . . . . . . . . . . . . 330 Rationalizing numerators and denominators . . . 332 Irrational exponents . . . . . . . . . . . . . . . . . . . . . . 333 Graphs of exponential functions . . . . . . . . . . . . 336 Exponential growth and decay . . . . . . . . . . . . . . 339 The number e and its exponential function . . . . 341 Other exponential functions (logistic models) . . 341 Compound interest . . . . . . . . . . . . . . . . . . . . . . . 345 Continuous compounding . . . . . . . . . . . . . . . . . 348 Constructing exponential growth functions . . . . 349 Constructing exponential decay functions . . . . . 351 Radioactive decay . . . . . . . . . . . . . . . . . . . . . . . . 352 Common logarithms . . . . . . . . . . . . . . . . . . . . . . 356 Natural logarithms . . . . . . . . . . . . . . . . . . . . . . . 358 Graphing logarithmic functions . . . . . . . . . . . . . 359 Solving logarithmic equations by graphing . . . . 360 Basic properties of logarithms . . . . . . . . . . . . . . 364 Product Law of Logarithms . . . . . . . . . . . . . . . . 365 Quotient Law of Logarithms . . . . . . . . . . . . . . . 366 Power Law of Logarithms . . . . . . . . . . . . . . . . . 366 Richter scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 Logarithms to base b . . . . . . . . . . . . . . . . . . . . . 371 Basic properties of logarithms to base b . . . . . . 372 Laws of logarithms to base b . . . . . . . . . . . . . . . 373 Change-of-base formula . . . . . . . . . . . . . . . . . . . 374 Exponential equations . . . . . . . . . . . . . . . . . . . . 379 Logarithmic equations . . . . . . . . . . . . . . . . . . . . 384 Exponential models . . . . . . . . . . . . . . . . . . . . . . 389 Logistic models . . . . . . . . . . . . . . . . . . . . . . . . . 391 Power models . . . . . . . . . . . . . . . . . . . . . . . . . . 392 Logarithmic models . . . . . . . . . . . . . . . . . . . . . . 394 401 402 Chapter Review Important Facts and Formulas • Rational Exponents: 1 n 2n c c 1 k ct • Laws of Exponents: crcs crs cr cs crs s crs cr 2 1 a cd 1 2 r cr dr r cr c dr db r 1 cr 10x log x is the inverse function of f 10log v v for all v 7 0 and log 10u u for all u ln x is the inverse function of f ex: x 2 1 eln v v for all v 7 0 and ln eu u for all u logb x is the inverse function of blogb v v for all v 7 0 and logb v, w 7 0 and any k: x k 1 2 bx: • Logarithm Laws: For all u for all u bu 1 2 vw ln 1 2 ln v ln w logb vw 2 1 logb v logb w logb a logb v wb vk 2 1 logb v logb w k logb v v wb vk ln a ln ln v ln w k ln v 1 2 • Exponential Growth Functions: P 1 Pax Pekx x x x f f f • Exponential Decay Functions: P 1 Pax Pekx • Logistic Function: x f 1 2 a 1 be kx • Compound Interest Formula: • Continuous Compounding: 1 r A P 1 A Pert P • Radioactive Decay Function: 1 • Change of Base Formula: logb v ln v ln b x f 1 2 0.5 t 2 x h 2 Review Exercises In Exercises 1–6, simplify the expression. Section 5.1 1. 323 c12 4. 3 5 1 5 2 2 1 1 3c 4c 1 1 2d 2d 2 4c 1 2 2 3 2 1 2c 2 2 4 1 2 23 4c3 d2 2. A c2d 3 A B 2 B 5 Chapter Review 403 3. A 2 a 2 3 b 5 a3 b6 B A 4 3 B 6. 3 2 c A 1 2c 2 3c 3 2 B In Exercises 7 and 8, simplify and write the expression without radicals or negative exponents. 7. 23 6c4 d14 2 d2 23 48c 8u5 2 8. 1 3 1u 1 4 2 2u8 9. Rationalize the numerator and simplify: 22x 2h 1 22x 1 h 10. Rationalize the denominator: 5 2x 3 Section 5.2 In Exercises 11–16, list the transformations needed to transform the graph of f(x) 5x into the graph of the given function. 11. 14 5x 52x 12. 15. h h x x 1 1 2 2 53x 5x 4 13. 16 2x 5x2 In Exercises 17 and 18, find a viewing window (or windows) that shows a complete graph of the function. 17. x f 1 2 2x 3x2 18. g x 1 2 850 1 5e 0.4x 19. Compunote offers a starting salary of $60,000 with $1000 yearly raises. Calcuplay offers a starting salary of $30,000 with a 6% raise each year. a. Complete the following table for each company. Year Compunote Year Calcuplay 1 2 3 4 5 $60,000 $61,000 $30,000 $31,800 1 2 3 4 5 b. For each company write a function that gives your salary in terms of years employed. c. If you plan on staying with the company for only five years, which job should you take to earn the most money? d. If you plan on staying with the company for 20 years, which is your best choice? 404 Chapter Review 20. A computer software company claims that the following function models the “learning curve” for their software. P t 2 1 100 1 48.2e 0.52t where t is measured in months and P(t) is the average percent of the software program’s capabilities mastered after t months. a. Initially what percent of the program is mastered? b. After 6 months what percent of the program is mastered? c. Roughly, when can a person expect to “learn the most in the least amount of time”? d. If the company’s claim is true, how many months will it take to have completely mastered the program? Section 5.3 21. Phil borrows $800 at 9% annual interest, compounded annually. a. How much does he owe after 6 years? b. If he pays off the loan at the end of 6 years, how much interest will he owe? 22. If you invest $5000 for 5 years at 9% annual interest, how much more will you make if interest is compounded continuously than if it is compounded quarterly? 23. Mary Karen invests $2000 at 5.5% annual interest, compounded monthly. a. How much is her investment worth in 3 years? b. When will her investment be worth $12,000? 24. If a $2000 investment grows to $5000 in 14 years, with interest compounded annually,
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what is the interest rate? 25. Company sales are increasing at 6.5% per year. If sales this year are $56,000, write the rule of a function that gives the sales in year x (where x 0 corresponds to the present year). 26. The population of Potterville is decreasing at an annual rate of 1.5%. If the population is 38,500 now, what will be the population x years from now? 27. The half-life of carbon-14 is 5730 years. How much carbon-14 remains from an original 16 grams after 12,000 years? 28. How long will it take for 4 grams of carbon-14 to decay to 1 gram? Section 5.4 In Exercises 29–34, translate the given exponential statement into an equivalent logarithmic one. 29. e6.628 756 32. eab c 30. e5.8972 364 31. er 21 u v 33. 102.8785 756 34. 10cd t In Exercises 35–38, translate the given logarithmic statement into an equivalent exponential one. 35. ln 1234 7.118 38. log 1234 3.0913 39. Find log 0.01 . 2 1 36. ln 1 ax b y 2 37. ln rs 2 1 t Chapter Review 405 In Exercises 40–43, describe the transformation from to the given function. Give the domain and range of the given function. or f (x) log x g(x) ln x 40. h 42 log 1 x 3 2 ln 3x 1 2 41. k 43. k x x 2 2 1 1 log 4 x 1 2 3 ln x 5 44. You are conducting an experiment about memory. The people who 2 1 t M 91 14 ln participate agree to take a test at the end of your course and every month thereafter for a period of two years. The average score for the group is given by the model months after the first test. a. What is the average score on the initial exam? b. What is the average score after three months? c. When will the average drop below 50%? d. Is the magnitude of the rate of memory loss greater in the first month t 1 t 12 ) or after the first year (from 0 t 24, where t is time in t 1 t 0 to , 1 2 after the course (from to t 13 )? e. Hypothetically, if the model could be extended past would it be possible for the average score to be 0%? t 24 months, Section 5.5 In Exercises 45–48, evaluate the given expression without using a calculator. 45. ln e3 46. ln e 47. eln 3 4 48. eln 1 x2y 2 49. Simplify: 3 ln 2x 1 2 ln x 50. Simplify: ln 1 4e e4e 1 2 In Exercises 51 and 52, write the given expression as a single logarithm. 51. ln 3x 3 ln x ln 3y 52. 4 ln x 2 ln x3 4 ln x 2 1 53. Which of the following statements is true? a. ln 10 c. ln 1 7b a ln 2 ln 5 1 21 ln 7 0 2 b. ln d. ln e a 6b e 1 2 ln e ln 6 1 e. None of the above is true. 54. Which of the following statements is false? log 50 a. c. log 5 10 1 2 log 1 ln 1 b. d. log 100 3 log 105 log 6 log 3 log 2 e. All of the above are false. 55. What is the domain of the function f ln x x 1b ? a x 1 2 Section 5.5.A In Exercises 56 and 57, translate the given logarithmic statement into an equivalent exponential one. 56. log5 cd k u 2 1 57. logd uv 1 2 w 58. Write log7 7x log7 y 1 as a single logarithm. 59. log20 400 ? 60. If log3 9x2 4, what is x? 406 Chapter Review Use the following six graphs for Exercises 61 and 62. y 3 2 1 −2 −1 −1 1 2 Figure 2 −1 −1 1 2 Figure II 2 −1 −1 1 2 Figure III y 1 −2 −1 −1 1 2 −2 −3 −2 −1 −1 1 2 −2 −1 −1 1 2 Figure IV Figure V Figure VI 61. If b 7 1, then the graph of a. I b. IV c. V d. VI x logb x f 1 e. none of these 2 could possibly be 62. If 0 6 b 6 1 a. II b. III then the graph of c. IV d. VI x g b x 1 2 e. none of these 1 could possibly be Section Section 5.6 5.6 In Exercises 63–71, solve the equation for x. 63. 8x 4x 23 64. e3x 4 65. 2 4x 5 4 66. 725e 4x 1500 67. u c d ln x 68. 2x 3x3 69. ln x ln 3x 5 ln 2 1 x2 1 2 2 log 2 x 1 1 2 71. log 1 70. ln x 8 2 1 ln x 1 72. At a small community college the spread of a rumor through the population of 500 faculty and students can be modeled by ln n ln 1 1000 2n 2 0.65t ln 998, where n is the number of people who have heard the rumor after t days. a. How many people know the rumor initially (at b. How many people have heard the rumor after four days? c. Roughly, in how many weeks will the entire population have heard the t 0 )? rumor? d. Use the properties of logarithms to write n as a function of t; in other words solve the model above for n in terms of t. Chapter Review 407 e. Enter the function you found in part d into your calculator and use the table feature to check your answers to parts a, b, and c. Do they agree? f. Graph the function. Over what time interval does the rumor seem to “spread” the fastest? 73. The half-life of polonium milligrams, how much will be left at the end of a year? 210Po 1 2 is 140 days. If you start with 10 74. An insect colony grows exponentially from 200 to 2000 in 3 months. How long will it take for the insect population to reach 50,000? 75. Hydrogen-3 decays at a rate of 5.59% per year. Find its half-life. 76. The half-life of radium-88 is 1590 years. How long will it take for 10 grams to decay to 1 gram? 77. How much money should be invested at 8% per year, compounded quarterly, in order to have $1000 in 10 years? 78. At what annual interest rate should you invest your money if you want to double it in 6 years? 79. One earthquake measures 4.6 on the Richter scale. A second earthquake is 1000 times more intense than the first. What does it measure on the Richter scale? Section 5.7 80. The table below gives the population of Austin, Texas. Year 1950 1970 1980 1990 2000 Population 132,459 253,539 345,890 465,622 656,562 a. Sketch a scatter plot of the data, with b. Find an exponential model for the data. c. Use the model to estimate the population of Austin in 1960 and 2005. corresponding to 1950. x 0 81. The wind-chill factor is the temperature that would produce the same cooling effect on a person’s skin if there were no wind. The table shows the wind-chill factors for various wind speeds when the temperature is 25 °F. Wind speed (mph) 0 5 10 15 20 25 30 35 40 45 Wind chill temperature (°F) 25 19 15 13 11 9 8 7 6 5 [Source: National Weather Service] a. What does a 20-mph wind make 25 °F b. Sketch a scatter plot of the data, with c. Explain why an exponential model would be appropriate. d. Find an exponential model for the data. e. According to the model, what is the wind-chill factor for a feel like? x 0 corresponding to 0 mph. 23-mph wind 12 y f (x, f(x)) x (b, f(b)) b Figure 5.C-1 Tangents to Exponential Functions ctions Tangent lines to a curve are important in calculus—where they are used to approximate function values close to a specific point and used for finding the zeros of general functions. The procedure developed in the Can Do Calculus for Chapter 3 will be used here to develop the equations of the tangent lines to exponential functions. Slopes of Secant Lines and Tangent Lines Recall that the difference quotient of f at the specific value x b is given by , where h is the amount of change in the x values from one point to another. The difference quotient can be interpreted as the slope of the secant line As h gets very b that passes through the points b h, f b h and . 1 22 approaches the value of the slope of 22 1 small, the value of f 1 the tangent line at b, f . b 1 22 1 1 f b Also, if f is any function, then the slope of the secant line through and any other point on the graph of f is given by x, f x b, f b 1 22 1 1 1 22 . As the point x, f x 1 22 approaches the point 1 b, f b 1 22 1 , the value of f 1 approaches the value of the slope of the tangent line to the curve at Figure 5.C-1 shows four secant lines that pass through the point The tangent line to f at is shown in red. b, f b b b, f 1 1 b 22 22 2 . . 1 1 22 f Consider the function 0 0, e that pass through Tangent Lines to the Exponential Function e x 2 1 that is Find the values of e 0 and the values of the slopes of secant lines when x is near 0. x e x 1 x 0.01 0.001 0.001 0.99502 0.9995 1.0005 0.01 1.005 408 y 4 2 −4 −2 0 2 4 x −2 −4 Figure 5.C-2 The table suggests that the slope of the tangent line to is 1. The tangent line to the curve at has slope x 0 at m 1. can be found by using the point-slope form of a line. x 0 Therefore, the equation of the tangent line to x 0 x 2 contains the point (0, 1) and e x e at x f f 1 1 2 x m y y0 y 1 1 y 1 x x x02 1 x 0 2 1 Point-slope form of a line y x 1 Equation of tangent line e x and the tangent line to the curve at 0, 1 1 2 is shown x f The graph of 1 in Figure 5.C-2. 2 Example 1 Tangent Line to the Exponential Function Find the tangent line to line. x f 1 2 Solution e x when x 1. Graph f and the tangent x 1, f e 1 e, 1 When is the point where the tangent line will touch the graph. To find the slope of the tangent line, look at values of the difference quotient near 2 1 x 1 . so the point 1 Alternately, you may use the numerical derivative feature of your calculator to find an approximate value of the slope of the tangent line for f You should find that the value of the slope of the tangent line is approximately 2.718282282. x 1. e at x 1 2 x Recall that e f at 1 tangent line’s equation is e 2.71828. 1, e x 2 1 2 x It appears that the slope of the tangent line to is e, which can be proved using calculus. Therefore, the Figure 5.C-3a 10 y e e x 1 1 2 or equivalently y ex and the graphs of f and the tangent line are in Figure 5.C-3b. ■ Calculator Exploration –5 5 –10 Figure 5.C-3b 2, 3, 4. Plot the points Find the slope of the tangent lines to 1, line, along with the corresponding points for tion would represent the graph of these points? 2, where y is the slope of the tangent x 0, 1. What func- x 3, when e x 409 In the Calculator Exploration, you should have found the values and plotted the points shown in Figure 5.C-4. Also shown in Figure 5.C-4 are all the points on the graph of f e x. x 1 2 Slope of the Tangent line to y e x The slope of the tangent line at any point on same value as the y coordinate, at that point. x, e Figure 5.C-4 y e x has the The exponential function is the only function with this characteristic. Example 2 Slope of y e x a. Find the x-value where the slope of b. Write the equation of the tangent line at c. Graph and the tangent line at y e x x 3 y ex 3. e is x 3. on the same screen
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. Solution y e x a. The slope of b. Using the point-slope form of a line with x 3 tion of the tangent line to is when y e e x x 3. 3 3 m e is and 3 3, e 2 1 , the equa- 5 y e 3 e y e at x 3 3 2 1 3x 2e 3 e x 2 3 1 2 c. The graphs of y e x and y e 3 x 2 2 1 are shown in Figure 5.C-5. ■ Exponential Functions with Bases Other Than e The procedure for finding the equation of the tangent line at a specific value of x for exponential functions with bases other than e is the same as that for finding the equation of the tangent line at a specific value of x for any function. 1. Find the values of slopes of secant lines by using the difference quo- tient and several values of x near the point in question. 2. Find the slope of the tangent line by determining the value of the slope suggested by the values found in Step 1. 3. Write the equation of the tangent line using the point-slope form of a linear equation. 4. Confirm your finding by graphing the function and the tangent line. 50 –5 Figure 5.C-5 –5 410 Calculator Exploration x 2, 1, Find the values of the slope of the tangent line for 0, 1, 2, 3. Graph the ordered pairs (x, slope of tangent) and find an equation to represent the graph by using exponential regression. y 2 at x The points to be graphed in the Calculator Exploration are 2, 0.17329 1, 0.34657 1 1 1 2 and the regression equation is 2 0, 0.69315 1, 1.3863 2, 2.7726 3, 5.5452 2 2 1 2 1 2 1 y 0.6931471806 x 2 . 2 1 ln 2 0.6931471806. c, f on the graph of c In fact, the slope of the tangent line at or is given by ln 2 f 2 x f c , x 1 22 1 2 1 2 Notice that any point 2 ln 2 . 1 c 1 2 Example 3 Tangent Line of y 2 x Find the equation of the tangent line to the curve confirm your result by graphing. y 2 x at x 4, and 30 Solution –1 6 –5 Figure 5.C-6 is 4 , 4, 2 1 11.09, 2 or (4, 16). The slope of the tanso the equation of the tangent The point on the curve at gent line at that point is line of x 4 ln 2 16 at (4, 16) is 2 21 x f 1 x 2 1 2 y 16 11.09 y 11.09 16 The graph of ure 5.C-6. x f 1 2 2 x and y 11.09 16 2 are shown in Fig- ■ Exercises In Exercises 1–4, write the equation of the tangent line f (x) ex. at the following values of x for the function Graph the function along with the tangent at each point. 5. x 0 7. x 2 6. x 1 8. x 2 1. x 0 3. x 2 2. x 1 4. x 2 In Exercises 9–12, write the equation of the tangent line at the following values of x for the function f (x) 3 Graph the function along with the tangent at each point. x. In Exercises 5–8, write the equation of the tangent line at the following values of x for the function f (x) e Graph the function along with the tangent at each point. x 2. 9. x 0 11. x 2 10. x 1 12. x 2 411 C H A P T E R 6 Trigonometry Where are we? Navigators at sea must determine their location. Surveyors need to determine the height of a mountain or the width of a canyon when direct measurement is not feasible. A fighter plane’s computer must set the course of a missile so that it will hit a moving target. Many phenomena such as the tides, seasonal change, and radio waves, have cycles that repeat. All of the situations can be described mathematically using trigonometry. 412 Chapter Outline 6.1 Right-Triangle Trigonometry 6.2 Trigonometric Applications 6.3 Angles and Radian Measure 6.4 Trigonometric Functions Interdependence of Sections 6.1 > 6.2 > 6.3 > 6.4 > > 6.5 6.5 Basic Trigonometric Identities Chapter Review can do calculus Optimization with Trigonometry T rigonometry, which means “triangle measurement,” was developed in ancient times for determining the angles and sides of triangles in order to solve problems in astronomy, navigation, and surveying. With the development of calculus and physics in the 17th century, a different viewpoint toward trigonometry arose, and trigonometry was used to model all kinds of periodic behavior, such as sound waves, vibrations, and planetary orbits. In this chapter, you will be introduced to both types of trigonometry, beginning with right-triangle trigonometry. 6.1 Right-Triangle Trigonometry Objectives Angles and Degree Measure • Define the six trigonometric ratios of an acute angle in terms of a right triangle • Evaluate trigonometric ratios, using triangles and on a calculator Recall from geometry that an angle is a figure formed by two rays with a common endpoint, called the vertex. The rays are called the sides of the angle. An angle may be labeled by the angle symbol and the vertex. The angle in Figure 6.1-1 may be labeled A. 1 2 side angle vertex A side Figure 6.1-1 413 414 Chapter 6 Trigonometry Angles may be measured in degrees, where 1 degree 2 angle is an entire circle, a 1 90° 360° angle is a quarter of a circle. A (See Figure 6.1-2.) Thus, a angle is half angle is also called of a circle, and a a right angle. A right angle is indicated on a diagram by a small square, is called an acute angle. as shown in Figure 6.1-3. An angle of less than The measure of an angle is indicated by the letter m in front of the angle symbol, such as mA 36°. 180° 90° 90° is 1 360 of a circle. 1° right-angle symbol Figure 6.1-2 Figure 6.1-3 Minutes and Seconds Fractional parts of a degree are usually expressed in decimal form or in 1 60 minutes and seconds. A minute of a degree, and a second 1 60 is is 2 2 1 1 of a minute, or 1 3600 of a degree. This form is often called DMS form, for degrees, minutes, seconds. Example 1 Converting Between Decimal Form and DMS Form a. Write b. Write 35° 15¿ 27– 48.3625° in decimal form. in DMS form. Solution a. 35° 15¿ 27– 35° ° a 15 27 3600b 60b 35° 0.25° 0.0075° 35.2575° a ° b. First, convert the entire decimal part to minutes by writing it in terms of 1 60 of a degree. 48.3625° 48° 0.3625° 48° a 0.3625° 60 60b 21.75 a 60 b 48° ° 48° 21.75¿ Second, convert the decimal part of the minutes to seconds by writing it in terms of 1 60 of a minute. Section 6.1 Right-Triangle Trigonometry 415 48° 21.75¿ 48° 21¿ 48° 21¿ 48° 21¿ 45– 0.75¿ ¿ 60 60b 45 60b a a Similar Triangles and Trigonometric Ratios Examine the following right triangles. B a C c 34° b f e 34° D Figure 6.1- Since all right triangles that contain a angle are similar, the corresponding ratios would be the same. Thus, the ratio is dependent only on the measure of the angle. These ratios, which can be determined for any angle between are the basis of trigonometry. and 90°, 34° 0° The hypotenuse (hyp) of a right triangle is the side across from the right angle. The hypotenuse is always the longest side of the triangle. The remaining sides are labeled by their relationship to the given angle, as shown in Figure 6.1-5. The adjacent (adj) side is the side of the given angle that is not the hypotenuse, and the opposite (opp) side is the side of the triangle that is across from the given angle. C opp ∠B adj ∠A b opp ∠A adj ∠B a A c hypotenuse Figure 6.1-5 B In the figure, if site side is a. If side is b. A B is the given angle, the adjacent side is b and the oppois the given angle, the adjacent side is a and the opposite There are six possible ratios for the three sides of a triangle. These ratios are called trigonometric ratios. 416 Chapter 6 Trigonometry Trigonometric Ratios For a given acute angle U in a right triangle: The sine of written as sin U, is the ratio U, sin U opposite hypotenuse hypotenuse opposite adjacent Figure 6.1-6 NOTE The Greek letter (theta) is commonly used u to label the measure of an angle in trigonometry. 13 12 Figure 6.1-7 U, The cosine of written as cos is the ratio U, cos U adjacent hypotenuse U, The tangent of written as tan U, is the ratio tan U opposite adjacent In addition, the reciprocal of each ratio above is also a trigonometric ratio. cosecant of U csc U hypotenuse opposite secant of U sec U hypotenuse adjacent cotangent of U cot U adjacent opposite 1 sin U 1 cos U 1 tan U Example 2 Evaluating Trigonometric Ratios 5 Evaluate the six trigonometric ratios of the angle u shown in Figure 6.1-7. Solution The opposite side has length 5, the adjacent side has length 12, and the hypotenuse has length 13. sin u opposite hypotenuse cos u adjacent hypotenuse 5 13 12 13 13 5 13 12 1.0833 0.9231 0.3846 2.6 csc u hypotenuse opposite sec u hypotenuse adjacent cot u adjacent opposite tan u opposite adjacent 5 12 0.4167 12 5 2.4 ■ Example 3 Evaluating Trigonometric Ratios Evaluate the six trigonometric ratios of ure 6.1-8. (Side lengths given are approximate.) 62° by using the triangle in Fig- 3 3.4 Figure 6.1-8 1.6 62° Section 6.1 Right-Triangle Trigonometry 417 Solution sin 62° opposite hypotenuse cos 62° adjacent hypotenuse 3 3.4 1.6 3.4 0.8824 0.4706 tan 62° opposite adjacent 3 1.6 1.8750 csc 62° hypotenuse opposite sec 62° hypotenuse adjacent cot 62° adjacent opposite 3.4 3 3.4 1.6 1.1333 2.1250 1.6 3 0.5333 ■ Evaluating Trigonometric Ratios Using a Calculator If the measure of an angle is given without a corresponding triangle, it may be difficult to accurately evaluate the trigonometric ratios of that angle. For example, to find sin it would be possible to draw a right and measure its sides. However, there may triangle with an angle of be inaccuracies in drawing and measuring the triangle. Tables of trigonometric ratios are available, but it is usually most convenient to use a calculator. 20°, 20° Technology Tip The following facts will be helpful in evaluating trigonometric ratios on a calculator. • Scientific and graphing calculators have modes for different units of angle measurements. When using degrees, make sure that your calculator is set in degree mode. • The functions on a calculator do not indicate the reciprocal functions. These functions will be discussed in Section 8.2. and tan sin 1, cos 1, 1 • Some calculators automatically insert an opening parenthesis “(” after sin, cos, or tan. Be sure to place the closing parenthesis “)” in the appropriate place. In trigonometry, many of the values used are approximate, and answers are usually rounded to 4 decimal places. However, in calculations involving trigonometric ratios, the values
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should not be rounded until the end of the problem. Example 4 Evaluating Trigonometric Ratios on a Calculator Evaluate the six trigonometric ratios of 20°. Solution Your calculator should have buttons for sine, cosine, and tangent. To find the cosecant, secant, and cotangent, take the reciprocal of each answer. 418 Chapter 6 Trigonometry sin 20° 0.3420 cos 20° 0.9397 tan 20° 0.3640 Figure 6.1-9 Special Angles csc 20° 1 sec 20° cot 20° sin 20° 1 cos 20° 1 tan 20° 1 0.3420 1 0.9397 1 0.3640 2.9238 1.0642 2.7475 ■ Properties of 30-60-90 and 45-45-90 triangles can be used to find exact val30°, 60°, These angles are called ues of the trigonometric ratios for special angles. and 45°. NOTE For a review of the properties of 30-60-90 and 45-45-90 triangles, see the Geometry Review in the Appendix. Example 5 Evaluating Trigonometric Ratios of Special Angles Evaluate the six trigonometric ratios of 30°, 60°, and 45°. Solution A 30-60-90 and a 45-45-90 triangle are shown below. 60° 1 2 3 30° Figure 6.1-10 45° 1 2 45° 1 In the 30-60-90 triangle: In the 45-45-90 triangle: • the hypotenuse is 2 • for the angle, 30° 1 is opposite and • for the 60° angle, 23 23 • the hypotenuse is 22 • for either 45° angle, is adjacent 1 is both opposite and adjacent is opposite and 1 is adjacent The following table summarizes the values of the trigonometric ratios for the special angles. NOTE The exact values of the trigonometric ratios of the special angles will be needed regularly. You should memorize the sine and cosine values for all three angles. The other ratios are easily derived from the sine and cosine. Section 6.1 Right-Triangle Trigonometry 419 U sin u opposite hypotenuse cos u adjacent hypotenuse 30° 1 2 23 2 tan u opposite adjacent 1 23 23 3 csc u hypotenuse opposite 2 2 1 sec u hypotenuse adjacent cot u adjacent opposite 2 23 223 3 23 1 23 45 1 22 1 22 22 2 22 2 1 1 1 22 1 22 1 22 22 1 1 1 60 23 2 1 2 23 1 23 2 23 223 3 2 2 1 1 23 23 3 ■ Exercises 6.1 In Exercises 1–4, write the DMS degree measurement in decimal form. 10. 1. 47° 15¿ 36– 3. 15° 24¿ 45– 2. 38° 33¿ 9– 4. 20° 51¿ 54– In Exercises 5–8, write the decimal degree measurement in DMS form. 5. 23.16° 7. 4.2075° 6. 50.3625° 8. 85.655° In Exercises 9–14, find the six trigonometric ratios for U. 9. 11 3 θ 2 11. 12 17 15 θ 8 420 13. 14. Chapter 6 Trigonometry In Exercises 15–20, use a calculator in degree mode to find the following. Round your answers to four decimal places. 15. sin 32° 16. cos 68° 17. tan 6° 18. csc 25° 19. sec 47° 20. cot 39° In Exercises 21–26, use the exact values of the trigonoU metric ratios for the special angles to find a value of that is a solution of the given equation. (See Example 5.) 21. sin u 1 2 22. tan u 1 23. csc u 22 24. cot u 23 25. cos u 1 2 26. sec u 2 In Exercises 27–32, refer to the figure below. Find the exact value of the trigonometric ratio for the given values of a, b, and c. A b c C a B 27. a 4, b 2, tan B ? 28. a 5, c 7, sin A ? 29. b 3, c 8, cos A ? 30. a 12, b 15, cot A ? 31. a 7, c 16, sec B ? 32. b 2, c 3, csc B ? In Exercises 33–38, use a calculator in degree mode to determine whether the equation is true or false, and explain your answer. 33. sin 50° 2 sin 25° 34. sin 50° 2 sin 25° cos 25° 35. 36. 37. 38. cos 28° 2 1 2 1 sin 28° 1 cos 28° 2 1 sin 28° 2 1 1 tan 75° tan 30° tan 45° 2 2 2 2 tan 75° tan 30° tan 45° 1 tan 30° tan 45° 39. Critical Thinking Complete the table below. U 1° 0.1° 0.01° 0.001° sin U cos U ? ? ? ? ? ? ? ? Based on the values in the table, what do you think would be a reasonable value for sin Verify your answers with a calculator. cos Why can’t these values be found by using the definition on page 416? 0°? 0° and 40. Critical Thinking Complete the table below. U 89° 89.9° 89.99° 89.999° sin U cos U ? ? ? ? ? ? ? ? Based on the values in the table, what do you 90° think would be a reasonable value for sin Verify your answers with a calculator. cos Why can’t these values be found by using the definition on page 416? 90°? and Section 6.2 Trigonometric Applications 421 41. Critical Thinking Use the diagram below to show that the area of a triangle with acute angle that has sides a and b is u A 1 2 ab sin u. b θ h a In Exercises 42–45, use the result of Exercise 41 to find the area of the given triangle. 42. 10 25° 14 43. 44. 45. 59° 72 140 44 30° 20 12 38° 9 6.2 Trigonometric Applications Objectives • Solve triangles using trigonometric ratios • Solve applications using triangles Solving Right Triangles Many applications in trigonometry involve solving a triangle. This means finding the lengths of all three sides and the measures of all three angles when only some of these quantities are known. Solving right triangles by using trigonometric ratios involves two theorems from geometry: Triangle Sum Theorem: The sum of the measures of the angles in a triangle is Pythagorean Theorem: In a right triangle with legs a and b and hypotenuse c, a2 b2 c 2. 180. If the measures of two angles are known, the Triangle Sum Theorem can be used to find the measure of the third. If the lengths of two sides of a right triangle are known, the Pythagorean Theorem can be used to find the length of the third. Trigonometric ratios are used to solve right triangles when the measure of an angle and the length of a side or when the lengths of two sides are given. The underlying idea is that the definition of each trigonometric ratio involves three quantities: • the measure of an angle • the lengths of two sides of the triangle 422 Chapter 6 Trigonometry When two of the three quantities are known, the third can always be found, as illustrated in the next two examples. Example 1 Finding a Side of a Triangle Find side x of the right triangle in Figure 6.2-1. 8 65° x Figure 6.2-1 Solution 65° The be found, so use the cosine ratio. angle and the hypotenuse are known. The adjacent side x must cos 65° adjacent hypotenuse x 8 Solve this equation for x and then use a calculator to evaluate cos 65°. cos 65° x 8 x 8 cos 65° 3.3809 Multiply both sides by 8 Use a calculator ■ Example 2 Finding an Angle of a Triangle Find the measure of angle u in Figure 6.2-2. 3 4 θ 5 Figure 6.2-2 Solution Note that sin u opposite hypotenuse 3 5 0.6. Section 6.2 Trigonometric Applications 423 u Before calculators were available, was found by using a table of sine values. You can do the same thing by having your calculator generate a table for by using the settings shown in Figure 6.2-3a and Figure 6.2-3b. sin Y1 X 1 2 View the table, shown in Figure 6.2-3c, and look through the column of sine values for the closest one to 0.6. Then look in the first column for the corresponding value of The closest entry to 0.6 in the sine column is 0.60042, which corresponds to an angle of Hence, u 36.9°. 36.9°. u. Figure 6.2-3a Figure 6.2-3b Figure 6.2-3c A faster and more accurate method of finding SIN1 on your calculator. SIN1 key in angle whose sine is 0.6, namely, key is labeled ASIN on some models.) When you (0.6), as in Figure 6.2-3d, the calculator produces an acute u 36.8699°. is to use the 1 u SIN1 NOTE Make sure your calculator is in degree mode. Figure 6.2-3d SIN1 Thus, the sine table, without actually having to construct the table. key provides the electronic equivalent of searching the ■ NOTE In this chapter, the and TAN1 u. keys will be used as they were in Example 2 to find an angle The other uses of these keys are discussed in Section 8.2, which deals with inverse trigonometric functions. key and the analogous COS1 SIN1 Here is a summary of how these techniques can be used to solve any right triangle. 424 Chapter 6 Trigonometry Solving Right Triangles A right triangle can be solved if the following information is given. Case 1: an acute angle and a side Case 2: two sides Sketch the triangle and label the acute angle, the right angle, and the given side. Sketch the triangle and label the right angle and the two given sides. Find the remaining acute angle by subtracting the 180°. known angles from Write a trigonometric equation that has an unknown side as the variable, and solve it with a calculator to evaluate the trigonometric ratio of the angle. Repeat the previous step or use the Pythagorean Theorem to find the third side. Find the third side by using the Pythagorean Theorem. Write a trigonometric equation that has an unknown angle as the variable. If the angle is a special angle, you can solve it by recognizing the value of the trigonometric ratio. If the angle is not one of the special angles, use the technique explained in Example 2. Example 3 Solving a Right Triangle θ Solve the right triangle in Figure 6.2-4. Solution 17 a 75° b Figure 6.2-4 One side and one angle are given, so use the first case above. To solve the triangle, it is necessary to find a, and b. u, To find u, subtract the measures of the given angles from 180°. u 180° 75° 90° 15° Write a trigonometric equation that has a as the variable. Since a is opposite the given angle and the hypotenuse is given, the sine is used. opposite hypotenuse sin 75° a 17 a 17 sin 75° a 17 0.9659 1 Next, write a trigonometric equation that has b as the variable. Since b is adjacent to the given angle, cosine is used. Evaluate cos by using a calculator, and solve. 16.42 75° 2 θ a 12 β 6 Figure 6.2-5 Section 6.2 Trigonometric Applications 425 cos 75° b 17 b 17 cos 75° b 4.40 adjacent hypotenuse ■ Example 4 Solving a Right Triangle Solve the right triangle in Figure 6.2-5. Solution Two sides are given, so use the second case above. To solve the triangle, it is necessary to find a, and b. u, To find a, use the Pythagorean Theorem. a2 62 122 a2 108 a 2108 623 b, To find use trigonometric ratios. The adjacent side and the hypotenuse are given, so cosine is used. cos b 6 12 1 2 adjacent hypotenuse From the table of trigonometric ratios of special angles on page 419, cos 60° 1 2 is the only acute angle with a cosine of Since 60° 1 2 , . b 60° and u 180° 60° 90° 30° ■ Applications The following examples illustrate a variet
