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dollars, to produce x bottles of 100% All-Natural Certiﬁed Free-Trade Organic Sasquatch Tonic is C(x) = 10x + 100, x ≥ 0 and the price-demand function, in dollars per bottle, is p(x) = 35 − x, 0 ≤ x ≤ 35. 12. The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is C(x) = 18x + 240, x ≥ 0 and the price-demand function, in cents per cup, is p(x) = 90 − 3x, 0 ≤ x ≤ 30. 13. The daily cost, in dollars, to produce x Sasquatch Berry Pies is C(x) = 3x + 36, x ≥ 0 and the price-demand function, in dollars per pie, is p(x) = 12 − 0.5x, 0 ≤ x ≤ 24. 14. The monthly cost, in hundreds of dollars, to produce x custom built electric scooters is C(x) = 20x + 1000, x ≥ 0 and the price-demand function, in hundreds of dollars per scooter, is p(x) = 140 − 2x, 0 ≤ x ≤ 70. 10We have already seen the graph of this function. It was used as an example in Section 1.6 to show how the graphing calculator can be misleading. 2.3 Quadratic Functions 201 15. The International Silver Strings Submarine Band holds a bake sale each year to fund their trip to the National Sasquatch Convention. It has been determined that the cost in dollars of baking x cookies is C(x) = 0.1x + 25 and that the demand function for their cookies is p = 10 −.01x. How many cookies should they bake in order to maximize their proﬁt? 16. Using data from Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = −0.0076t2 + 0.45t + 16, 0 ≤ t ≤ 28, where t is the number of years since 1980. Find and interpret the coordinates of the vertex of the graph of y = F (t). 17. The temperature T, in degrees Fahrenheit, t hours after 6 AM is given by: T (t) = − 1 2 t2 + 8t + 32,
0 ≤ t ≤ 12 What is the warmest temperature of the day? When does this happen? 18. Suppose C(x) = x2 − 10x + 27 represents the costs, in hundreds, to produce x thousand pens. How many pens should be produced to minimize the cost? What is this minimum cost? 19. Skippy wishes to plant a vegetable garden along one side of his house. In his garage, he found 32 linear feet of fencing. Since one side of the garden will border the house, Skippy doesn’t need fencing along that side. What are the dimensions of the garden which will maximize the area of the garden? What is the maximum area of the garden? 20. In the situation of Example 2.3.4, Donnie has a nightmare that one of his alpaca herd fell into the river and drowned. To avoid this, he wants to move his rectangular pasture away from the river. This means that all four sides of the pasture require fencing. If the total amount of fencing available is still 200 linear feet, what dimensions maximize the area of the pasture now? What is the maximum area? Assuming an average alpaca requires 25 square feet of pasture, how many alpaca can he raise now? 21. What is the largest rectangular area one can enclose with 14 inches of string? 22. The height of an object dropped from the roof of an eight story building is modeled by h(t) = −16t2 + 64, 0 ≤ t ≤ 2. Here, h is the height of the object oﬀ the ground, in feet, t seconds after the object is dropped. How long before the object hits the ground? 23. The height h in feet of a model rocket above the ground t seconds after lift-oﬀ is given by h(t) = −5t2 + 100t, for 0 ≤ t ≤ 20. When does the rocket reach its maximum height above the ground? What is its maximum height? 24. Carl’s friend Jason participates in the Highland Games. In one event, the hammer throw, the height h in feet of the hammer above the ground t seconds after Jason lets it go is modeled by h(t) = −16t2 + 22.08t + 6. What is the hammer’s maximum height? What is the hammer’s total time in the air? Round your answers to two decimal places. 202 Linear and Quadratic Functions 25. Assuming