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y of applications of the trigonometric ratios. Example 5 Height Above Sea Level A straight road leads from an ocean beach at a constant upward angle of How high above sea level is the road at a point 1 mile from the beach? 3°. ocean 5 2 8 0 f t 3° r o a d Figure 6.2-6 h = height above sea level sea level 426 Chapter 6 Trigonometry Technology Tip When using a calculator to evaluate trigonometric ratios, do not round your answer until the end of the problem. The rounding error can be increased significantly as other operations are performed. It may help to store the value in the calculator’s memory, or to use the entire trigonometric expression in each calculation. Solution Figure 6.2-6 shows a right triangle with the road as the hypotenuse 1 mi 5280 ft and the opposite side h as the height above sea level. 1 Write a trigonometric equation that uses sine. 2 sin 3° h 5280 h 5280 sin 3° h 5280 0.0523 1 h 276.33 ft 2 opposite hypotenuse Solve for h Use a calculator to evaluate sin 3° Simplify At one mile, the road is about 276 feet above sea level. ■ Example 6 Ladder Safety According to the safety sticker on a 20-foot ladder, the distance from the bottom of the ladder to the base of the wall on which it leans should be one-fourth of the length of the ladder: 5 feet. a. How high up the wall will the ladder reach? b. If the ladder is in this position, what angle does it make with the ladder 20 ft wall h ground? Solution θ ground 5 ft Figure 6.2-7 Draw the right triangle formed by the ladder, the wall, and the ground. Label the sides and angles as shown in Figure 6.2-7. a. Since the length of the ladder and the distance from the wall are known, find the third side by using the Pythagorean Theorem. h2 52 202 h2 375 h 19.36 The ladder will safely reach a height of a little more than 19 feet up the wall. b. The hypotenuse and the side adjacent to angle are known, so use u the cosine ratio. Figure 6.2-8 cos u adjacent hypotenuse 5 20 1 4 Use the COS1 key to find that u 75.5°, as shown in Figure 6.2-8. ■ Angles of Elevation and Depression In many applications the angle between a horizontal line and another line is used, such as the line of sight from an observer to a distant object. If the line is above the horizontal, the angle is called the angle of elevation. Section 6.2 Trigonometric Applications 427 If the line is below the horizontal, the angle is called the angle of depression. Horizontal Angle of elevation Angle of depression Figure 6.2-9 Example 7 Indirect Measurement A flagpole casts a 60-foot shadow when the angle of elevation of the sun as shown in Figure 6.2-10. Find the height of the flagpole. is 35°, h 35° 60 ft Figure 6.2-10 Solution A right triangle is formed by the flagpole and its shadow. The opposite side is unknown and the adjacent side is given, so the tangent is used. tan 35° h 60 h 60 tan 35° h 42.012 opposite adjacent Solve for h Use a calculator Thus, the flagpole is about 42 feet high. ■ Example 8 Indirect Measurement A wire needs to reach from the top of a building to a point on the ground. The building is 10 m tall, and the angle of depression from the top of the How long should the wire be? building to the point on the ground is 22°. 428 Chapter 6 Trigonometry α 22° Angle of depression 10 Wire Figure 6.2-11 Solution Figure 6.2-11 shows that the sum of the angle of depression and the angle a formed by the wall of the building and the wire is 90°. a 90° 22° 68° The wall, wire, and ground form a right triangle where the wall is the side adjacent to and the wire is the hypotenuse. Thus, a adjacent hypotenuse cos 68° 10 w w 10 cos 68° w 26.7 m Thus, the wire should be about 27 m long. ■ Example 9 Indirect Measurement A person on the edge of a canal observes a lamp post on the other side to the top of the lamp post and an angle with an angle of elevation of of depression of to the bottom of the lamp post from eye level. The person’s eye level is 152 cm (about 5 ft). 12° 7° a. Find the width of the canal. b. Find the height of the lamp post. 152 12° 7° Figure 6.2-12 Section 6.2 Trigonometric Applications 429 Solution The essential information is shown in Figure 6.2-13 below. Note that is parallel to so CD is also 152 cm. DE, AC 152 A E 12° 7° Figure 6.2-13 B C D 152 a. The width of the canal AC is adjacent to the 7° angle, and 152 is opposite the 7° angle. tan 7° 152 AC opposite adjacent AC 152 tan 7° 1237.94 cm, or about 12.38 m wide b. The height BC can be represented in terms of the width of the canal found in part a. tan 12° BC AC opposite adjacent BC AC tan 12° tan 12° 2 2 1 263.13 cm 1 1237.94 BC CD. The height of the lamp post is BC CD 263.13 152 415.13 cm ■ Exercises 6.2 In Exercises 1–6, find side c in the figure below by using the given conditions. b c A C a B 2. 3. sin C 3 4 tan A 5 12 4. sec A 2 5. cot A 6 6. csc C 1.5 b 12 a 15 b 8 a 1.4 b 4.5 1. cos A 12 13 b 39 In Exercises 7–12, find the exact value of h without using a calculator. 430 7. 8. 9. 10. 11. 12. Chapter 6 Trigonometry 25 h h 45° 30° h 60° 72 150 h 45° 12 h 30° 100 20 h 60° Use the figure below for Exercises 13–24. A c B b a C In Exercises 13–16, find the indicated value without using a calculator. 13. a 4 14. c 5 15. c 10 16. a 12 mA 60° mA 60° mA 30° mA 30° Find c. Find a. Find a. Find c. In Exercises 17–24, solve the triangle with the given conditions. 17. b 10 18. c 12 19. a 6 20. a 8 21. c 5 22. c 4 23. b 3.5 24. a 4.2 mC 50° mC 37° mA 14° mA 40° mA 65° mC 28° mA 72° mC 33° In Exercises 25–28, find angle U. 25. 26. 27. 4 3 θ 12 θ 10 2 θ 3 28. θ 200 144 In Exercises 29–36, use the figure for Exercises 13–24 to find angles A and C under the given conditions. 29. a 4 and c 6 30. b 14 and c 5 31. a 7 and b 10 32. a 5 and c 3 33. b 18 and c 12 34. a 4 and b 9 35. a 2.5 and c 1.4 36. b 3.7 and c 2.2 37. A 24-ft ladder positioned against a wall forms an 75° with the ground. angle of a. How high up the wall does the ladder reach? b. How far is the base of the ladder from the wall? 38. A guy wire stretches from the top of an antenna tower to a point on level ground 18 feet from the base of the tower. The angle between the wire and the ground is How high is the tower? 63°. 39. A plane takes off at an angle of 5°. After traveling 1 mile along this flight path, how high (in feet) is the plane above the ground? 1 mi 5280 ft 1 2 5° 40. A plane takes off at an angle of 6° traveling at the rate of 200 feet/second. If it continues on this flight path at the same speed, how many minutes will it take to reach an altitude of 8000 feet? 41. The Ohio Turnpike has a maximum uphill slope 3°. of How long must a straight uphill segment of the road be in order to allow a vertical rise of 450 feet? 42. Ruth is flying a kite. Her hand is 3 feet above ground level and is holding the end of a 300-ft Section 6.2 Trigonometric Applications 431 kite string, which makes an angle of horizontal. How high is the kite above the ground? 57° with the 43. Suppose that a person with a reach of 27 inches and a shoulder height of 5 feet is standing upright on a mountainside that makes a the horizontal, as shown in the figure below. Can the person touch the mountain? angle with 62° 62° 44. A swimming pool is 3 feet deep in the shallow end. The bottom of the pool has a steady downward drop of 12° the pool is 50 feet long, how deep is the deep end? toward the deep end. If 45. A wire from the top of a TV tower makes an angle 49.5° of with the ground and touches the ground 225 feet from the base of the tower. How high is the tower? 46. A plane flies a straight course. On the ground directly below the flight path, observers 2 miles apart spot the plane at the same time. The plane’s angle of elevation is point and plane? from one observation from the other. How high is the 71° 46° 71° 46° 2 miles 2 miles 432 Chapter 6 Trigonometry 47. A buoy in the ocean is observed from the top of a 40-meter-high radar tower on shore. The angle of depression from the top of the tower to the base How far is the buoy from the of the buoy is base of the radar tower? 6.5°. 48. A plane passes directly over your head at an 33° QP altitude of 500 feet. Two seconds later you observe that its angle of elevation is plane travel during those 2 seconds? How far did the 42°. 49. A man stands 12 feet from a statue. The angle of elevation from eye level to the top of the statue is and the angle of depression to the base of the 30°, statue is How tall is the statue? 15°. 50. Two boats lie on a straight line with the base of a lighthouse. From the top of the lighthouse, 21 meters above water level, it is observed that the angle of depression of the nearest boat is 53° the angle of depression of the farthest boat is How far apart are the boats? and 27°. 53° 27° 54. A drinking glass 5 inches tall has a 2.5-inch diameter base. Its sides slope outward at a angle as shown. What is the diameter of the top of the glass? 4° 4° 4° 55. In aerial navigation, directions are given in 90°, degrees clockwise from north, called headings. Thus east is shown below. A plane travels from an airport for 200 miles at a heading of the airport is the plane? How far west of and so on, as south is 180°, 300°. 51. A rocket shoots straight up from the launch pad. Five seconds after lift-off, an observer 2 miles away notes that the rocket’s angle of elevation is 3.5° is 4 seconds? . Four seconds after that, the angle of elevation How far did the rocket rise during those 41°. 0° North 300° 270° West 90° East 52. From a window 35 meters high, the angle of depression to the top of a nearby streetlight is The angle of depression to the base of the streetlight is . How tall is the streetlight? 57.8° 55°. Distance west of airport 180° South 53. A 60-foot drawbridge is 24 feet above water level when closed. When open, the bridge makes an with the horizontal. angle of a. How high is the tip P of the open bridge above 33° the water? b. When the bridge is open, what is the distance from P to Q? 56. A plane travels from an airport at a const
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ant 300 65° mph at a heading of a. How far east of the airport is the plane after . (See Exercise 55.) half an hour? b. How far north of the airport is the plane after 2 hours and 24 minutes? 57. A car on a straight road passes under a bridge. Two seconds later an observer on the bridge, 20 feet above the road, notes that the angle of How fast, in miles depression to the car is per hour, is the car traveling? (Note: 60 mph is equivalent to 88 feet/second.) 7.4°. 58. A pedestrian overpass is shown in the figure below. If you walk on the overpass from one end to the other, how far have you walked? 15° 18 ft 21° 200 ft Section 6.3 Angles and Radian Measure 433 59. Critical Thinking A 50-ft flagpole stands on top of a building. From a point on the ground the angle of elevation to the top of the pole is and the angle of elevation to the bottom of the pole is How high is the building? 40°. 43° 60. Critical Thinking Two points on level ground are 500 meters apart. The angles of elevation from and these points to the top of a nearby hill are 52° respectively. The two points and the ground67°, level point directly below the top of the hill lie on a straight line. How high is the hill? 6.3 Angles and Radian Measure Objectives Extending Angle Measure • Use a rotating ray to extend the definition of angle measure to negative angles and angles greater than 180° • Define radian measure and convert angle measures between degrees and radians In geometry and triangle trigonometry, an angle is a static figure consisting of two rays that meet at a point. But in modern trigonometry, which will be introduced in the next section, an angle is thought of as being formed dynamically by rotating a ray around its endpoint, the vertex. The starting position of the ray is called the initial side and its final position after the rotation is called the terminal side. The amount the ray is rotated is the measure of the angle. Counterclockwise rotations have positive measure and clockwise rotations have negative measure. T er m in al Vertex T e r m i n a l Initial Initial Initial 40° 830° Figure 6.3-1 T e r m i n a l 43° Terminal Initial 312° 434 Chapter 6 Trigonometry An angle in the coordinate plane is said to be in standard position if its vertex is at the origin and its initial side is on the positive x-axis. y y Positive angle x Figure 6.3-2 x Negative angle Angles formed by different rotations that have the same initial and terminal sides are called coterminal. (See Figure 6.3-3.) For example, and 360° are coterminal angles. 0° Example 1 Coterminal Angles Find three angles coterminal with an angle of 60° in standard position. Solution To find an angle that is coterminal with a given angle, add or subtract a complete revolution, or To find additional angles, add or subtract any multiple of Three possible angles are shown below. 360°. 360°. 60° 360° 420° 60° 360° 300° 60° 2 360° 2 1 780° y y y 60° 60° x x 420° −300° Figure 6.3-4 Arc Length 60° 780° x ■ Recall from geometry that an arc is a part of a circle and that a central angle is an angle whose vertex is the center of the circle. The length of an arc depends on the radius of the circle and the measure of the central angle that it intercepts, as shown in Figure 6.3-5 Initial Figure 6.3-3 arc θ r center Figure 6.3-5 Section 6.3 Angles and Radian Measure 435 Arc length can be calculated by considering an arc as a fraction of the entire circle. Suppose an arc in a circle of radius r has a central angle meas- ure of u . Since there are 360° in a full circle, the arc is The circumference of the circle is 2pr , so, the length of the arc is / u 360 of the circle. / u 360 2pr upr 180 Example 2 Finding an Angle Given an Arc Length / An arc in a circle has an arc length which is equal to the radius r. Find the measure of the central angle that the arc intercepts. Solution r / r upr 180 180r upr 180 up 180 p u The central angle measure is 180 p b ° , a or about 57.3°. ■ Radian Measure The angle found in Example 2 leads to another unit used in finding angle measure called a radian. Because it simplifies many formulas in calculus and physics, radians are used as a unit of angle measurement in mathematical and scientific applications. Angle measurement in radians can be described in terms of the unit circle, which is the circle of radius 1 centered at the origin, whose equation When an angle is in standard position, its initial side lies is on the x-axis and passes through Its terminal side intersects the unit circle at some point P, as shown in Figure 6.3-6. x2 y2 1. 1, 0 . 2 1 y P x (1, 0) Figure 6.3-6 436 Chapter 6 Trigonometry Definition of Radian Measure NOTE Generally, measurements in radians are not labeled with units, although the word radian or the abbreviation rad may sometimes be used for clarity. The radian measure of an angle is the distance traveled along the unit circle in a counterclockwise direction by the point P, as it moves from its starting position on the initial side to its final position on the terminal side of the angle. 1 radian 180 P b a 57.3 If the vertex of an angle is the center of a circle of radius r, then an angle of 1 radian intercepts an arc of length r. Movement along the unit circle is counterclockwise for positive measure and clockwise for negative measure. y distance = 3.75 y 3.75 radians x (1, 0) P (1, 0) x −2 radians P distance = 2 Figure 6.3-7 Consider an angle in standard position, with its terminal side rotating 360° around the origin. In degree measure, one full revolution produces a angle. The radian measure of this angle is the circumference of the unit circle, namely Other angles can be considered as a fraction of a full revolution, as shown in Figure 6.3-8. 2p. 1 revolution 2p radians y 1 3 4 3/4 revolution 2p 3p 2 y radians 1/2 revolution 2p p radians 1 2 −1 x 1 −1 −1 1 −1 x 1 −1 Figure 6.3-8 y 1 −1 1 4 1/4 revolution 2p p 2 y radians x 1 −1 1 −1 x 1 Section 6.3 Angles and Radian Measure 437 Radian Measure of Special Angles The special angles of tion of a full revolution. Note that and 30°, 60°, 45° can also be considered as a frac- • • • 360 12 360 6 360 8 30, so 30° is 1 12 of a complete revolution: 1 6 60, so 60° 45, so 45° is of a complete revolution: is of a complete revolution: 1 8 rad 2p p 6 30° 1 12 2p p 3 2p p 4 60° 1 6 45° 1 8 rad rad NOTE p; Radian measurements are usually given in terms of however, it is useful to know the decimal equivalents for common measurements when using a calculator. p 3.14 2p 6.28 p 2 p 6 p 1.57 4 p 0.52 3 1.05 0.79 y x 16π 3 = 2π + 2π + 4π 3 Figure 6.3-10 Figure 6.3-9 shows a unit circle with radian and degree measures for important values. The radian measures for the angles shown in the first quadrant and on the x- and y-axes should be memorized. 90° = π 2 2π 3 120° = 3π 4 135° = 150° = 5π 6 60° = π 3 45° = π 4 30° = π 6 180° = ππ 0° = 360° = 2ππ 210° = 7π 6 225° = 5π 4 240° = 4π 3 11π 6 330° = 7π 4 315° = 5π 3 300° = 270° = 3π 2 Figure 6.3-9 As shown in Figure 6.3-9, radians corresponds to a full revolution of the terminal side of an angle in standard position. So an angle of radian measure t is coterminal with the angles whose radian measures are t ± 2p, t ± 4p, and so on, as shown in Figure 6.3-10. 2p Increasing or decreasing the radian measure of an angle by an integer multiple of 2P results in a coterminal angle. Converting Between Degrees and Radians As shown in Figure 6.3-9, p radians 180° . 438 Chapter 6 Trigonometry y Dividing both sides by p shows that 1 1 radian 180 p b a ° 57.3°, 1 rad x which agrees with the definition of radian. (1, 0) Similarly, both sides of the original equation can be divided by 180. Figure 6.3-11 Radian/Degree Conversion p 180 ˛ radians 1° These two equations give the conversion factors for radians to degrees and degrees to radians. To convert radians to degrees, multiply by To convert degrees to radians, multiply by 180 P . P 180 . Example 3 Converting From Radians to Degrees Convert the following radian measurements to degrees. a. p 5 Solution b. 4p 9 c. 6p a. p 5 180 p 36° b. 4p 9 180 p 80° c. 6p 180 p 1080° Example 4 Converting From Degrees to Radians Convert the following degree measurements to radians. a. 75° Solution b. 220° c. 400° a. 75° p 180 5p 12 b. 220° p 180 11p 9 c. 400° p 180 20p 9 Arc Length and Angular Speed The formula for arc length can also be written in terms of radians. ■ ■ NOTE One radian, which is illustrated in Figure 6.3-11, is close to 60°. There are about 2p 6.28 6 radians complete circle. 1 2 in a NOTE To help you remember which conversion factor to use, it may be helpful to notice that radians are usually written in terms of p. p Degrees to radians: to get in final answer, multiply by p 180 Radians to degrees: to cancel in final answer, p multiply by 180 p Section 6.3 Angles and Radian Measure 439 Arc Length An arc with central angle measure rU U radians has length In other words, the arc length is the radius times the radian measure of the central angle of the arc. 10 9 8 12 11 7 6 1 π 2 5 2 4 Figure 6.3-12 Example 5 Arc Length The second hand on a clock is 6 inches long. How far does the tip of the second hand move in 15 seconds? 3 Solution The second hand makes a full revolution every 60 seconds, that is, it moves through an angle of radians. During a 15-second interval it will make 15 60 (Figure 6.3-12), so the tip of the second hand travels along an arc with a of a revolution, moving through an angle of p 2 1 4 radians 1 4 ˛1 2p 2p 2 central angle measure of p 2 15 seconds is the arc length . Therefore, the distance that the tip moves in ru 6˛a p 2 b 3p 9.4 inches. ■ Example 6 Central Angle Measure Find the central angle measure (in radians) of an arc of length 5 cm on a circle with a radius of 3 cm. y 5 cm 3 cm Solution x Solve the arc length formula for u . / ru / r 5 3 u radians Figure 6.3-13 This is a little more than one-quarter of a complete revolution, as shown in Figure 6.3-13. ■ Linear and Angular Speed Suppose that a wheel is rota
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ting at a constant rate around its center, O, and P is a point on the outer edge of the wheel. There are two ways to measure the speed of point P, in terms of the distance traveled or in terms of the angle of rotation. The two measures of speed are called linear speed and angular speed. 440 Chapter 6 Trigonometry P Recall that the speed of a moving object is distance time . If the object is trav- O eling in a circular path with radius r, the linear speed is given by linear speed arc length ru t time and the angular speed is given by angular speed angle time u t Figure 6.3-14 u where is the radian measure of the angle through which the object travels in time t. Notice the relationship between linear speed and angular speed: NOTE The angular speed of an object traveling in a circular path is the same, regardless of its distance from the center of the circle. When the angular speed of the object stays the same, the linear speed increases as the object moves farther from the center. linear speed r angular speed Example 7 Linear and Angular Speed A merry-go-round makes 8 revolutions per minute. a. What is the angular speed of the merry-go-round in radians per minute? b. How fast is a horse 12 feet from the center traveling? c. How fast is a horse 4 feet from the center traveling? Solution a. Each revolution of the merry-go-round corresponds to a central angle of of 2p 8 2p 16p radians in one minute. radians, so the merry-go-round travels through an angle angular speed u t 16p 1 16p radians per minute b. The horse 12 feet from the center travels along a circle of radius 12. From part a, linear speed r angular speed 12 16p 192p ftmin which is about 6.9 mph. c. The horse 4 feet from the center travels along a circle of radius 4. From part a, linear speed r angular speed 4 16p 64p ftmin which is about 2.3 mph. ■ NOTE follows: The units ft/min in Example 7 can be converted to mph as 192p ft 1 min 60 min 1 hr 1 mi 5280 ft 6.9 mi 1 hr Section 6.3 Angles and Radian Measure 441 Exercises 6.3 In Exercises 1–10, find the degree and radian measure of the angle in standard position formed by rotating the terminal side by the given amount. In Exercises 47 – 52, determine the positive radian measure of the angle that the second hand of a clock travels through in the given time. of a circle 10. 55. If the radius of the circle in the figure is 20 cm 47. 40 seconds 48. 50 seconds 49. 35 seconds 50. 2 minutes 15 seconds 51. 3 minutes 25 seconds 52. 1 minute 55 seconds 53. The second hand on a clock is 6 cm long. How far does its tip travel in 40 seconds? 54. The second hand on a clock is 5 cm long. How far does its tip travel in 2 minutes and 15 seconds? 85 cm, and angle u? what is the radian measure of the y θ x 1 9 of a circle 1 18 1 36 of a circle of a circle of a circle 1. 3. 5. 7. 9. 2 3 4 5 2. 4. 6. 8. 1 24 1 72 of a circle of a circle 1 5 of a circle 7 12 5 36 of a circle of a circle In Exercises 11–22, convert the given radian measure to degrees. 11. 15. p 5 3p 4 12. p 6 16. 5p 3 19. 5p 12 20. 7p 15 13. p 10 14. 2p 5 17. p 45 21. 27p 5 18. p 60 22. 41p 6 In Exercises 23–34, convert the given degree measure to radians. Write your answer in terms of P. 23. 6° 27. 75° 24. 10° 25. 12° 26. 36° 28. 105° 29. 135° 30. 165° 31. 225° 32. 252° 33. 930° 34. 585° 56. Find the radian measure of the angle u in the preceding figure if the diameter of the circle is 150 cm and 360 cm. In Exercises 35–42, state the radian measure of an angle in standard position between 0 and that is coterminal with the given angle in standard position. 2P In Exercises 57–60, assume that a wheel on a car has radius 36 cm. Find the angle (in radians) that the wheel turns while the car travels the given distance. 35. p 3 39. 7p 5 36. 3p 4 40. 45p 8 37. 19p 4 38. 16p 3 41. 7 42. 18.5 In Exercises 43–46, find the radian measure of four angles in standard position that are coterminal with the given angle in standard position. 43. p 4 44. 7p 5 45. p 6 46. 9p 7 57. 2 meters (200 cm) 58. 5 meters 59. 720 meters 60. 1 kilometer (1000 meters) In Exercises 61–64, find the length of the circular arc with the central angle whose radian measure is given. Assume that the circle has diameter 10. 61. 1 radian 62. 2 radians 63. 1.75 radians 64. 2.2 radians 442 Chapter 6 Trigonometry The latitude of a point P on Earth is the degree measure of the angle between the point and the plane of the equator, with Earth’s center as the vertex, as shown in the figure below. U P θ Equator In Exercises 65–68, the latitudes of a pair of cities are given. Assume that one city is directly south of the other and that the earth is a perfect sphere of radius 4000 miles. Use the arc length formula in terms of degrees to find the distance between the two cities. 65. The North Pole: latitude 90° Springfield, Illinois: latitude north 40° north 66. San Antonio, Texas: latitude Mexico City, Mexico: latitude 29.5° 20° north north 67. Cleveland, Ohio: latitude Tampa, Florida: latitude 41.5° north 28° north 68. Rome, Italy: latitude 42° north Copenhagen, Denmark: latitude 54.3° north In Exercises 69–76, a wheel is rotating around its axle. Find the angle (in radians) through which the wheel turns in the given time when it rotates at the given number of revolutions per minute (rpm). Assume t 77 0 k 77 0. and 69. 3.5 minutes, 1 rpm 70. t minutes, 1 rpm 78. A circular saw blade has an angular speed of 15,000 radians per minute. a. How many revolutions per minute does the saw make? b. How long will it take the saw to make 6000 revolutions? 79. A circular gear rotates at the rate of 200 revolutions per minute (rpm). a. What is the angular speed of the gear in radians per minute? b. What is the linear speed of a point on the gear 2 inches from the center in inches per minute? in feet per minute? 80. A wheel in a large machine is 2.8 feet in diameter and rotates at 1200 rpm. a. What is the angular speed of the wheel? b. How fast is a point on the circumference of the wheel traveling in feet per minute? in miles per hour? 81. A riding lawn mower has wheels that are 15 inches in diameter, which are turning at 2.5 revolutions per second. a. What is the angular speed of a wheel? b. How fast is the lawn mower traveling in miles 71. 1 minute, 2 rpm 72. 3.5 minutes, 2 rpm per hour? 73. 4.25 minutes, 5 rpm 74. t minutes, 5 rpm 82. A bicycle has wheels that are 26 inches in 75. 1 minute, k rpm 76. t minutes, k rpm 77. One end of a rope is attached to a circular drum of radius 2 feet and the other to a steel beam. When the drum is rotated, the rope wraps around it and pulls the object upward (see figure). Through what angle must the drum be rotated in order to raise the beam 6 feet? diameter. If the bike is traveling at 14 mph, what is the angular speed of each wheel? 83. A merry-go-round horse is traveling at 10 feet per second when the merry-go-round is making 6 revolutions per minute. How far is the horse from the center of the merry-go-round? 84. The pedal sprocket of a bicycle has radius 4.5 inches and the rear wheel sprocket has radius 1.5 inches (see figure). If the rear wheel has a radius of 13.5 inches and the cyclist is pedaling at the rate of 80 rpm, how fast is the bicycle traveling in feet per minute? in miles per hour? Section 6.4 Trigonometric Functions 443 away on the western horizon. (The figure is not to scale.) Assuming that the radius of the earth is 3950 miles, how high was the plane when the picture was taken? Hint: The sight lines from the plane to the horizons are tangent to the earth and a tangent line to a circle is perpendicular to the radius at that point. The arc of the earth between St. Louis and Cleveland is 520 miles long. Use this fact and the arc length formula to find angle Your answers will be in radians. Note that a u 2 why? u. . 2 1 85. A spy plane on a practice run over the Midwest takes a picture that shows Cleveland, Ohio, on the eastern horizon and St. Louis, Missouri, 520 miles St. Louis θ α Cleveland 6.4 Trigonometric Functions Objectives Extending the Trigonometric Ratios • Define the trigonometric ratios in the coordinate plane • Define the trigonometric functions in terms of the unit circle NOTE P can be any point on the terminal side of the angle, except for the origin, since different choices for P generate similar right triangles. Thus, the value of a trigonometric ratio depends only on the angle. Trigonometric ratios were defined for acute angles in Section 6.1. The next step is to develop a definition of these ratios that applies to angles of any measure. u To do this, first consider an acute angle in standard position. Choose a point P, with coordinates (x, y), on the terminal side, and draw a right triangle, as shown in Figure 6.4-1. The side adjacent to has length x and the side opposite has length y. The length of the hypotenuse, r, is the distance from the origin, which may be the Pythagorean Theorem. found by using u u x2 y2 r2 r 2x2 y2 y hypotenuse r θ adjacent x opposite y x Figure 6.4-1 444 Chapter 6 Trigonometry The trigonometric ratios can now be written in terms of x, y, and r. For example, sin u opposite hypotenuse y r and cos u adjacent hypotenuse x r Thus, the trigonometric ratios can be described without triangles by using a point on the terminal side of the angle. More importantly, this process can be carried out for any angle, not just acute angles. Therefore, the following definition applies to any angle and agrees with the previous definition when the angle is acute. Trigonometric Ratios in the Coordinate Plane y θ (−3, −2) x Figure 6.4-2 U Let be an angle in standard position and let point on the terminal side of Let r be the distance from (x, y) to the origin: P (x, y) U. be any r 2x2 y2 U Then the trigonometric ratios of are defined as follows: sin U y r cos U x r tan U y x csc U r y sec U r x cot U x y P(x, y) θ y r x Example 1 Trigonometric Ratios in the Coordinate Plane Find the sine, cosine, and tangent of the angle whose terminal side passes th
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rough the point 3, 2 . u, 1 2 Solution Using the values sin u x 3, y 2, 2 213 cos u and 3 213 r 2 3 2 2 1 1 tan u 2 2 2 3 2 213, 2 3 ■ Trigonometric Functions Trigonometric ratios have been defined for all angles. But modern applications of trigonometry deal with functions whose domains consist of real numbers. The basic idea is quite simple: If t is a real number, then sin t is defined to be the sine of an angle of t radians; cos t is defined to be the cosine of an angle of t radians; Section 6.4 Trigonometric Functions 445 and so on. Instead of starting with angles, as was done up until now, this new approach starts with a number and only then moves to angles, as summarized below. Trigonometric Functions of Real Numbers ⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭ Begin with a number t > Form an angle of t radians > Determine sin t, cos t, tan t ⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭ Trigonometric Ratios of Angles Adapting earlier definitions of ratios to this new viewpoint produces the following definition of trigonometric functions of real numbers. Use Figure 6.4-3 for reference. Trigonometric Functions of a Real Variable Let t be a real number. Choose any point side of an angle of t radians in standard position. Then cos t x r sec t r x tan t y x cot t x y sin t y r csc t r y (x, y) on the terminal y where r 2x 2 y 2 is the distance from (x, y) to the origin. (x, y) r t x y x Although this definition is essential for developing various facts about the trigonometric functions, the values of these functions are usually approximated by a calculator in radian mode, as shown in Figure 6.4-4. Figure 6.4-3 NOTE Unless stated otherwise, use radian mode when evaluating trigonometric functions of real numbers. Figure 6.4-4 Trigonometric Functions and the Unit Circle Recall that the unit circle is the circle of radius 1 centered at the origin, whose equation is The unit circle is the basis for the most useful description of trigonometric functions of real numbers. x2 y2 1. Let t be any real number. Construct an angle of t radians in standard posibe the point where the terminal side of this angle meets tion. Let the unit circle, as shown in Figure 6.4-5. x, y P 2 1 446 Chapter 6 Trigonometry NOTE arc from 1 The length of the to P is t. 1, 0 2 y 1 t P(x, y) −1 x 1 −1 Figure 6.4-5 P, r 1, The distance from P to the origin is 1 because the unit circle has radius 1. Using the point and the definition of trigonometric functions of real numbers shows the following: sin t y r y and cos t x r x y 1 x 1 Unit Circle Description of Trigonometric Functions Let t be a real number and let P be the point where the terminal side of an angle of t radians in standard position meets the unit circle. Then P has coordinates (cos t, sin t) and tan t y x sec t 1 x sin t cos t 1 cos t cot t x y csc t 1 y cos t sin t 1 sin t Graphing Exploration With your calculator in radian mode and parametric graphing mode, set the range values as follows: 0 t 2p 1.8 x 1.8 1.2 y 1.2 Then, graph the curve given by these parametric equations: x cos t y sin t The graph is the unit circle. Use the trace to move around the circle. At each point, the screen will display three numbers: the values of t, x, and y. For each t, the cursor is on the point where the terminal side of an angle of t radians meets the unit circle, so the corresponding x is the number cos t and the corresponding y is the number sin t. Section 6.4 Trigonometric Functions 447 The coordinates of points on a circle of radius r that is centered at the origin can be written by using r and t with the definition of the trigonometric ratios in the coordinate plane. See Exercise 61. Domain and Range By the domain convention in Section 3.1, the domain of a function is all real numbers for which the function is defined. For any real number t, an appropriate angle of t radians and its intersection point with the unit circle are always defined, so the domain of the sine function and of the cosine function is the set of all real numbers. The range of a function is the set of all possible outputs. Because sin t and cos t are the coordinates of a point on the unit circle, they take on all values between and 1 and no other values. Thus, 1 the range of the sine function and of the cosine function is the set of all real numbers between and 1, that is, the interval tan t y x, all points on the unit circle except (0, 1) and The tangent function is defined as whenever 0, 1 x 0, . 1 [1, 1]. that is, for 1 2 The point (0, 1) is on the terminal side of an angle of p 2 radians or any angle obtained by adding integer multiples of 2p p (a complete circle) to it, that is, , 7p 2 , 3p 2 9p 2 5p 2 , p p 2 , , , The point (0, 1) is on the terminal side of an angle of radians or any 3p 2 to it, that is, p angle obtained by adding integer multiples of 2 p , 5p 2 , p 2 11p 2 3p 2 7p 2 , , , , p Combining these facts shows that the domain of the tangent function consists of all real numbers except P 2 2kP, where k 0, 1, 2, 3, p . In contrast to sine and cosine, the range of the tangent function is the set of all real numbers. A proof of this fact is found in Exercise 60. Figure 6.4-6 shows that values of the tangent can be very large positives, very large negatives, or in between. Signs of the Trigonometric Functions It is often important to know whether the value of a trigonometric function is positive or negative. For any real number t, the point (cos t, sin t) is on the terminal side of an angle of t radians in standard position. The Figure 6.4-6 448 Chapter 6 Trigonometry π 2 < t < π sin t + cos t − tan t − 3π 2 π < t < sin t − cos t − tan t + y π 0 < t < 2 sin t + cos t + tan t + 3π 2 < t < 2π sin t − cos t + tan t − Figure 6.4-7 quadrant in which this point lies determines the signs of sine and cosine, as well as those of the other trigonometric functions, as summarized in Figure 6.4-7. x Exact Values of Trigonometric Functions Although a calculator is used to evaluate trigonometric functions approximately, there are a few special numbers for which exact values can be found. Recall that 30°, 45°, and 60° are the same as p 6 , p 4 , and p 3 , respec- tively. Therefore, the chart on page 419 can be translated as follows. t sin t cos t tan t csc t sec t cot t P 6 1 2 23 2 1 23 23 3 2 2 1 2 23 223 3 23 1 23 P 4 22 2 1 22 1 22 22 2 1 22 1 22 1 22 22 P 3 23 2 1 2 23 2 23 223 3 2 2 1 1 1 1 1 23 23 3 The exact values of the trigonometric functions can also be found for any number that is an integer multiple of p 6 , p 4 , and p 3 . The technique for doing this depends on the concept of a reference angle. Example 2 Exact Values of Trigonometric Functions Find the exact value of the sine, cosine, and tangent functions when p 2 3p 2 and , p, 2p. , t 0, Solution If the measure of an angle is a multiple of p 2 , then its terminal side lies on an axis. Thus, the only possible values of the sine and cosine functions NOTE The diagram of of such angles are 0 and 1. The following chart shows the value of the sine, cosine, and tangent functions for these angles between 0 and 2p. 1, Section 6.4 Trigonometric Functions 449 the unit circle shown in Figure 6.4-8 will help you memorize the values of sin t p 2 and cos t for multiples of . Definition of Reference Angle t 0 p 2 p 3p 2 2p sin t cos t tan undefined 0 undefined 0 Reference Angles y (0, 1) (−1, 0) x (1, 0) (0, −1) Figure 6.4-8 ■ U For an angle in standard position, the reference angle is the positive acute angle formed by the terminal side of and the x-axis. U y t′ = t y In the following figure, the reference angle standard position is shown in two ways. t¿ for an angle of t radians in Definition of Reference Angle y t t = t′ x t′ y t x y t x t′ x t′ t′ = − tπ t′ = t − π t′ = 2 − t π Unit Circle with Reference Angle Placed in Quadrant I P(x, y) t = t′ x Q(−x, y) t′ y t P(x, y) t′ x y t P(x, y) t′ x t′ Q(−x, −y) Figure 6.4-9 y t t′ t′ P(x, y) x Q(x, −y) 450 Chapter 6 Trigonometry In every case, the figure that references the unit circle illustrates the following fact that can be proved by using congruent triangles: x-coordinate of Q ± y-coordinate of Q ± x-coordinate of P y-coordinate of P 2 2 1 1 By definition, the values of the trigonometric functions for t are given by the coordinates of Q and the values of these functions for are given by the coordinates of P. So, these values will be the same, except for a plus or minus sign. The correct sign is determined by the quadrant in which the terminal side of an angle of t radians lies, as shown in Figure 6.4-7 on page 448. t¿ Finding Trigonometric Function Values To find the sine, cosine, or tangent of t radians, • Sketch an angle of t radians in standard position and determine the quadrant in which the terminal side lies. • Find the reference angle, which has measure • Find the sine, cosine, and tangent of t appropriate sign. t and append the radians. Example 3 Using Reference Angles Use reference angles to find the exact value of sin t, cos t, and tan t. a. t 3p 4 b. t 4p 3 c. t 11p 6 x Solution a. Sketch the angle, as shown in Figure 6.4-10. The terminal side is in the second quadrant, so the reference angle is p 3p 4 p 4 p t . y 3π 4 π 4 Figure 6.4-10 Because the terminal side of the angle of 3p 4 radians lies in the second quadrant, sin 3p 4 is positive, and cos 3p 4 and tan 3p 4 are negative. sin tan 3p 4 4p 3 sin p 4 22 2 cos 3p 4 cos p 4 22 2 tan p 4 1 b. Sketch the angle, as shown in Figure 6.4-11. The terminal side is in quadrant III, so the reference angle is t p . 4p 3 p p 3 y 4π 3 x π 3 Figure 6.4-11 Section 6.4 Trigonometric Functions 451 Thus, the sine, cosine, and tangent functions are sin tan 4p 3 3p 4 sin p 3 23 2 cos 4p 3 cos p 3 1 2 tan p 3 23 c. Sketch the angle, as shown in Figure 6.4-12. The terminal side is in quadrant IV, so the reference angle is 2p 11p 6 p 6 2p t . Thus, the sine, cosine, and tangent functions are sin tan 11p 6 11p 6 sin tan p 6 p 6 11p 6 cos p 6 23 2 cos 1 2 23 3 ■ y 11π 6 x π 6 Figure 6.4-12 , it is possible to f
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ind a coterIf an angle is less than 0 or greater than 2p . minal angle between 0 and Thus, the trigonometric functions of a real variable have the following property. by adding or subtracting multiples of 2p 2p Trigonometric Ratios of Coterminal Angles Any trigonometric function of a real number t is equal to the same trigonometric function of all numbers where k is an integer. t 2kP, Example 4 Trigonometric Functions Where t 7 2P Find the sine, cosine, and tangent of 7p 3 . y 7π 3 π 3 x Figure 6.4-13 Solution 7p 3 can be written as p 3 2p. Therefore, is coterminal with p 3 . 7p 3 23 2 1 2 sin cos tan 7p 3 7p 3 7p 3 sin cos tan p 3 p 3 p 3 23 ■ 452 Chapter 6 Trigonometry Exercises 6.4 Note: Unless stated otherwise, all angles are in standard position. In Exercises 1–6, find sin t, cos t, and tan t when the terminal side of an angle of t radians passes through the given point. 1. 4. 2, 7 2 4, 3 1 1 2. 5. 2 3, 2 2 23, 10 1 1 3. 6. 5, 6 1 2 p, 2 2 1 2 In Exercises 7–10, find sin t, cos t, and tan t when the terminal side of an angle of t radians passes through the given point on the unit circle. 7. 9. 2 25 a , 1 25b 3 5 , 4 5b a 8. 1 210 a , 3 210b 10. 1 0.6, 0.8 2 In Exercises 11–14, identify an angle that is coterminal with the given angle, and find the sine and cosine of the given angle. 0 t P 11. 13p 6 12. 9p 2 13. 16p 14. 7p 4 In Exercises 15–23, a. Use a calculator in radian mode to find the sine, cosine, and tangent of each number. Round your answers to four decimal places. b. Use the signs of the functions to identify the quadrant of the terminal side of an angle of t radians. If the terminal side lies on an axis, identify which axis and whether it is on the positive or negative side of the axis. Explain your reasoning. 15. 7p 5 18. 23p 21. 9.5p 16. 11 19. 10p 3 22. p 17 17. 14p 9 20. 6.4p 23. 17 In Exercises 24 – 29, sketch each angle whose radian measure is given and find its reference angle. 24. 7p 3 25. 17p 6 26. 6p 5 27. 1.75p 28. 3p 4 29. p 7 In Exercises 30–47, find the exact value of the sine, cosine, and tangent of the number without using a calculator. 30. 34. 7p 6 5p 4 31. 7p 3 35. 3p 2 32. 17p 3 33. 11p 4 36. 3p 37. 23p 6 38. 11p 6 39. 19p 3 40. 10p 3 41. 15p 4 42. 25p 4 46. p 43. 5p 6 47. 4p 44. 17p 2 45. 9p 2 In Exercises 48 – 53, write the expression as a single real number. Do not use decimal approximations. 48. sin p 6 b a cos p 2 b a cos 49. cos p 2 b a cos p 4 b a sin p 6 b a p 2 b a sin sin p 2 b a p 4 b a 50. cos 2p 3 b a cos p sin 2p 3 b a sin p 51. sin 3p 4 b a cos 5p 6 b a cos 3p 4 b a sin 5p 6 b a 52. sin 7p 3 b a cos 5p 4 b a cos 7p 3 b a sin 5p 4 b a 53. sin p 3 b a cos p sin p cos p 3 b a In Exercises 54–59, the terminal side of an angle of t radians lies in the given quadrant on the given line. Find sin t, cos t, and tan t. (Hint: Find a point on the terminal side of the angle.) 54. Quadrant III; line 2y 4x 0 55. Quadrant IV; line through 3, 5 1 and 1 2 9, 15 2 56. Quadrant III; line through the origin parallel to 7x 2y 6 57. Quadrant II; line through the origin parallel to 2y x 6 58. Quadrant I; line through the origin perpendicular to 3y x 6 59. Quadrant IV; line y 3x 60. The terminal side of an angle of t radians lies on a straight line through the origin, and therefore, has an equation of the form slope of the line. where m is the y mx, y y = mx t x m tan t. a. Prove that Hint: a point on the terminal side of the angle has coordinates x, mx b. Explain why tan t approaches infinity as t 1 2 approaches p 2 from below. Hint: What happens to the slope of the terminal side when t is close to p 2 ? c. Explain why tan t approaches negative infinity as t approaches from above. Hint: When t is p 2 p 2 Section 6.4 Trigonometric Functions 453 61. The figure below shows an angle of t radians. Use trigonometric functions to write the coordinates of point P in terms of r and t. y r P t x 62. Complete the following table by writing each value as a fraction with denominator 2 and a radical in the numerator. You may find the resulting pattern an easy way to remember these function values. t sin t cos t 0 2? 2 2? 2 P 6 2? 2 2? 2 P 4 2? 2 2? 2 P 3 2? 2 2? 2 P 2 2? 2 2? 2 63. Find the domain and range of the cosecant function. 64. Find the domain and range of the secant function. 65. Find the domain and range of the cotangent a bit larger than , is the slope of its terminal function. side positive or negative? d. Use parts b and c to show that the range of the tangent function is the set of all real numbers. 66. Critical Thinking Using only the definition and no calculator, determine which number is larger: sin(cos 0) or cos(sin 0). 454 Chapter 6 Trigonometry 6.5 Basic Trigonometric Identities Objectives • Develop basic trigonometric identities The algebra of trigonometric functions is just like that of other functions. They may be added, subtracted, composed, etc. However, two notational conventions are normally used with trigonometric functions. NOTE Most calculators automatically insert an opening parenthesis when a trigonometric function key is pushed. The display cos is interpreted as 1 cos . If you want cos 1 5 3, parenthesis after the 5: cos 1 you must insert Technology Tip Calculators do not use the convention of writing an exponent between the trigonometric function and its argument. In order to obtain you must enter sin sin3 4, 4 ^3. 1 2 Parentheses can be omitted whenever no confusion can result. Figure 6.5-1 shows, however, that parentheses are needed to distinguish cos 1 t 3 2 and cos t 3. Figure 6.5-1 When dealing with powers of trigonometric functions, exponents (other than 1 ) are written between the function symbol and the variable. For example, Furthermore, cos t 2 1 3 is written cos3t. sin t3 means sin 1 t3 2 not sin t 2 1 3 or sin3 t, as illustrated in Figure 6.5-2. Figure 6.5-2 Identities Trigonometric functions have numerous relationships that can be expressed as identities. An identity is an equation that is true for all val- Section 6.5 Basic Trigonometric Identities 455 ues of the variables for which every term of the equation is defined. For example, a b 1 2 2 a2 2ab b2 is an identity because it is true for all possible values of a and b. The unit circle description of trigonometric functions (see the box on page 446) leads to the following quotient identities. Quotient Identities tan t sin t cos t cott cos t sin t Example 1 Quotient Identities Simplify the expression below. tan t cos t Solution By the quotient identity, tan t cos t sin t cos t ˛cos t sin t ■ Reciprocal Identities The reciprocal identities follow immediately from the definitions of the trigonometric functions. sin t 1 csc t csc t 1 sin t cos t 1 sec t sec t 1 cos t tan t 1 cot t cot t 1 tan t Example 2 Reciprocal Identities Given that sin t 0.28 and cos t 0.96, find csc t and sec t. Solution By the reciprocal identities, csc t 1 sin t 1 0.28 3.57 sec t 1 cos t 1 0.96 1.04 ■ Reciprocal Identities CAUTION An identity may not be true for a value of the variable that makes a term of the equation undefined. For example, if while cot t is undefined. tan t 1 cot t tan t 0 t 0, then for Thus, t 0. 456 Chapter 6 Trigonometry y 1 P(cos t, sin t) Pythagorean Identities t x 1 −1 0 −1 Figure 6.5-3 Pythagorean Identities For any real number t, the coordinates of the point P where the terminal side of an angle of t radians meets the unit circle are (cos t, sin t), as shown in Figure 6.5-3. Since P is on the unit circle, its coordinates must satisfy x2 y2 1, which is the equation of the unit circle. That is, cos2 t sin2 t 1 This identity, which is usually written is called the Pythagorean identity. It can be used as follows to derive two other identities, which are also called Pythagorean identities. sin2 t cos2 t 1, sin2 t cos2 t sin2 t cos2 t 1 1 cos2 t cos2 t cos2 t tan2 t 1 sec2 t Divide by cos2 t Simplify Similarly, dividing both sides of sin2 t cos2 t 1 by sin2 t shows that 1 cot 2 t csc2 t sin2 t cos2 t 1 tan2 t 1 sec2 t 1 cot2 t csc2 t In addition to the version shown above, the following forms of the Pythagorean identity are also commonly used. sin2 t 1 cos2 t cos2 t 1 sin2 t Example 3 Pythagorean Identities Simplify the expression below. tan2 t cos2 t cos2 t Solution By the quotient and Pythagorean identities, tan2 t cos2 t cos2 t sin2 t cos2 t ˛ cos2 t cos2 t sin2 t cos2 t 1 ■ Periodicity Identities Let t be any real number. Construct two angles in standard position of measure t and radians, as shown in Figure 6.5-4. Since both of t 2p Section 6.5 Basic Trigonometric Identities 457 these angles have the same terminal side, the point P where the terminal side intersects the unit circle is the same for both angles. y y P (cos t, sin t) P (cos (t + 2π), sin (t + 2π)) t x t + 2π x Figure 6.5-4 In both cases, the sine is the y-coordinate of P, so sin t sin 1 t 2p . 2 In addition, the terminal side of the angle is the same for measures of t, t ± 2p, t ± 6p, t ± 4p, sin t sin 1 and so on. Thus, t ± 4p sin 1 2 t ± 2p sin 1 2 t ± 6p 2 p Similarly in both cases, the cosine is the x-coordinate of P, so cos 1 cos t cos 1 cos 1 t ± 2p t ± 4p t ± 6p 2 2 2 p The identities above show that sine and cosine functions repeat their values at regular intervals. Such functions are called periodic. A function is said to be periodic if there exists some constant k such that 1 for every number t in the domain of f. The smallest value of k that has this property is called the period of the function f. 2 2 1 f t f t k Since the tangent function is the quotient of the sine and cosine functions, . However, there is a number it must also be true that 2 2p that has this property. Figure 6.5-5 shows the angles t and smaller than t p. A rotation of so the image of the point (x, y) is radians is the same as a rotation of tan t tan 1 x, y t 2p Thus, 180°, p y (x, y) t x t + π (−x, −y) Figure 6.5-5 tan tan t Calculator Exploration Use your calculator to verify the following: 3 4p sin sin 3 sin 2 1 1 4 6p cos cos 4 cos 1 2 1 tan 1 tan 1 5p
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tan 1 1 3 2p 4 2p 1 p 2 2 2 2 458 Chapter 6 Trigonometry Periodicity Identities The sine and cosine functions are periodic with period For every real number t, 2P. sin (t 2P) sin t and cos (t 2P) cos t P. The tangent function is periodic with period number t in the domain of the tangent function, For every tan (t P) tan t Example 4 Periodicity Identities Find the exact value of sin 13p 6 . Solution By the periodicity identity for sine, sin 13p 6 sin p 6 a 12p 6 b sin p 6 a 2p b sin p 6 1 2 ■ Negative Angle Identities Let t be any real number and construct two angles in standard position of measure t and radians, as shown in Figure 6.5-6. t y (cos t, sin t) 1 P −1 t −t x 1 (cos (−t), sin (−t)) −1 Q Figure 6.5-6 Since the point Q is the reflection of the point P across the x-axis, the x-coordinates of P and Q are the same, and the y-coordinates are opposites of each other. Thus, cos t cos and sin t sin t t 1 2 1 2 Also, tan t 2 1 sin cos t 2 t 2 1 1 sin t cos t sin t cos t tan t Section 6.5 Basic Trigonometric Identities 459 Negative Angle Identities sin(t) sin t cos(t) cos t tan(t) tan t Example 5 Negative Angle Identities Find the exact value of sin p 6 b a and of cos p 6 b a . Solution By the negative angle identities, sin p a 6 b sin p 6 1 2 and cos p a 6 b cos p 6 23 2 ■ y Other Identities Q(−x, y) P(x, y) π − t t t x p t Let t be any real number. Figure 6.5-7 shows the angles of t and radians in standard position. The terminal side of the angle of t radians meets the unit circle at P, and the terminal side of the angle of radians meets the unit circle at Q. Congruent triangles can be used to prove what the figure illustrates: p t The y-coordinates of P and Q are the same, and their x-coordinates are opposites. Figure 6.5-7 This leads to the following identities. Identities Involving P t sin t sin(P t) cos t cos(P t) tan t tan(P t) Example 6 Identities Involving P t Find the exact value of sin 5p 6 b . a NOTE The identity sin t sin p t is used in solving basic trigonometric equations. (See Section 8.3.) 1 2 Solution By the identity p t sin t, sin 1 sin 2 6p 6 a sin a 5p 6 b p 6 b sin p p 6 b a sin p 6 b a 1 2 ■ 460 Chapter 6 Trigonometry Summary of Identities Quotient Identities: tan t sin t cos t cot t cos t sin t Reciprocal Identities: sin t 1 csc t csc t 1 sin t cos t 1 sec t sec t 1 cos t tan t 1 cot t cot t 1 tan t Pythagorean Identities: sin2 t cos2 t 1 tan2 t 1 sec2 t 1 cot2 t csc2 t Periodicity Identities: sin (t 2P) sin t cos(t 2P) cos t tan(t P) tan t Negative Angle Identities: sin (t) sin t cos(t) cos t tan (t) tan t Identities Involving sin t sin(P t) : P t cos t cos(P t) tan t tan(P t) Exercises 6.5 In Exercises 1–4, use the quotient and reciprocal identities to simplify the given expression. remaining five trigonometric functions. Round your answers to four decimal places. 1. cot t sin t 3. csc t sin t 2. tan t cot t 4. cot t sec t In Exercises 5–8, use the Pythagorean identities to simplify the given expression. 5. sin 2 ˛t cot 2 ˛t sin 2 t 6. 1 sec2 ˛ t 7. 8. csc 2 ˛t cot 2 sin 2 t ˛t sin 2 ˛t cos 2 sin 2 t ˛t sin 2 t In Exercises 9–14, the value of one trigonometric func- tion is given for 0 6 t 6 p 2 and Pythagorean identities to find the values of the . Use quotient, reciprocal, 9. sin t 0.3251 10. cos t 0.4167 11. tan t 3.6294 12. sec t 2.5846 13. csc t 6.2474 14. cot t 1.8479 In Exercises 15–25, use basic identities and algebra to simplify the expression. Assume all denominators are nonzero. 15. 16. 17. sin t cos t 1 sin t cos t 1 2 sin t cos t 21 2 2 sin t tan t 18. 1 tan t 2 tan t 3 1 2 21 6 tan t 2 2 tan t 19. 4 cos2 t sin2 t b a a sin t 4 cos tb 2 20. 21. 22. 5 cos t sin2 t sin2 t sin t cos t sin2 t cos2 t cos2 t 4 cos t 4 cos t 2 sin2 t 2 sin t 1 sin t 1 23. 1 cos t sin t tan t 24. 1 tan2 t 1 tan2 t 2 sin2 t 25. 2sin3 t cos t 2cos t Recall that a function is even if f(x) f(x) and a function is odd if Section 6.5 Basic Trigonometric Identities 461 39. sin 2p t 2 1 41. tan t 43. tan 2p t 2 1 40. cos t 42. cos 44. sin t 2 1 p t 1 2 In Exercises 45–50, cos t 2 5 and p 66 t 66 3p 2 . Use basic identities and the signs of the trigonometric functions in each quadrant to find each value. 45. sin t 46. tan t 47. cos 49. sin 2p t 1 4p t 2 2 1 48. cos 50. tan t 2 4p t 1 1 2 In Exercises 51–54, it is given that sin p 8 32 22 2 f(x) f(x) Use basic identities to find each value. for every value of x in the domain of f. In Exercises 26–32, use the negative angle identities to determine whether the function is even, odd, or neither. 26. 28. 30. 32 sin t tan t t sin t t cos t 27. 29. 31 cos t sec t t tan t In Exercises 33–36, use the Pythagorean identities to find sin t for the given value of cos t. Make sure that the sign is correct for the given quadrant. 33. cos t 0.5 34. cos t 3 210 35. cos t 1 2 36. cos t 2 25 p 6 t 6 3p 3p 2 6 t 6 2p In Exercises 37–44, sin t 3 5 identities and the signs of the trigonometric functions in each quadrant to find each value. 0 66 t 66 p 2 . Use basic and 37. sin t 2 1 38. sin 1 t 10p 2 51. cos p 8 53. sin 17p 8 52. tan p 8 54. tan 15p 8 In Exercises 55–60, use the Pythagorean identities to determine if it is possible for a number t to satisfy the given conditions. 55. sin t 5 13 and cos t 12 13 56. sin t 2 and cos t 1 57. sin t 1 and cos t 1 58. sin t 1 22 and cos t 1 22 59. sin t 1 and tan t 1 60. cos t 8 17 and tan t 15 8 61. Use the periodicity identities for sine, cosine, and tangent to write periodicity identities for cosecant, secant, and cotangent. 62. Use the negative angle identities for sine, cosine, and tangent to write negative angle identities for cosecant, secant, and cotangent Important Concepts Section 6.1 Section 6.2 Section 6.3 Section 6.4 462 Angle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 Vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414 Hypotenuse. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 Opposite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 Adjacent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 Trigonometric ratios . . . . . . . . . . . . . . . . . . . . . . 416 Sine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 Tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 Cosecant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 Secant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 Cotangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 Special angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 Solving a triangle . . . . . . . . . . . . . . . . . . . . . . . . 421 Triangle Sum Theorem . . . . . . . . . . . . . . . . . . . . 421 Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . 421 Angle of elevation. . . . . . . . . . . . . . . . . . . . . . . . 426 Angle of depression . . . . . . . . . . . . . . . . . . . . . . 426 Initial side of an angle . . . . . . . . . . . . . . . . . . . . 433 Terminal side of an angle . . . . . . . . . . . . . . . . . . 433 Standard position of an angle . . . . . . . . . . . . . . . 434 Coterminal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 Arc length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 Radian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 Unit circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 Trigonometric ratios in the coordinate plane . . . 444 Trigonometric functions of a real variable . . . . . 445 Unit circle descriptions of trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446 Reference angle . . . . . . . . . . . . . . . . . . . . . . . . . . 449 Trigonometric ratios of coterminal angles. . . . . . 451 Chapter Review 463 Section 6.5 Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 Quotient identities . . . . . . . . . . . . . . . . . . . . . . . 455 Reciprocal identities . . . . . . . . . . . . . . . . . . . . . . 455 Pythagorean identities. . . . . . . . . . . . . . . . . . . . . 456 Period of a function . . . . . . . . . . . . . . . . . . . . . . 457 Periodicity identities . . . . . . . . . . . . . . . . . . . . . . 458 Negative angle identities . . . . . . . . . . . . . . . . . . 459 Important Facts and Formulas For a given acute angle u: sin u csc u opposite hypotenuse hypotenuse opposite cos u sec u adjacent hypotenuse hypotenuse adjacent tan u cot u opposite adjacent adjacent opposite U sin U cos U tan U csc U sec U cot U 30° 60° 45° 1 2 23 2 22 2 23 2 1 2 22 2 23 3 23 2 223 3 223 3 2 1 22 22 23 23 3 1 To convert radians to degrees, multiply by To convert degrees to radians, multiply by 180 p . p 180 . Quotient Identities: tan t sin t cos t cot t cos t sin t Reciprocal Identities: sin t 1 csc t csc t 1 sin t cos t 1 sec t sec t 1 cos t tan t 1 cot t cot t 1 tan t Pythagorean Identities: sin2 t cos2 t 1 tan2 1 sec2 t 1 cot2 t sec2 t 464 Chapter Review Important Facts and Formulas Periodicity Identities: sin t ± 2p sin t 1 2 Negative Angle Identities: cos t ± 2p 1 2 cos t tan 1 t ± p 2 tan t sin t sin t 1 2 Identities Involving p t : cos t 2 1 cos t tan t 2 1 tan t 2 1 sin t sin p t 2 1 cos t cos p t 2 1 tan t tan p t 2 1 Review Exercises Section 6.1 1. Write 41° 6¿ 54– in decimal form. 2. Write 10.5625° in DMS form. 3. Which of the following statements about the angle u is true? a. c. e. sin u 3 4 tan u 3 5 sin u 4 3 b. d. cos u 5 4 sin u 4 5 4 5 3 θ In Exercises 4–9, use the right triangle in the figure to find each ratio. 4. sin u 7. csc u 5. cos u 8. sec u 6. tan u 9. cot u 7 θ 4 10. Find the length of side h in the triangle, given that angle A measures 40° and the distance from C to A is 25. B h C A Section 6.2 In Exercises 11–14, solve triangle ABC. Chapter Revi
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ew 465 11. A 40° b 10 12. C 35° a 12 13. A 56° a 11 14 15. From a point on level ground 145 feet from the base of a tower, the angle of elevation to the top of the tower is 57.3°. How high is the tower? 16. A pilot in a plane at an altitude of 22,000 feet observes that the angle of depression to a nearby airport is the point on the ground directly below the plane? 26°. How many miles is the airport from 17. A lighthouse keeper 100 feet above the water sees a boat sailing in a straight line directly toward her. As she watches, the angle of depression to the boat changes from How far has the boat traveled during this time? 40°. 25° to Section 6.3 18. 9p 5 radians ___ ? degrees 19. 17p 12 radians ___ ? degrees 20. 11p 4 radians ___ ? degrees 21. 36° ___ ? radians 22. 220° ___ ? radians 23. 135° ___ ? radians 24. Find a number between 0 and u 2p in standard position is coterminal with an angle of such that an angle of 23p 3 u radians radians in standard position. 25. Through how many radians does the second hand of a clock move in 2 minutes and 40 seconds? 26. 10 revolutions per minute ___ ? radians per minute 27. 4p radians per minute ___ ? revolutions per minute Section 6.4 28. If the terminal side of an angle of t radians in standard position passes through the point 1 2, 3 , then 2 tan t ___ ? . 29. If the terminal side of an angle of t radians in standard position passes through the point 1 6, 8 , then 2 cos t ___ ? . 30. If the terminal side of an angle of t radians in standard position passes through the point (1.2, 3.5), then sin t ___ ? . In Exercises 31–42, give the exact values. 31. cos 34. cos 47p 2 3p 4 32. sin 13p 1 2 35. tan 8p 3 33. sin 7p 6 36. sin 7p 4 b a 466 Chapter Review 37. cot 4p 3 38. cos p 6 b a 40. sin 11p 6 b a 41. sin p 3 39. sec 42. csc 2p 3 5p 2 43. The value of cos t is negative when the terminal side of an angle of t radians in standard position lies in which quadrants? In Exercises 44–46, express as a single real number (no decimal approximations allowed). 44. cos 3p 4 sin 5p 6 sin 3p 4 cos 5p 6 45. sin a p 6 2 1 b 46. sin p 2 b a sin 0 cos 0 Section 6.5 47. Write tan t cot t entirely in terms of sin t and cos t, then simplify. 48. 3 sin S 2 p 5500b T a S 3 cos 2 p 5500b T a ? 49. Which of the following could possibly be a true statement about a real number t? a. and cos t sin t 2 and cos t 1 22 sin t 1 2 2 sin t 1 and cos t 1 sin t p 2 sin t 3 5 and cos t 4 5 and cos t 1 p 2 b. c. d. e. 50. If p 2 6 t 6 p and sin t 5 13 , then cos t ? 51. If sin t 4 5 and the terminal side of an angle of t radians in standard position lies in the third quadrant, then cos t ___ ? . 52. Simplify 53. If sin tan sin t p 2 1 t 2p 2 . 1 101p a 2 b 1, then sin 105p 2 b a ? 54. Which of the statements (i)–(iii) are true? sin x cos x tan x x 2 x 2 x 2 (i) sin 1 (ii) cos 1 (iii) tan 1 (i) and (ii) only (ii) only (i) and (iii) only a. b. c. d. all of them e. none of them 55. Suppose is a real number. Consider the right triangle with sides as shown Chapter Review 467 u in the figure. Then: a. b. c. cos u sin u d. e. none of the above sin θ 2 cos θ 56. Determine the following segment lengths in terms of a single trigonometric function of t. a. OR b. PR c. SQ d. OQ Figure 6.C-1 Optimization with Trigonometry Optimization problems involve finding a solution that is either a maximum or a minimum value of a function. Calculus is needed to find exact solutions to most optimization problems, but tables or graphs can often be used to find approximate solutions. Example 1 Maximum Area A gutter is to be made from a strip of metal 24 inches wide by bending up the sides to form a trapezoid, as shown in Figure 6.C-1. a. Express the area of the cross-section of the gutter as a function of the angle t. b. For what value of t will this area be as large as possible? Solution a. The cross-section of the gutter is a trapezoid, shown in Figure 6.C-2. b2 8 h x t t 8 The bases are parallel, so these alternate interior angles are equal. = 8 b1 Figure 6.C-2 The area of a trapezoid is b1 h 1 b22 2 . The top base b2 8 2x, where The height is h, where . cos t x 8 So x 8 cos t. So h 8 sin t. sin t h 8 . Thus, the area of the cross-section is A 8 sin t 1 8 8 2 2 1 8 cos t 22 4 sin t 1 16 16 cos t 64 sin t 1 cos t . 2 1 2 b. To find the value of t that makes the area be as large as possible, first notice that t must be between 0 and p 2 1.57. By examining a table of values, it is possible to estimate the maximum value of A over this interval. 468 NOTE Trigonometric functions often have maxima and minima at p. fractional multiples of It is a good idea to try these values as increments for a table. 6 ft t d1 – t π 2 d2 8 ft A good starting interval for the table is p 12 0.26. The table in Figure 6.C-3 shows the highest value at about 1.05, or 4p 12 p 3 . Figure 6.C-3 Figure 6.C-4 It is possible to confirm this value or get a better estimate by using a table with a smaller step size, such as firms that p 3 area, about p 144 appears to be the value of t that corresponds to the largest 83.1 in2. Figure 6.C-4 con- 0.02. ■ Example 2 Maximum Length Two corridors meet at a right angle. One corridor is 6 ft wide, and the other is 8 ft wide. A ladder is being carried horizontally along the corridor. What is the maximum length of a ladder that can fit around the corner? Solution The length of the longest ladder that fits around the corner is the same as the shortest length of the red segment in Figure 6.C-5 as it pivots about the corner. Let the part of the segment from the corner to the opposite and the part to the wall of the 8-ft corridor wall of the 6-ft corridor be be Then the desired length is d1, d2. d2. d1 Figure 6.C-5 For the angle t in Figure 6.C-5, sin t 6 d1 6 d1 sin t sin p 2 a t b 8 d2 d2 8 p 2 Q t R sin The function that describes the desired length is L 6 sin t 8 p 2 Q . t R sin 469 To find the minimum of the function, note that t is between 0 and p 2 . Construct a table with an increment of p 12 to construct a table with an increment of , then use the minimum value p 144 . The minimum appears to be at t 0.74, which corresponds to a length of about 19.7 ft. Figure 6.C-6 ■ In Examples 1 and 2, an area and a length were represented in terms of an angle to find the optimal solution. In the following example, the angle is the quantity to be maximized. Example 3 Maximum Viewing Angle The best view of a statue is where the viewing angle is a maximum. In Figure 6.C-7, the height of the statue is 24 ft and the height of the pedestal is 8 ft. Find the distance from the statue where the viewing angle is optimal. t eye level 5 ft Figure 6.C-7 Solution In Figure 6.C-8, the angle The important quantities are opposite and adjacent to the angles, so the tangent function is used to describe the relationship. t t1 t2. tan t1 27 d tan t2 3 d tan˛ t1 1 27 d t2 tan˛ 1 3 d t tan˛ 1 27 d tan˛ 1 3 d statue 24 ft pedestal 8 ft t2 t1 t d Figure 6.C-8 eye level 5 ft 470 d 7 0. To create a table, first notice that Start with a large increment, such as 5 ft, and narrow the increment to refine your estimate, as shown in Figure 6.C-9. Figure 6.C-9 The best distance to view the statue is at about 9 ft away. The viewing angle at this distance is about 0.93 radians, which is about 53°. ■ Exercises Estimate the maximum value of the given function between 0 and P 2 by using tables with increments of P 12 1. 3. and P 144 . f t 2 1 f t 2 1 sin t cos t 2. f t 2 1 sin t 2 cos t 3 sin t sin p 2 a t b 4. f t 2 1 2 cos t 1 1 sin t 5. The cross section of a tunnel is a semicircle with radius 10 meters. The interior walls of the tunnel form a rectangle. y t 10 10 x a. Express the area of the rectangular cross-section of the tunnel opening as a function of angle t. b. For what value of t is the cross-sectional area of the tunnel opening as large as possible? What are the dimensions of the tunnel opening in this case? 6. A 30-ft statue stands on a 10-ft pedestal. Find the best distance to view the statue, assuming eye level is 5 ft (see Example 3). 7. Two towns lie 10 miles apart on opposite sides of a mile-wide straight river, as shown. A road is to be built along one side of the river from town A to point X, then across the river to town B. The cost of building on land is $10,000 per mile, and the cost of building over the water is $20,000 per mile. a. Express the cost of building the road as a function of the angle t. b. Find the minimum cost of the road. A 1 mi road X t 10 mi B 471 C H A P T E R 7 Trigonometric Graphs Stay tuned for more! Radio stations transmit by sending out a signal in the form of an electromagnetic wave that can be described by a trigonometric function. The shape of this signal is modified by the sounds being transmitted. AM radio signals are modified by varying the “height,” or amplitude, of the waves, whereas FM signals are modified by varying the frequency of the waves. The signal displayed in the photo is from an AM radio station found at 900 on the broadcast dial. See Exercise 65 of Section 7.3. 472 Chapter Outline 7.1 Graphs of the Sine, Cosine, and Tangent Functions 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions 7.3 Periodic Graphs and Amplitude 7.4 Periodic Graphs and Phase Shifts 7.4.A Excursion: Other Trigonometric Graphs Chapter Review can do calculus Approximations with Infinite Series Interdependence of Sections > 7.2 > 7.3 7.1 > 7.4 G raphs of trigonometric functions often make it very easy to see the essential properties of these functions, particularly the fact that they repeat their values at regular intervals. Because of the repeating, or peri- odic, nature of trigonometric functions, they are used to model a variety of phenomena that involve cyclic behavior, such as sound waves, elec- tron orbitals, planetary orbits, radio transmissions, vibrating strings, pendulums, and many more.• 7.1 Graphs of the Sine, Cosine, and Tangent Functions Objectives • Graph the sine, cosine, and tangent functions • State all values in the do
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main of a basic trigonometric function that correspond to a given value of the range • Graph transformations of the sine, cosine, and tangent graphs Although a graphing calculator will quickly sketch the graphs of the sine, cosine, and tangent functions, it will not give you much insight into why these graphs have the shapes they do and why these shapes are important. So the emphasis in this section is the connection between the functions’ definitions and their graphs. Using radians and the unit circle, you learned in Chapter 6 that trigonometric functions can be defined as functions of real numbers. Using this definition, you will see that the graphs of trigonometric functions are directly related to angles in the unit circle. Graph of the Sine Function Consider an angle of t radians in standard position. Let P be the point where the terminal side of the angle meets the unit circle. Then the sin t y-coordinate of P is the number sin t. As t increases, the graph of can be sketched from the corresponding y-coordinates of P. t f 1 2 473 474 Chapter 7 Trigonometric Graphs Change in t from 0 to p 2 Movement of point P sint (y-coordinate of P) Corresponding graph from (1, 0) to (0, 1) increases from 0 to 1 (0, 1) P t (1, 0) y y y y 1 −1 1 −1 1 −1 1 −1 t t t t π 2 π 2π 3π 2 π 2 π 2π 3π 2 π 2 π 2 π 2π 3π 2 π 2π 3π 2 from p 2 to p from 0, 1 to 1 2 1 1, 0 2 decreases from 1 to 0 (0, 1) t P (−1, 0) from p to 3p 2 from 1, 0 1 to 1 2 0, 1 2 decreases from 0 to 1 (−1, 0) t P (0, −1) from 3p 2 to 2p from 0, 1 1 to 1 2 1, 0 2 increases from 1 to 0 t (1, 0) P (0, −1) CAUTION Throughout this chapter, the independent variable used for trigonometric functions will be t to avoid any confusion with the x and y that are part of the definition of these functions. However, using a graphing calculator in function mode, you must enter x as the independent variable. Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 475 Your graphing calculator can provide a dynamic view of the graph of the sine function and its relationship to points on the unit circle. Graphing Exploration With your graphing calculator in parametric mode, set the viewing window as shown below, with a t-step of 0.1. 0 t 2p p 3 x 2p 2.5 y 2.5 On the same screen, graph the two functions given below. sin t cos t, Y1 t, Y2 sin t X1 X2 Use the trace feature to move the cursor along the first graph, which is the unit circle. Stop at a point, and note the values of t and y. Use the up or down key to move the cursor to the second graph, which is the graph of the sine function. The value of t will remain the same. What are the x- and y-coordinates of this point? How does the y-coordinate of the new point compare with the y-coordinate of the original point on the unit circle? To complete the graph of the sine function, note that as t goes from 4p, the point P on the unit circle retraces the path it took from 0 to 2p the same curve will repeat on the graph. This repetition occurs each along the horizontal axis, therefore the sine function has a period of That is, for any real number t, 2p to 2p, so units 2p. t ± 2p sin 1 sin t. 2 y 1 h(t) = sin t t −4π −3π −2π −π 0 π 2π 3π 4π −1 Graph of the Cosine Function Graph of the Sine Function Let P be the point where the terminal side of an angle of t radians in standard position meets the unit circle. Then the x-coordinate of P is the the same process as number To obtain the graph of cos cos t. t t f , 1 2 1 2 476 Chapter 7 Trigonometric Graphs that used for the sine function is followed, except the x-coordinate is observed. The following chart illustrates the graph of the cosine function. Movement of point P cos t (x-coordinate of P) Corresponding graph from (1, 0) to (0, 1) decreases from 1 to 0 y Change in t from 0 to p 2 from p 2 to p (0, 1) P t (1, 0) from (0, 1) to 1, 0 2 decreases from 0 to 1 1 (0, 1) P t (−1, 0) from p to 3p 2 from 1, 0 1 to 1 2 0, 1 2 increases from 1 to 0 (−1, 0) t P (0, −1) from 3p 2 to 2p from 0, 1 1 2 to (1, 0) increases from 0 to 1 t (1, 0) P (0, −1) 1 −1 1 −1 1 −1 1 − 3π 2 2π π 2 π 2 π 2 π 3π 2 2π π 3π 2 2π π 3π 2 2π As the value of t increases, the point P on the unit circle retraces its path t along the unit circle, so the graph of repeats the same curve 2 Because the cosine function also has a period of at intervals of length 2p, for any number t, cos 2p. t f 1 2 1 cos 1 t ± 2p 2 cos t. Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 477 Graph of the Cosine Function y 1 0 −1 h(t) = cos t π 2π 3π 4π t −4π −3π −2π −π The graphs of the sine and cosine functions visually illustrate two basic facts about these functions. Because the graphs extend infinitely to the right and to the left, the domain of the sine and cosine functions is the set of all real numbers. Also, the y-coordinate of every point on these graphs lies between and 1 (inclusive), so that 1 the range of the sine and cosine functions is the interval [1, 1]. You can use the period of the function to state all values of t for which sin t is a given number, as shown in Examples 1 and 2. cos t or Example 1 Finding All t-values State all values of t for which sin t is 1. Solution (i.e., and 1 and has a period of units on the horizontal axis), so there are an These points occur every units on the horizontal axis, and a few are highlighted in red on the graph The sine function oscillates between it repeats the pattern every infinite number of t-values for which sin t is 2p y sin t shown in Figure 7.1.1. 1. 2p 2p 1 y 1 0 −1 3π 2 t π 2π 3π 4π −4π −3π −2π −π 2π 2π 2π Figure 7.1-1 478 Chapter 7 Trigonometric Graphs On the interval y sin t 3 2 has only one point, 0, 2p , highlighted in red on the graph above, the graph of 1. t 3p 2 Therefore, all values of t for which 2kp, where k is any integer. 3p 2 a , 1 b sin t , at which the y-coordinate is is 1 can be expressed as ■ Example 2 Finding All t-values State all values of t for which cos t is 1 2 . Solution The cosine function repeats its pattern of y-values at intervals of 2p, so there are an infinite number of t-values for which y cos t 1 2 of . shown in Figure 7.1-2 highlights a few points with a y-coordinate cos t is The graph of . 1 2 1 −4π −3π −2π −π 0 −1 y π 3 y = 1 2 π 2π 3π 4π t 5π 3 On the interval y cos t of 0, 2p 2 3 has two points, Figure 7.1-2 , highlighted in red on the graph above, the graph p 3 , 1 2 b a and 5p 3 , 1 2 b a , at which the y-coordinate 1 . is 2 t p 3 Therefore, all values of t for which cos t is 1 2 can be expressed as 2kp or 5p 3 2kp, where k is any integer. ■ Graph of the Tangent Function f a connection between To determine the shape of the graph of the tangent function and slope can be used. As shown in Figure 7.1-3, the point P where the terminal side of an angle of t radians in standard position meets the unit circle has coordinates This point and the point (0, 0) can be used to compute the slope of the line containing the terminal side. cos t, sin t t . 2 1 2 1 tan t, y 1 P (cos t, sin t) −1 t x 1 −1 Figure 7.1-3 Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 479 slope sin t 0 cos t 0 sin t cos t tan t The graph of minal side of an angle of t radians, as t takes different values. can be sketched by noting the slope of the ter- f t 2 1 tan t Change in t from 0 to p 2 Movement of terminal side from horizontal upward toward vertical tan t (terminal side slope) increases from 0 in the positive direction Corresponding graph t from 0 to p 2 from horizontal downward toward vertical decreases from 0 in the negative direction t π− 2 π− 2 π 2 π 2 When t ± p 2 , the terminal side of the angle is vertical, so its slope is not defined. The graph of the tangent function has vertical asymptotes at the values of t for which the function is undefined. To complete the graph of the tangent function, note that as t goes from p 2 the terminal side goes from almost vertical with negative slope 3p 2 to , to almost vertical with positive slope, exactly as it does from So the graph repeats this pattern at intervals of length p. p 2 to p 2 . 480 Chapter 7 Trigonometric Graphs Graph of the Tangent Function −2π −π − 3π 2 − π 2 h(t) = tan t π 2 π 3π 2 t 2π y 4 2 0 −2 −4 Notice that the domain of the tangent function is all real numbers except odd multiples of p 2 . The range of the tangent function is all real numbers. Because the tangent function has a period of domain, p, for any number t in its tan 1 t ± p 2 tan t. Example 3 Finding All t-values State all values of t for which tan t is 1. Solution p, The tangent function repeats its pattern of y-values at intervals of so there y tan t are an infinite number t-values for which The graph of shown in Figure 7.1-4 highlights a few points with a y-coordinate of 1. tan t is 1. y y = tan t − 5π 4 −2π −π 2 − π 4 0 −2 3π 4 π 7π 4 t 2π y = −1 π π π Figure 7.1-4 p 2 , p 2 b S On the interval , highlighted in red on the graph above, the graph of y tan t has only one point, p a 4 , 1 b , at which the y-coordinate is Technology Tip p Most calculators have a window setting that automatically rescales the horizontal axis in fractional units of when in radian mode. On TI models, select ZTRIG in the ZOOM menu, and on Casio, select F3 (V-Window) then F2 (TRIG) from GRAPH mode. Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 481 1. t p 4 Therefore, all values of t for which kp, where k is any integer. tan t is 1 can be expressed as ■ h(t) = 4 cos t f(t) = cos t 4 −2π 2π −4 Figure 7.1-5 Basic Transformations of Sine, Cosine, and Tangent The graphical transformations (such as shifting and stretching) that were considered in Section 3.4 also apply to trigonometric graphs. Example 4 Vertical Stretch List the transformation needed to change the graph of the graph of Graph both equations on the same screen. 4 cos t. h t f t 1 2 cos t into 1 2 Solution h Because by a factor of 4. Both graphs are identified in Figure 7.1-5. the graph of h is the graph of f after a vertical stretch t t , 2 1 2 1 4 f ■ Example 5 Reflect