no air resistance or forces other than the Earth’s gravity, the height above the ground at time t of a falling object is given by s(t) = −4.9t2 + v0t + s0 where s is in meters, t is in seconds, v0 is the object’s initial velocity in meters per second and s0 is its initial position in meters. (a) What is the applied domain of this function? (b) Discuss with your classmates what each of v0 > 0, v0 = 0 and v0 < 0 would mean. (c) Come up with a scenario in which s0 < 0. (d) Let’s say a slingshot is used to shoot a marble straight up from the ground (s0 = 0) with an initial velocity of 15 meters per second. What is the marble’s maximum height above the ground? At what time will it hit the ground? (e) Now shoot the marble from the top of a tower which is 25 meters tall. When does it hit the ground? (f) What would the height function be if instead of shooting the marble up oﬀ of the tower, you were to shoot it straight DOWN from the top of the tower? 26. The two towers of a suspension bridge are 400 feet apart. The parabolic cable11 attached to the tops of the towers is 10 feet above the point on the bridge deck that is midway between the towers. If the towers are 100 feet tall, ﬁnd the height of the cable directly above a point of the bridge deck that is 50 feet to the right of the left-hand tower. 27. Graph f (x) = \|1 − x2\| 28. Find all of the points on the line y = 1 − x which are 2 units from (1, −1). 29. Let L be the line y = 2x + 1. Find a function D(x) which measures the distance squared from a point on L to (0, 0). Use this to ﬁnd the point on L closest to (0, 0). 30. With the help of your classmates, show that if a quadratic function f (x) = ax2 + bx + c has two real zeros then the x-coordinate of the vertex is the midpoint of the zeros. In Exercises 31 - 36, solve the quadratic equation for the indicated variable. 31
. x2 − 10y2 = 0 for x 32. y2 − 4y = x2 − 4 for x 33. x2 − mx = 1 for x 34. y2 − 3y = 4x for y 35. y2 − 4y = x2 − 4 for y 36. −gt2 + v0t + s0 = 0 for t (Assume g = 0.) 11The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power line does not form a parabola. We shall see in Exercise 35 in Section 6.5 what shape a free hanging cable makes. 2.3 Quadratic Functions 203 2.3.2 Answers 1. f (x) = x2 + 2 (this is both forms!) No x-intercepts y-intercept (0, 2) Domain: (−∞, ∞) Range: [2, ∞) Decreasing on (−∞, 0] Increasing on [0, ∞) Vertex (0, 2) is a minimum Axis of symmetry x = 0 2. f (x) = −(x + 2)2 = −x2 − 4x − 4 x-intercept (−2, 0) y-intercept (0, −4) Domain: (−∞, ∞) Range: (−∞, 0] Increasing on (−∞, −2] Decreasing on [−2, ∞) Vertex (−2, 0) is a maximum Axis of symmetry x = −2 3. f (x) = x2 − 2x − 8 = (x − 1)2 − 9 x-intercepts (−2, 0) and (4, 0) y-intercept (0, −8) Domain: (−∞, ∞) Range: [−9, ∞) Decreasing on (−∞, 1] Increasing on [1, ∞) Vertex (1, −9) is a minimum Axis of symmetry x = 1 4. f (x) = −2(x + 1)2 + 4 = −2x2 − 4x + 2 √ 2, 0) 2, 0) and (−1 + √ x-intercepts (−1 − y-intercept (0, 2) Domain: (−∞, ∞) Range: (−∞, 4] Increasing on (−�
�, −1] Decreasing on [−1, ∞) Vertex (−1, 4) is a maximum Axis of symmetry x = −1 y 10 2 −1 1 2 x −4 −3 −2 −1 −1 y x −2 −3 −4 −5 −6 −7 −8 y 2 1 −2 −1 −1 1 2 3 4 x −2 −3 −4 −5 −6 −7 −8 −9 y 4 3 2 1 −1 −2 −3 −4 1 x −3 −2 −1 204 Linear and Quadratic Functions 5. f (x) = 2x2 − 4x − 1 = 2(x − 1)2 − 3 √ 6 √ 6 2+ 2, 0 2− 2, 0 and x-intercepts y-intercept (0, −1) Domain: (−∞, ∞) Range: [−3, ∞) Increasing on [1, ∞) Decreasing on (−∞, 1] Vertex (1, −3) is a minimum Axis of symmetry x = 1 6. f (x) = −3x2 + 4x − 7 = −3 x − 2 3 2 − 17 3 No x-intercepts y-intercept (0, −7) Domain: (−∞, ∞) Range: −∞, − 17 3 Increasing on −∞, 2 3 Decreasing on 2 3, ∞ Vertex 2 is a maximum 3, − 17 Axis of symmetry 1 1 2 3 x 1 2 x −1 −2 −3 y −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 −14 7. f (x) = x2 + No x-intercepts y-intercept (0, 1) Domain: (−∞, ∞) 4, ∞ Range: 3 Increasing on − 1 2, ∞ Decreasing on −∞, − 1 2 Vertex − 1 is a minimum 2, 3 4 Axis of symmetry 2 −1 1 x 2.3 Quadratic Functions 205 8. f (x) = −3x2 + 5x + 4 = −3 x − 5 6 5− 73, 0 5+ and, 0 73 √ √ 2 + 73 12 6 x-intercepts 6 y-intercept (0, 4) Domain