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ion and Vertical Stretch Graph g 1 2 t 2 1 sin t on the interval 2p, 2p . 4 3 Solution The graph of g is the graph of t f 1 2 sin t 1 2 compressed vertically by a factor of , as shown in Figure 7.1-6. reflected across the x-axis and y 1 −1 1 g(t) = − sin t 2 t 2π π f(t) = sin t −2π −π Figure 7.1-6 ■ Example 6 Vertical Shift tan t 5 on the interval 3p, 3p . 4 3 Graph h t 2 1 Solution The graph of h is the graph of tan t shifted up 5 units. t f 1 2 482 Chapter 7 Trigonometric Graphs NOTE For a complete discussion of symmetry and odd and even functions, see Excursion 3.4.A. y 8 6 4 2 0 −2 −3π −2π −π t π 2π 3π Figure 7.1-7 ■ Even and Odd Functions Trigonometric functions can be classified as odd or even as determined by their symmetry. Even Functions A graph is symmetric with respect to the y-axis if the part of the graph on the right side of the y-axis is the mirror image of the part on the left side of the y-axis. Graphing Exploration 1. For each pair of functions f and g below, answer the following questions. cos t sin t tan and and and g g g t cos t 2 1 2 1 t sin t 2 1 2 1 t tan t 2 1 2 1 • Is f symmetric with respect to the y-axis? • Does the graph of g appear to coincide with the graph of f? 2. If a graph is symmetric with respect to the y-axis, describe the graph after a reflection across the y-axis. A function f whose graph is symmetric with respect to the y-axis is called an even function. Even Function A function f is even if f (x) f(x) for every x in the domain of f. The graph of an even function is symmetric with respect to the y-axis. Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 483 f For example, t cos 1 1 2 cos t t 2 cos t is an even function because for every t in the domain of f cos t. t 2 1 Odd Functions If a graph is symmetric with respect to the origin, then whenever (x, y) is on is also on the graph. A function f whose graph is symthe graph, metric with respect to the origin is called an odd function. x, y 2 1 Odd Function A function f is odd if f(x) f(x) for every x in the domain of f. The graph of an odd function is symmetric with respect to the origin. f For example, t 1 t 1 sin tan 2 2 1 sin t t 2 sin t tan t and g t 2 1 tan t are odd functions because for every t in the domain of f for every t in the domain of g sin t tan t. t t 2 2 1 1 Summary of the Properties of Sine, Cosine, and Tangent Functions Function Symbol Domain Range Period Even/Odd sine cosine tangent sin t all real numbers cos t all real numbers all real numbers from to 1, inclusive all real numbers from to 1, inclusive 1 1 tan t all real numbers except all real numbers odd multiples of p 2 2p 2p p odd even odd Exercises 7.1 In Exercises 1 – 6, graph each function on the given interval. 6. h t 2 1 tan t; 5p 3 S , 3p T 1. 2. h 4. f t t 2 2 1 1 sin t; cos t; tan t; sin t; 3 2p, 6p 4 p, 3p 4 3 3 p, 2p 4 3 5p, 3p 5. g t 2 1 cos t; 7p 6 , 7p 2 T S 4 7. For what values of t on the interval sin t 1? 8. For what values of t on the interval cos t 0? 2p, 2p 2p, 2p 3 3 is is 4 4 9. What is the maximum value of g cos t? t 2 1 10. What is the minimum value of f sin t? t 2 1 484 Chapter 7 Trigonometric Graphs 11. For what values of t on the interval tan t 3 have vertical 4 2p, 2p does the graph of asymptotes? h t 2 1 12. What is the y-intercept of the graph of sin t? f t 2 1 13. What is the y-intercept of the graph of g t 2 1 cos t? 14. What is the y-intercept of the graph of tan t? h t 2 1 15. For what values of t on the interval p, p 3 is f t 2 1 4 sin t increasing? 16. For what values of t on the interval cos t decreasing? g t 2 1 17. For what values of t on the interval tan t greater than 1? 18. For what values of t on the interval tan t less than 0? 19. For what values of t on the interval tan t increasing? h t 2 1 3p, p is 4 2p, 2p 2p, 2p is is 4 4 p, 2p is 4 3 3 3 3 In Exercises 20–33, find all the exact t-values for which the given statement is true. 20. tan t 0 22. sin t 0 24. tan t 1 26. cos t 0 28. sin t 1 21. sin t 22 2 23. cos t 1 2 25. sin t 23 2 27. cos t 23 2 29. sin t 1 2 30. tan t 23 3 31. cos t 22 2 32. cos t 1 33. tan t 23 In Exercises 34–43, list the transformations that change the graph of f into the graph of g. State the domain and range of g. 34. 35. f t 2 1 f t 2 1 cos t; g cos t 2 t 2 1 cos t; g cos t t 2 1 36. 37. 38. 39. 40. 41. 42. 43 sin t; g 1 tan t; g tan t; g t 1 2 3 sin t t 2 t 1 tan t 5 2 tan t cos t; g t 1 sin t; g t 1 2 3 cos t 2 2 sin t sin t; g 1 cos t; g sin t; g t 1 2 3 sin t 2 t 2 t 1 5 cos t 3 2 sin t 3 In Exercises 44– 48, sketch the graph of each function. 44. 46 cos t 4 tan t 48. f t 2 1 3 sin t 1 2 45. f t 2 1 5 sin t 1 47. f t 2 1 1 4 cos t In Exercises 49–54, match a graph to a function. Only one graph is possible for each function 49. 51. 53. a. 2π b. 2π 2 tan t sin t 1 3 tan t 1 50. 52. 54.5 cos t 2.5 sin t cos t 1 3 3 3 3 2π 2π Section 7.1 Graphs of the Sine, Cosine, and Tangent Functions 485 c. 3 2π 2π 60. Scientists theorize that the average temperature at a specific location fluctuates from cooler to warmer and then to cooler again over a long period of time. The graph shows a theoretical prediction of the average summer temperature for the last 150,000 years for a location in Alaska. d. e. f. 3 3 π π 3 3 2π 2π 3 3 π π 3 55. Fill the blanks with “even” or “odd” so that the resulting statement is true. Then prove the statement by using an appropriate identity. Excursion 3.4.A may be helpful. a. b. c. d. e. is an ___ function. is an ___ function. is an ___ function. sin t cos t tan t t sin t t tan t is an ___ function. is an ___ function In Exercises 56–59, find tan t, where the terminal side of an angle of t radians lies on the given line. 56. y 1.5x 58. y 0.32x 57. y 1.4x 59. y 11x 85 80 75 70 65 60 55 150000 −100000 −50000 Years Ago a. Find the highest and lowest temperature represented. b. Over what time interval does the temperature repeat the cycle? c. What is the estimated average summer temperature at the present time? 61. A rotating beacon is located at point P, 5 yards from a wall. The distance d, as measured along the wall, where the light shines is given by d 5 tan 2pt where t is time measured in seconds since the beacon began to rotate. When the light is aimed at point A. When the beacon is aimed to the right of A, the distance d is positive, and when it is aimed to the left of A, the value of d is negative. t 0, d A 5 yds P Graph the function and estimate the value of d for the following times. b. a. c. d. e. Determine the position of the beacon when t 0 t 0.7 t 0.5 t 1.4 t 0.25 of d for that value of t. and discuss the corresponding value 486 Chapter 7 Trigonometric Graphs 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions Objectives • Graph the cosecant, secant, and cotangent functions • Graph transformations of the cosecant, secant, and cotangent graphs y sin t, y cos t, and y tan t The graphs of that were developed in Section 7.1 are closely related to the graphs of the reciprocal functions y csc t, y sec t, and y cot t that are studied in this section. Graph of the Cosecant Function The general shape of the graph of the graph of the sine function and the fact that csc t 1 sin t f t 1 2 . csc t can be determined by using Graphing Exploration Graph the two functions below on the same screen in a viewing window with 2p t 2p 4 y 4 and sint g f t 2 1 1 sint t 2 1 How are the graphs alike and how are they different? sin t csc t and are reciprocals, csc t Because that is, when t is an integer multiple of t f 1 of csc t t csc t is all real numbers except integer multiples of has vertical asymptotes at integer multiples of p. is not defined when p. sin t 0; Therefore, the domain of and the graph p, 2 f 1 2 Graph of the Cosecant Function −2π −π y = csc t y = sin t t π 2π y 4 2 −2 −4 Notice that as the graph of csc t of decreases to a height of 1, f 1 t 2 decreases to a height of 1, and as the graph of increases to a height of 1, the graph y sin t increases to a height csc t f t y sin t the graph of 1 2 Section 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions 487 1. The range of of to 1 or less than or equal to f t 1 2 csc t 1. is all real numbers greater than or equal 2p. The period of the cosecant function is Example 1 Reflection and Vertical Stretch 3 csc t. Graph h t 2 1 Solution y 3 sin t, y sin t First consider the graph of stretched vertically by a factor of 3 and reflected across the horizontal and that of axis. The relationship between the graph of 3 csc t h as shown in Figure 7.2-1. y csc t and y sin t, is similar to that between which is the graph of y 3 sin t t 1 2 y 8 4 0 −4 −8 h(t) = −3 csc t y = −3sin t t π 2 π 2π 3π 2 −2π − 3π 2 −π − π 2 Figure 7.2-1 ■ Graph of the Secant Function t The graph of 2 that the graph of f 1 sec t f 1 t 2 csc t is related to the cosine graph in the same way is related to the sine graph. Graphing Exploration Graph the two functions below on the same screen in a viewing window with 2p t 2p 4 y 4. and cos t g f t 2 1 1 cos t t 2 1 How are the graphs alike and how are they different? Because cos t and sec t are reciprocals, sec t that is, when t is an odd multiple of is not defined when p 2 . Therefore, the domain of cos t 0; f t sec t 2 f 1 1 t of 2 or equal to sec t 1. is all real numbers except odd multiples of p 2 , and the range is all real numbers greater than or equal to 1 or less than The period of the secant function is 2p. 488 Chapter 7 Trigonometric Graphs Graph of the Secant Function −2π −π g(t) = sec t y = cos t π t 2π y 4 2 −2 −4 Example 2 Vertical Stretch and Vertical Shift 2 sec t 3. Graph g t 2 1 Solution y 2 cos t 3, stretched First graph vertically by a factor of 2 and shifted down 3 units. The graphs of g are related in the same way as the 1 graphs of y 2 cos t 3 y cos t, as shown in Figure 7.2-2. which is the graph of 2 sec t 3 y sec t and and t 2 y cos t −2π −π y g(t) = 2 sec t − 3 π 2π t y = 2 cos t − 3 6 4 2 0 −2 −4 −6 −8 −10 Figure 7.2-2 ■ Graph of the Cotangent Function Becaus
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e cot t cost sint ing the quotient , the graph of cot t t f 1 2 can be obtained by graph- y cos t sin t . The cotangent function is not defined when whenever t is an integer multiple of f(t) the range of f(t) has vertical asymptotes at integer multiples of p. p. sin t 0 , and this occurs Therefore the domain of p, and cot t t f 1 2 cot t consists of all real numbers except integer multiples of cot t is the set of real numbers. The graph of Section 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions 489 Graph of the Cotangent Function y 4 2 0 −2 −4 y = tan t f(t) = cot t t π 2 π 2π 3π 2 −2π − 3π 2 −π − π 2 Notice that as the graph of decreases, and as the graph of increases. The period of the cotangent function is increases, the graph of decreases, the graph of p. y tan t y tan t cot t cot t t t f 1 f 1 2 2 Example 3 Reflection, Vertical Stretch, and Horizontal Shift Graph k t 2 1 3 cot t p 4 b . a Solution The graph of k zontal shift of 2 t 1 p 4 3 cot t p 4 b a is the graph of y cot t after a hori- units to the right, a reflection across the horizontal axis, and a vertical stretch by a factor of 3. The graph of k 3 cot t 2 1 t p 4 b a is shown with the graph of y cot t −2π −π y 8 6 4 2 0 −2 −4 −6 −8 in Figure 7.2-3 below. k(t) = −3 cot ( y = cot t t − π 4 ) π t 2π Figure 7.2-3 ■ Even and Odd Functions The fact that the cosecant, secant, and cotangent functions are reciprocals of the sine, cosine, and tangent functions, respectively, can be used to determine whether the functions are even or odd. 490 Chapter 7 Trigonometric Graphs The secant function is an even function, as shown below. sec t 2 1 1 t cos 1 2 1 cos t sec t The cosecant and cotangent functions are odd functions. csc t 1 cot t 2 1 1 t 2 sin 1 t cos 1 t sin 1 2 2 2 1 sin t 1 sin t csc t cos t sin t cos t sin t cot t Summary of the Properties of Secant, Cosecant, and Cotangent Functions Function Symbol Domain Range Period Even/Odd secant sec t t f 1 2 all real numbers except odd multiples of p 2 all real numbers less 1 than or equal to greater than or equal to 1 or cosecant csc t t f 1 2 all real numbers except multiples of p all real numbers less 1 than or equal to greater than or equal to 1 or cotangent cot t t f 1 2 all real numbers except multiples of p all real numbers 2p 2p p even odd odd Exercises 7.2 11. 12. t sec t f 1 and shifted 1 unit to the left 2 stretched vertically by a factor of 3 t csc t f 1 and shifted 1 unit up 2 compressed vertically by a factor of 0.5 In Exercises 1–10, describe the transformations that change or h(t) cot t f(t) csc t, g(t) sec t, into the graph of the given function. the graph of 1. s t 1 3. m 1 2 t 3 sec t 2 csc t 2 1 2 4 5. p t 2 1 1 2 sec t 1 2. 4. k t 2 1 t 3 2 5 cot 1 2 cot t 2 1 13. t f 1 2 sec t reflected across the horizontal axis and 2 csc t compressed vertically by a factor of 1 4 7. q t 2 1 sec t 2 1 8 8. s t 2 1 5 cot t 2 1 2 9. v t 2 1 p csc t 10. j t 2 1 1 4 sec t 2 1 14. 15. t cot t f 1 shifted down 2 units 2 reflected across the vertical axis and csc t t f 1 2 down shifted p 2 units to the right and 5 units In Exercises 11–17, state the rule of a function g whose graph is the given transformation of the graph of f. 16. csc t f t 2 1 of 0.75 compressed vertically by a factor Section 7.2 Graphs of the Cosecant, Secant, and Cotangent Functions 491 17. t cot t f 1 across the horizontal axis 2 reflected across the vertical axis and In Exercises 18–21, match graph a, b, c, or d with each function. a. b. c. d. −1π − π 2 −1π − π 2 −1π − π 2 −1π − 2 −4 −6 6 4 2 −2 −4 −6 6 4 2 −2 −4 −6 6 4 2 −2 −4 −6 0 0 0 0 t 1π t 1π t 1π t 1π π 2 π 2 π 2 π 2 18. f t 2 1 1 2 cot t 1 19. f t 2 1 1 2 cot t 1 20. t f 1 2 2 cot t 1 21. t f 1 2 2 cot t 1 In Exercises 22 – 25, match graph a or b with each function. a. b. −1π − 3π 4 − π 2 − π 4 −1π − 3π 1 −2 −3 −4 −5 5 4 3 2 1 −1 −2 −3 −4 −5 0 0 t t π 4 π 2 3π 4 1π π 4 π 2 3π 4 1π 22. 24. f t 2 1 f t 2 1 csc t csc t 2 1 23. 25. f t 2 1 f t 2 1 csc t csc t 2 1 In Exercises 26 – 29, match graph a or b with each function. a1 −2 −3 −4 −5 0 π 4 π 2 3π 4 1π 5π 4 3π 2 t 492 b. Chapter 7 Trigonometric Graphs y 5 4 3 2 1 −1 −2 −3 −4 −5 0 π 4 π 2 3π 4 1π 5π 4 3π 2 − π 2 − π 4 26. 28. f f t t 2 2 1 1 sec t 2 1 sec t 2 1 27. 29 sec 1 sec t 1 2 2 In Exercises 30 – 33, match graph a or b with each function. a. b. −1π − 3π 4 − π 2 − π 4 −1π − 3π 1 −2 −3 −4 −5 5 4 3 2 1 −1 −2 −3 −4 −5 0 0 t t π 4 π 2 3π 4 1π π 4 π 2 3π 4 1π 30. 32. f f t t 2 2 1 1 cot t 2 1 cot t 2 1 31. 33. f f t t 2 2 1 1 cot t 2 1 cot t 2 1 t In Exercises 34–38, graph at least one cycle of the given function. 34. 36. 38 sec 2t 35. f t 2 1 cot 3t 4 5 csc t a p 2 b 37. f t 2 1 3 4 csc t 2 3 sec t p 2 1 39. Critical Thinking Show graphically that the equation sec t t but none between has infinitely many solutions, p 2 and p 2 . 40. Critical Thinking A rotating beacon is positioned 5 yards from a wall at P. In the figure, the distance a is given by a 5 sec 2pt , 0 where t is the number of seconds since the beacon began to rotate. 0 d A 5 yds a. Use the graph of a as a function of t to find a for the following times. t 0 t 0.75 t 1 b. For what values of t is c. How fast is the beacon rotating? a 5? Section 7.3 Periodic Graphs and Amplitude 493 7.3 Periodic Graphs and Amplitude Objectives • State the period and amplitude (if any) given the function rule or the graph of a sine, cosine, or tangent function • Use the period and amplitude (if any) to sketch the graph of a sine, cosine, or tangent function A surprisingly large number of physical phenomena can be described by functions like the following: 5 sin and g 4 cos 0.5t 1 3t In this section and in the next, the graphs of such functions will be analyzed. All of these functions are periodic and their graphs consist of a series of identical waves. A single wave of the graph is called a cycle. The length of each cycle is the period of the function. 2 0 2 sine cycle 2π Figure 7.3-1 Every cycle for the sine function resembles the graph of 0 to 2p, as shown in Figure 7.3-1. t f 1 2 sin t from • beginning at a point midway between its maximum and minimum value • rising to its maximum value • falling to its minimum value • returning to the beginning point Every cycle repeats the same pattern. 2 0 2 cosine cycle 2π Figure 7.3-2 Similarly, every cycle for the cosine function resembles the graph of g as shown in Figure 7.3-2. from 0 to cos t 2p, t 1 2 494 Chapter 7 Trigonometric Graphs • beginning at its maximum value • falling to its minimum value • returning to the beginning point Again, every cycle repeats the same pattern. Period Before proceeding to the discussion about functions that have different periods, it will be helpful to consider functions of the form t f 1 2 sin bt and g 1 t 2 cos bt where b is a constant. The constant b changes the period of the sine or cosine function. Its effect on the graph is to increase or decrease the length of each cycle. Graphing Exploration Graph each function below, one at a time, in a viewing window Answer the questions that follow for each funcwith tion. 0 t 2p. t f 1 2 cos 4t h 1 t 2 sin 5t Determine the number of complete cycles between 0 and 2p. Find the period, or length of one complete cycle. Hint: Use division. The exploration above suggests the following rule. Period of sin bt and cos bt If b 77 0, then the graph of either f(t) sin bt or g(t) cos bt makes b complete cycles between 2P b has a period of . 0 and 2P, and each function y 1 0 −1 g(t) = sin 3t t π 3 2π 3 π 4π 3 5π 3 2π t h The graph of 0 to 2p. on values from 1 g Similarly, the graph of 0 to 2p. 2 sin t 1 2 completes one cycle as t takes on values from completes one cycle as 3t takes sin 3t t When 3t 0, t must be 0. When 3t 2p, t must be 2p 3 . Figure 7.3-3 ues from 0 to 2p 3 , as shown in Figure 7.3-3. Therefore, the graph of g sin 3t t 2 1 completes one cycle as t takes on val- Section 7.3 Periodic Graphs and Amplitude 495 Example 1 Determining Period Determine the period of each function. a. b. k t 2 1 f t 2 1 cos 3t sin t 2 Solution a. The function Figure 7.3-4. k t 2 1 cos 3t has a period of 2p b 2p 3 , as shown in y 1 −1 k(t) = cos 3t π 3 2π 3 π 4π 3 5π 3 t 2π 1 cycle 1 cycle 1 cycle Figure 7.3-4 b. Rewrite f sin t 2 1 has a period of 2p b t as f t 1 2 2p 1 2 sin 2 4p, 1 2 a . t b The function f sin t 2 1 1 2 a t b as shown in Figure 7.3-5. −2π −1π f(t) = sin t 2 1π t 2π y 1 −1 1 cycle Figure 7.3-5 ■ 496 Chapter 7 Trigonometric Graphs CAUTION t A calculator may not produce an accurate graph of g cosbt for large values of b. For instance, the graph of sin 50t sinbt or has 50 complete cycles between 0 and your calculator will show. (try it!) 2p, but that is not what Graphing Exploration Graph each function below, one at a time, in a viewing window with p 2 t p 2 . Answer the questions that follow for each function. f t 2 1 tan 3t g tan 4t t 2 1 Determine the number of complete cycles between p 2 and p 2 . Find the period, that is, the length of one complete cycle. The exploration above suggests the following rule. Period of tan bt If b 77 0, then the graph of f(t) tan bt makes b complete cycles between function has a period of p b . p 2 and p 2 , and the y 4 2 0 −2 −4 − π 2 − π 4 Figure 7.3-6 π 4 π 2 k(t) = tan 2t Example 2 Determining Period Determine the period of each function. t a. k t 2 1 tan 2t b. f t 2 1 tan t 3 Solution a. The function k cycle between t 2 1 p 4 tan 2t has a period of p b p 2 . It completes one and p 4 , as shown in Figure 7.3-6. Section 7.3 Periodic Graphs and Amplitude 497 y f(t) = tan 4 2 t 3 t 0 −π 3π − 2 − π 2 π 2 π 3π 2 2 1 3p, of p b p 1 3 Amplitude b. Rewrite as f t tan t 3 as f t 2 1 tan 1 3 a . t b The function has a period as shown in Figure 7.3-7. ■ −4 Figure 7.3-7 Recall from Section 3.4 that multiplying the rule of a function by a positive constant has the effect of stretching or compressing its graph vertically. Example 3 Vertical and Horizontal Stretches or Compressions y g(t) = 7 cos 3t Graph each function. k(t) = cos 3t a.
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g t 2 1 7 cos 3t b. h t 2 1 1 3 sin t 2 7 1 −π − 2π 3 − π 3 π 3 2π 3 t π Solution a. The function is the function by 7. Consequently, the graph of g is the graph of k (see Example 1a) stretched vertically by a factor of 7. multiplied g k t t 2 2 1 1 7 cos 3t cos 3t −7 Figure 7.3-8 y f(t) = sin t 2 t π 2π 1 h(t) = sin 3 t 2 1 1 3 −2π −π 0 − 1 3 −1 Figure 7.3-9 As Figure 7.3-8 shows, stretching the graph affects only the height of the waves in the graph, not the period of the function. So the period cos 3t, of g is the same as that of namely k t . 2p b 2p 3 1 2 b. The function h 1 3 t 2 1 sin t 2 is the function f sin t 2 1 t 2 multiplied by 1 3 . Consequently, the graph of h is the graph of f (see Example 1b) vertically compressed by a factor of 2p 1 2 as the period of f, namely 2p b The period of h is the same . 1 3 4p. ■ As the graphs in Example 3 illustrate, vertically stretching or compressing the graph affects only the height, not the period of the function. 7 cos 3t 1 2 t g The graph of in Example 3 reaches a maximum value of 7 units above the horizontal axis and a minimum value of 7 units below a cos bt the horizontal axis. In general, the graph of units above and below the horizontal axis, and reaches a distance of The graph of g(t) 7 cos 3t has an is said to have an amplitude of amplitude of 7. a sin bt g or 498 Chapter 7 Trigonometric Graphs Amplitude and Period If a 0 and b 77 0, f(t) a sin bt then each of the functions or g(t) a cos bt has an amplitude of and a period of a 00 00 2p b . Example 4 Determining Amplitude and Period Determine the amplitude and period of p 2 T the interval p 2 , . S Solution 2 sin 4t. Then graph f on f t 2 1 The amplitude of 2 sin 4t is t f 1 2 t f 1 2p b 2 sin 4t 2 p 2p 2 4 . is a 0 0 2 0 0 2, and the period of So the graph of f consists of cycles that are p 2 long and rise and fall between the heights of 2 and 2. To graph this function, be sure to notice that its graph is the reflection of h across the horizontal axis, as shown in Figure 7.3-10. 2 sin 4t t 1 2 ■ Although the graph of any function can be vertically stretched or compressed, amplitude only applies to bounded periodic functions. y f(t) = −2 sin 4t 2 h(t) = 2 sin 4t Figure 7.3-10 Exercises 7.3 In Exercises 1–15, state the amplitude (if any) and period of each function. 1. 3. 5. 7. 9 cos t cos 3t 4 cos t 5 tan 2t 0.3 sin t 3 2. 4. 6. 8. 10 sin 2t 2.5 tan t 3 sin t 1.2 cos 0.5t tan 0.4t 11. f t 2 1 1 2 sin 3t 12. f t 2 1 1 2 tan 3t 13. f t 2 1 5 cos 1.7t 15. f t 2 1 1 3 tan pt 4 14. f t 2 1 2 sin 2pt 3 16. a. What is the period of f b. For what values of t (with t 2 1 sin 2pt? 0 t 1 ) is c. For what values of t (with d. For what values of t (with 0 t 1 ) is 0 t 1 ) is 17. a. What is the period of f b. For what values of t (with t 2 1 cos pt? 0 t 2 ) is c. For what values of t (with d. For what values of t (with 0 t 2 ) is 0 t 2 ) is 0? 1? 1? 0? 1? 1 18. a. What is the period of f 2 b. For what values of t (with 1 t tan pt? 1 2 t 1 2 ) is 0? f t 2 1 c. For what values of t (with 1? f t 2 1 d. For what values of t (with 1 ) is 1 2 t 1 2 ) is In Exercises 19–38, describe the transformations that change the graph of f into the graph of g. State the amplitude (if any) and the period of g. Section 7.3 Periodic Graphs and Amplitude 499 In Exercises 39 – 44, sketch at least one cycle of the graph of each function. 39. 41. 43 cos t 2 2 tan 3t 3.5 sin 2pt 40. 42. 44. f t 2 1 2 3 sin 2t f t 2 1 f t 2 1 0.8 cos pt tan pt 2 sin 5t t 2 In Exercises 45–50, match a graph to a function. Only one graph is possible for each function. a. 5 2 19. 20. 21. 22. 23. 24. 25. 26. 27. 28 sin t; g 1 tan t; g cos t; g 1 cos t; g 1 tan 3t cos 8t t t 2 2 cos t 2 1 tan t; g tan t 2 1 sin t; g t 1 sin t; g 1 cos t; g 2 2 t sin 2 1 sin 1.6t cos 2.6t t 2 1 sin t; g cos t; g t 2 1 t 2 1 3 sin t 1 2 cos t 29. f t 2 1 tan t; g 1 3 t 2 1 tan t 30. 31. 32. 33. 34. 35 sin t; g t 2 1 4 sin t 2 sin t; g tan t; g t 2 1 t 2 1 tan t; g cos t; g cos t sin 2t 2 tan t 2 2 tan 0.2t 3 cos 6t 2 5 cos 8t 36. f t 2 1 sin t; g t 2 1 2 sin 3pt 5 37. f t 2 1 tan t; g 1 3 t 2 1 tan pt 38. f t 2 1 cos t; g 5 3 t 2 1 cos pt 3 2π 2π 2π 2π b. c. d. 2π 2π 2π 2π 5 5 5 10 10 5 5 500 e. 2π f. Chapter 7 Trigonometric Graphs 2π 5 5 10 2π 2π 58. y 3 0 −3 59. y 5 0 −5 t t 2π 3 2π 5 In Exercises 60–64, state all local minima and maxima of the function on the given interval. 10 45. 47 sin 2t 3 cos t 2 49. f t 2 1 5 tan t 3 46. 48 cos 2t 3 sin t 2 50. f t 2 1 3 tan 2t In Exercises 51–56, write an equation for a sine function with the given information. 60. 61. 62. 63. 64 sin 2t; 0 t p cos 3t; 0 t p cos t 2 ; 2p t p sin t 3 ; 2p t p 3 sin 2pt; 1.5 t 1.5 65. The current generated by an AM radio transmitter 2 1 t A sin 2000 pmt, is given by a function of the form f the location on the broadcast dial and t is measured in seconds. For example, a station at 900 on the AM dial has a function of the form 550 m 1600 where is f t A sin 2000p 900 t A sin 1,800,000pt 1 2 2 1 Sound information is added to this signal by modulating A, that is, by changing the amplitude of the waves being transmitted. AM means amplitude modulation. For a station at 900 on the dial, what is the period of function f ? 66. Find the function f, its period, and its frequency for a radio station at 1440 on the dial. (See Exercise 65.) 51. amplitude 2, period 4p 52. amplitude 1 2 , period p 2 53. amplitude 1.8, period 3p 2 54. amplitude 1, period 2 55. amplitude 3 2 , period 4 56. amplitude 6, period 1 2 In Exercises 57–59, state the rule of a sine function whose graph appears to be identical to the given graph. 57. y 2 0 −2 t π 2 Section 7.4 Periodic Graphs and Phase Shifts 501 7.4 Periodic Graphs and Phase Shifts Objectives In Section 7.3, you studied graphs of functions of the form • State the period, amplitude, vertical shift, and phase shift given the function rule or graph of a sine or cosine function • Use graphs to determine whether an equation could possibly be an identity a sin bt and f t 2 1 a cos bt g t 2 1 and learned how the constants a and b affect the amplitudes and periods of the functions. In this section, you will consider functions of the form f t 2 1 a sin bt c d 2 1 and g t 2 1 a cos bt c d 2 1 where a, b, c, and d are constants, and you will determine how these constants affect the graphs of the functions. Vertical Shifts Recall from Section 3.4 that adding a constant to the rule of a function shifts the graph vertically. Example 1 illustrates a vertical shift in combination with a reflection and a change in amplitude. Example 1 Reflection, Vertical Stretch, and Vertical Shift Describe the graph of 2p, 2p . 3 4 2 cos t 3. Then graph k on the interval k t 2 1 Solution k reflected The graph of across the horizontal axis, vertically stretched by a factor of 2, and shifted 3 units upward, as shown in Figure 7.4-1. is the graph of g t t 2 2 1 1 2 cos t 3 cos t y k(t) = −2 cos t + 3 5 4 3 2 1 −2π −π −1 Figure 7.4-1 t π y = cos t 2π After the vertical shift, the graph of tered on the horizontal line y 3. k t 2 1 2 cos t 3 is vertically cen- ■ 502 Chapter 7 Trigonometric Graphs Phase Shifts Recall from Section 3.4 that when the independent variable t in the rule where c is a constant, the graph of a function is replaced by is shifted horizontally. For periodic functions, the number c is the phase shift associated with the graph. t c, t c or Example 2 Phase Shift Describe the graph of each function. t p 2 b a sin cos b. a 2p 3 b a Solution a. The graph of g sin t 2 1 t p a 2 b is the graph of f t 2 1 sin t shifted to the left p 2 units, as shown in Figure 7.4-2. y f(t) = sin t 1 0 −1 π 2 −2π −π g(t) = sin( t + π 2 ) π 2 Figure 7.4-2 t π 2π When the graph of g t 2 1 sin t p 2 b a , of g that begins at sin t t f 1 2 is shifted to become the graph of t 0 becomes a cycle the cycle of f that begins at t p 2 t 2p a 3 b is the graph of . cos Thus, g has a phase shift of p 2 . cos t shifted to f t 2 1 b. The graph of h t 2 1 the right 2p 3 units, as shown in Figure 7.4-3. y 1 0 −1 h(t) = cos( t − 2π 3 ) t 2π π 2π 3 Figure 7.4-3 −2π −π f(t) = cos t Section 7.4 Periodic Graphs and Phase Shifts 503 The cycle of cos h t 1 2 t f 2 1 t 2p 3 b a 2p 3 . phase shift of cos t that begins at t 0 becomes a cycle of that begins at t 2p 3 . Thus, the function h has a ■ Combined Transformations Now that you are familiar with the effects of various transformations on the sine and cosine functions, you are ready for some examples that simultaneously include changes in amplitude, period, and phase shift. Example 3 Combined Transformations State the amplitude, period, and phase shift of f t 2 1 3 sin 2t 5 . 2 1 Solution Rewrite the function rule. f t 2 1 3 sin 2t 5 1 2 3 sin t 5 a 2 b d 2 c When the rule of f is written in this form, you can see that it is obtained from the rule of k t 2 1 3 sin 2t by replacing t with t 5 2 . Therefore, the graph of f can be obtained by horizontally shifting the graph of k to the left units, as shown in Figure 7.4-4. 5 2 f(t) = 3 sin (2t + 5) −6 −4 − 5 2 y k(t) = 3 sin 2t t 2 4 6 4 2 0 −2 −4 Figure 7.4-4 t 0 becomes a cycle of f that begins at The cycle of k that begins at t 5 2 ; so the function f has a phase shift of The amplitude of f is 3 and its period is 2p 2 5 2 p. . ■ 504 Chapter 7 Trigonometric Graphs The procedure that is used in Examples 1–3 can be used to analyze any function whose rule is of the form a sin bt c d. t f 1 2 1 2 First rewrite the function rule as follows. f t 2 1 a sin bt c 2 1 d a sin t c bb d b a c d Thus, the graph of f is obtained from the graph of k t 2 units horizontally. The cycle of k that 1 a sin bt by shift- ing it d units vertically and c b begins at t 0 becomes the cycle of f that begins at t c b , so f has phase shift c b . The amplitude of both f and k is A similar analysis applies to the function g a 0 t 2 0 1 and both have period bt c a cos d . 1 2 2p b . If and b 77 0, a 0 f(t) a sin(bt c) d then each of the functions and g(t) a cos(bt c) d has the follow
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ing characteristics: amplitude a 00 00 phase shift c b period 2p b vertical shift d Example 4 Combined Transformations Describe the graph of g t 2 1 2 cos 3t 4 1 2 1. Solution Identify the amplitude, period, vertical shift and phase shift. 1 2 cos 3t 4 g t 1 2 1 amplitude a 0 0 3t 4 1 2 cos 1 2 2 period 2p b 3 2 4 2p 3 1 2 phase shift c b 2 cos 4 a 3 b 1, 3t 4 t shown in Figure 7.4-5, is vertically The graph of 2 2 1 y 1. centered on the horizontal line The waves reach a maximum of 2 units above that horizontal line and a minimum of 2 units below that vertical shift 1 4 3 g 1 Combined Transformations 2.5 2π 2π horizontal line. The graph begins a cosine wave at t 4 3 and completes 3.5 Figure 7.4-5 one cycle in 2 p 3 units. ■ Section 7.4 Periodic Graphs and Phase Shifts 505 Example 5 Combined Transformations Identify the amplitude, period, vertical shift, and phase shift of 4 sin . Then graph at least one complete cycle of f. Solution The function rule c 1, d 3. and f t 2 1 f(t) = −4 sin y ( t 2 + 1) 3+ is of the form a sin bt c 2 1 d, with a 48 −6 −4 0 −2 2 4 6 8 10 y = 4 sin ( Figure 7.4-6 t 2 + 1) 3+ amplitude a 4 0 0 0 phase shift c b 4 period 2p b 2p 1 2 4p 2 vertical shift 3 0 1 1 2 The waves of the graph are vertically centered on the horizontal line reaching a maximum of 7 and a minimum of wave at y 3 The graph begins a sine units. The graph of 1. 4p 12.6 t 2 and completes one cycle in 1 y 4 sin 3 t 2 a b f is the graph of y 3, as shown in Figure 7.4-6. reflected across the horizontal line ■ −2π 4 2 −2 −4 3π 4 π 4 Figure 7.4-7 Example 6 Identifying Graphs Find a sine function and a cosine function whose graphs look like the graph shown in Figure 7.4-7. Solution 2π This graph appears to have an amplitude of 2 and to be centered vertib 1. so cally on the horizontal axis. The period appears to be 2 cos t Therefore, the graph looks like the graph of shifted horizontally. 2 sin t 2p, g t or f 1 t 2 1 2 The graph of 2 Figure 7.4-7 intercepts the x-axis at intercepts the x-axis at t p 4 , t f 1 2 sin t t 0. The graph in so it looks like the graph of 2 sin t shifted t f 1 2 p 4 units to the right. Therefore, this graph closely resembles the graph of h 1 At t 0, the graph of g t 2 1 t 2 sin 2 2 cos t t p 4 b a . reaches its maximum of 2. The graph in Figure 7.4-7 reaches its maximum of 2 at t 3p 4 , so it looks like the graph of g t 2 1 2 cos t shifted 3p 4 units to the right. Therefore, this graph closely resembles the graph of k 2 cos t 2 1 t 3p 4 b . a ■ 506 Chapter 7 Trigonometric Graphs NOTE Identities are proved algebraically in Chapter 9. Graphs and Identities Graphing calculators can be used to determine equations that could possibly be identities. A calculator cannot prove that such an equation is an identity, but it can provide evidence that it might be one. On the other hand, a calculator can prove that a particular equation is not an identity. Example 7 Possible Identities Which of the following equations could possibly be an identity? 4 g(t) = sin t a. cos p 2 a t b sin t b. cos p 2 a t b sin t −2π 2π π 2 f(t) = cos ( + t) −4 Figure 7.4-8 Solution a. If cos p 2 Q t R sin t is an identity, then cos f t 2 1 p 2 Q t R and g t 2 1 sin t are equivalent functions and have the same graph. The graphs of f and g, shown in Figure 7.4-8 are obviously different. Therefore, cos p 2 Q t R sin t is not an identity. b. In Figure 7.4-9, the graphs of cos f t 2 1 p 2 Q t g and t 1 R 2p, 2p sin t 2 appear to coincide on the interval values for f and g, shown in Figure 7.4-10, also supports the idea Comparing a table of . 3 4 that f t 2 1 cos p 2 Q t R and g t 2 1 sin t are equivalent functions. −2π 4 −4 2π Figure 7.4-9 Figure 7.4-10 This evidence strongly suggests that the equation cos is an identity, but does not prove it. Therefore, cos could possibly be an identity. Q p 2 Q p t 2 R sin t t R sin t ■ Section 7.4 Periodic Graphs and Phase Shifts 507 Example 8 Possible Identities Which of the following equations could possibly be an identity? a. cot t cos t sin t b. sin t tan t cos t f(t) = cot t cos t 1.5 g(t) = sin t Solution 2π 2π a. Rewrite f as sin t. g t 1 2 1 tan t cos t f t 2 1 and compare its graph with the graph of The graphs of f and g, shown in Figure 7.4-11 are obviously different. Therefore, cot t cos t sin t is not an identity. 1.5 Figure 7.4-11 However, it does appear from the graph that cot t cos t csc t could possibly be an identity. b. In Figure 7.4-12a, the graphs of sin t tan t t f 1 2 and g t 2 1 cos t appear to coincide. Comparing a table of values for f and g, shown in Figure 7.4-12b, also supports the idea that sin t tan t t f 1 2 and cos t t g 1 defined; that is, all values of t except those that make tan t 0. are equivalent functions for all values of t for which f is 2 1.5 2π 2π 1.5 Figure 7.4-12a Figure 7.4-12b Therefore, sin t tan t cos t could possibly be an identity for tan t 0. ■ CAUTION Do not assume that two graphs that look the same on a calculator screen actually are the same. Depending on the viewing window, two graphs that are actually quite different may appear identical. 508 Chapter 7 Trigonometric Graphs Exercises 7.4 In Exercises 1–20, state the amplitude, period, phase shift, and vertical shift of the function. 1. h t 2 1 cos t 1 1 2 2. m t 2 1 7 cos t 3 2 1 3. f t 2 1 5. k 7. p t t 2 2 1 1 8. f t 2 1 9. h t 2 1 4. k t 2 1 cos 2pt a 3 b 6. g t 2 1 3 sin 2t p 2 1 5 sin 2t t p 4 2 3p t 1 sin 1 6 cos 1 4.5 sin 2 12t 6 1 2 3t p 6 b a 5 1 4 cos 10. p t 2 1 5 sin t 4 a 3 b 1 11. q t 2 1 7 sin 7t 1 7 B A 12. h t 2 1 16 sin 2t 3 a 4 b 13. d t 2 1 3 sin 2t 5p a 4 b 14. c t 2 1 cos 3t 2 a p 3 b 5 15. 17. g t 2 1 s t 2 1 97 cos 14t 5 1 2 7 cos 2pt 16. 18. f t 2 1 m t 2 1 19. k t 2 1 3 cos p t 3 a 1 b 5 20. h t 2 1 4 sin t 3 a p 4 b 3 2 cos 4t 1 1 t 5 2 2 2 4 cos 1 In Exercises 21–30, state the rule of a sine function with the given amplitude, period, phase shift, and vertical shift, respectively. 21. 3, p 4 , p 5 , 0 22. 1, 2, 3, 4 23. 2 3 , 3p, 2p 3 , 2 24. 8, 1 2 , 2 3 , 4 25. 0.5, 2.5, 1.5, 0.6 26. 1, 5, 0, 3 27. 6, 5p 3 , 0, 1 29. 5 2 , 1.8, 0.2, 0 28. 2, 8p, 1, 1 30. 1, 1, 1, 1 In Exercises 31–40, a. State the rule of a function of the form t a sin bt c f identical to the given graph. d 1 1 2 2 whose graph appears to be b. State the rule of a function of the form t a cos bt c f identical to the given graph. d 1 1 2 2 whose graph appears to be 31. 32. 33. 34. 35. 36. y 12 0 −12 18 0 −18 1 −1 1 −1 1 2 1 2 − 3.5 −3. 3π 2 π 2 3π 4 t t π 8 π 4 5π 16 π 3 π 37. 38. 39. 40. t 3π t π 2 4π 3 π 4 2π 2 −4 y 0 −2 −4 −6 −8 −10 10 8 6 4 2 0 −π y −2 −4 −6 −8 −10 Section 7.4 Periodic Graphs and Phase Shifts 509 47. f t 2 1 sin 2t 3 1 2 1 48. g t 2 1 5 cos t p a 3 b 2 In Exercises 49–52, graph the function over the interval and determine the location of all local maxima and minima. (0, 2p), 49. f t 2 1 1 2 sin t p 3 b a 50. g t 2 1 2 sin 2t 3 a p 9 b 51. f t 2 1 2 sin 1 52. h t 2 1 1 2 cos p 2 a 3t p 2 t p 8 b 1 53. Describe the graph of f sin2 t cos2 t. t 2 1 In Exercises 54–57, use graphs to determine whether the equation could possibly be an identity or is definitely not an identity. 54. cos t t p a 2 b cos 56. sec t csc t 1 tan t cot t 55. sin t 1 cos t cot t csc t 57. tan t cot p 2 a t b 2p t 2p. In Exercises 58 – 61, graph f in a viewing window Use the trace feature to determine with constants a, b, and c such that the graph of f appears to coincide with the graph of a sin bt c g t . 1 2 1 2 58. 59. 60. 61 sin t 2 cos t 3 sin 4t 2 1 2 2 cos 4t 1 2 sin 3t 5 3 cos 1 2 sin t 5 cos t 2 1 2 1 3t 2 2 In Exercises 41–48, sketch the graph of at least one cycle of the function. 41. k t 2 1 3 sin t 43. p t 2 1 1 2 sin 2t 42. y t 2 1 2 cos 3t 44. q t 2 1 2 3 cos 3 2 t 45. h t 2 1 3 sin 2t p 2 b a 46. p t 2 1 3 cos 3t p 1 2 In Exercises 62–63, explain why there could not possibly be constants a, b, and c such that the graph of g coincides with the graph of f. a sin bt c t 1 2 62. 63 sin 2 3t 1 1 sin 2t cos 3t 2 3 cos 4t 1 1 2 510 Chapter 7 Trigonometric Graphs 7.4.A Excursion: Other Trigonometric Graphs Objectives • Write a sine function whose graph looks like the graph of another given sinusoidal function • Find viewing windows for the graphs of other trigonometric functions A graphing calculator enables you to explore with ease a wide variety of trigonometric functions. Graphing Exploration Graph each function on the same screen in a viewing window with 0 t 2p. cos t f g t 2 1 sin t 2 1 t p 2 b a How do the two graphs compare? Do they appear to coincide? The exploration above suggests that the equation cos t sin t p a 2 b is an identity and that the graph of the cosine function can be obtained by horizontally shifting the graph of the sine function. This is true, and it will be proved in Section 9.2. Consequently, every cosine function, such 6, t g as can be expressed as a sine function of the form 2 1 a sin t f The shape of the graph of such a function is called a sinusoid. 1 bt c 3 cos 4t 5 2 d. 2 2 1 1 Other Sinusoidal Graphs The exploration below suggests that other trigonometric functions can be bt c expressed in the form f a sin t . 1 2 1 2 Graphing Exploration 1. Graph with g 2 sin 2p t 2p 3 cos t 7 in a viewing window 2 1 . Does the function appear to be periodic? t 2 t 2 1 2 1 2. Using the calculator’s minimum and maximum finders, deter- mine the approximate amplitude of this function. 3. Using the calculator’s zero finder, estimate the period of this function (find the length of a complete cycle). 4. What is the smallest positive t-value at which a sine cycle begins? 5. Use the information from Questions 2–4 to write a function of the a sin t whose graph looks very much like the form 2 3 cos g t graph of . Graph the new function on the same screen with g. Do the graphs appear to coincide? 1 2 sin 2 t 7 bt Section 7.4.A Excursion: Other Trigonometric Graphs 511 The results of the preceding graphing exploration suggest that the graph of looks like the graph of 4.95 f 2 sin 1 t 2.60 . 3 cos t g
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2 1 sin These results illustrate the following facts. Sinusoidal Graphs If b, d, k, r, and s are constants, then the graph of the function g(t) d sin(bt r) k cos(bt s) is a sinusoid and there are constants a and c such that d sin(bt r) k cos(bt s) a sin(bt c). Example 1 Sinusoidal Graphs Find a sine function whose graph looks like the graph of g t 2 1 4 sin 3t 2 1 2 2 cos 3t 4 . 2 1 5 0.84 3.94 Solution The graph of 7.4.A-1. g t 2 1 4 sin 3t 2 2 1 2 cos 3t 4 1 2 is shown in Figure 2π 2π 5 Figure 7.4.A-1 By using a graphing calculator’s minimum and maximum finders with 2 cos 3t 2 t you see that g has an the graph of 2 2 amplitude of approximately 3.94. 4 sin 3t 4 g , 2 1 1 1 The function g has period 2p 3 because this is the period of both sin 3t 4 The function f . 2 t 2 1 a sin bt c 2 1 has period and cos 1 b 3. So 3t 2 1 2p b 2 . 2p 3 By using a graphing calculator’s zero finder, you can see that a sine cycle in the graph of g begins at approximately so the phase shift c b is approximately t 0.84, 0.84. Find c. 0.84 c b c 3 c 2.52 c 2.52 0.84 Substitute 3 for b Therefore, a 3.94, b 3, c 2.52, looks like the graph of f t 2 1 and the graph of 3t 2.52 2 3.94 sin 1 g t 2 1 4 sin 3t 2 2 1 2 cos 3t 4 . 2 1 ■ 512 Chapter 7 Trigonometric Graphs Other Trigonometric Graphs In Example 1, the variable t has the same coefficient b in both the sine and cosine terms of the function’s rule. When this is not the case, the graph will consist of waves of varying size and shape, as shown in Figure 7.4.A-2. sin 3t cos sin 3 t 5 1 2 4 cos t 2 2 1 h t 2 1 2 sin 2t 3 cos 3t 6 6 6 6 Figure 7.4.A-3 Figure 7.4.A-4 2π 2π 6 0 6 2π 2π 6 6 Figure 7.4.A-2 2π 2π 6 6 2π Example 2 Finding a Viewing Window Find a viewing window for one complete cycle of 2π Solution 4 sin 100pt 2 cos 40pt. f t 2 1 A graph of f in a viewing window with includes so many cycles that the calculator cannot display an accurate graph, as shown in Figure 7.4.A-3. 2p t 2p Instead, find the period of f by using the following method: Let h t 2 1 4 sin 100pt and g 2 cos 40pt. t 2 1 The period of h is 0.10 2p 100p 1 50 0.02. The period of g is 2p 40p 1 20 0.05. The period of f is the least common multiple of 0.02 and 0.05, which is 0.10. Therefore, the viewing window with one complete cycle of t f 4 sin 100pt 2 cos 40pt. 0 t 0.10 in Figure 7.4.A-4 shows 1 2 ■ Damped and Compressed Trigonometric Graphs Suppose a weight hanging from a spring is set in motion by an upward push. No spring is perfectly elastic, and friction acts to slow the motion of the weight as time goes on. Consequently, the graph showing the height of the weight above or below its equilibrium point at time t will consist of waves that get smaller and smaller as t gets larger. Many other physical situations can be described by functions whose graphs consist of waves of diminishing or increasing heights. Other situations, such as Section 7.4.A Excursion: Other Trigonometric Graphs 513 sound waves in FM radio transmission, are modeled by functions whose graphs consist of waves of uniform height and varying frequency. 35 Example 3 Analyzing a Damped Graph −35 35 Solution Analyze the graph of f t cos t. t 2 1 −35 Figure 7.4.A-5 y y = 0.5t t y = −0.5t not to scale Figure 7.4.A-6 Graph f in a viewing window with shown in Figure 7.4.A-5. 35 t 35 and 35 y 35, as Recall that sidering the cases t 6 0. ) sign when 1 cos t 1. t 0 and Multiply each term of the inequality by t con(Remember to reverse the inequality t 6 0. when when t t cos t t t t cos t t t 0 t 6 0 In graphical terms, this means that the graph of f(t) t cos t lies between the straight lines y t, with the waves growing larger or t and y smaller to fit the space between the lines. The graph touches the lines y t and y 1. This occurs when t k , where k is an integer. t exactly when t cos t t, that is, when cos t p ± ± 0 Therefore, the graph of f(t) t cos t consists of waves that diminish in amplitude as t approaches 0 from both negative and positive values, and the waves are bounded by the lines y ± t. ■ Graphing Exploration Illustrate the analysis of the graph f(t) t, and y f(t) t cos t t on the same screen. t cos t, y by graphing Example 4 Analyzing a Damped Graph Analyze the graph of g(t) 0.5t sin t. Solution No single viewing window gives a completely readable graph of g. To the left of the y-axis, the graph gets quite large; but to the right, it almost coincides with the t-axis. To get a better mental picture, note that multiply each term of the known inequality for every t. To find the bounds of by 0.5t sin t, 0.5t. 0.5t 7 0 1 sin t 1 sin t 1 1 0.5t 0.5t sin t 0.5t for every t Therefore, the graph of g lies between the graphs of the exponential funcThe graph of g will consist of sine waves tions y 0.5t y 0.5t. and 514 Chapter 7 Trigonometric Graphs rising and falling between the graph of the exponential functions 0.5t y as indicated in Figure 7.4.A-6 (which is not to scale). ■ y 0.5t, and Graphing Exploration Find viewing window ranges that clearly show the graph in Example 4 when t is in the following domains. 2p t 0 0 t 2p 2p t 4p Example 5 Oscillating Behavior Analyze the graph of f(t) sin p t b . a Solution Using a wide viewing window, it is clear that the t-axis is an asymptote of the graph of f(t) sin p t b a , as shown in Figure 7.4.A-7a. Near the ori- gin, however, the graph is not readable, even in a very narrow viewing window like Figure 7.4.A-7b. 1.5 1.5 −76 76 −0.5 0.5 −1.5 Figure 7.4.A-7a −1.5 Figure 7.4.A-7b Consider what happens to the graph between t , and t 0. 1 2 As t goes from 1 2 to 1 4 , sin p t b a goes from sin p 1 2 b a to sin a , that is p 1 4 b from sin wave for 2p 1 4 4p. to sin 1 t 2 Therefore, the graph of f makes one complete sine . Similarly, for t 1 6 1 4 , the graph of f makes another complete sine wave. The same pattern continues so that the graph of f 1 6 makes a complete wave for , and so on. A sim- t t , for 1 10 1 8 ilar phenomenon occurs as t takes values between and 0. Conse- 1 8 1 2 Section 7.4.A Excursion: Other Trigonometric Graphs 515 quently, the graph of f near 0 oscillates infinitely often between and 1, with the waves becoming more and more compressed as t gets closer to 0, as indicated in Figure 7.4.A-8. Because the function is not defined at t 0, the left and right halves of the graph are not connected. 1 Graph oscillates infinitely often here y 1 −1 1 −1 Figure 7.4.A-8 f(t) = sin (π/t) t ■ 2 Exercises 7.4.A In Exercises 1 – 6, find a sine function whose graph looks like the graph of the given function f. 1. 2. 3. 4. 5. 6 sin t 2 cos t 3 sin t 2 cos t 2 sin 4t 5 cos 4t 3 sin 1 5 sin 0.3 sin 2t 1 4 cos 2t 3 2 3t 2 1 2t 4 1 2 2 1 2 cos 1 0.4 cos 2 3t 1 2 2t 3 1 2 In Exercises 7–16, find a viewing window that shows a complete graph of the function. 7. g(t) (5 sin 2t)(cos 5t) 8. h(t) esin t 9. f(t) t 2 cos 2t 10. g t 2 1 sin t 3 2 2 cos t 4 2 11. h t 2 1 sin 300t cos 500t ¢ ≤ ¢ ≤ 12. f 13. g t t 2 2 1 1 3 sin 1 5 sin 14. h t 2 1 4 sin 1 200t 1 2 cos 300t 2 250pt 5 1 600pt 3 2 1 2 cos 2 400pt 7 2 6 cos 1 500pt 3 1 2 2 15. f 16. g t t 2 2 1 1 4 sin 0.2pt 5 cos 0.4pt 6 sin 0.05pt 2 cos 0.04pt In Exercises 1724, describe the graph of the function verbally, including such features as asymptotes, undefined points, amplitude and number of waves between 0 and 2 . Find viewing windows that illustrate the main features of the graph. P 17. g(t) sin et 19. f(t) 21. h(t) 2 0 t 0 cos t 1 t sin t 18. h(t) 20. g(t) cos 2t 1 t2 t2 8 e sin 2pt 22. f(t) t sin 1 t 23. h(t) ln 0 cos t 0 24. h(t) ln 0 sin Important Concepts Section 7.1 Section 7.2 Section 7.3 Section 7.4 Graph of the sine function . . . . . . . . . . . . . . . . . 475 Graph of the cosine function. . . . . . . . . . . . . . . . 477 Domain and range of the sine and cosine functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 Graph of the tangent function . . . . . . . . . . . . . . 480 Domain and range of the tangent function. . . . . 480 Even function . . . . . . . . . . . . . . . . . . . . . . . . . . . 482 Odd function. . . . . . . . . . . . . . . . . . . . . . . . . . . . 483 Graph of the cosecant function . . . . . . . . . . . . . . 486 Domain and range of the cosecant function . . . . 486 Graph of the secant function. . . . . . . . . . . . . . . . 488 Domain and range of the secant function. . . . . . 487 Domain and range of the cotangent function . . . 488 Graph of the cotangent function . . . . . . . . . . . . . 489 Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494 Amplitude. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497 Phase shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 Combined transformations of sine and cosine graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 503 Graphs and identities . . . . . . . . . . . . . . . . . . . . . 506 Section 7.4.A Sinusoidal graphs . . . . . . . . . . . . . . . . . . . . . . . . 510 Other trigonometric graphs . . . . . . . . . . . . . . . . 512 Important Facts and Formulas If and , and a cos then each of the functions has: d bt c 1 2 f t 2 1 a sin bt c d 2 1 , 0 a amplitude 0 phase shift c b period 2p b , vertical shift d If b 7 0, then the function h tan bt has period t 2 1 p b . 516 Chapter Review 517 Review Exercises Section 7.1 1. Which of the following is not true about the graph of f It has no sharp corners. a. b. It crosses the horizontal axis more than once. c. It rises higher and higher as t gets larger. d. It is periodic. e. It has no vertical asymptotes. sin t? t 2 1 In Exercises 2–4, graph each function on the given interval. 2. f t 2 1 sin t 4. h t 2 1 tan t 7p 2 , 7p T 2p, 3p 4 S 3 3. g t 2 1 cos t 5p, 7p 2 T S In Exercises 5–7, find all the exact t-values for which the given statement is true. 5. cos t 1 7. tan t 23 6. sin t 1 2 In Exercises 8–10, list the transformations that change the graph of f into the graph of g. State the domain and range of g. 8. f t 2 1 sin t g 1 2 t 2 1 sin t 9. f t 2 1 tan t g t 2
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1 tan 2t 10. f t 2 1 cos t g cos t 2 1 1 2 a t b 1 In Exercises 11–13, sketch the graph of each function. 11. 13 cos t 2 sin t 3 12. h t 2 1 tan t 4 14. Which of the following functions has the graph shown below between p and p? a. b. c. d. e sin x, cos x, cos x 1 sin x, sin x e if x 0 if x 6 0 if x 0 if x 6 0 , −π 2 1 cos x 0 21 sin2 x 0 y 1 −1 − π 2 t π π 2 15. Between (and including) 0 and 2p, the function h a. 3 zeros and is undefined at 2 places b. 2 zeros and is undefined at 3 places c. 2 zeros and is undefined at 2 places d. 3 zeros and is defined everywhere e. no zeros and is undefined at 3 places tan t has ? . t 2 1 518 Chapter Review 16. Which of the statements i–iii are true? y 2 π 2 −2 −π − t π π 2 Section 7.2 The sine function is an odd function. The cosine function is an odd function. i. ii. iii. The tangent function is an odd function. i and ii only a. b. ii only c. d. all of them e. none of them i and iii only 17. Which of the following functions has the graph shown at left? a. b. c. d. e tan t t p tan a 2 b 1 tan t 3 tan t tan t 18. Which of the following is true about sec t ? a. b. 0 0 sec 2 1 sec t 1 sin t Its graph has no asymptotes. c. d. It is a periodic function. e. It is never negative. In Exercises 19–21, sketch the graph of each function. 19 g t 2 1 20. f t 2 1 cot t 2 3 sec t 2 21. h t 2 1 csc 1 2 a t b In Exercises 22–23, complete the statement with “odd” or “even.” 22. The cosecant function is an function. 23. The secant function is an function. Section 7.3 24. Let f 3 2 t 2 1 sin 5t. a. What is the largest possible value of b. Find the smallest positive number t such that ? t f 1 2 0. f t 2 1 25. Sketch the graph of g 26. Sketch the graph of f 27. Sketch the graph of cost. 1 2 sin 2t on the interval 2p t 2p. sin 4t on the interval 0 t 2p. 28. What is the period of the function g sin 4pt? t 2 1 29. If t g 2 1 that g 20 sin 1? t 1 2 , for how many values of t with 0 t 2p is it true 200t 2 1 Chapter Review 519 30. What is the period of f tan t 2 1 t 2 b ? a 31. Which of the following statements is true? 3 sin 2t 1 is 2. a. The amplitude of b. The period of g t 2 1 c. The period of 2 d. The amplitude of tan 2t cos 2t is 4p. p is 2 3 tan t . is 3 13 cos 14t 15 1 Section 7.4 32. What are the amplitude, period, and phase shift of the function 33. State the rule of a sine function with amplitude 8, period 5, and phase shift 14. 34. State the rule of a sine function with amplitude 3, period p, and phase shift p 3 . t 2π 5 π 4π 5 6π 5 8π 5 2π 35. State the rule of a periodic function whose graph from t 0 to t 2p closely resembles the graph at left. y 2 1 −1 −2 In Exercises 36–38, sketch the graph of at least one cycle of each function. 36. f t 2 1 1 2 cos 1 2t p 3 2 37. g t 2 1 sin 1 3 a t p b 38. g t 2 1 4 cos 2t 3 b a 5 In Exercises 39–42, determine graphically whether the given equation could possibly be an identity. 39. cos t sin t p 2 b a 41. sin t sin 3t cos t cos 3t tan t 40. tan t 2 sin t 1 cos t 42. cos 2t 1 1 2 sin2 t Section 7.4.A In Exercises 43 and 44, find a sine function whose graph looks like the graph of the given function. 43. 44 sin 1 5 sin 4t 7 5 cos 4t 8 2 5t 3 1 2 cos 2 5t 2 1 2 1 2 2 In Exercises 45 and 46, find a viewing window that shows a complete graph of the function. 45. f 46. g t t 2 2 1 1 3 sin 1 5 sin 300t 5 2 cos 500t 8 400pt 1 2 1 1 2 cos 2 150pt Approximations with Infinite Series In the Chapter 1 Can Do Calculus, it was shown that the infinite geo- metric series a ar ar 2 ar3 . . . a 1 r when r 0 0 6 1. This section will investigate certain functions that can be represented by an infinite series, a topic considered in depth in calculus. Example 1 Representing a Function as a Series Write f 1 1 x x 1 2 as an infinite series. Solution The expression 1 1 x a ar ar is in the form a 1 r 2 ar3 . . . when r 1 x x2 x3 . . . , when x 0 0 a 1 r 1 1 x where a 1 and r x. , Because 0 6 1, 0 6 1. ■ 5 To confirm that 1 1 x 1 x x2 x3 . . . when 6 1, graph x 0 0 and 1 x y 1 x x2 x3 x4 y 1 Figure 7.C-1 where is drawn with a heavy line. but The graphs of the function and the series are very close when they diverge when When more terms are used in graphing the series, the series approximates the function more closely The set of all values of x for which the series converges to when x 6 1 and when x 7 1. y 1 x x2 x3 x4 on the same screen, as shown in 6 1, 6 1. x x 0 0 the function is called the interval of convergence. The function 1 6 x 6 1. 1 1 x x 1, the infinite series does not converge to a single value. interval of convergence when It is not defined when x 1 has and 0 0 Other Types of Series Many interesting functions that can be represented by a series include the product of all the integers from 1 to n. Such a product is written as n!, which is read “n factorial.” n 1 n! 1 2 3 4 n 2 0! is defined to be the number 1. n 2 21 1 p 3 3 5 Figure 7.C-1 Technology Tip The factorial feature is found in the PROB (or PRB) submenu of the MATH or OPTN menu on most calculators. 520 10 Example 2 A Series that Approximates a Function 10 10 10 Figure 7.C-2a 10 10 10 10 Figure 7.C-2b 10 Find a function that is approximated by the following series in the interval p x p. x x3 3! x5 5! x7 7! x9 9! p n1 1 1 2 x2n1 2n 1 1 ! 2 Solution Begin by graphing the function formed by the first five terms of the series, as shown in Figure 7.C-2a. Next graph the functions formed by first six terms of the series and then the first seven terms, as shown in Figures 7.C-2b and 7.C-2c. The graph of the series is beginning to resemble the graph of the sine function. To test the hypothesis that the series converges to the sine function, graph both the sine function and the function formed by series on the same screen using several terms of the series. In Figure 7.C-2d, the is shown as the series is displayed with the heavier line, and lighter line. In calculus it will be shown that the infinite series converges to the sine function, and the interval of convergence is the entire set of real numbers. y sin x 1 10 10 15 15 10 Figure 7.C-2c Exercises 1 Figure 7.C-2d ■ Find an infinite geometric series that represents the given function, and state the interval of convergence. Find a function that is approximated by the following series. State the interval of convergence. 1. y 2 1 3x 3. y 2 1 x 2. y 3 1 2x 4. y 3 1 2x 5. 1 x2 2! x4 4! x6 6! p x 1 2 6. 1 x 1 2 1 7. 1 x x2 2! 1 x3 3! 2 x4 4 521 C H A P T E R 8 Solving Trigonometric Equations Round and round we go! Trigonometric functions are used to analyze periodic phenomena, because simple harmonic motion models circular motion or any phenomenon that is “back and forth.’’ Some examples of simple harmonic motion include a vibrating prong of a tuning fork, a buoy bobbing up and down in water, seismic and ocean waves, spring-mass systems, a piston in a running engine, a particle of air during the passage of a simple sound wave, or a turning Ferris wheel. See Exercise 1 of Section 8.4. 522 Chapter Outline 8.1 Graphical Solutions to Trigonometric Equations 8.2 Inverse Trigonometric Functions 8.3 Algebraic Solutions of Trigonometric Equations 8.4 Simple Harmonic Motion and Modeling 8.4.A Excursion: Sound Waves Chapter Review can do calculus Limits of Trigonometric Functions Interdependence of Sections 8.1 8.2 > 8.3 > 8.4 There are two kinds of trigonometric equations. Identities, which will be studied more in Chapter 9, are equations that are valid for all val- ues of the variable for which the equation is defined, such as sin2 x cos2 x 1 and cot x 1 tan x . In this chapter, conditional equations will be studied. Conditional equations are valid only for certain values of the variable, such as sin x 0, cos x 1 2 , and 3 sin2 x sin x 2. If a trigonometric equation is conditional, solutions are found by using techniques similar to those used to solve algebraic equations. Graphs were used to solve some simple trigonometric equations in Chapter 7. This chapter will extend graphical solution techniques and introduce analytic solution methods. Graphical solution methods are presented in 8.1. Inverse trigonometric functions are discussed in Section 8.2. Methods that use inverse functions, basic identities, and algebra to solve trigonometric equations are considered in Section 8.3. Skills from the Sections 8.1 through 8.3 are applied to problem-solving and real-world applications in Section 8.4. NOTE In Chapter 7, the variable t was used for trigonometric functions to avoid confusion with the x’s and y’s that appear in their definitions. Now that you are comfortable with these functions, the letter x, or occasionally y, will be used for the variable. Unless otherwise stated, all trigonometric functions in this chapter are considered as functions of real numbers, rather than functions of angles in degree measure. 523 524 Chapter 8 Solving Trigonometric Equations 8.1 Graphical Solutions to Trigonometric Equations Objectives • Solve trigonometric equations graphically • State the complete solution of a trigonometric equation Any equation involving trigonometric functions can be solved graphically. To solve trigonometric equations graphically, the same methods of graphical solutions are used here as have been used previously to solve polynomial equations, except that trigonometric equations typically have an infinite number of solutions. These solutions are systematically determined by using the periodicity of the function. Basic Trigonometric Equations An equation that involves a single trigonometric function set equal to a number is called a basic equation. Some examples include the following: sin x 0.39, cos x 0.5, and tan x 3 Examples 1 and 2 show how they can be solved graphically. Example 1 The Intersection Method Solve tan x 2. Solution The equation can be solved by graphing on the same screen and finding intersection points. The x-coordinate of every such point is a number whose tangent is 2; or a solution of Figure 8.1-1 indicates that there are infinitely many intersectio
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n points, so the equation has an infinite number of solutions. tan x 2. and Y1 tan x Y2 2 y 4 2 −2π −3π 2 −π − π 2 −2 π 2 π 2π 3π 2 5π 2 3π 7π 2 4π x One period Figure 8.1-1 5 π 2 π 2 5 Figure 8.1-2 NOTE Solutions in this chapter are often rounded, but the full decimal expansion given by the calculator is used in all computations. The symbol even though these calculator solutions are approximations of the actual solutions. is used rather than Section 8.1 Graphical Solutions to Trigonometric Equations 525 1 f x tan x The function p 2 b in this interval. Using the intersecand there is one solution of tion finder on a graphing calculator gives the approximate solution in this interval. completes one cycle on the interval tan .1071 tan x f x repeats its pattern to the left and to the Because the graph of right, the other solutions will differ from this first solution by multiples of the period of the tangent function. The other solutions are 1.1071 ± p, 1.1071 ± 2p, and 1.1071 ± 3p, p, 1 2 and so on. All solutions can be expressed as 1.1071 kp, where k is any integer. Example 2 The x-Intercept Method Solve sin x 0.75. Solution Rewrite the equation sinx 0.75 as sin x 0.75 0. ■ 3 Recall from Section 2.1 that the solutions of this equation are the x-intercepts of the graph of sin x 0.75. f x 1 2 2π 2π 1 Figure 8.1-3 2p The graph of f is shown in Figure 8.1--3. The function has a period of and the viewing window can be modified to show one period of sin x 0.75. f Figures 8.1-4a and 8.1-4b show that there are two zeros of f on the interval so the equation has two solutions on that interval. 0, 2p x , 1 2 4 3 The calculator’s zero finder calculates the zeros: x 3.9897 x 5.4351 3 0 1 3 2π 0 2π Figure 8.1-4a Figure 8.1-4b 1 526 Chapter 8 Solving Trigonometric Equations Because the graph repeats its pattern every differ from these two by multiples of Therefore, all solutions of 2p, sin x 0.75 2p, the period of are 1 2 the other solutions will sin x 0.75. x f x 3.9897 2kp and x 5.4351 2kp, where k is any integer. ■ Other Trigonometric Equations The procedures in Examples 1 and 2 can be used to solve any trigonometric equation graphically. Example 3 The x-intercept Method Solve 3 sin2 x cos x 2 0. Solution Technology Tip 3 sin2 x Enter culator as on a cal- 2 1 sine and 3 sin2 x cos x 2 cosine have period is at most the period of Both The graph of f, which is shown x f in two viewing windows in Figure 8.1-5, does not repeat its pattern over so you can conclude that f has a period any interval of less than of 2p. 2p, 2p. so 2p, sin x 3 3 1 or 1 sin 2 2 2 2. x 1 22 2π 4π 0 4 Figure 8.1-5a 2 4 2π Figure 8.1-5b The function f makes one complete period on the interval as shown 2p, in Figure 8.1-5b. The equation has four solutions between 0 and namely, the four x-intercepts of the graph in that interval. A graphical zero finder shows these four solutions. , 2 3 0, 2p x 1.1216 x 2.4459 x 3.8373 x 5.1616 Because the graph repeats its pattern every tion are given by 2p, all solutions of the equa- x 1.1216 2kp, x 3.8373 2kp, x 2.4459 2kp, x 5.1616 2kp, where k is any integer. ■ Section 8.1 Graphical Solutions to Trigonometric Equations 527 The solution methods in Examples 1 through 3 depend only on knowing the period of a function and all the solutions of the equation in one period. A similar procedure can be used to solve any trigonometric equation graphically. Solving Trigonometric Equations Graphically 1. Write the equation in the form f(x) 0. 2. Determine the period p of f. 3. Graph f over an interval of length p. 4. Use a calculator’s zero finder to determine the x-intercepts of the graph in this interval. 5. For each x-intercept u, all of the numbers u kp where k is any integer are solutions of the equation. In Example 1, for example, p was p. In Examples 2 and 3, p was 2p. Example 4 Solving Any Trigonometric Equation 3 π 2 π 2 Solve tan x 3 sin 2x. Solution First rewrite the equation tan x 3 sin 2x 0 3 Figure 8.1-6 Next, determine the period of y 3 sin 2x f x 1 2 tan x 3 sin 2x. p, Recall from Section which is also the period of 7.3 that y tan x. has a period of 2p 2 p. Therefore, the period of f is p 2 b p 2 , a , f on the interval an interval of length p. Figure 8.1-6 shows the graph of Even without the graph, it can be easily verified that there is an x-intercept at 0. f 0 1 2 tan 0 3 sin 2 0 2 1 0 Using the calculator’s zero finder gives the other x-intercepts of the graph of f on this interval. x 1.1503 and x 1.1503 Because f has a period of are p, all solutions of the equation tan x 3 sin 2x x 1.1503 kp, x 0 kp, and x 1.1503 kp, where k is any integer. ■ 528 Chapter 8 Solving Trigonometric Equations Trigonometric Equations in Degree Measure Some real-world applications of trigonometric equations require solutions to be expressed as angles in degree measure. The graphical solution procedure is the same, except that you must set the mode of your calculator to “degree.” Example 5 Trigonometric Equations in Degree Measure Solve 2 sin2u 3 sin u 3 0. Solution 360 The period of the function ure 8.1-7 shows the graph of f on the interval 2 sin2u 3 sin u 3 . 0°, 360° u f 2 1 3 2 is 360°, and Fig- A graphical zero finder determines the approximate x-intercepts. u 223.33° and u 316.67° Figure 8.1-7 Using the fact that the period of f is 360°, all solutions of the equation are u 223.33° 360°k and u 316.67° 360°k, where k is any integer. ■ 3 0 5 Exercises 8.1 In Exercises 1–12, solve the equation graphically. 13. Use the graph of the sine function to show the 1. 4 sin 2x 3 cos 2x 2 2. 5 sin 3x 6 cos 3x 1 3. 3 sin32x 2 cos x 4. sin2 2x 3 cos 2x 2 0 5. tan x 5 sin x 1 6. 2 cos2x sin x 1 0 7. cos3x 3 cos x 1 0 8. tan x 3 cos x 9. cos4x 3 cos3x cos x 1 10. sec x tan x 3 11. sin3x 2 sin2 x 3 cos x 2 0 12. csc2x sec x 1 following. a. The solutions of sin x 1 are 5p x p , 2 2 3p 2 x , 9p , 2 7p 2 , p and 11p 2 , , . . . . b. The solutions of sin x 1 are 11p 2 x 3p , 2 p 2 x 7p , 2 5p 2 , , p and 9p 2 , . . . . , 14. Use the graph of the cosine function to show the following. a. The solutions of ± 4p, x 0, ± 2p, b. The solutions of x ± p, ± 3p, cos x 1 are ± 6p, . . . . cos x 1 ± 5p, . . . . are Section 8.2 Inverse Trigonometric Functions 529 In Exercises 15–18, approximate all solutions of the given equation in (0, 2p) . 15. sin x 0.119 16. cos x 0.958 17. tan x 5 18. tan x 17.65 At the instant you hear a sonic boom from an airplane overhead, your angle of elevation to the plane is given by the equation A sin A 1 m In Exercises 19–28, find all angles with that are solutions of the given equation. U 0 U 66 360 19. tan u 7.95 20. tan u 69.4 21. cos u 0.42 22. cot u 2.4 23. 2 sin2 u 3 sin u 1 0 24. 4 cos2u 4 cos u 3 0 25. tan2u 3 0 26. 2 sin2u 1 27. 4 cos2u 4 cos u 1 0 28. sin2u 3 sin u 10 where m is the Mach number for the speed of the plane (Mach 1 is the speed of sound, Mach 2.5 is 2.5 times the speed of sound, etc.). In Exercises 29–32, find the angle of elevation (in degrees) for the given Mach number. Remember that an angle of elevation must be between 90. and 0 29. m 1.1 31. m 2 30. m 1.6 32. m 2.4 33. Critical Thinking Under what conditions (on the constant) does a basic equation involving the sine and cosine function have no solutions? 34. Critical Thinking Under what conditions (on the constant) does a basic equation involving the secant and cosecant function have no solutions? 8.2 Inverse Trigonometric Functions Objectives • Define the domain and range of the inverse trigonometric functions • Use inverse trigonometric function notation NOTE Other ways of restricting the domains of trigonometric functions are possible. Those presented here for sine, cosine, and tangent are the ones universally agreed upon by mathematicians. Many trigonometric equations can be solved without graphing. Nongraphical solution methods make use of the inverse trigonometric functions that are introduced in this section. Recall from Section 3.6 that a function cannot have an inverse function unless its graph has the following property. No horizontal line intersects the graph more than once. You have seen that the graphs of trigonometric functions do not have this property. However, restricting their domains can modify the trigonometric functions so that they do have inverse functions. Inverse Sine Function , the interval The restricted sine function is p 2 p 2 T ber v in the interval 3 p 2 S , such that sin u v. p 2 T , S 4 sin x, f x 1 2 when its domain is restricted to Its graph in Figure 8.2-1 shows that for each num. 1, 1 there is exactly one number u in the interval 530 Chapter 8 Solving Trigonometric Equations y 1 v (u, v) = (u, sin u) x u π 2 − π 2 −1 Figure 8.2-1 Because the graph of the restricted sine function passes the Horizontal Line Test, it has an inverse function. This inverse function is called the inverse sine (or arcsine) function and is denoted by g(x) sin1 x g(x) arcsin x. or It is convenient to think of a value of an inverse trigonometric function as an angle; sin 1 23 2 sine is 23 2 . Since sin represents an angle in the interval 23 2 p 3 23 2 then 1 sin whose The graph of the inverse sine function, shown in Figure 8.2-2, is readily obtained from a calculator. Because is the inverse of the restricted sine function, its graph is the reflection of the graph of the restricted sine function across the line 2 1 y x . sin Figure 8.2-2 x 2 sin 1 x is the interval 1, 1 3 4 , and its range is the g The domain of p 2 T interval p 2 , S 1 . Inverse Sine Function For each v with 1 v 1, sin1 v is the unique number u in the interval sine is v; that is, P 2 , S P 2 T whose sin1 v u exactly when sin u v. Section 8.2 Inverse Trigonometric Functions 531 Technology Tip The inverse sine function can be evaluated by using the times labeled ASIN) on a calculator. For example, SIN1 key (some- Unless otherwise noted, make sure your calculator is in radian mode. 1 sin 1 0.67 2 0.7342 and sin 1 0.42 0.4334. For many special values, however, you can evaluate the
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inverse sine function without using a calculator. Technology Tip If you attempt to use a calculator to evaluate the inverse sine function at a number not in its domain, such as sin you will get an error message. 1 2 , 2 1 Example 1 Special Values Evaluate: a. sin 1 1 2 Solution b. 1 sin 22 2 b a a. 1 2 sin 1 is the number in the interval p 2 T your study of special values, you know that p 2 , S whose sine is 1 2 . From sin p 6 1 2 . Because p 6 is in the interval b. 1 sin 22 a 2 b p 2 S interval , p 2 S p 4 sin because sin p 4 b a 22 2 and p 4 is in the , p 2 T . ■ CAUTION 1 x sin The notation 1 1 sinx sin x or 1 2 . is not exponential notation. It does not mean For instance, Example 1 shows that sin 1 1 2 p 6 0.5236, but this is not equivalent to sin a 1 2 b 1 1 sin 1 2 1 0.4794 2.0858. Suppose sine function, 1 v u. 1 v 1 p 2 sin and sin u p 2 u and sin sin u v. sin u and 1 v 2 1 2 1 sin 1 Therefore, sin 1 v 1 sin u 2 1 2 v. Then by definition of the inverse This shows that the restricted sine function and the inverse sine function have the usual composition properties of other inverse functions. 532 Chapter 8 Solving Trigonometric Equations Properties of Inverse Sine sin1(sin u) u if p 2 u p 2 sin(sin1 v) v if 1 v 1 Example 2 Composition of Inverse Functions Explain why sin 1 sin a p 6 b p 6 is true but sin 1 sin a 5p 6 b 5p 6 is not true. Solution You know that sin p 6 1 2 , so by substitution 1 sin sin a p 6 b sin 1 1 2 b a p 6 because p 6 is in the interval p 2 S , p 2 T . Although sin 5p 6 is also 1 2 , by substitution 1 sin sin a 5p 6 b sin 1 1 2 b a p 6 , not 5p 6 , because 5p 6 is not in the interval p 2 , S p 2 T . ■ Inverse Cosine Function The restricted cosine function is 0, p . to the interval 4 3 1, 1 v in the interval such that cos u v. 3 4 cos x, f x when its domain is restricted Its graph in Figure 8.2-3 shows that for each number 0, p , there is exactly one number u in the interval 1 2 3 4 y 1 v −1 u x π (u, v) = (u, cos u) Figure 8.2-3 Section 8.2 Inverse Trigonometric Functions 533 Because the graph of the restricted cosine function passes the horizontal line test, it has an inverse function. This inverse function is called the inverse cosine (or arccosine) function and is denoted by g(x) cos1 x or g(x) arccos x. The graph of the inverse cosine function, which is the reflection of the y x, graph of the restricted cosine function (Figure 8.2-3) across the line is shown in Figure 8.2-4. 2 π π 2 Figure 8.2-4 2 The domain of g x 1 2 cos 1 x is the interval 1, 1 3 4 and its range is 0, p . 4 3 Inverse Cosine Function For each v with 1 v 1, cos1 v cosine is v; that is, is the unique number u in the interval [0, P] whose cos1v u exactly when cos u v. The properties of the inverse cosine function are similar to the properties of the inverse sine function. Properties of Inverse Cosine cos1(cos u) u if 0 u p cos(cos1v) v if 1 v 1 Example 3 Evaluating Inverse Cosine Expressions Evaluate the following. a. cos 1 1 2 b. cos 1 0 c. cos 1 0.63 2 1 534 Chapter 8 Solving Trigonometric Equations CAUTION 1 cos cos x 1 does not mean 1 1 or cosx . 2 Solution a. cos 1 1 2 p 3 because p 3 whose cosine is 1 2 . is the unique number in the interval 0, p 4 3 b. cos 1 0 p 2 COS1 c. The 1 0.63 cos 1 2 because cos p 2 0 and 0 p 2 p. command on a calculator shows that 2.2523. ■ 1−v2 1 u v Figure 8.2-5 Example 4 Equivalent Algebraic Expressions Write sin cos 1 v 2 1 Solution as an algebraic expression in v. Let cos 1 v u , where 0 u p . Construct a right triangle containing an angle of u radians where cos u adjacent hypotenuse 8.2-5. By the Pythagorean Theorem, the length of the side opposite u is 21 v2 , as shown in Figure v . By the definition of sine, sin u opposite hypotenuse 21 v2. 1 v cos sin 1 2 21 v2 1 Therefore, 21 v2 ■ Inverse Tangent Function The restricted tangent function is p to the interval 2 Q p 2 R , . tan x, f x 1 2 when its domain is restricted Its graph in Figure 8.2-6 shows that for every real number v, there is exactly one number u between tan u v. p 2 and p 2 such that y v (u, v) = (u, tan u) x − π 2 u π 2 Figure 8.2-6 Section 8.2 Inverse Trigonometric Functions 535 Because the graph of the restricted tangent function passes the horizontal line test, it has an inverse function. This inverse function is called the inverse tangent (or arctangent) function and is denoted g(x) tan1 x g(x) arctan x. or The graph of the inverse tangent function, which is the reflection of the y x, graph of the restricted tangent function (Figure 8.2-6) across the line is shown in Figure 8.2-7. π 2 10 10 π 2 Figure 8.2-7 The domain of is the interval x g 2 1 p 2 is the set of all real numbers and its range 1 x tan p 2 b . , a Inverse Tangent Function For each real number v, tan1 v is the unique number u in the interval P 2 a , P 2 b whose tangent is v; that is, tan1v u exactly when tan u v. The properties of the inverse tangent function are similar to the properties of the inverse sine and inverse cosine functions. Properties of Inverse Tangent tan1(tan u) u if P 2 6 u 6 P 2 tan(tan1v) v for every real number v. CAUTION Example 5 Evaluating Inverse Tangent Expressions 1 tan tan x 1 does not mean 1 1 or tan x . 2 Evaluate: 1 1 tan a. b. tan 1 136 536 Chapter 8 Solving Trigonometric Equations Solution a. tan 1 1 p 4 because p 4 such that tan p 4 1. is the unique number in the interval p 2 Q , p 2 R b. The TAN1 key on a calculator shows that tan 1 136 1 2 1.5634. Example 6 Exact Values Find the exact value of cos 1 tan a 25 2 b . Solution Let tan 1 25 2 u. Then tan u 25 2 and p 2 6 u 6 p 2 . Because tan u 25 2 is positive, u must be between 0 and p 2 . Draw a right triangle containing an angle of u radians whose tangent is 25 2 . tan u opposite adjacent 25 2 5 3 u 2 Figure 8.2-8 The hypotenuse has length 1 25 cos tan a 2 b 1 2 cos u adjacent hypotenuse 222 15 2 24 5 3. Therefore, 2 3 . ■ Exercises 8.2 In Exercises 1–14, find the exact value without using a calculator. 11. 1 tan 23 A B 2. cos 10 5. cos 11 3. 6. 1 tan tan 1 1 11 2 13. 1 cos 1 a 2 b 12. 1 cos a 22 2 b 14. sin 1 1 a 2 b 8. cos 1 23 2 10. sin 1 23 2 In Exercises 15–24, use a calculator in radian mode to approximate the functional value. 15. sin 10.35 16. cos 10.76 17. 1 tan 3.256 1 2 18. sin 1 0.795 1 2 1. sin 11 4. sin 7. tan 1 1 2 1 1 23 3 9. sin 1 a 22 2 b 19. 20. 22. 24. 1 1 1 sin cos sin 7 2 Hint: the answer is not 7. cos 3.5 2 1 sin tan 1 1 2 sin C 1 2D tan 12.4 1 tan cos 1 tan C 1 1 cos C 1 4 2D 8.5 2D 21. 23. 2 25. Given that u sin 1 of cos u and tan u. a 23 2 b , find the exact value Section 8.2 Inverse Trigonometric Functions 537 45. tan sin 1v A B 46. sin A 2 sin 1v B In Exercises 47–50, graph the function. 47. f 48. g 49 cos 1 x 1 A B tan 1 x p sin 1 sin x A B 50. k x 1 2 sin A sin 1 x B 26. Given that u tan 1 4 3 b a sin u and sec u. , find the exact value of 51. A model plane 40 feet above the ground is flying away from an observer. In Exercises 27–42, find the exact functional value without using a calculator. 28. 1 cos sin a p 6 b 30. 1 tan cos p 2 1 32. 1 cos tan a 7p 4 b (See Exercise 19.) 27. 1 sin cos 0 1 2 29. 1 cos sin a 4p 3 b 31. 1 sin cos a 7p 6 b 33. 1 sin sin a 2p 3 b 34. 1 cos cos a 5p 4 b x 40 θ Observer a. Express the angle of elevation of the plane as a function of the distance x from the observer to the plane. u b. What is when the plane is 250 feet from the u observer? 35. 1 cos 36. 1 tan cos tan S S p a 6 b T 4p a 3 b T 37. sin 1 cos S 3 5 b T a (See Example 6.) 38. tan 40. cos 1 sin 1 sin S S 42. sin 1 cos S 3 5 b T 23 5 b T 23 13 b T a a a 39. cos 41. tan 1 tan 3 a 4 b T 1 sin 5 13 b T a S S In Exercises 43–46, write the expression as an algebraic expression in v, as in Example 4. 43. cos sin 1 v A B 44. cot cos 1 v A B 52. Show that the restricted secant function, whose domain consists of all numbers x such that 0 x p , has an inverse function. and x p 2 Sketch its graph. 53. Show that the restricted cosecant function, whose domain consists of all numbers x such that p x 0, 2 has an inverse function. x p 2 and Sketch its graph. 54. Show that the restricted cotangent function, whose domain is the interval function. Sketch its graph. 1 0, p 2 , has an inverse 55. a. Show that the inverse cosine function actually has the two properties listed in the box on page 533. b. Show that the inverse tangent function actually has the two properties listed in the box on page 535. 538 Chapter 8 Solving Trigonometric Equations 56. Critical Thinking A 15-foot-wide highway sign is placed 10 feet from a road, perpendicular to the road. A spotlight at the edge of the road is aimed at the sign, as shown in the figure below. 57. Critical Thinking A camera on a 5-foot-high tripod is placed in front of a 6-foot-high picture that is mounted 3 feet above the floor, as shown in figure below. 10 A Sign θ Spotlight θ 5 ft x 6 ft 3 ft a. Express angle as a function of the distance x u from the camera to the wall. b. The photographer wants to use a particular u 36° lens, for which How far . p 5 a radians b a. Express as a function of the distance x from u point A to the spotlight. b. How far from point A should the spotlight be placed so that the angle possible? u is as large as should she place the camera from the wall to be sure the entire picture will show in the photograph? 8.3 Algebraic Solutions of Trigonometric Equations Objective • Solve trigonometric equations algebraically Trigonometric equations were solved graphically in Section 8.1. In this section you will learn how to use algebra with inverse trigonometric functions and identities to solve trigonometric equations. Recall from Section 8.1 that equations such as sin x 0.75, cos x 0.6, and tan x 3 are called basic equations. Algebraic solution methods for basic equations are illustrated in Examples 1 through 3. Example 1 Solving Basic Cosine Equations Solve cos x 0.6. Section 8.3 Algebraic Solutions of Trigonometric Equations 539 2 Solution π π 2 Figure 8.3-1 1.5 Y1 cos x The graphs of and just
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two solutions (intersection points) on the interval one full period of the cosine function. in Figure 8.3-1 show that there are which is p, p , 3 4 Y2 0.6 The definition of the inverse cosine function states that cos 10.6 Using the inverse cosine function, of by using the identity 0, p is the number in the interval 3 10.6 0.9273 x cos p, p . 4 3 cos x, is one solution The second solution can be found with whose cosine is 0.6. x 0.9273. cos x 0.6 cos 4 on the interval x 2 1 0.9273 cos Therefore, the solutions of cos 0.9273 0.6 1 2 cos x 0.6 10.6 0.9273 and x cos on the interval p, p are 10.6 0.9273 3 4 x cos Because the interval tion, all solutions of p, p 3 cos x 0.6 4 are given by is one complete period of the cosine func- x 0.9273 2kp and x 0.9273 2kp, where k is any integer. ■ Example 2 Solving Basic Sine Equations Solve sin x 0.75. Solution The definition of the inverse sine function states that 1 sin 1 0.75 2 is the number in the interval S , p 2 1 0.75 1 p p on the interval , . S 2 T 2 sin x, p x x sin 2 sin 1 0.8481 2 sin 3.9897 whose sine is 0.75. p 2 T 0.8481 is the solu- A second solution can be x 0.8481. with 0.75 2 4 3.9897 1 2 is also a solution of sin x 0.75, Using the inverse sine function, tion of sinx 0.75 found by using the identity p sin 1 3 0.8481 x p Therefore, 2 and all solutions are given by 1 π π where k is any integer. x 0.8481 2kp and x 3.9897 2kp, 1.5 Figure 8.3-2 Recall that there are an infinite number of solutions to many trigonometric equations. Figure 8.3-2 indicates that there are two solutions in the interx 0.8481 val can in the solution be found by letting . The solution x 3.9897 2kp x 2.2935 : x 2.2935 k 1 p, p and . 4 3 ■ 540 Chapter 8 Solving Trigonometric Equations Example 3 Solving Basic Tangent Equations 5 Solve tan x 3. Solution The definition of the inverse tangent function states that π π 5 Figure 8.3-3 tan 1 3 is the number in the interval p p a 2 b 2 1 3 1.2490 x tan p a 2 of the tangent function, all solutions are given by Using the inverse tangent function, tan x 3 on the interval p a 2 Because p 2 b p 2 b , , , . whose tangent is 3. is the solution of is one full period where k is any integer. x 1.2490 kp, ■ The solution method used in Examples 1–3 is summarized in the following table, where k is any integer. Solutions of Basic Trigonometric Equations Equation Possible values of c Solutions sin x c 1 6 c 6 1 x sin 1 c 2kp c 1 c 1 and p sin 1 c 2kp 2 x 1 x p 2 x p 2 2kp 2kp c 7 1 or c 6 1 no solution cos cos 1 c 2kp and 1 c 2kp x cos x 0 2kp 2kp x p 2kp c 7 1 or c 6 1 no solution tan x c all real numbers x tan 1 c kp Section 8.3 Algebraic Solutions of Trigonometric Equations 541 Example 4 Using the Solution Algorithm Solve 8 cos x 1 0. Solution First rewrite the equation as an equivalent basic equation. 8 cos x 1 0 cos x 1 8 Then solve the basic equation using the inverse cosine function. One solution is in Quadrant I. The other solution on the interval is in Quadrant IV. x cos 1.4455 1 1 8 p, p x cos 3 1 1 8 4 1.4455 All solutions are given by x 1.4455 2kp and x 1.4455 2kp, where k is any integer. ■ Example 5 Solving Basic Equations with Special Values Solve sin u 22 2 Solution exactly, without using a calculator. Because terval p sin 4 p, p 3 4 22 2 , u p 4 is one solution of sin u 22 2 on the in- . Another solution is in Quadrant II. u p p 4 3p 4 Therefore, the exact solution is given by u p 4 2kp and u 3p 4 2kp, where k is any integer. ■ Sometimes trigonometric equations can be solved by using substitution to make them into basic equations. 542 Chapter 8 Solving Trigonometric Equations Example 6 Using Substitution and Basic Equations Solve sin 2x 22 2 Solution exactly, without using a calculator. First, let u 2x, and solve the basic equation From Exam- ple 5, you know the complete exact solution of is given by sin u 22 2 . sin u 22 2 u p 4 2kp and u 3p 4 2kp, each of these solutions leads to a solution of the original for u, and solve for x. where k is any integer. u 2x, Because 2 x equation. Substitute u p 4 2x p 4 x p 8 2kp 2kp kp Therefore, all solutions of x p 8 where k is any integer. and u 3p 4 2x 3p 4 x 3p 8 2kp 2kp kp are given by sin 2x 22 2 kp and x 3p 8 kp, ■ Algebraic Techniques Many trigonometric equations can be solved algebraically — by using factoring, the quadratic formula, and basic identities to write an equivalent equation that involves only basic equations, as shown in the following examples. Example 7 Factoring Trigonometric Equations Find the solutions of 3 sin2 x sin x 2 0 in the interval p, p . 4 3 Solution Let u sin x . 3 sin2 x sin x 2 0 3u2 u 2 0 Substitution Section 8.3 Algebraic Solutions of Trigonometric Equations 543 This quadratic equation can be solved by factoring. 3u2 u 2 0 u 1 0 3u 2 2 1 2 or u 1 1 u 2 3 Substituting sin x for u results in two basic equations. sin x 2 3 or sin x 1 If sin x 2 3 b a , then 2π x sin 1 2 a 3 b 0.7297 2kp x p 2 2kp , then If sin x 1 x p sin 1 2 a 3 b 3.8713 2kp or Therefore, the solutions of x 0.7297 2kp, where k is any integer. 3 sin2 x sin x 2 0 x p 2 2kp, and are x 3.8713 2kp, p, p , Figure 8.3-4 indicates that there are three solutions in the interval is outside which is marked with vertical lines. The solution the interval, but the corresponding solution within the interval can be x 3.8713 2kp. found by letting , the solutions are x 3.8713 k 1 p, p Within in 4 3 4 3 x 3.8713 2p 2.4119, x 0.7297, and x p 2 . ■ Example 8 Factoring Trigonometric Equations Solve tan x cos2x tan x. Solution Write an equivalent equation as an expression equal to zero, and factor. tan x cos2x tan x 0 0 cos2 x 1 0 cos2 x 1 2cos2x 21 cos x ± 1 x 0 kp tan x or cos2x 1 tan x 0 1 2 x 0 2kp or x p 2kp −2π 3 −3 Figure 8.3-4 CAUTION cos2 x 1 sin2 x 544 Chapter 8 Solving Trigonometric Equations 2 0 2 More simply stated, the solution of tan x cos2 x tan x is x 0 kp kp, 2π where k is any integer. The graphs of shown in Figure 8.3-5. Y1 tan x cos2 x and Y2 tan x are ■ Figure 8.3-5 Many trigonometric equations can be solved if trigonometric identities are used to rewrite the original equation, as shown in Examples 9 and 10. Example 9 Identities and Factoring Solve 10 cos2 x 3 sin x 9 0. Solution Use the Pythagorean identity to rewrite the equation in terms of the sine function. 10 cos2x 3 sin x 9 0 3 sin x 9 0 1 sin2x 10 10 10 sin2x 3 sin x 9 0 10 sin2x 3 sin x 1 0 1 2 Factor the left side and solve. 2 sin x 1 0 sin x 1 2 2 sin x 1 1 2 1 5 sin x 1 0 2 5 sin x 1 0 or sin x 1 5 x sin 1 1 2 b a x sin 1 1 a 5 b or x 0.2014 2kp or 2kp x p 6 x p p 6 5p 6 2kp 2kp x p 0.2014 1 2 2kp 3.3430 2kp 15 5 π 2 Figure 8.3-6 Therefore, all solutions of x p 6 10 cos2 x 3 sin x 9 0 2 kp, x 5p 6 2kp, are x 0.2014 2kp, and x 3.3430 2kp, where k is any integer. The graph of in Figure 8.3-6 confirms the solution. Y1 10 cos2x 3 sin x 9 shown ■ 3π 2 Example 10 Identities and Quadratic Formula Solve sec2 x 5 tan x 2. Section 8.3 Algebraic Solutions of Trigonometric Equations 545 Solution Use a Pythagorean identity to rewrite the equation in terms of the tangent function. sec2x 5 tan x 2 1 tan2x sec2x 5 tan x 2 0 5 tan x 2 0 tan2x 5 tan x 3 0 2 1 Use the quadratic formula to solve for tan x. tan x 5 ± 252 4 2 1 1 2 1 3 2 21 1 5 ± 213 2 0.6972 or tan x 4.3028 tan x 5 213 2 0.6972 1 1 0.6088 kp 2 x tan Therefore, the solution set of x 0.6089 pk and sec2x 5 tan x 2 5 213 2 4.3028 1 1 1.3424 kp 2 x tan is x 1.3424 pk, sec2 x 5 tan x and 20 π 2 3π 2 10 Figure 8.3-7 where k is any integer. The graphs of in Figure 8.3-7 confirm the solution. Y1 Y2 2 ■ Exercises 8.3 In Exercises 1–8, find the exact solutions. 1. sin x 23 2 3. tan x 23 5. 2 cos x 23 2. 2 cos x 22 4. tan x 1 6. sin x 0 7. 2 sin x 1 0 8. csc x 22 In the following exercises, find exact solutions if possible and approximate solutions otherwise. When a calculator is used, round to four decimal places. Use the following information in Exercises 9–12. When a light beam passes from one medium to another (for instance, from air to water), it changes both its speed and direction. According to Snell’s Law of Refraction, sin u1 sin u2 v1 v2 , v1 where speed in the second medium, and is the speed of light in the first medium, its u1 the angle of incidence, the angle of refraction, as shown in the figure. v2 u2 The number v1 v2 is called the index of refraction. Angle of incidence Incident ray, speed v 1 θ 1 Refracted ray, speed v 2 θ 2 Angle of refraction 9. The index of refraction of light passing from air to , find water is 1.33. If the angle of incidence is the angle of refraction. 38° 546 Chapter 8 Solving Trigonometric Equations 10. The index of refraction of light passing from air to 40. tan x sec x 3 tan x 0 ordinary glass is 1.52. If the angle of incidence is 17°, find the angle of refraction. 11. The index of refraction of light passing from air to dense glass is 1.66. If the angle of incidence is , 24° find the angle of refraction. 12. The index of refraction of light passing from air to , find quartz is 1.46. If the angle of incidence is the angle of refraction. 50° In Exercises 13–32, find all the solutions of each equation. 13. sin x 0.465 14. sin x 0.682 15. cos x 0.564 16. cos x 0.371 17. tan x 0.237 18. tan x 12.45 19. Hint: cot x 1 cot x 2.3 S tan x . T 20. cot x 3.5 21. sec x 2.65 22. csc x 5.27 23. sin 2x 23 2 24. cos 2 x 22 2 26. 2 sin x 3 1 25. 2 cos x 2 22 27. tan 3 x 23 28. 5 sin 2x 2 29. 5 cos 3x 3 30. 2 tan 4x 16 32. 5 sin x 4 4 31. 4 tan x 2 8 41. 4 sin x tan x 3 tan x 20 sin x 15 0 Hint: One factor is tan x 5. 42. 25 sin x cos x 5 sin x 20 cos x 4 43. sin2x 2 sin x 2 0 44. cos2x 5 cos x 1 45. tan2x 1 3 tan x 46. 4 cos2x 2 cos x 1 47. 2 tan2x 1 3 tan x 48. 6 sin2x 4 sin x 1 49. sec2x 2 tan2x 0 50. 9 12 sin x 4 cos2x 51. sec2 x tan x 3 52. cos2x sin2x sin x 0 53. 2 tan2x tan x 5 sec2x 54. The number of hours of daylight in Detroit on t 0 day t of a non-leap year (with 1) is given by the following function. being January 3 sin d t 2 1 2p 365 1 S t 80 2 T 12 a. On what days of the yea
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r are there exactly 11 hours of daylight? b. What day has the maximum amount of daylight? 55. A weight hanging from a spring is set into motion moving up and down. Its distance d (in centimeters) above or below the equilibrium point at time t seconds is given by d 5 1 sin 6 t 4 cos 6 t . 2 At what times during the first 2 seconds is the weight at the equilibrium position d 0 ? 1 2 In Exercises 33–53, use factoring, the quadratic formula, or identities to solve the equation. Find all solutions in the interval [0, 2p). 33. 3 sin2x 8 sin x 3 0 34. 5 cos2x 6 cos x 8 35. 2 tan2x 5 tan x 3 0 36. 3 sin2x 2 sin x 5 37. cot x cos x cos x 38. tan x cos x cos x 39. cos x csc x 2 cos x In Exercises 56–59, use the following information. When a projectile (such as a ball or a bullet) leaves its starting point at angle of elevation with velocity v, the horizontal distance d it travels is given by the equation u d v2 32 sin 2u, where d is measured in feet and v in feet per second. Note that the horizontal distance traveled may be the same for two different angles of elevation, so that some of these exercises may have more than one correct answer. Section 8.4 Simple Harmonic Motion and Modeling 547 Vmax is the maximum voltage, f is the where frequency (in cycles per second), t is the time in seconds, and is the phase angle. a. If the phase angle is 0, solve the voltage f equation for t. f 0, Vmax b. If 20, V 8.5, and f 120, find the smallest positive value of t. 61. Critical Thinking Find all solutions of sin2x 3 cos2x 0 in the interval 0, 2p . 2 3 62. Critical Thinking What is wrong with this “solution’’? sin x tan x sin x tan x 1 or 5p 4 x p 4 Hint: Solve the original equation by moving all terms to one side and factoring. Compare your answers with the ones above. 63. Critical Thinking Let n be a fixed positive integer. Describe all solutions of the equation sin nx 1 2 . θ d a d b 56. If muzzle velocity of a rifle is 300 feet per second, at what angle of elevation (in radians) should it be aimed in order for the bullet to hit a target 2500 feet away? 57. Is it possible for the rifle in Exercise 56 to hit a target that is 3000 feet away? At what angle of elevation would it have to be aimed? 58. A fly ball leaves the bat at a velocity of 98 miles per hour and is caught by an outfielder 288 feet away. At what angle of elevation (in degrees) did the ball leave the bat? 59. An outfielder throws the ball at a speed of 75 miles per hour to the catcher who is 200 feet away. At what angle of elevation was the ball thrown? 60. In an alternating current circuit, the voltage is given by the formula V Vmax sin 1 2pft f , 2 8.4 Simple Harmonic Motion and Modeling Objective In Section 7.4, graphs of functions of the form • Write a sinusoidal function whose graph resembles a given graph • Write a sinusoidal function to represent a given simple harmonic motion, and use the function to solve problems • Find a sinusoidal model for a set of data, and use the model to make predictions f t 2 1 a sin bt c 2 1 d and g t 2 1 a cos bt c 2 1 d, were studied; and the constants a, b, c, and d were examined to see how they affect the graphs of the functions. In this section, trigonometric functions of this form are used to model real-world phenomena. Recall that if a 0 and b 7 0, bt c a sin 1 2 f t 2 1 then each of the functions d and g t 2 1 a cos bt c d 2 1 has the following characteristics. a amplitude 0 phase shift c b 0 period 2p b vertical shift d 548 Chapter 8 Solving Trigonometric Equations Recall the shapes of sine and cosine waves. 2 0 2 2π 0 2π 2 Figure 8.4-1 2 Figure 8.4-2 The wave shape of the graphs of these functions is called a sinusoid and the functions are called sinusoidal functions. a sin a 1 2 f t bt , is the number Recall that the amplitude of the function 0 1 units above the horand that its graph consists of waves that rise to 0 units below the horizontal axis. In other izontal axis and fall to words, the amplitude is half the distance from the maximum value to the minimum value of the function. This number remains the same when the graph is shifted vertically or horizontally. Thus the amplitude of the sinusoidal function is the number d or g a cos a sin bt c bt fmax fmin2 , where fmax and fmin denote the maximum and minimum values of f. The vertical shift d of a sinusoidal function can be determined by averaging the maximum and minimum values as shown below. vertical shift d fmax fmin 2 Example 1 Constructing Sinusoidal Functions Write a sine function and a cosine function whose graph resembles the sinusoidal graph below. y 4 2 0 −2 Figure 8.4-3 −2π −1π t 2π 1π 3π 4 NOTE The function that 2 1 t t represents the graph, 3 cos 2t 1, g is not unique because infinitely many functions can name the graph, such as 1. 3 cos h 2 1 1 They can be written using different phase shifts. The first function above is the representative answer because it uses the phase shift that is closest to zero. t p 2 2 2 1 Section 8.4 Simple Harmonic Motion and Modeling 549 Solution The graph shows that the function has a maximum of 4 and a minimum of Therefore, the amplitude is 2. a 1 2 1 fmax fmin2 1 2 1 4 2 1 22 3. The graph shows one complete cosine cycle between 0 and is Therefore, p. p, so the period p 2p b b 2 The vertical shift d is one unit. d fmax fmin 2 4 2 1 2 2 1 The variable c depends on whether the function is described in terms of g is used, there is no phase shift because a sine or cosine. If cosine wave’s maximum occurs at So the phase shift is 0, and a function for the sinusoidal graph in Figure 8.4-3 has a value of 0 for c. cos t t 0. t 2 1 3 cos 2t 1 g t 2 1 The graph indicates that a sine wave begins at for a function using f sin t is t 2 1 3p 4 . t 3p 4 , so the phase shift 3p 4 3p 4 c b c 2 c 3p 2 phase shift is c b b 2 Solve for c. Therefore, a function for the sinusoidal graph in Figure 8.4-3 is 3 sin f t 2 1 2t 3p 2 b a 1. ■ Simple Harmonic Motion Motion that can be described by a function of the form f t 2 1 a sin bt c 1 2 d or g t 2 1 a cos bt c d 2 1 is called simple harmonic motion. Many kinds of physical motion are simple harmonic motions. 550 Chapter 8 Solving Trigonometric Equations Example 2 Rotating Wheel P(2, 0) A wheel with a radius of 2 centimeters is rotating counterclockwise at 3 radians per second. A free-hanging rod 10 centimeters long is connected to the edge of the wheel at point P and remains vertical as the wheel rotates (Figure 8.4-4). a. Assuming that the center of the wheel is at the origin and that P is find a function that describes the y-coordinate at (2, 0) at time of the tip E of the rod at time t. t 0, b. What is the first time that the tip E of the rod will be at a height of 9 centimeters? Solution a. The wheel is rotating at 3 radians per second. After t seconds, the point P has moved through an angle of from the origin. 3t radians and is 2 units E Figure 8.4-4 To find the time t that it takes to complete one revolution (i.e., 2p), solve 3t 2p. t 2p 3 −2 −4 −6 −8 −10 −12 y 0 t π 6 π 3 π 2 2π 3 5π 6 π 1 4 After of a revolution, or 1 4 maximum of 2 centimeters, so the y-coordinate of E is 1 2 After of a revolution, or 2p 3 p 3 2p 3 p 6 1 2 , , the height of P reaches its 2 10 8. the height of P is at 0, so the y-coordinate of E is 10. Continuing this process, it can be found that the y-coordinate of E is 12 and at 2p 3 the y-coordinate of E is p at 2 10. Figure 8.4-5 Plotting these key points shows the main features of the graph. The amplitude is a 1 2 1 fmax fmin2 1 2 1 8 12 1 22 2. Use the period to find b. 2p 2p 3 b b 3 so there is no phase shift and c 0. The sine wave begins at The vertical shift, d, is d fmax t 0, 10 units. fmin 2 8 1 2 12 2 10 Thus, the function giving the y-coordinate of E at time t is 2 sin 3t 10. f t 2 1 Section 8.4 Simple Harmonic Motion and Modeling 551 b. To find the first time that the tip E of the rod will be at a height of 2 sin 3t 10 9 centimeters, solve for t. 9 2 sin 3t 10 9 2 sin 3t 1 sin 3t 1 2 3t sin 1 1 2 b a 3t p 6 t p 18 2kp 1 2kp 3 k is any integer 2 0.1745 2kp 3 The first time that the tip E of the rod will be at a height of 9 cm is when k 0, that is, t p 18 0.1745 seconds. ■ Example 3 Bouncing Spring Suppose that a weight hanging from a spring is set in motion by an upward push (Figure 8.4-6). It takes 5 seconds for it to complete one cycle of moving from its equilibrium position to 8 centimeters above, then dropping to 8 centimeters below, and finally returning to its equilibrium position. (This is an idealized situation in which the spring has perfect elasticity, and friction, air resistance, etc., are negligible.) Equilibrium position Figure 8.4-6 a. Find a sinusoidal function to represent the motion of the moving weight. b. Sketch a graph of the function you wrote in part a. c. Use the function from part a to predict the height of the weight after 3 seconds. d. In the first 5 seconds, when will the height of the weight be 6 centimeters below the equilibrium position? 552 Chapter 8 Solving Trigonometric Equations Solution a. Let h t 1 2 denote the distance of the weight above or below t h equilibrium position at time t. Then from 0 to 5, increases from 0 to 8, decreases to again to 0. In the next 5 seconds it repeats the same pattern, and so on. Therefore, the graph of h is periodic and has some kind of wave shape. Two possibilities are shown in Figures 8.4-7a and 8.4-7b. is 0 when t is 0. As t increases and increases h t 1 2 1 2 1 2 8, 1 2 its h(t) 8 h(t) 8 −8 t or −8 t Figure 8.4-7a Figure 8.4-7b Careful physical experiment suggests that the curve in Figure 8.4-7a, which resembles the sine graphs you have studied, is a reasonably accurate model of this process. Facts from physics and calculus show that the rule of the function h is the form for some constants a, b, and c. a sin bt c h t 2 2 1 1 The function h has an amplitude of 8, a period of 5, and a phase shift 0, so the constants a, b, and c must satisfy 8, 2p b 5, and c b 0, a 0 0 y 8 4 0 −4 −8 or equi
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valently, a 8, b 2p 5 , and c 0. 1 2 3 4 5 t Therefore, the motion of the moving weight can be described by this function: 8 sin h t 2 1 2p 5 t Figure 8.4-8 b. The graph of h 8 sin t 2 1 2p 5 t is shown in Figure 8.4-8. c. The value of h 3 1 2 gives the height of the weight after 3 seconds. 8 sin h 3 1 2 2p 5 a 3 b 8 sin 6p 5 4.7 The height of the weight after 3 seconds is approximately 4.7 centimeters below the equilibrium point. d. To find the times in the first 5 seconds when the weight is 6 centimeters below the equilibrium, you must solve the equation 10 0 10 10 0 Figure 8.4-9 5 5 10 Figure 8.4-10 Section 8.4 Simple Harmonic Motion and Modeling 553 8 sin 2p 2p 5 Y2 6 Y1 8 sin a t b and are shown in Figure 8.4-9 and 8.4-10. The points of The graphs of 0 t 5 intersection show that the weight will be 6 centimeters below the equilibrium at two times between 0 and 5 seconds. t 3.1749 and t 4.3251 in a window with ■ Modeling Trigonometric Data Periodic data that appears to resemble a sinusoidal curve when plotted can often be modeled by a sine function, as shown in Example 4. Example 4 Temperature Data The following table shows the average monthly temperature in Cleveland, Ohio, based on data from 1971 to 2000. Since average temperatures are not likely to vary much from year to year, the data essentially repeats the same pattern in subsequent years. So, a periodic model is appropriate. Month Temperature °F 1 2 Month Temperature °F 2 1 Jan. Feb. Mar. Apr. May June 25.7 28.4 37.5 47.6 58.5 67.5 July Aug. Sep. Oct. Nov. Dec. 71.9 70.2 63.3 52.2 41.8 31.1 [Source: National Climatic Data Center] a. Make a scatter plot of the data. b. Find a sinusoidal function that models the temperature data. c. Use the sine regression feature on a calculator to find another sinusoidal model for the data. d. How do the models in parts b and c differ from one another? e. Use one of the models to predict time(s) of year in which the average temperature is 45°F. 100 0 0 13 Solution a. Let t 1 represent January. Enter 1 through 12 in List 1 and the Figure 8.4-11a temperatures in List 2. The scatter plot is shown in Figure 8.4-11a. 554 Chapter 8 Solving Trigonometric Equations b. To find a sinusoidal function to represent the temperature data, examine the properties of the scatter plot. The minimum value is 25.7 and the maximum value is 71.9, so the amplitude is a 1 2 1 fmax fmin2 1 2 1 71.9 25.7 23.1. 2 One complete cycle is 12 months, so the period, 2p b , is 12. 2p b 12 b p 6 12 Figure 8.4-11b The vertical shift is d fmax fmin 2 71.9 25.7 2 48.8. A sine wave begins close to the data point (4, 47.6), as shown in Figure 8.4-11b, so the phase shift c b is approximately 4. Find c. 4 4 c b c p 6 c 2p 3 Therefore, a sinusoidal function to represent the temperature data is approximately 23.1 sin f t 2 1 p 6 t 2p 3 b a 48.8. The graph of this function is shown with the scatter plot of the data in Figure 8.4-12. c. Using the 12 given data points, the regression feature on a calculator 13 Figure 8.4-12 produces the following model. 23.1202 sin t f 1 2 0.5018t 2.0490 48.6927 2 1 The period of this function is approximately 2p 0.5018 12.52, which is not a very good approximation for a 12-month cycle. Because the data repeats the same pattern from year to year, a more accurate model can be obtained by using the same data repeated for 100 0 0 100 0 0 Section 8.4 Simple Harmonic Motion and Modeling 555 the second year. The data for a two-year period is plotted in Figure 8.4-13a. The sine regression feature on a calculator produces this model from the 24 data points: 22.7000 sin 0.5219t 2.1842 49.5731 t f 2 1 The period of this function is 2 1 2p 0.5219 12.04, which is slightly off from the expected 12-month period. However, its graph in Figure 8.4-13b appears to fit the data well. d. Using the decimal approximation of p, the rule of the function found in part b becomes f t 2 1 23.1 sin 1 0.5236 t 2.0944 48.8. 2 This model differs only slightly from the second model in part c. f t 2 1 22.7000 sin 1 0.5219 t 2.1842 49.5731 2 Visually, however, the model shown in Figure 8.4-12 does not seem to fit the data points quite as well as the model shown in Figure 8.4-13b. Nevertheless, considering that the model found in part b can be obtained without technology, it is remarkedly close. e. The model from part c will be used to predict the times of year in which the average temperature is 45°F . There are two points of intersection of the graphs of 0.5219 t 2.1842 22.7000 sin y 45 and t f 1 2 1 49.5731, 2 as shown in Figure 8.4-14. Their approximate coordinates are (3.8, 45) and (10.6, 45). Therefore, according to this model, the temperature would be around in late March and late October. 45°F ■ 80 0 0 80 0 0 100 0 0 25 25 13 Figure 8.4-13a Figure 8.4-13b Figure 8.4-14 Exercises 8.4 1. The original Ferris wheel, built by George Ferris for the Columbian Exposition of 1893, was much larger and slower than its modern counterparts: It had a diameter of 250 feet and contained 36 cars, each of which held 40 people; it made one revolution every 10 minutes. Suppose that the Ferris wheel revolves counterclockwise in the x-y plane with its center at the origin. Car D in the figure had coordinates (125, 0) at time the rule of a function that gives the y-coordinate of car D at time t. t 0. Find y 125 −125 125 x D 556 Chapter 8 Solving Trigonometric Equations 2. Do Exercise 1 if the wheel turns at 2 radians per minute and car D is at 1 0, 125 at time 2 t 0. 3. A circular wheel with a radius of 1 foot rotates counterclockwise. A 4-foot rod has one end attached to the edge of this wheel and the other end to the base of a piston (see figure). It transfers the rotary motion of the wheel into a back-andforth linear motion of the piston. If the wheel is rotating at 10 revolutions per second, point W is and point P is always on at x-axis, find the rule of a function that gives the x-coordinate of P at time t. at time t 0, 1, 0 1 2 from the pendulum to the (dashed) center line at time t seconds (with distances to the right of the line measured by positive numbers and distances to the left by negative ones). Assume that the pendulum is on the center line at time moving to the right. Assume the motion of the pendulum is simple harmonic motion. Find the rule of the function t 0 and 1 −1 A B 4. Do Exercise 3 if the wheel has a radius of 2 feet, rotates at 50 revolutions per second, and is at 2, 0 t 0. when 1 2 In Exercises 5–8, suppose a weight is hanging from a spring (under the same conditions as in Example 3). The weight is pushed to start it moving. At time t, let h(t) be the distance of the weight above or below its equilibrium point. Assume the maximum distance the weight moves in either direction from the equilibrium point is 6 cm and that it moves through a complete cycle every 4 seconds. Express h(t) in terms of the sine or cosine function under the given conditions. 5. There is an initial push upward from the equilibrium point. 6. There is an initial pull downward from the equilibrium point. Hint: What does the graph of y a sin bt look like when a 6 0 ? 7. The weight is pulled 6 cm above the equilibrium ) is point, and the initial movement (at downward. Hint: Think of the cosine graph. t 0 8. The weight is pulled 6 cm below its equilibrium point, and the initial movement is upward. 9. A pendulum swings uniformly back and forth, taking 2 seconds to move from the position directly above point A to the position directly above point B, as shown in the figure. The distance from A to B is 20 centimeters. Let be the horizontal distance d t 1 2 10. The following figure shows a diagram of a merrygo-round that is turning counterclockwise at a constant rate, making 2 revolutions in 1 minute. On the merry-go-round are horses A, B, C, and D 4 meters from the center and horses E, F, and G t a 8 meters from the center. There is a function 1 that gives the distance the horse A is from the y-axis (this is the x-coordinate of A’s position) as a function of time t measured in minutes. Similarly, gives the x-coordinate for B as a function of b time, and so on. Assume the diagram shows the situation at time t 0. Which of the following function rules does a have? 4 cos t, 4 cos pt, 4 cos 2t, 4 cos 2pt, t 2 1 1 2 4 cos , 4 cos t t b b b. Describe the function a a p 2 , 4 cos 4pt t b 2 using the cosine function and g t 2 1 Section 8.4 Simple Harmonic Motion and Modeling 557 c. Suppose the x-coordinate of a horse S is given 4 cos by the function and the s t 4pt 5p 6 b a 1 2 x-coordinate of another horse R is given by 2 . t r 8 cos Where are these horses 4pt p a 1 3 b located in relation to the rest of the horses? Copy the diagram and mark the positions of R and S at t 0 . 11. The following table shows the number, in millions, of unemployed people in the labor force for 1991–2002. Year Unemployed Year Unemployed 1991 1992 1993 1994 1995 1996 8.628 9.613 8.940 7.996 7.404 7.236 1997 1998 1999 2000 2001 2002 6.739 6.210 5.880 5.655 6.742 8.234 a. Sketch a scatter plot of the data, with x 0 corresponding to 1990. b. Does the data appear to be periodic? If so, find an appropriate model. c. Do you think this model is likely to be accurate much beyond the year 2002? Why? In Exercises 12 and 13, do the following: a. Use 12 data points (with x 1 corresponding to January) to find a periodic model of the data. b. What is the period of the function found in part a? Is this reasonable? c. Plot 24 data points (two years) and graph the function from part a on the same screen. Is the function a good model in the second year? d. Use the 24 data points in part c to find another periodic model for the data. e. What is the period of the function in part d? Does its graph fit the data well? 12. The table shows the average monthly temperature in Chicago, IL, based on data from 1971 to 2000. Month Temperature °F 2 1 Jan. Feb. Mar. Apr. May June July Aug. Sep. Oct. Nov. Dec. 22.0 27.0 37.3 47.8 58.7 68.2 73.3 71.7 63.8 52
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.1 39.3 27.4 [Source: National Climatic Data Center] 13. The table shows the average monthly precipitation, in inches, in San Francisco, CA, based on data from 1971 to 2000. Month Precipitation Jan. Feb. Mar. Apr. May June July Aug. Sep. Oct. Nov. Dec. 4.45 4.01 3.26 1.17 0.38 0.11 0.03 0.07 0.20 1.04 2.49 2.89 [Source: National Climatic Data Center] 558 Chapter 8 Solving Trigonometric Equations 14. Critical Thinking A grandfather clock has a pendulum length of k meters. Its swing is given (as in Exercise 9) by the function f t 0.25 sin vt , where v 2 1 a. Find k such that the period of the pendulum is B 1 2 9.8 k . b. The temperature in the summer months causes the pendulum to increase its length by 0.01%. How much time will the clock lose in June, July, and August? Hint: These three months have a total of 92 days (7,948,800 seconds). If k is increased by 0.01%, what is f 2 ? 1 2 2 seconds. 8.4.A Excursion: Sound Waves Objectives • Find the frequency of a sound wave using a tuning fork and a data collection device • Model sound wave data with sinusoidal functions and graphs Sound travels through the air like small ripples traveling across a body of water. Throwing a rock into a calm body of water causes the water to begin to move up and down around the entry point. The movement of the water causes ripples to move outward. If a flower is floating in this body of water, the ripples will cause the flower to move up and down on the water. Sound is produced by a vibrating object that disturbs the surrounding air molecules and causes them to vibrate. These vibrations cause a periodic change in air pressure that travels through the air much like the ripples in a body of water. When the air pressure waves reach the eardrum, they cause it to vibrate at the same frequency as the source. These air pressure waves are more commonly called sound waves. The voices and sounds heard each day are generally a combination of many different sound waves. The sound from a tuning fork is a single tone that can be described mathematically using a sinusoidal function, or d. g a cos a sin bt c bt Recall from Sections 7.4 and 8.4 that if functions a 0 and b 7 0, then each of the f t 2 1 a sin bt c 2 1 d and g t 2 1 a cos bt c d 2 1 has the following characteristics: amplitude 0 phase shift c b a period 2p b 0 vertical shift d The period of a sound wave determines the sound’s frequency. The frequency f of a sound wave is the reciprocal of its period. f b 2p Section 8.4.A Excursion: Sound Waves 559 Technology Tip Data obtained may differ from one collection to another, and specific values produced by a calculator may vary. Frequency gives the number of cycles (periods) that the sound wave completes in one second. Frequencies are measured in units called Hertz (Hz), where one Hertz is one cycle per second. Most tuning forks have their frequency and corresponding musical note listed on them. A tuning fork, a microphone connected to a data collection device (such as a CBL2), and a calculator with a tuning program can be used to simulate the sounds heard through the eardrum. NOTE A tuning program can be obtained by downloading or entering the DataMate application available from Texas Instruments. data collection device calculator tuning fork (-) microphone Figure 8.4.A-1 Example 1 Frequency Confirm with a sinusoid graph that the frequency of a sound wave formed by striking a C tuning fork is 262 Hz. Solution Connect the calculator to the data collection device, connect the microphone to the data collection device, and run the program needed to calculate the pressure for the sound waves. Follow the directions on the screen of your calculator to obtain a graph like the one shown in Figure 8.4.A-2. 0.05 0 −0.05 0.05 Figure 8.4.A-2a 0.05 0 −0.05 0.05 0 Figure 8.4.A-2b 0.05 0.05 −0.05 Figure 8.4.A-2c The period of the data can be found by dividing the differences between the x-value of the first maximum and the x-value for the last maximum by the number of complete cycles between the maximums. For the data 560 Chapter 8 Solving Trigonometric Equations shown in Figure 8.4.A-2a, Figures 8.4.A-2b and 8.4.A-2c show that the first maximum occurs when , the last maximum occurs when x 0.0208608 , and there are five complete cycles between the first and last maximum. Therefore, the period of the data is given by the following. x 0.0020768 0.0208608 0.0020768 5 0.0037568 cycle length The frequency is the reciprocal of the period. 1 0.0037568 266.184 Therefore, the graph indicates a frequency of about 266.184, which is very close to the actual frequency of 262 Hz. ■ 0.06 0.0001 Example 2 Sinusoidal Model Find a sinusoidal model to fit the sound waves produced by a tuning fork with the note G. 0.024 Solution −0.06 Figure 8.4.A-3 Use a data collection device, a G tuning fork, and a tuning program to obtain a graph similar to the one shown in Figure 8.4.A-3. Find the amplitude a by finding half the difference between the maximum value and minimum value. The data graphed in Figure 8.4-3 has a maximum value of 0.043942 and a minimum value of 0.045618 . a 0.043942 1 2 0.045618 2 0.04478 amplitude To find the period, find the x-value of the first maximum of the graph, find the x-value of the last maximum of the graph, and divide the difference between the two x-values by 8 (the number of cycles between the first and last maximum). The first maximum shown on the graph occurs x 0.0221088 . when , and the last maximum occurs when x 0.0020768 p 0.0221088 0.0020768 8 0.002504 period Use the period to find b. 2p b 0.002504 The graph has a maximum at tion using cosine is 0.0020768. , so the phase shift for a func- b 2p 0.002504 x 0.0020768 Section 8.4.A Excursion: Sound Waves 561 Use the phase shift and b to find c. c b phase shift c 0.0020768 2p 0.002504 b a 1.658786p Find the vertical shift d by finding the average of the maximum and minimum values. In this example, d 0. Therefore, the sinusoidal model to fit the sound waves produced by a tuning fork with the note G is y 0.04478 cos 2p 0.002504 a x 1.658786p b . Notice that the frequency, the reciprocal of the period, is 399.361, which is very close to the actual frequency of 392 Hz for the note G. ■ When two sounds of slightly different frequency are produced simultaneously, a beat is heard. A beat is a single sound that gets louder and softer at periodic intervals. By using more than one tuning fork, a beat can be displayed. Example 3 Chord Frequency Place tuning forks with notes of C and G close to a microphone. a. Find a graphical representation for the sound waves produced by playing the two notes simultaneously. b. Find the period of the function. c. How does the frequency of this sound compare with the actual frequencies of C and G? Solution a. Using a data collection device, the graph in Figure 8.4.A-4 is obtained. b. The graph appears to be periodic. By finding the length of one complete cycle, the period of this graph appears to be approximately 0.0077632. c. The frequency of this sound is the reciprocal of the period of the Figure 8.4.A-4 function. 1 0.0077632 128.8128607 Hz The frequency of this sound is very close to the difference between the actual frequency of C, 262 Hz, and the actual frequency of G, 392 Hz. 392 262 130 ■ 562 Chapter 8 Solving Trigonometric Equations NOTE The intensity of the strike on the tuning fork determines the amplitude. Graphing Exploration Find a sinusoidal function to model the graph produced by the C tuning fork. Add the sinusoidal models for C and G notes. (Use the model from Example 2.) Graph this sum. What is the frequency of the graph? How does the frequency of this sound compare with the sum of the actual frequencies of C and G? Most sounds are more complex than those produced by tuning forks. A tuning fork produces a graph of a single note. Most musical instruments produce a sound that is a combination of several different sounds. A C-G chord was produced with the tuning forks in Example 3. The exploration above indicates that this sound can be modeled by the sum of the models for the C note and the G note. Exercises 8.4.A Bottles of water can be tuned using a microphone and data collection device. Place some water in the bottle and blow air over the top of the bottle to produce a sound. Use the frequency of the graph formed from the sound to approximate the frequency. If a higher note is needed, place more water in the bottle and calculate the frequency again. Display the graph of the following notes using the chart below. Note Frequency in Hz G# or Ab A A# or Bb B 415 440 466 494 524 Note Frequency in Hz C (next octave) C C# or Db D D# or Eb E F F# or Gb G 262 277 294 311 330 349 370 392 1. D 2. B 3. A 4. C# 5. Using only one tuning fork at a time and the sum of three functions, sketch a graph of the C-major chord C E G . 1 2 6. Using three tuning forks for the notes C, E, and G at one time, find a graph of the C-major chord Important Concepts Section 8.1 Section 8.2 Section 8.3 Section 8.4 Basic trigonometric equations . . . . . . . . . . . . . . . 524 Intersection method . . . . . . . . . . . . . . . . . . . . . . 524 x-intercept method . . . . . . . . . . . . . . . . . . . . . . . 525 Inverse sine function. . . . . . . . . . . . . . . . . . . . . . 530 Inverse cosine function . . . . . . . . . . . . . . . . . . . . 533 Inverse tangent function . . . . . . . . . . . . . . . . . . . 535 Solution algorithm for basic trigonometric equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540 Writing sinusoidal functions . . . . . . . . . . . . . . . . 548 Simple harmonic motion. . . . . . . . . . . . . . . . . . . 549 Modeling trigonometric data . . . . . . . . . . . . . . . 553 Section 8.4.A Frequency of a sound . . . . . . . . . . . . . . . . . . . . . 558 Sound waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558 Important Facts and Formulas sin cos 1 v u exactly when sin u v a 1 v u exactly when cos , 1 v 1 , 1 v 1 b tan 1 v u exactly whe
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n tan real number b a 0 Let teristics. and b 7 0. The following functions have the given charac- f t 2 1 a sin bt c 2 1 d and g t a cos bt c 0 2 1 a amplitude fmax 1 2 1 1 fmin2 0 , phase shift c b fmin2 fmax period 2p b vertical shift d 1 2 1 d 2 563 564 Chapter Review Review Exercises In Exercises 1–6, solve each equation graphically. Section 8.1 1. 5 tan x 2 sin 2x 2. sin3 x cos2 x tan x 2 3. sin x sec2 x 3 4. cos2 x csc2 x tan x p 2 R Q 5 0 5. cos 2x sin x 6. 3 sin 2 x cos x 7. A weight hanging from a spring is set into motion, moving up and down. (See Figure 8.4-6 in Example 3 of Section 8.4.) Its distance in centimeters above or below the equilibrium point at time t seconds is given by d 5 sin 3t 3 cos 3t. weight at the equilibrium position At what times during the first two seconds is the d 0 ? 1 2 In Exercises 8–17, find the exact value without using a calculator. Section 8.2 8. 1 cos 22 2 b a ? 9. 1 sin 23 2 b a ? 10. tan 123 ? 12. 1 cos sin a 5p 3 b ? 14. 1 sin sin 0.75 1 ? 2 16. 1 sin sin a 8p 3 b ? 11. 1 sin cos a 11p 6 b ? 13. 1 tan cos a 7p 2 b ? 15. 1 cos cos 2 1 2 ? 17. 1 cos cos a 13p 4 b ? 18. Sketch the graph of f 19. Sketch the graph of g x 2 1 x 2 1 tan 1 x p. sin 1 x 2 . 2 1 20. Find the exact value of sin 1 cos a 1 4 b . 21. Find the exact value of sin 1 tan a 1 2 cos 1 4 5 b . Section 8.3 22. Find all angles with u 0° u 360° such that sin u 0.7133. 23. Find all angles with u 0° u 360° such that tan u 3.7321. In Exercises 24–38, solve the equation by any means. Find exact solutions when possible and approximate solutions otherwise. 24. 2 sin x 1 26. tan x 1 28. sin x 0.7 30. tan x 13 25. cos x 23 2 27. sin 3x 23 2 29. cos x 0.8 31. cot x 0.4 32. 2 sin2 x 5 sin x 3 33. 4 cos2 x 2 0 Chapter Review 565 34. 2 sin2 x 3 sin x 2 35. sec2 x 3 tan2 x 13 36. sec2 x 4 tan x 2 37. 2 sin2 x sin x 2 0 38. cos2 x 3 cos x 2 0 39. A cannon has a muzzle velocity of 600 feet per second. At what angle of elevation should it be fired in order to hit a target 3500 feet away? Hint: Use the projectile equation given for Exercises 56–59 of Section 8.3. Section 8.4 40. The following table gives the average population, in thousands, of a southern town for each month throughout the year. The population is greater in the winter and smaller in the summer, and it repeats this pattern from year to year. Month Population Month Population Jan. Feb. Mar. Apr. May June 10.5 9.3 7.8 6.0 4.9 4.5 July Aug. Sep. Oct. Nov. Dec. 4.7 5.8 7.6 9.4 10.6 10.9 a. Make a scatter plot of the data. b. Find a sinusoidal function to represent the population data. c. Use the sine regression feature on a calculator to find a periodic model for the data. d. Use the model from part c to predict time(s) of year in which the average population is 6200. 41. The paddle wheel of a steamboat is 22 feet in diameter and is turning at 3 revolutions per minute. The axle of the wheel is 8 feet above the surface of the water. Assume that the center of the wheel is at the origin and that a point P on the edge of the paddle wheel is at (0, 19) at time a. What maximum height above the water does point P reach? b. How far below the water does point P reach? c. d. Write a cosine function for the height of point P at time t. e. Write a sine function for the height of point P at time t. f. Use the function from part d or e to find the time(s) at which point P In how many seconds does the wheel complete one revolution? t 0 seconds. will be at a height of 10 feet. Section 8.4.A 42. Confirm with a sinusoid graph that the frequency of a sound wave formed by striking an F tuning fork is 349 Hz. 43. Find a sinusoidal model to fit the sound waves produced by striking an E tuning fork(x) k lim f(x) = k x n x n x Figure 8.C-1 Limits of Trigonometric Functions A main focus of calculus is the behavior of the output of a function as the input approaches a specific value. The value that the function approaches, if it exists, is called a limit. In this section an informal description of a limit is illustrated with some interesting trigonometric functions, but the discussion is not intended to be complete. See Chapter 14 for a detailed discussion of limits. x If the output of a function approaches a single real number k as the input approaches the real number n, then the function is said to have a limit of k as the input approaches n. This is written as lim xSn f 1 x 2 k, and is read “the limit of f x 1 2 as x approaches n is k”. See Figure 8.C-1. If the outputs of the function do not approach a single real number as the inputs approach n, the limit does not exist. Calculus is needed to find limits analytically, but a calculator’s table feature and a graph can approximate a limit, if it exists. A table or a graph will also indicate when a limit does not exist. The following two limits are very important in calculus and are used in future can do calculus features. Example 1 Limit of sin x x Find lim xS0 sin x x , if it exists, by using a table and a graph. Solution Enter Y1 sin x x into the function editor of a calculator and produce the Figure 8.C-2a table shown in Figure 8.C-2a and the graph shown in Figure 8.C-2b. 2 The table confirms that sin 0 0 0 0 is undefined, but it also suggests that the values of itive and negative. sin x x are approaching 1 for x-values near 0, both pos- –2π 2π proaches 0 from the positive side and from the negative side. Therefore, The graph confirms that the values of sin x x are approaching 1 as x ap- –0.5 Figure 8.C-2b 566 sin x x lim xS0 1 ■ There is another trigonometric limit that is often used in calculus. Example 2 Limit of cos x 1 x Find lim xS0 cos x 1 x Solution , if it exists, by using a table and a graph. Figure 8.C-3a Figure 8.C-3a confirms that cos x 1 x 0 0 is undefined, but that the val- –2π 2 –2 ues of cos x 1 x are approaching 0 when x is near 0. The graph shown in Figure 8.C-3b also illustrates this. Therefore, 2π cos x 1 x lim xS0 0. ■ There are two ways in which a limit may not exist, as shown in the next examples. The first example illustrates a function that does not have a Figure 8.C-3b limit as x approaches p 2 because function values on either side of p 2 do not approach a single real number. The second example illustrates a function that does not have a limit as x approaches 0 because the function values oscillate wildly. NOTE p 2 1.57. when x 7 p 2 Example 3 Determining the Behavior of a Function Near an x-Value Discuss the behavior of f 1 cos x x 2 1 as x approaches p 2 . Find lim xSp 2 1 cos x, if it exists, by using a table and a graph. Solution As shown in Figure 8.C-4a, the values of f 1 cos x x 1 2 when x 6 p 2 are large positive numbers, and the values are large negative numbers . Figure 8.C-4b shows that the graph of f 1 cos x x 1 2 has a p 2 vertical asymptote at . Because the values of f do not approach a x 1 2 single real number as x approaches p 2 , lim xSp 2 1 cos x does not exist. 567 10 –2π 2π Figure 8.C-4a –10 Figure 8.C-4b ■ Example 4 Oscillating Function Values Discuss the behavior of f cos 1 xb a x 2 1 near x 0 and find lim xS0 cos 1 xb , a if it exists. Solution A table of values of f cos 1 xb a x 1 2 near x 0 is shown in Figure 8.C-5a. 1.1 –2π 2π Figure 8.C-5a –1.1 Figure 8.C-5b The table suggests that the function values near may be near 0.86232, but using the trace feature on the graph shown in Figure 8.C-5b indicates that the function value is near 0.365 when x is near 0. x 0 Using a window where lates wildly around x 0. 0.01 x 0.01, you can see that the graph oscil- See Figure 8.C-5c. 1 –0.01 0.01 –1 Figure 8.C-5c Because the values of x f 1 2 cos 1 xb a oscillate as x approaches 0, lim xS0 cos 1 xb a does not exist. ■ 568 Exercises Discuss the behavior of the function around the given x-value by using a table and a graph. Find the limit of each function, if it exists. 1. lim xS0 1 cos x 3. lim xSp 2 tan x x 5. lim xS0 sin 3x 3x 7. lim xS0 x tan2 x 9. lim xS0 3x sin 3x 11. lim xS0 sin 3x sin 4x 2. lim xSp 2 x tan x 4. lim xS0 tan x x 6. lim xS0 sin 2x 2x 8. lim xS0 x sin2 x 10. lim xS0 x sin x 12. lim xS0 sin 8x sin 7x 13. lim xS0 sin a 1 xb 15. lim xSp 2 x cos x 17. lim xS2 3x 19. lim xS1 x 1 x 14. lim xS0 2x sin x x 16. lim xS0 tan 1 xb a 18. 2x lim xS2 20. lim xS1 x 1 x 21. lim xS3 x 2 x 6 x 3 22. lim xS1 x 2 2x 3 x 1 23. lim xS1 3 1 x x 1 24. lim xS1 x 3 x2 x 1 x 1 25. Make a conjecture about the and c are real numbers. lim xS0 sin bx sin cx , where b 569 C H A P T E R 9 Trigonometric Identities and Proof Time. To find the exact period of the oscillations of a simple pendulum, a trigonometric expression must be written in an alternate form, which is obtained by using trigonometric identities. See Exercise 75 in Section 9.3. 570 Chapter Outline 9.1 Identities and Proofs 9.2 Addition and Subtraction Identities 9.2.A Excursion: Lines and Angles 9.3 Other Identities 9.4 Using Trigonometric Identities Chapter Review can do calculus Rates of Change in Trigonometry Interdependence of Sections 9.3 9.1 9.4 9.2 > > > T he basic trigonometric identities, which were discussed in Chapter 6 and used in Chapter 8, are not the only identities that are useful in rewriting trigonometric expressions and in solving trigonometric equa- tions. This chapter presents many widely used trigonometric identities and specific methods for solving particular forms of trigonometric equa- tions. 9.1 Identities and Proofs Objectives • Identify possible identities by using graphs • Apply strategies to prove identities Recall that an identity is an equation that is true for all values of the variable for which every term of the equation is defined. Several trigonometric identities have been discussed in previous sections. This section will introduce other identities and discuss techniques used to verify that an equation is an identity. Trigonometric identities can be used for simplifying expressions, rewriting the rules of trigonometric functions, and performing numerical calculations. There are no hard and fast rules for dea
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ling with identities, but some suggestions follow. The phrases “prove the identity” and “verify the identity” mean “prove that the given equation is an identity.” 571 572 Chapter 9 Trigonometric Identities and Proof 4 Graphical Testing 2π 2π 4 Figure 9.1-1 When presented with a trigonometric equation that might be an identity, it is a good idea to determine graphically whether or not this is possible. x For instance, the equation can be tested to determine sin t cos p 2 a b if it is possibly an identity by graphing Y1 cos p 2 a x b and Y2 sin x on the same screen, as shown in Figure 9.1-1 where the graph of darker than the graph of concluded that the equation is not an identity. is . Because the graphs are different, it can be Y2 Y1 Any equation can be tested by simultaneously graphing the two functions whose rules are given by the left and right sides of the equation. If the graphs are different, the equation is not an identity. If the graphs appear to be the same, then it is possible that the equation is an identity. However, The fact that the graphs of both sides of an equation appear identical does not prove that the equation is an identity, as the following exploration demonstrates. Graphing Exploration In the viewing window with both sides of the equation p x p and 2 y 2, graph cos x 1 x2 2 x4 24 x6 720 x8 40,320 Do the graphs appear identical? Now change the viewing window so that Is the equation an identity? 2p x 2p. Example 1 Graphical Identity Testing Is either of the following equations an identity? a. b. 2 sin2x cos x 2 cos2x sin x 1 sin x sin2x cos x cos x tan x 4 Solution Test each equation graphically to see if it might be an identity by graphing each side of the equation. 2 sin2x cos x 2 cos2x sin x a. Graph and Y2 Y1 on the same is shown darker screen, as shown in Figure 9.1-2a. The graph of Y2. than 2 sin2x Because the graphs are not the same, the equation cos x 2 cos2x sin x is not an identity. Y1 2π 2π 4 Figure 9.1-2a Section 9.1 Identities and Proofs 573 b. The graph shown in Figure 9.1-2b suggests that 1 sin x sin2x cos x cos x tan x may be an identity, but the proof that it actually is an identity must be done algebraically. 4 π π Figure 9.1-2b 4 ■ Example 2 Finding an Identity Find an equation involving 2 sin x cos x that could possibly be an identity. Solution y 2 sin x cos x is shown in Figure 9.1-3a. Does it look familThe graph of iar? At first it looks like the graph of but there is an important difference. The function graphed in Figure 9.1-3a has a period of As was shown in Section 7.3, the graph of looks like the sine graph but has a period of y sin 2x y sin x, p. Y1 The graphs and 9.1-3b. Because the graphs appear identical, may be an identity. and Y2 sin 2x are shown in Figures 9.1-3a 2 sin x cos x sin 2x p. 2 sin x cos x ■ Proving Identities CAUTION Be sure to use parentheses correctly when entering each function to be graphed. 3 y = 2 sin x cos x 2π 2π 3 Figure 9.1-3a 3 y = sin 2x 2π 2π A useful feature of trigonometric functions is that they can be written in many ways. One form may be easier to use in one situation, and a different form of the same function may be more useful in another. 3 Figure 9.1-3b The elementary identities that were given in Section 6.5 are summarized for your reference on the following page. Memorizing these identities will benefit you greatly in the future. NOTE The definitions of the basic trigonometric ratios may help you remember the quotient and reciprocal identities. The shapes of the graphs of sine, cosine, and tangent may help you remember the periodicity and negative angle identities. Also, if you can remember the first of the Pythagorean identities, which is based on the Pythagorean Theorem, the other two can easily be derived from it. 574 Chapter 9 Trigonometric Identities and Proof Basic Trigonometry Identities Quotient Identities tan x sin x cos x cot x cos x sin x Reciprocal Identities sin x 1 csc x cos x 1 sec x csc x 1 sin x cot x 1 tan x sec x 1 cos x tan x 1 cot x Periodicity Identities sin(x 2P) sin x csc(x 2P) csc x tan(x P) tan x cos(x 2P) cos x sec(x 2P) sec x cot(x P) cot x sin2x cos2x 1 Pythagorean Identities tan2x 1 sec2x 1 cot2x csc2x sin( x) sin x Negative Angle Identities cos(x) cos x tan(x) tan x Just looking at the graphs of the two expressions that make up the equation is not enough to guarantee that it is an identity. Although there are no exact rules for simplifying trigonometric expressions or proving identities, there are some common strategies that are often helpful. Strategies for Proving Trigonometric Identities 1. Use algebra and previously proven identities to transform one side of the equation into the other. 2. If possible, write the entire equation in terms of one trigonometric function. 3. Express everything in terms of sine and cosine. 4. Deal separately with each side of the equation A B. First use identities and algebra to transform A into some expression C, then use (possibly different) identities and algebra to transform B into the same expression C. Conclude that 5. Prove that with B 0 and D 0. You can then A B. AD BC, C A B D . conclude that Section 9.1 Identities and Proofs 575 There are often a variety of ways to proceed, and it will take some practice before you can easily decide which strategies are likely to be the most efficient in a particular case. Keep these two purposes of working with trigonometric identities in mind: • to learn the relationships among the trigonometric functions • to simplify an expression by using an equivalent form CAUTION Proving identities is not the same as solving equations. Properties that apply to equations, such as adding the same value to both sides, are not valid when verifying identities because the beginning statement (to be verified) may not be true. In the following example, the Pythagorean identity is used to replace 1 sin2x Consider using one of the Pythagorean identities whenever a squared trigonometric function appears. cos2x. with replace sin2x cos2x tan2x with 1 cos2x 1 sin2x sec2x 1 replace csc2x sec2x cot2x with 1 cot2x tan2x 1 csc2x 1 Example 3 Transform One Side into the Other Side Verify that 1 sin x sin2x cos x cos x tan x. Solution The graph of each side of the equation is shown in Figure 9.1-2b of Example 1, where it was noted that the equation might be an identity. Begin with the left side of the equation. 1 sin x sin2x cos x sin x 1 sin2x 1 2 cos x cos2x sin x cos x cos2x sin x cos x cos x cos x sin x cos x cos x tan x regrouping terms Pythagorean identity a b c a c b c a2 a a quotient identity ■ Strategies for proving identities can also be used to simplify complex expressions. 576 Chapter 9 Trigonometric Identities and Proof Example 4 Write Everything in Terms of Sine and Cosine Simplify 1 csc x cot x 1 cos x . 2 2 1 Solution csc x cot x 1 2 1 2 cos x sin x b 2 1 cos x ˛ 1 1 cos x 2 2 2 ˛1 sin x 1 cos x 1 a sin x 1 cos x sin x 1 cos x 1 1 1 cos2x sin x sin2x sin x sin x 1 cos x 1 2 reciprocal and quotient identities ab c a b a b 2 2 ˛1 1 a2 b2 Pythagorean identity x2 x x ■ The strategies presented above and those to be considered are “plans of attack.” By themselves they are not much help unless you also have some techniques for carrying out these plans. In the previous examples, the techniques of basic algebra and the use of known identities were used to change trigonometric expressions into equivalent expressions. There is another technique that is often useful when dealing with fractions. Rewrite a fraction in equivalent form by multiplying its numerator and denominator by the same quantity. Example 5 Transform One Side into the Other Side Prove that sin x 1 cos x 1 cos x sin x . Solution Beginning with the left side, multiply the numerator and denominator by 1 cos x. sin x 1 cos x sin x 1 cos x sin x 1 cos x 1 1 cos x 1 cos x 1 cos x 2 1 cos x 1 sin x 1 2 1 1 cos x 2 1 cos2x 2 1 cos x sin x 1 sin2x 1 cos x sin x 2 Pythagorean identity ■ NOTE If a denominator 1 cos x, is of the form 1 cos x multiplying by 1 cos2x sin2x. gives Similarly, if a denominator is of the form multiplying by gives Compare this with earlier techniques used to rationalize denominators and simplify numbers with complex denominators. 1 sin x, 1 sin x 1 sin2x cos2x. Section 9.1 Identities and Proofs 577 Alternate Solution The numerators of the given equation, look similar to the Pythagorean identity—except the squares are missing. So begin with the left side and introduce some squares by multiplying it by sin x sin x 1. and sin x 1 cos x, sin x 1 cos x sin x sin x sin x 1 cos x sin x˛1 sin2x 1 cos x 2 1 cos2x 1 cos x sin x˛1 1 cos x 1 sin x˛1 1 cos x sin x 2 1 cos x 2 ˛1 1 cos x 2 Pythagorean identity 2 a2 b2 a b a b 2 21 1 Example 6 Dealing with Each Side Separately Prove that csc x cot x sin x 1 cos x . Solution Begin with the left side. csc x cot x 1 sin x 1 cos x sin x cos x sin x ■ [1] Example 5 shows that the right side of the identity to be proved can also be transformed into this same expression. sin x 1 cos x 1 cos x sin x Combining the equalities [1] and [2] proves the identity. csc x cot x 1 cos x sin x sin x 1 cos x [2] ■ Proving identities involving fractions can sometimes be quite complicated. It often helps to approach a fractional identity indirectly, as in the following example. 578 Chapter 9 Trigonometric Identities and Proof Example 7 Proving Identities that Involve Fractions Prove the first identity below, then use the first identity to prove the second identity. a. sec x˛1 sec x cos x tan2x 2 b. sec x tan x tan x sec x cos x Solution a. Begin by transforming the left side. sec x˛1 sec x cos x 2 cos x ˛cos x sec2x sec x cos x sec2x 1 sec2x 1 tan2x tan2x. reciprocal identity Pythagorean identity Therefore, sec x˛1 b. By part a, sec x cos x 2 sec x cos x sec x˛1 Divide both sides of this equation by sec x˛1 tan x˛1 sec x˛1 tan x˛1 sec x cos x 2 sec x cos x 2 sec x cos x 2 sec x cos x 2 sec x tan x . tan2x 2 sec x cos x tan x˛1 tan2x sec x cos x tan
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x tan x sec x cos x tan x˛1 2 tan x˛1 tan x sec x cos x 2 2 ■ Look carefully at how identity b was proved in Example 7. First prove AD BC B tan x, identity a, which is of the form C tan x, D sec x cos x Then divide both sides by BD, that . A sec x, (with and sec x cos x 2 is, by tan x˛1 to conclude that , 2 A B C D . This property pro- vides a useful strategy for dealing with identities involving fractions. Example 8 If AD BC, with B 0 and D 0, then A B C D . Prove that cot x 1 cot x 1 1 tan x 1 tan x . Solution Use the same strategy used in Example 7. First prove A cot x 1, B cot x 1, and D 1 tan x. AD BC, C 1 tan x, AD BC cot x 1 1 2 ˛1 1 tan x cot x 1 1 tan x 2 2 ˛1 2 1 with [3] Section 9.1 Identities and Proofs 579 Multiply out the left side of [3]. cot x 1 1 2 ˛1 1 tan x 2 cot x cot x tan x 1 tan x cot x 1 tan x cot x 1 1 tan x cot x tan x. tan x 1 tan x Similarly, multiply the right side of [3]. cot x 1 1 2 ˛1 1 tan x 2 cot x cot x tan x 1 tan x cot x 1 1 tan x cot x tan x. Because the left and right sides are equal to the same expression, [3] has been proven to be an identity. Therefore, conclude that cot x 1 cot x 1 1 tan x 1 tan x is also an identity. CAUTION ■ C D and cross multiply to eliminate the fractions. If you did Strategy 5 does not say that you begin with a fractional equation A B that, you would be assuming that the statement was true, which is what has to be proved. What the strategy says is that to prove an identity involving fractions, you need only prove a different identity that does not involve fractions. In other words, if you prove that then you can conclude whenever D 0, B 0 and that Note that you do not assume that AD BC; you use AD BC C A B D . some other strategy to prove this statement. It takes a good deal of practice, as well as much trial and error, to become proficient at proving identities. The more practice you have, the easier it will become. Because there are many correct methods, your proofs may be quite different from those of your classmates, instructor, or text answers. If you do not see what to do immediately, try something and see where it leads: multiply out, factor, or multiply numerator and denominator by the same nonzero quantity. Even if this does not lead anywhere, it may give you some ideas on other strategies to try. When you do obtain a proof, check to see if it can be done more efficiently. Do not include the “side trips” in your final proof—they may have given you some ideas, but they are not part of the proof. 580 Chapter 9 Trigonometric Identities and Proof Exercises 9.1 In Exercises 1–4, test the equation graphically to determine whether it might be an identity. You need not prove those equations that appear to be identities. 1. sec x cos x sec x sin2x 2. tan x cot x sin x cos x 3. 4. 1 cos 2x 2 sin2x tan x cot x csc x sec x In Exercises 5–8, insert one of a–f on the right of the equal sign so that the resulting equation appears to be an identity when you test it graphically. You need not prove the identity. a. cos x b. sec x c. sin2x d. sec2 x e. sin x cos x f. 1 sin x cos x 5. csc x tan x ____ 6. 7. sin x tan x ____ sin4x cos4 ˛x sin x cos x ____ 8. tan2 x 2 1 sin x 1 sin x ____ 2 In Exercises 9–18, prove the identity. 9. tan x cos x sin x 10. cot x sin x cos x 11. cos x sec x 1 12. sin x csc x 1 13. tan x csc x sec x 14. sec x cot x csc x 15. 17. 18. tan x sec x sin x 16. cot x csc x cos x 1 cos x csc x 1 1 cos x 2 1 csc x 1 sin2x 2 cot2x 2 2 1 1 1 In Exercises 19–48, state whether or not the equation is an identity. If it is an identity, prove it. 19. sin x 21 cos2x 20. cot x csc x sec x 21. sin cos x 2 x 2 1 1 tan x 22. tan x 2sec2x 1 23. cot x 2 1 cot x 24. 25. sec x sec x 1 2 1 sec2x tan2x 26. sec4 ˛x tan4 ˛x 1 2 tan2x 27. sec2x csc2x tan2x cot2x 28. sec2x csc2x sec2x csc2x sin2x 1 cos2x cot x 1 2 sec x 1 2 2 cos2x tan x 1 2 2 1 1 cos x 2 2 1 2 sin2x tan2x sin2x tan2x 1 29. 30. 31. 33. 34. 35. 36. 32. cot2x 1 csc2x tan2x tan2x 1 2 1 csc x sin x 2 cos2x 1 1 2 1 cos2x 1 tan x sec x csc x cos sin x 1 x 1 2 2 cot x 37. cos4x sin4x cos2x sin2x 38. cot2x cos2x cos2x cot2x 39. 40. 41. 42. 43. 44. sin x cos x 2 sin2x cos2x 1 1 1 tan x 2 2 2 sec2x sec x csc x sin x cos x 2 tan x 1 cos x sin x sin x 1 cos x 2 csc x sec x csc x 1 tan x csc x cot x 1 1 tan x csc x sec x Section 9.2 Addition and Subtraction Identities 581 54. sin x 1 cot x cos x 1 tan x cos x sin x 55. cos x 1 sin x sec x tan x 45. 1 csc x sin x sec x tan x 46. 47. 48. 1 csc x csc x cos2x 1 sin x sin x cos x tan x tan x sin x cos x cot x csc x 1 csc x 1 cot x In Exercises 49–52, half of an identity is given. Graph and this half in a viewing window with write a conjecture as to what the right side of the identity is. Then prove your conjecture. 2P x 2P 49. 50. 51. 52. 1 sin2x has a graph that looks like this? 1 cos x ? Hint: What familiar function 1 cos x cos2x sin x cot x ? sin x cos x sec x csc x cot x 2 ? 1 cos3x 21 1 tan4x sec4x 2 ? 2 1 In Exercises 53–66, prove the identity. 53. 1 sin x sec x cos3x 1 sin x 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 1 sec x tan x sin x csc x cos x cot x cot x cos x cot x cos x cos x cot x cos3x sin3x cos x sin x 1 sin x cos x 2 2 cot x sec x tan x cos x log101 log101 2 csc x cot x log101 log101 log101 log101 log101 log101 tan x tan y tan x tan y 2 sec x tan x 2 2 csc x cot x 2 sec x tan x cot x cot y 1 2 2 tan x tan y cot x cot y tan x tan y cos x sin y cos y sin x cos y sin x cos x sin y tan x tan y cot x cot y tan x tan y 1 1 cot x cot y 9.2 Addition and Subtraction Identities Objectives • Use the addition and subtraction identities for sine, cosine, and tangent functions • Use the cofunction identities Many times, the input, or argument, of the sine or cosine function is the sum or difference of two angles, and you may need to simplify the expression. Be careful not to make this common student error. x p 6 b is not sin x sin p 6 sin a Graphing Exploration Verify graphically that the expressions above do NOT form an iden- tity by graphing Y1 sin x p a 6 b and Y2 sin x sin p 6 . 582 Chapter 9 Trigonometric Identities and Proof The exploration shows that sin x y sin x sin y because it is false when y p 6 . 1 So, is there an identity involving sin 2 x y 1 ? 2 Graphing Exploration Graph Y1 sin x p a 6 b and Y2 23 2 ˛ sin x 1 2 ˛ cos x on the same screen. Do the graphs appear identical? The exploration suggests that sin x p 6 b a 23 2 sin x 1 2 cos x may be an identity. Furthermore, note that the coefficients on the right side can 23 2 be expressed in terms of In other words, 1 2 p 6 the following equation appears to be an identity. x p 6 b sin x cos cos x sin : cos˛ and sin ˛ p 6 p 6 p 6 p 6 sin a . Graphing Exploration 1 Y1 sin x 5 Graph on the same screen. Do the graphs appear identical? What identity does this suggest? Repeat the process with some other number in place of 5. Are the results the same? sin x cos 5 cos x sin 5 and Y2 2 The equations examined in the discussion and exploration above are examples of the first identity listed below. Each identity can be confirmed by assigning a constant value to y and then graphing each side of the equation, as in the Graphing Exploration above. Addition and Subtraction Identities for Sine and Cosine sin(x y) sin x cos y cos x sin y sin(x y) sin x cos y cos x sin y cos(x y) cos x cos y sin x sin y cos(x y) cos x cos y sin x sin y The addition and subtraction identities are important trigonometric identities. You should become familiar with the examples and special cases that follow. NOTE In order to use addition or subtraction identities to find exact values, first write the argument as a sum or difference of two terms for which exact values are known, such as , and p. Section 9.2 Addition and Subtraction Identities 583 Example 1 Addition Identities Use the addition identities to find the exact values of sin 5p 12 and cos 5p 12 . Solution Because x p 6 5p 12 2p 12 y p 4 . and 3p 12 p 6 p 4 , apply the addition identities with sin 5p 12 sin p a 6 22 2 p 4 b 23 2 p 6 22 2 1 2 p 4 22 sin cos cos p 6 sin p 4 A 1 23 4 B cos sin p 4 p 4 22 sin p 6 23 1 4 A B cos 5p 12 cos a 23 2 p 4 b p 6 22 2 1 2 cos p 6 22 2 Example 2 Subtraction Identity Find sin p y . 2 1 Solution Apply the subtraction identity for the sine function with 1 sin p cos y cos p sin y 0 cos y p y sin x p. sin y sin y 1 2 1 2 ■ ■ Example 3 Addition Identity Prove that cos x cos y 1 2 3 cos 1 x y 2 cos x y 1 . 2 4 Solution Begin with the more complicated right side and use the addition and subtraction identities for cosine to transform it into the left side. 1 2 3 cos x y cos x y 1 2 4 2 1 1 cos x cos y sin x sin y 2 2 3 1 cos x cos y sin x sin y 2 4 cos x cos y cos x cos y 1 2 1 cos x cos y 2 2 cos x cos y 2 ■ 1 1 2 1 584 Chapter 9 Trigonometric Identities and Proof NOTE Recall that the difference quotient of a function f is . Simplifying the Difference Quotient of a Trigonometric Function The difference quotient is very important in calculus, and the addition identities are needed to simplify difference quotients of trigonometric functions. Example 4 The Difference Quotient of f(x) sin x Show that for the function Solution sin x and any number h 0, x 2 1 sin x cos h 1 h a b cos x sin h a h b . Use the addition identity for sin sin y h. with sin sin x cos h cos x sin h sin x h cos x sin h sin x 1 cos h 1 2 h cos h 1 h a sin x cos x b sin h a h b ■ Addition and Subtraction Identities for the Tangent Function The addition and subtraction identities for sine and cosine can be used to obtain the addition and subtraction identities for the tangent function. Addition and Subtraction Identities for Tangent tan(x y) tan(x y) tan x tan y 1 tan x tan y tan x tan y 1 tan x tan y A proof of these identities is outlined in Exercise 36. Example 5 Addition and Subtraction Identities for Tangent Find the exact values of such that 0 6 x 6 p 2 , and x y sin 2 1 p 6 y 6 3p 2 , x y tan 1 sin x 3 4 , 2 and if x and y are numbers cos y 1 3 . Deter- NOTE See Figure 6.4-
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7 for the signs of the functions in each quadrant. y ( 7, 3) x x 4 3 x 42 − 32 = 7 sin x = 3 4 Figure 9.2-1a y 1 y 3 32 − 12 = 8 = 2 2 (−1, −2 2) cos y = − 1 3 Figure 9.2-1b Section 9.2 Addition and Subtraction Identities 585 mine in which of the following intervals x y lies: p 2 b , 0, a 3p 2 b , or 3p 2 a , 2p . b Solution Use the Pythagorean identity and the fact that cos x and tan x are positive in the first quadrant to obtain the following. See Figure 9.2-1a. 27 4 cos x 21 sin2x 1 9 16 3 4 b 7 16 1 B B B a 2 tan x sin x cos x 3 4 27 4 3 27 327 7 Because y lies between p and 3p 2 , its sine is negative. See Figure 9.2-1b. sin y 21 cos2y 28 3 222 3 tan y sin y cos y 222 3 1 3 222 3 3 1 222 The addition identities for sine and tangent show exact values. x y sin 1 x y tan 1 2 2 sin x cos y cos x sin y 222 1 27 3 3 3 4 4 tan x tan y 1 tan x tan y 3 12 2214 12 3 2214 12 222 327 7 327 222 327 1422 7 7 6214 7 B 1 7 bA a 327 1422 7 6214 2727 3222 65 must be in the interval Both the sine and tangent of 3p 2 intervals in which both sine and tangent are negative. , 2p b a x y are negative numbers. Therefore x y because it is the only one of the four ■ Cofunction Identities Special cases of the addition and subtraction identities are the cofunction identities. 586 Chapter 9 Trigonometric Identities and Proof Cofunction Identities sin x cos P 2 a x b tan x cot P 2 a x b sec x csc P 2 a x b cos x sin cot x tan csc x sec P 2 a x b The first cofunction identity is proved by using the identity for cos with p 2 in place of x and x in place of y. x y 2 1 cos p 2 a x b cos p 2 cos x sin p 2 sin x 0 cos x 1 sin x sin x Because the first cofunction identity is valid for every number x, it is also valid with the number p 2 x in place of x. sin p 2 a x b cos p 2 c p 2 a x b d cos x Thus, the second cofunction identity is proved. The others now follow from these previous two. For instance, Also, tan p 2 a x b sin a cos a p 2 p 2 x b x b cos x sin x cot x csc p 2 a x b 1 p 2 sin a x b 1 cos x sec x Example 6 Cofunction Identities cos x p 2 b a cos x tan x. Verify that Solution Beginning with the left side, the term cos x p 2 b a looks almost, but not quite, like the term cos . Therefore, in the cofunction identity. But note that Section 9.2 Addition and Subtraction Identities 587 cos x p a 2 b cos x cos c x p a cos x 2 b d cos x p a 2 cos x b sin x cos x tan x negative angle identity with x p 2 in place of x cofunction identity quotient identity ■ Exercises 9.2 In Exercises 1–12, find the exact value. 1. sin 4. sin 7. tan p 12 5p 12 7p 12 2. cos 5. cot p 12 5p 12 8. cos 11p 12 3. tan 6. cos p 12 7p 12 9. cot 11p 12 10. sin 75° Hint: 75° 45° 30°. 11. sin 105° 12. cos 165° 25. If sin x 1 3 and 0 6 x 6 p 2 , then sin p 4 a x b ? 26. If cos x 1 4 and p 2 6 x 6 p, then cos 27. If cos x 1 5 and p 6 x 6 3p 2 , then sin ? 28. If sin x 3 4 and 3p 2 6 x 6 2p, then cos p 4 a x b ? In Exercises 13–18, rewrite the given expression in terms of cos x. sin x and 13. sin p 2 a x b 15. cos x 3p 2 b a 17. sec x p 1 2 14. cos x p 2 b a 16. csc x p a 2 b 18. cot 1 x p 2 In Exercises 19–24, simplify the given expression. 19. sin 3 cos 5 cos 3 sin 5 20. sin 37° sin 53° cos 37° cos 53° x y x y 2 sin y 2 sin y 21. cos 1 sin 22. 23. 1 cos 1 24. sin 1 x y x y x y x y cos y sin 1 2 2 cos y cos 1 2 cos 1 x y 2 sin 1 x y 2 2 In Exercises 29–34, assume that sin y 20.75 and that x and y lie between 0 and sin x 0.8 and P 2 . Evaluate the given expressions. x y x y 2 2 30. sin 32. sin 34. tan 1 29. cos 31. cos 33. tan 1 1 1 35. If f x 1 2 x y 2 cos x prove that the difference quotient is and h is a fixed nonzero number cos x cos h 1 h b a sin x sin h a h b . 36. Prove the addition and subtraction identities for the tangent function. Hint: tan 1 x y 2 sin 1 cos 1 x y 2 x y 2 Use the addition identities on the numerator and denominator; then divide both numerator and denominator by , and simplify. cos x cos y x y 1 x y 1 cos 2 sin 2 1 x y 1 x y cos2 ˛x cos2 ˛y sin2 ˛x sin2 ˛y 2 sin2x cos2y cos2x sin2y 2 56. sin In Exercises 57–66, determine graphically whether the equation could not possibly be an identity (by choosing a numerical value for y and graphing both sides), or write a proof that it is. 588 Chapter 9 Trigonometric Identities and Proof 37. If x is in the first quadrant and y is in the second quadrant, sin x 24 25 x y value of and tan sin 2 x y quadrant in which and , 1 1 lies. sin y 4 5 x y and the , 2 find the exact 50. cos 51. tan x p x p 2 2 1 1 cos x tan x 38. If x and y are in the second quadrant, and , cos y 3 4 x y and lies. , 2 x y cos 1 which find the exact value of tan x y 2 1 and the quadrant in , sin x 1 3 x y sin 1 52. sin x cos y 1 2 3 sin 53. sin x sin y 1 2 3 cos 54. cos x sin y 1 2 3 sin sin 1 x y 2 4 cos x y 1 2 4 sin 1 x y 2 4 , 2 39. If x is in the first quadrant and y is in the second 55. cos quadrant, sin x 4 and , 5 x y 2 x y exact value of 1 quadrant in which cos cos y 12 , 13 x y tan and lies. find the and the 2 40. If x is in the fourth quadrant and y is in the first quadrant, cos x 1 3 x y and value of sin 2 x y quadrant in which , 1 and tan 1 lies. cos y 2 3 x y and the , find the exact 41. Express sin u v w 1 cosines of u, v, and w. Hint: First apply the addition identity with y w. and 2 x u v in terms of sines and 1 2 42. Express x y z cos cosines of x, y, and z. 1 in terms of sines and 2 60. cos 43. If x y p 2 , show that sin2x sin2y 1. 44. Prove that cot x y 1 cot x cot y 1 cot x cot y . 2 In Exercises 45–56, prove the identity. 61. 62. 63. 45. sin 1 46. cos 47. cos 1 1 48. tan 49. sin sin x cos x cos x tan x 2 2 2 sin x 2 57. 58. x y cos 2 sin x cos y 1 x y cos 2 sin x cos y 1 59. sin 1 x y 2 x y 2 cot x tan y cot x tan y sin x sin y cos x cos y 1 1 1 1 1 sin sin sin sin cos cos 1 1 x y x y 2 2 tan x tan y tan x tan cot y cot x cot y cot x cot x tan y cot x tan y 64. cos cos 65. tan 66. cot cot y tan x cot y tan x 2 2 tan x tan y 2 cot x cot y 2 Section 9.2.A Excursion: Lines and Angles 589 9.2.A Excursion: Lines and Angles Objectives • Find the angle of inclination of a line with a given slope • Find the angle between two lines Several interesting concepts dealing with lines are defined in terms of trigonometry. They lead to useful facts whose proofs are based on the addition and subtraction identities for sine, cosine, and tangent. If L is a nonhorizontal straight line, the angle of inclination of L is the formed by the part of L above the x-axis and the x-axis, positive angle as shown in Figure 9.2.A-1 Figure 9.2.A-1 The angle of inclination of a horizontal line is defined to be Thus, the radian measure of the angle of inclination of any line satisfies 0 u 6 p. Furthermore, u 0. Angle of Inclination Theorem If L is a nonvertical line with angle of inclination tan U slope of L. U, then Proof If L is horizontal, then L has slope 0 and angle of inclination tan u tan˛ 0 0, tan u slope L 0. so of u 0. Hence, If L is not horizontal, then it intersects the x-axis at some point shown for two possible cases in Figure 9.2.A-2. 1 x1, 0 , as 2 (x2, y2) (x2, y2) L y2 y2 L (x1, 0) θ x θθπ − x x2 − x1 Figure 9.2.A-2a x1 − x2 Figure 9.2.A-2b (x1, 0) 590 Chapter 9 Trigonometric Identities and Proof The right triangle in Figure 9.2.A-2a shows that slope of L y2 x2 0 x1 y2 x2 x1 opposite adjacent tan u. The right triangle in Figure 9.2.A-2b shows that slope of L 0 y2 x2 x1 y2 x1 x2 opposite adjacent tan p u . 2 1 [1] Use the fact that the tangent function has period and the negative angle identity for tangent to obtain tan tan p u u p tan u. 2 L tan ˛u shows that slope of tan u 2 1 1 2 Combining this fact with also. 1 1 3 4 in this case ■ Example 1 Angle of Inclination Find the angle of inclination of a line of slope 5 3 . Solution By the Angle of Inclination Theorem, culator shows that u 1.0304 radians, or . The TAN1 tan u 5 3 u 59.04° . key on a cal- ■ Example 2 Angle of Inclination Find the angle of inclination of a line L with slope 2. Solution Because line L has slope tan u 2 that lies between 1.1071. solution p, another solution is needed. tan˛ t tan Recall that 2, p 2 its angle of inclination is a solution of and A calculator gives the approximate p. Because an angle of inclination must be between 0 and 1 t p , for every t. So u 1.1071 p 2.0344 2 is the solution of angle of inclination is approximately 2.03 radians, or about in the interval from 0 to tan u 2 p. Therefore, the 116.57°. ■ θπ − θ θ θπ − Figure 9.2.A-3 Angles Between Two Lines If two lines intersect, then they determine four angles with vertices at the point of intersection, as shown in Figure 9.2.A-3. If one of these angles Section 9.2.A Excursion: Lines and Angles 591 u measures p u vertical angle theorem from plane geometry. radians, then each of the two angles adjacent to it measures radians by the radians. (Why?) The fourth angle also measures u The angles formed by intersecting lines can be determined from the angles a of inclination of the lines. Suppose L and M have angles of inclination Basic facts about parallel lines, as and respectively, such that b a is one angle between L and illustrated in Figure 9.2.A-4, show that M, and is the other. b a. b a p bβ − α) L α M β x Figure 9.2.A-4 The angle between two lines can also be found from their slopes by using the following fact. Angle Between Two Lines If two nonvertical, nonperpendicular lines have slopes m and k, then one angle between them satisfies U tan˛ U m k 1 mk ` ` . Proof Suppose L has slope k and angle of inclination and that M has slope m and angle of inclination a b. m k 1 mk By the definition of absolute value m k m k 1 mk 1 mk ` or ` m k 1 mk ` ` whichever is positive. It will be shown that one angle between L and M has tangent , and that the other has tangent . Thus, one m k 1 mk m k 1 mk of them necessarily has tangent . If b a, then b a is one m k 1 mk ` ` angle between L and M. By the subtraction identity for tangent, tan 1 b a 2 tan b tan a 1 tan b tan a p m k 1 mk b a The other angle between L
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and M is 2 negative angle identity, and the addition identity 1 and by periodicity, the 592 Chapter 9 Trigonometric Identities and Proof p tan tan 1 3 b a tan 1 tan b tan a 1 tan b tan a m k 1 mk . This completes the proof when similar. b a. The proof in the case a b is ■ Example 3 The Angle Between Two Lines If the slopes of lines L and M are 8 and angle between them. 3 , respectively, then find one Solution Substitute 8 for m and 3 ` tan u for k to find the tangent values. 3 2 3 2 1 11 23 8 1 1 8 11 23 ` ` u 0.4461 radians, or ` Solving the equation yields 25.56°. ■ Exercises 9.2.A In Exercises 1–6, find the angle of inclination of the straight line through the given points. 1. 1 1, 2 2 , (3, 5) 3. (1, 4), (6, 0) 5. 1 3, 7 2 , (3, 5) 2. (0, 4), 4. (4, 2), 6. (0, 0), 5, 1 2 3, 2 4, 5 1 1 1 2 2 In Exercises 7–12, find one of the angles between the straight lines L and M. 7. L has slope 3 2 and M has slope 1. 8. L has slope 1 and M has slope 3. 9. L has slope 10. L has slope 1 2 and M has slope 0. and M has slope 3. 11. (3, 2) and (5, 6) are on L; (0, 3) and (4, 0) are on M. 12. 1, 2 1 on M. 2 and 1 3, 3 2 are on L; 3, 3 2 1 and (6, 1) are Section 9.3 Other Identities 593 9.3 Other Identities Objectives • Use the following identities: double-angle power-reducing half-angle product-to-sum sum-to-product Double-Angle Identities A variety of identities that are special cases of the addition and subtraction identities of Section 9.2 are presented in this section. These identities include double-angle identities, power-reducing identities, half-angle identities, product-to-sum identities, and sum-to-product identities. Double-Angle Identities Special cases of the addition identities occur when two angles have the same measure. These identities are called the double-angle identities. sin ˛2x 2 sin x cos x cos˛ 2x cos2x sin2x tan 2x 2 tan x 1 tan2x Proof Substitute x for y in the addition identities. sin 2x sin 1 cos 2x cos 1 x x 2 x x 2 tan 2x tan x x 2 1 sin˛ x cos ˛x cos ˛x sin ˛x 2 sin˛ x ˛cos x cos ˛x ˛cos ˛x sin ˛x ˛sin˛ x cos2x sin2 tan ˛x tan ˛x 1 tan ˛x tan ˛x 2 tan x 1 tan2 ˛x ˛x Example 1 Use Double-Angle Identities If p 6 x 6 3p 2 and cos ˛x 8 17 , find sin 2x and cos 2x, and show that 5p 2 6 2x 6 3p. Solution In order to use the double-angle identities, first determine sin x, which can be found by using the Pythagorean identity. sin2 ˛x 1 cos2 ˛x 1 8 2 1 64 289 225 289 Thus, sin x ± B rant, and sin x is negative there. . Since 225 289 a 17 b p 6 x 6 3p 2 , x must be in the third quad- sin ˛x B 225 289 15 17 y x 8 x 15 17 (−8, −15) Figure 9.3-1 594 Chapter 9 Trigonometric Identities and Proof Now substitute these values in the double-angle identities to find sin 2x and cos 2x. sin ˛2x 2˛ sin˛ x cos ˛x 2 15 a 17 ba 8 17 b 240 289 0.8304 cos 2x cos2x sin2x 8 a 17 b 2 2 15 a 17 b 64 289 225 289 161 289 0.5571 You know that x lies in the third quadrant. Multiply the inequality p 6 x 6 3p 2 first or second quadrant. The calculations above show that at 2x, sine is positive and cosine is negative. This can occur only if 2x lies in the sec- That is, 2x is in either the 2p 6 2x 6 3p. by 2 to find that ond quadrant, so 5p 2 6 2x 6 3p. ■ Example 2 Use Double-Angle Identities Express the rule of the function and constants. f x 1 2 sin 3x in terms of powers of sin x Solution First use the addition identity for f x 1 2 sin 3x sin x 2x 1 2 2 sin x cos x 2 1 identity for sin 2x 1 2 sin with y 2x. x y sin x cos 2x cos x sin 2x sin x˛1 cos x cos2x sin2x 2 identity for cos 2x sin x cos2x sin3x 2 sin x cos2x 3 sin x cos2x sin3x 3 sin x sin3x 2 1 1 sin2x Pythagorean identity 3 sin x 3 sin3x sin3x 3 sin x 4 sin3x ■ Forms of cos 2x The double-angle identity for can be rewritten in several useful ways. For instance, we can use the Pythagorean identity in the form of cos2x 1 sin2x cos 2x cos 2x cos2x sin2x to obtain the following. 1 sin2x 1 2 sin2x 1 2 sin2x sin2x 1 cos2x to Similarly, use the Pythagorean identity in the form obtain the following. cos 2x cos2x sin2x cos2x 1 cos2x 2 1 2 cos2x 1 Forms of cos 2x cos 2x 1 2 sin2x cos 2x 2 cos2x 1 Section 9.3 Other Identities 595 Example 3 Use Forms of cos 2x Prove that 1 cos 2x sin 2x tan x. Solution Use the first identity in the preceding box and the double-angle identity for sine. 1 cos 2x sin 2x 1 1 2 sin2x 2 sin x cos x 1 2 2 sin2x 2 sin x cos x sin x cos x tan x ■ Power-Reducing Identities If the first equation in the preceding box is solved for ond one for new forms are called the power-reducing identities. and the secalternate forms for these identities are obtained. The cos2 x, sin2x Power-Reducing Identities sin2x 1 cos 2x 2 cos2x 1 cos 2x 2 Example 4 Use Power-Reducing Identities Express the function of cosine functions. f x 1 2 Solution sin4 x in terms of constants and first powers Begin by applying the power-reducing identity. f x 1 2 sin4x sin2x sin2x 1 cos 2 x 2 1 2 cos 2 x cos2 2 x 4 1 cos 2x 2 596 Chapter 9 Trigonometric Identities and Proof NOTE To write cos2 2x in terms of first powers of cosine functions, use 2x in place of x in the powerreducing identity for cosine. 2x cos2 2x 1 cos 2 1 2 1 cos 4x 2 2 Half-Angle Identities Next apply the power-reducing identity for cosine to (See Note.) 1 2 cos 2 x cos22 x 4 2 cos22x. 1 2 cos 2 x 1 cos 4x 4 cos 2x 1 8 ˛ cos 2x 1 8 1 2 1 2 cos 4x 1 1 4 3 8 1 cos 4 x 2 ■ Half-Angle Identities The power-reducing identities with the half-angle identities. x 2 in place of x can be used to obtain 1 cos2 x 2 b a 2 ± 1 cos x 2 B sin2 sin x 2 b a x 2 b a 1 cos x 2 This proves the first of the half-angle identities. sin x 2 1 cos x 2 B cos x 2 ±± 1 cos x 2 B tan x 2 ±± 1 cos x 1 cos x B The sign in front of the radical depends upon the quadrant in which x 2 lies. The half-angle identity for cosine is derived from a power-reducing identity, as was the half-angle identity for sine. The half-angle identity for tangent then follows immediately since tan x 2 b a sin x 2 b a cos x 2 b a . Example 5 Half-Angle Identities Find the exact value of a. cos 5p 8 b. sin p 12 Section 9.3 Other Identities 597 use the half-angle identity with x 5p 4 and Solution a. Because 5p 8 1 2 a , 5p 4 b 22 2 5p 4 5p 8 the fact that cos . The sign chart given in Section 6.4 shows that cos is negative because 5p 8 is in the second quadrant. So, use the negative sign in front of the radical. cos 5p 8 cos 5p 4 2 R 1 cos 5p 4 b a R R 2 12 2 b 1 a 2 2 12 A B 2 2 2 12 4 B 22 12 2 p 12 1 2 a p 6 b and p 12 b. Because positive, is in the first quadrant, where sine is sin p 12 sin p 6 2 R S 1 cos p 6 b a 2 1 23 2 2 2 23 2 2 2 23 4 S C 32 23 2 ■ The problem of determining signs in the half-angle formulas can be eliminated for the tangent by using the following formulas. 598 Chapter 9 Trigonometric Identities and Proof Half-Angle Identities for Tangent tan tan x 2 x 2 1 cos x sin x sin x 1 cos x Proof The proof of the first of these identities follows from the identity tan x 1 cos 2 x this identity. which was proved in Example 3. Replace x by sin 2x in x 2 , tan x 2 1 cos 2 x 2 b a sin 2 x 2 b a 1 cos x sin x The second identity in the box is proved in Exercise 71. Example 6 Use Half-Angle Identity for Tangent If tan x 3 2 and p 6 x 6 3p 2 , find tan˛ x 2 . Solution The terminal side of an angle of x radians in standard position lies in the third quadrant, as shown in Figure 9.3-2. The tangent of the angle in standard position whose terminal side passes through the point is 3 2 Because there is only one angle in the third quadrant with tan- 2, 3 3 2 . 2 1 gent 3 2 , the point 2, 3 1 2 radians. must lie on the terminal side of the angle of x The distance from triangle. 1 2, 3 2 to the origin is the hypotenuse of the Therefore 213 sin x 3 213 cos x 2 213 By the first of the half-angle identities for tangent, tan ˛ x 2 1 cos x sin x 1 2 213 b a 3 213 213 2 213 3 213 213 2 3 ■ y x 2 x 3 13 (−2, −3) Figure 9.3-2 Section 9.3 Other Identities 599 Product-to-Sum Identities Use the addition and subtraction identities to rewrite x y . 2 sin x cos y cos x sin y sin x cos y cos x sin y 2 sin x cos y sin x y sin 2 1 1 sin 1 x y sin x y 1 2 2 Dividing both sides of the equation by 2 produces the first of the following identities. Product-to-Sum Identities sin x cos y 1 sin x sin y 1 cos x cos y 1 2 cos x sin y 1 2 2 ˛ [sin(x y) sin(x y)] 2 ˛ [cos(x y) cos(x y)] [cos(x y) cos(x y)] [sin(x y) sin(x y)] The proofs of the second and fourth product-to-sum identities are similar to the proof of the first. The third product-to-sum identity was proved in Example 3 of Section 9.2. Sum-to-Product Identities Use the first product-to-sum identity with x y 2 in place of x and ˛ x y 2 ˛ in place of y to obtain the first sum-to-product identity sin ˛ cos sin sin ˛ sin x sin y 2 Multiplying both sides of the equation by 2 produces the identity. Sum-to-Product Identities sin x sin y 2 sin sin x sin y 2 cos cos x cos y 2 cos ˛ cos ˛ sin ˛ cos cos x cos y 2 sin ˛ sin 600 Chapter 9 Trigonometric Identities and Proof The other sum-to-product identities are proved the same way as the first. Example 7 Use Sum-to-Product Identities Prove the identity below. sin t sin 3t cos t cos 3t tan 2t Solution Use the first sum-to-product identity with sin t sin 3t 2 sin t 3t 2 a b ˛ cos a Similarly, x t t 3t 2 b and y 3t . 2 sin 2t cos t 2 1 cos t cos 3t 2 cos ˛a t 3t 2 b ˛ cos a t 3t 2 b 2 cos 2t cos t . 2 1 Therefore, sin t sin 3t cos t cos 3t 2 sin 2t cos 2 cos 2t cos t 2 t 2 1 1 sin 2t cos 2t tan 2t. ■ Exercises 9.3 In Exercises 1–12, use the half-angle identities to evaluate the given expression exactly. In Exercises 23–30, find under the given conditions. sin 2x, cos 2x, and tan 2 x 1. cos 5. tan p 8 p 12 2. tan p 8 6. sin 5p 8 9. sin 7p 8 10. cos 7p 8 11. tan 3. sin 3p 8 7. cos p 12 7p 8 4. cos 8. tan 12. cot 3p 8 5p 8 p 8 In Exercises 13–18, write each expression as a sum or difference. 13. sin 4x cos 6x 14. sin 5x sin 7x 15. cos 2x cos 4x 16. sin 3x cos 5x 17. sin 17x sin 3x 1 2 18. cos 13x cos 5x 2 1 In Exercises 19–2
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2, write each expression as a product. 19. sin 3x sin 5x 20. cos 2x cos 6x 21. sin 9x sin 5x 22. cos 5x cos 7x 23. sin x 5 13 , for 0 6 x 6 p 2 24. sin x 4 5 , for p 6 x 6 3p 2 25. cos x 3 5 , for p 6 x 6 3p 2 26. cos x 1 3 , for p 2 6 x 6 p 27. tan x 3 4 , for p 6 x 6 3p 2 28. tan x 3 2 , for p 2 6 x 6 p 29. csc x 4, for 0 6 x 6 p 2 30. sec x 5, for p 6 x 6 3p 2 5910ac09_570-613 9/21/05 2:30 PM Page 601 In Exercises 31–36, find sin the given conditions. x 2 , cos x 2 , and tan x 2 under 31. cos x 0.4, for 0 6 x 6 p 2 32. sin x 0.6, for p 2 6 x 6 p 33. sin x 3 5 , for 3p 2 6 x 6 2p 34. cos x 0.8, for 3p 2 6 x 6 2p 35. tan x 1 2 , for p 6 x 6 3p 2 36. cot x 1, for p 6 x 6 p 2 In Exercises 37–42, assume sin x 0.6 and evaluate the given expression. Section 9.3 Other Identities 601 53. cos4x sin4x cos 2x 54. sec 2x 1 1 2 sin2x 55. cos 4x 2 cos 2x 1 56. sin2x cos2x 2 sin x 57. 1 cos 2x sin 2x cot x 58. sin 2x 2 cot x csc2x 59. sin 3x sin x 1 2 ˛1 60. 61. 4 cos x sin x sin 4x 1 cos 2x 2 tan x sec2x 2 ˛1 3 4 sin2x 2 1 2 sin2x 2 and 0 66 x 66 P 2 62. cos 3x cos x 1 2 ˛1 3 4 cos2x 2 37. sin 2x 38. cos 4x 39. cos 2x 40. sin 4x 41. sin x 2 42. cos x 2 63. csc2 ˛a x 2 b 2 1 cos x 64. sec2 x 2 b ˛a 2 1 cos x 43. Express cos 3x in terms of cos x. In Exercises 65–70, prove the identity. 44. a. Express the function f cos3x in terms of x 1 2 constants and first powers of the cosine function, as in Example 4. b. Do the same for f x 2 1 cos4x. In Exercises 45–50, simplify the given expression. 45. sin 2x 2 sin x 47. 2 cos 2y sin 2y 46. 1 2 sin2 ˛a x 2 b 48. 49. 50. cos2 ˛a x 2 b sin2 ˛a x 2 b sin x cos x 2 sin 2x 2 1 2 sin x cos3x 2 sin3x cos x In Exercises 51–64, determine graphically whether the equation could not possibly be an identity, or write a proof showing that it is. 51. sin 16x 2 sin 8x cos 8x 52. cos 8x cos24x sin24x 65. 66. 67. 68. 69. 70. sin x sin 3x cos x cos 3x tan x sin x sin 3x cos x cos 3x cot 2x sin 4x sin 6x cos 4x cos 6x cot x cos 8x cos 4x cos 8x cos 4x cot 6x cot 2x sin x sin y cos x cos y cot x y 2 b a sin x sin y cos x cos y tan x y 2 b a 71. a. Prove that 1 cos x sin x sin x 1 cos x . b. Use part a and the half-angle identity proved in the text to prove that tan x 2 sin x 1 cos x . 72. If x cos 2t and y sin t, find the relation between x and y by eliminating t. 602 Chapter 9 Trigonometric Identities and Proof 73. The horizontal range of a projectile R is given by the equation R vt cos a, where v is the initial velocity of the projectile, t is is the angle between the the time of flight, and t 2v sin a line of fire and the horizontal. If a , g where g is acceleration due to gravity, show that R v2 sin 2a . g 74. The expression sin p 2 of reflection of light waves. Show that this expression can be written as 1 2 sin2u. 2u a b occurs in the theory 75. The expression 1 2 the theory of the motion of a pendulum. Show that this equation can be written as a sin2 sin2 a cos u cos a. is used in 1 2 u 2 b 76. A batter hits a baseball that is caught by a fielder. If the ball leaves the bat at an angle of radians to the horizontal, with an initial velocity of v feet per second, then the approximate horizontal distance d traveled by the ball is given by u d v2 sin u cos u 16 . a. If the initial velocity is 90 ft/sec, find the horizontal distance traveled by the ball when u 0.5 radian. b. Use an identity to show that d v2 sin 2u radian and when u 0.75 . 32 9.4 Using Trigonometric Identities Objectives • Use identities to solve trigonometric equations Recall that the basic identities are used to simplify expressions and to algebraically solve trigonometric equations. The trigonometric identities introduced in Section 9.3 can also be used with the techniques shown in Section 8.3, where equations were rewritten into a basic form and then solved. Example 1 Use Double-Angle Identities Solve 5 cos x 3 cos 2x 3. Solution cos2x. 5 cos x 3 Use a double-angle identity to rewrite cos 2x in terms of 5 cos x 3 cos 2x 3 2 cos2x 1 3 5 cos x 6 cos2x 3 3 6 cos2x 5 cos x 6 0 0 or 2 1 2 cos x 3 0 3 cos x 2 2 cos x 3 2 1 2 1 double-angle identity factor the quadratic expression cos x 3 2 3 cos x 2 0 cos x 2 3 Section 9.4 Using Trigonometric Identities 603 The equation cos x 3 2 has no solutions because cos x always lies between 1 and 1. A calculator shows that the solutions of cos x 2 3 are x 0.8411 2kp and x 0.8411 2kp for any integer k. ■ Example 2 Use Double-Angle Identities Solve the equation sin x cos x 1. Solution Use the double-angle identity to rewrite sin x cos x. 2 sin x cos x sin 2x sin x cos x 1 2 sin 2x Replace sin x cos x with 1 2 sin 2x and multiply both sides by 2. 1 2 sin 2x 1 sin 2x 2 Because the sine of any number must be between and 1, there is no solution to the last equation. Therefore, there is no solution to the original equation. 1 ■ Example 3 Use Double-Angle Identities Find exact solutions of cos2x sin2x 1 2 Solution Because cos2x sin2x cos 2x, the equation can be rewritten cos 2x 1 2 2x cos 2x p 3 x p 6 2pk or pk 1 1 2 2x 5p 3 x 5p 6 2pk pk for any integer k. ■ 604 Chapter 9 Trigonometric Identities and Proof Example 4 Use Addition Identities Find the exact solutions of sin 2x cos x cos 2x sin x 1. Solution The left side of the equation is similar to the right side of the addition identity for sine. sin 1 x y 2 sin x cos y cos x sin y. Substitute 2x for x and x for y. sin 2x cos x cos 2x sin x sin 1 2x x 1 2 sin 3x 1 For any integer k, 3x sin 3x p 2 x p 6 11 2pk 2pk 3 Example 5 Use Half-Angle Identities Find the solutions of sin x sin , where 0 x 2p. x 2 Solution CAUTION Squaring both sides of an equation may introduce extraneous solutions. Be sure to check all solutions in the original equation. 1 sin x sin x 2 1 cos x sin x ± B 2 sin2x 1 cos x 1 cos2x 1 cos x 2 2 cos2x 1 cos x 2 2 half-angle identity square both sides Pythagorean identity 2 cos2x cos x 1 0 0 cos x 1 2 cos x 1 2 1 2 cos x 1 0 2 or cos x 1 0 cos x 1 cos x 1 2 For any integer k, x 2p 3 2pk x 4p 3 2pk x 0 2pk For 0 x 2p, x 0, 2p 3 , or 2p. ■ ■ Section 9.4 Using Trigonometric Identities 605 Solving a sin x b cos x c (Optional) Equations of the form a sin x b cos x c occur often. For the case when c 0, the equations can be rewritten as a sin x b cos x sin x b a cos x tan x b a The last equation can be solved by methods discussed in Section 8.3. For the case when solutions to c 0, a very different approach is needed to find the The procedure involves rewriting the equation as a sin x b cos x c 2 is the angle whose terminal side contains the point (a, b), and where then using the addition identity for the sine function. a 1 sin x a k, a To find and b, where a lies on the positive x-axis and tex at the origin. , construct a right triangle in the coordinate plane with sides a is the angle with its ver- a Begin by writing the equation so that the coefficient of sin x, a, is positive. The position of the point (a, b) depends on whether b is positive or negative. If b is positive, the point is in the first quadrant; if b is negative, the point is in the fourth quadrant. Both possibilities are shown in Figures 9.4-1 and 9.4-2. a 2a2 b2 2a2 b2 Divide each side of the original equation by to obtain a sin x a 2a2 b2 sin x b cos x b 2a2 b2 cos x c c 2a2 b2 [2] Substitute the equivalent expressions from [1] into equation [2]. cos a sin x sin a cos x c 2a2 b2 Use the addition identity for sine to rewrite the left side of the equation. sin 1 x a 2 c 2a2 b2 The last equation can be solved by using the methods from Section 8.3. The steps of the procedure are summarized in the following box. Positive b In both cases, sin a b 2a2 b2 and cos a . [1] y y x x (a, b) b a2 + b2 α a Figure 9.4-1 a α a2 + b2 b (a, b) Negative b Figure 9.4-2 606 Chapter 9 Trigonometric Identities and Proof a sin x Solving b cos x c, where c 0 Let a, b, and c be nonzero real numbers. To solve a sin x b cos x c 1. Multiply by 1, if needed, to make a positive. 2. Plot the point (a, b) position that contains A and let be the angle in standard on its terminal side. (a, b) 3. Find • the length of the hypotenuse of the reference triangle • expressions that represent • the measure of A sin A and cos A 4. Divide each side of the equation by 2a2 b2 a 2a2 b2 sin x b 2a2 b2 cos x yielding c 2a2 b2 5. Use the addition identity for sine to rewrite the equation. sin(x A) c 2a2 b2 6. Solve the equation using techniques previously discussed. Example 6 Solve a sin x b cos x c Solve the equation 23 sin x cos x 23. Solution Step 1 Make the coefficient of 1. of the equation by sin x positive by multiplying both sides 23 sin x cos x 23 so a 23 Step 2 Sketch a diagram of the angle a that has nal side. See Figure 9.4-3. A b 1 and 23, 1 on its termi- B Step 3 The length of the hypotenuse is a Find sin and cos 3 A from the figure. 23 2 B 1 1 2 2 2 . a cos a 23 2 and sin a 1 2 Find a . a p 6 or a 11p 6 Step 4 Divide both sides of the equation in Step 1 by the hypotenuse, 23 3 A B 2 1 1 2 2. 2 23 2 sin x 1 2 cos x 23 2 y 1 −1 0 −1 −2 3 α 2 x 1 ( 3, −1) Figure 9.4-3 Section 9.4 Using Trigonometric Identities 607 Step 5 Rewrite the equation by substituting cos 11p 6 for 23 2 and sin 11p 6 for 1 2 and then use the addition identity for sine. CAUTION When substituting sin for , the 11p 6 1 2 sign between the terms changes to . cos 11p 6 sin x sin 11p 6 x 11p 6 b cos x 23 2 23 2 sin a Step 6 Solve the equation. x 11p 6 sin p x 11p 3 6 x p 3 x 3p 2 2pk 11p 6 2pk or 2pk x p 2 2pk for any integer k. Maxima and Minima of 23 1 2 b a x 11p 2p 3 6 x 2p 3 x 7p 6 x 5p 6 2pk 11p 6 2pk 2pk 2pk ■ f(x) a sin x b cos x a sin x b cos x, maximum and miniFor functions of the form mum values can be found by using a technique similar to that described in the algorithm to solve equations of the form a sin x b cos x c. x f 1 2 Example 7 Maximum and Minimum of f(x) a sin x b cos x Find the maximum and minimum of the function 3 sin x 4 cos x. x f 1 2 Solution Note that 232 42 225 5. Let a cos 1 or equivalently a sin 1 4 5 , and write the func
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tion in the form sin a NOTE The value of not needed to find the maximum or minimum of the function. is x f 1 2 5 3 5 sin x 4 5 a cos x b 5 cos a sin x sin a cos x 2 1 Because the sine function varies between 5 sin is 5 and the minimum is x a 5. 1 2 x a 1 2 and 1, the maximum of ■ 1 5 sin 608 Chapter 9 Trigonometric Identities and Proof To find the values of x that produce the maximum or minimum values of the function, solve the equation a sin x b cos x c, where c is the maximum or minimum value. In Example 7, the maximum value of 5 occurs when 3 sin x 4 cos x 5 3 sin x 4 5 5 x a cos x 1 1 sin 1 2 x a sin 11 p 2 x p 2 a 1.5708 0.9273 a cos 1 3 5 0.9273 0.6435 In one revolution, the maximum occurs at approximately 0.6435. Where the minimum value occurs is found in a similar manner. Exercises 9.4 In Exercises 1–27, find all solutions of the equation in the interval [0, 2P]. 1. sin2x 3 cos2x 0 2. sin 2x cos x 0 3. cos 2x sin x 1 4. sin x 2 1 cos x 5. 4 sin2 x 2 b a cos2x 2 6. sin 4x sin 2x 0 7. sin x sin x 1 cos x 1 2 8. sin 2t cos t cos 2t sin t 0 9. sin 2x sin x cos x 0 10. cos x cos 1 2 x (Check for extraneous solutions.) 11. sin 2x cos 2x 0 12. cos 4x cos x sin 4x sin x 0 13. sin 4x cos 2x 14. cos 2x sin2 x 0 15. 2 cos2x 2 cos 2x 1 16. cos x p 2 b a sin x 1 17. sin x p 2 b a cos x 1 18. sin x 23 cos x 0 19. sin x cos x 0 20. sin 2x cos x 0 21. cos 2x cos x 0 22. 1 sin x cos x 2 1 2 23. sin x cos x 1 2 0 24. sin2 x 2 b a cos x 0 25. csc2 x 2 2 sec x 26. 22 sin x 22 cos x 1 27. 2 sin x 2 cos x 22 In Exercises 28–31, find the solution to each equation in the interval [P, P]. 28. sin x cos x 1 29. sin2x cos 2x 1 30. sin 2x cos x 0 31. cos x 23 sin x 1 In Exercises 32–35, solve each equation in [0, 2P). 32. sin 4x sin 2x sin x Section 9.4 Using Trigonometric Identities 609 33. sin 2x sin 1 2 x 34. cos 3x cos x 0 35. sin 3x sin x 0 In Exercises 36–40, 2 1 f x k sin a. Express each function in the form x a b. Find the maximum value that c. Find all values of x in x maximum value of 0, 2p . can assume. that give the 36. 37. 38. 39. 40 23 sin x cos x sin x 23 cos x 2 sin x 2 cos x sin x cos x 4 sin x 3 cos Important Concepts Section 9.1 Basic trigonometry identities . . . . . . . . . . . . . . . . 574 Strategies for proving trigonometric identities. . . 574 Section 9.2 Addition and subtraction identities • for sine and cosine . . . . . . . . . . . . . . . . . . . . . 582 • for tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . 584 Cofunction identities . . . . . . . . . . . . . . . . . . . . . . 586 Section 9.2.A Angle of inclination . . . . . . . . . . . . . . . . . . . . . . . 589 Angle between two lines . . . . . . . . . . . . . . . . . . . 591 Section 9.3 Section 9.4 cos 2x Double-angle identities. . . . . . . . . . . . . . . . . . . . . 593 Forms of . . . . . . . . . . . . . . . . . . . . . . . . . . . 595 Power-reducing identities. . . . . . . . . . . . . . . . . . . 595 Half-angle identities . . . . . . . . . . . . . . . . . . . 596–598 Product-to-sum identities . . . . . . . . . . . . . . . . . . . 599 Sum-to-product identities. . . . . . . . . . . . . . . . . . . 599 a sin x b cos x c Solving Maxima and minima of a sin x b cos x x f 1 2 . . . . . . . . . . . . . . . . . 606 . . . . . . . . . . . . . . . . . . . 607 Important Facts and Formulas Addition and Subtraction Identities sin sin cos cos tan tan sin x cos y cos x sin y sin x cos y cos x sin y cos x cos y sin x sin y cos x cos y sin x sin y tan x tan y 1 tan x tan y tan x tan y 1 tan x tan y 610 Chapter Review 611 Cofunction Identities sin x cos p 2 a x b cos x sin p 2 a x b tan x cot sec x csc cot x tan a csc x sec a p 2 p 2 x b x b Double-Angle Identities sin 2x 2 sin x cos x cos 2x cos2x sin2x 2 cos2x 1 1 2 sin2x tan 2x 2 tan x 1 tan2x Half-Angle Identities sin cos tan cos x 2 1 cos x B 2 1 cos x sin x sin x 1 cos x Review Exercises In Exercises 1–4, simplify the given expression. Section 9.1 1. 2. sin2t 1 tan2t 2 tan t 4 3 tan2t 3 tan t 2 cos2t sec2t csc t csc2t sec t 3. tan2x sin2x sec2x sin x cos x 4. 1 sin x cos x 21 sin2x 2 1 In Exercises 5–11, determine graphically whether the equation could not possibly be an identity, or write a proof showing that it is. 5. sin4t cos4t 2 sin2t 1 6. 1 2 cos2t cos4t sin4t 7. 9. sin t 1 cos t 1 cos t sin t 8. sin2t cos2t 1 1 cos2t cos2 sin2 sin2t 10. tan x cot x sec x csc x 11. 1 sin x cos x 2 2 sin 2x 1 612 Chapter Review In Exercises 12–16, prove the given identity. 12. 14. sec x 1 tan x tan x sec x 1 1 tan2x tan2x csc2x 16. tan2x sec2x cot2x csc2x 13. cos4x sin4x 1 tan4x cos4x 15. sec x cos x sin x tan x 17. B a. c. ? 1 cos2x 1 sin2x tan x 0 B 0 1 sin2x 1 cos2x e. undefined 18. 1 2 1 csc x ? 2 sec2x 1 cos2x 1 1 tan2x 21 2 sin x 1 sin x 1 1 tan3x 21 1 a. c. e. b. 0 cot x 0 d. sec x b. sin x sin3x d. sin x 1 1 tan2x 2 Section 9.2 In Exercises 19–20, prove the given identity. 19. 20. cos x y 1 2 x y cos 2 cos x cos y 1 cos x y 2 1 cos2x sin2y 1 tan x tan y 21. Evaluate the following in exact form, where the angles and a b satisfy the conditions: sin a 4 5 for p 2 6 a 6 p tan b 7 24 a b b a c. cos 2 2 1 cot y 5 12 b. 2 tan 1 and p 6 x 6 3p 2 , and for p 6 b 6 3p 2 , with 3p 2 6 y 6 2p, find with p 6 x 6 3p 2 , and sec y 13 12 with 3p 2 6 y 6 2p, find a. 22. If sin 23. If cos 24. If b a sin 1 tan x 4 3 x y . 2 1 sin x 12 13 x y . 2 1 sin x 1 4 and 0 6 x 6 p 2 , then sin p 3 a x b ? 25. If sin x 2 5 and 3p 2 6 x 6 2p, then cos p 4 a x b ? 26. Find the exact value of sin 5p 12 . Chapter Review 613 27. Express sec x p 2 1 in terms of sin x and cos x. Section 9.2.A 28. Find the angle of inclination of the straight line through the points (2, 6) and 1 2, 2 . 2 29. Find one of the angles between the line L through the points (5, 1) and the line M, which has slope 2. 3, 2 2 1 and Section 9.3 30. Evaluate the following in exact form, where the angles and a b satisfy the conditions: sin a 44 125 for p 2 6 a 6 p tan b 15 112 for 3p 2 6 b 6 2p a. sin b 2 c. tan 2a b. cos 2a d. cos a b cos 1 2 1 a b 2 In Exercises 31–34, prove the given identity. 31. 1 cos 2x tan x sin 2x 32. tan x sin x 2 tan x sin2 x 2 33. 2 cos x 2 cos3 x sin x sin 2x 34. sin 2x 1 tan x cot 2x 35. If tan x 5 12 and sin x 7 0, find sin 2x. 36. If cos x 15 17 and 0 6 x 6 p 2 , find sin x 2 . 37. If sin x 0, is it true that sin 2x 0? Justify your answer. 38. If cos x 0, is it true that cos 2x 0? Justify your answer. 39. Show 32 23 22 26 2 by computing cos p 12 in two ways, using the half-angle identity and the subtraction identity for cosine. 40. True or false: 2 sin x sin 2x. Justify your answer. 41. If sin x 0.6 and 0 6 x 6 p 2 , find sin 2x. 42. If sin x 0.6 and 0 6 x 6 p 2 , find sin x 2 . Section 9.4 Solve the equation. Find exact solutions when possible and approximate ones otherwise. 43. 5 tan x 2 sin 2x 44. cos 2x cos x 45. 2 cos x sin x 0 46. sin 2x cos Rates of Change in Trigonometry As discussed in Section 3.7, the difference quotient represents the average rate of change of a function over the interval from x to x h average rate of change As the value of h becomes smaller and smaller, the average rate of change approaches the instantaneous rate of change at x as discussed in the Chapter 3 Can Do Calculus. Another way to represent the instantaneous rate of change is with limit notation, shown in the Can Do Calculus in Chapter 8. That is, instantaneous rate of change of a function f at x is the limit of the difference quotient. f 1 x h 2 h lim hS0 f x 1 2 instantaneous rate of change The instantaneous rate of change of a particular function is given by the expression in x that is the simplified form of the limit given above. Example 1 Instantaneous Rate of Change of f (x) sin x Find an expression for the instantaneous rate of change of f sin x. x 1 2 Solution The difference quotient of addition identity for the sine function. x f 1 2 sin x can be simplified by using the Let y h. sin sin 1 1 x y x h 2 2 sin x cos y cos x sin y sin x cos h cos x sin h Therefore, the difference quotient for the sine function can be simplified as follows. (See Example 4 of Section 9.2 for details of the simplification.) sin 1 x h 2 h sin x sin x cos h 1 h a b cos x sin h a h b The expression that represents the instantaneous rate of change is found by finding the limit of sin x a cos h 1 h cos x b sin h a h b as h approaches 0. lim hS0 a sin x cos h 1 h a b cos x sin h a h bb instantaneous rate of change 614 NOTE See Chapter 14 for a complete discussion of limits. The two fractional expressions in the limit above were evaluated in the Chapter 8 Can Do Calculus as follows: lim hS0 sin h h 1 and lim hS0 cos h 1 h 0 Substitute the values above into the expression for instantaneous rate of change to find a simpler expression. lim hS0 a sin x cos h 1 h a b cos x sin h a h bb sin x 1 cos x 0 2 21 cos x 1 2 21 1 instantaneous rate of change ■ The expression for the instantaneous rate of change for can be used to find the instantaneous rate of change at any particular value of x. x f 1 2 sin x Example 2 Finding Instantaneous Rate of Change at x k Find the instantaneous rate of change of f sin x when x 1 2 x p 3 . Inter- pret the result. Solution Because the instantaneous rate of change of cos x, x g 1 2 1 2 the instantaneous rate of change at x p 3 1 2 , or When . sin x is changing unit per unit increase in x. sin x x f 2 1 x p 3 is given by p cos 3 b p 3 , is g a Exercises 1 2 x sin x f each result. x 0 a. c. x p 2 1. What is the instantaneous rate of change of 3. What is the instantaneous rate of change of for the following values of x. Interpret for the following values of x? x cos x f Interpret each result. 1 2 b. d. x p 4 x p 6 a. c. x 0 x p 2 b. x p 4 d. x p 6 2. Find the expression for the instantaneous rate of change of f x 1 2 cos x. ■ 615 C H A P T E R 10 Trigonometric Applications A Bridge over Troubled Waters When planning a bridge or building, architects and engineers must determine the stress on cables and other parts of the structure to be sure that al
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