: (−∞, ∞) Range: −∞, 73 12 Increasing on −∞, 5 6 Decreasing on 5 6, ∞ Vertex 5 6, 73 12 Axis of symmetry x = 5 6 is a maximum 100 x − 1 = x − 1 200 1− 1+ 40001 2 − 40001 40000 √ 40001 and 200 √ 9. f (x) = x2 − 1 x-intercepts 200 y-intercept (0, −1) Domain: (−∞, ∞) Range: − 40001 40000, ∞ Decreasing on −∞, 1 200 200, ∞ Increasing on 1 Vertex 1 200, − 40001 Axis of symmetry x = 1 200 40000 is a minimum12 y 6 5 4 3 2 1 −1 1 2 3 x −1 −2 −2 −1 1 2 x 10. P (x) = −2x2 + 28x − 26, for 0 ≤ x ≤ 15. 7 T-shirts should be made and sold to maximize proﬁt. The maximum proﬁt is $72. The price per T-shirt should be set at $16 to maximize proﬁt. The break even points are x = 1 and x = 13, so to make a proﬁt, between 1 and 13 T-shirts need to be made and sold. 11. P (x) = −x2 + 25x − 100, for 0 ≤ x ≤ 35 Since the vertex occurs at x = 12.5, and it is impossible to make or sell 12.5 bottles of tonic, maximum proﬁt occurs when either 12 or 13 bottles of tonic are made and sold. The maximum proﬁt is $56. The price per bottle can be either $23 (to sell 12 bottles) or $22 (to sell 13 bottles.) Both will result in the maximum proﬁt. The break even points are x = 5 and x = 20, so to make a proﬁt, between 5 and 20 bottles of tonic need to be made and sold. 12You’ll need to use your calculator to zoom in far enough to see that the vertex is not the y-intercept. 206 12. 13. 14. Linear and Quadratic Functions P (x) = −3x2 + 72x − 240, for 0 ≤ x ≤ 30
12 cups of lemonade need to be made and sold to maximize proﬁt. The maximum proﬁt is 192¢ or $1.92. The price per cup should be set at 54¢ per cup to maximize proﬁt. The break even points are x = 4 and x = 20, so to make a proﬁt, between 4 and 20 cups of lemonade need to be made and sold. P (x) = −0.5x2 + 9x − 36, for 0 ≤ x ≤ 24 9 pies should be made and sold to maximize the daily proﬁt. The maximum daily proﬁt is $4.50. The price per pie should be set at $7.50 to maximize proﬁt. The break even points are x = 6 and x = 12, so to make a proﬁt, between 6 and 12 pies need to be made and sold daily. P (x) = −2x2 + 120x − 1000, for 0 ≤ x ≤ 70 30 scooters need to be made and sold to maximize proﬁt. The maximum monthly proﬁt is 800 hundred dollars, or $80,000. The price per scooter should be set at 80 hundred dollars, or $8000 per scooter. The break even points are x = 10 and x = 50, so to make a proﬁt, between 10 and 50 scooters need to be made and sold monthly. 15. 495 cookies 16. The vertex is (approximately) (29.60, 22.66), which corresponds to a maximum fuel economy of 22.66 miles per gallon, reached sometime between 2009 and 2010 (29 – 30 years after 1980.) Unfortunately, the model is only valid up until 2008 (28 years after 1908.) So, at this point, we are using the model to predict the maximum fuel economy. 17. 64◦ at 2 PM (8 hours after 6 AM.) 18. 5000 pens should be produced for a cost of $200. 19. 8 feet by 16 feet; maximum area is 128 square feet. 20. 50 feet by 50 feet; maximum area is 2500 feet; he can raise 100 average alpacas. 21. The largest rectangle has area 12.25 square inches. 22. 2 seconds. 23. The rocket reaches its maximum height of 500 feet 10 seconds after lift-oﬀ. 24. The
hammer reaches a maximum height of approximately 13.62 feet. The hammer is in the air approximately 1.61 seconds. 2.3 Quadratic Functions 207 25. (a) The applied domain is [0, ∞). (d) The height function is this case is s(t) = −4.9t2 + 15t. The vertex of this parabola is approximately (1.53, 11.48) so the maximum height reached by the marble is 11.48 meters. It hits the ground again when t ≈ 3.06 seconds. (e) The revised height function is s(t) = −4.9t2 + 15t + 25 which has zeros at t ≈ −1.20 and t ≈ 4.26. We ignore the negative value and claim that the marble will hit the ground after 4.26 seconds. (f) Shooting down means the initial velocity is negative so the height functions becomes s(t) = −4.9t2 − 15t + 25. 26. Make the vertex of the parabola (0, 10) so that the point on the top of the left-hand tower where the cable connects is (−200, 100) and the point on the top of the right-hand tower is (200, 100). Then the parabola is given by p(x) = 9 4000 x2 + 10. Standing 50 feet to the right of the left-hand tower means you’re standing at x = −150 and p(−150) = 60.625. So the cable is 60.625 feet above the bridge deck there. √ 3 − 2 7, −1 − 2 √ 7 28. 27. y = \|1 − x22 −1 1 2 x 29. D(x) = x2 +(2x+1)2 = 5x2 +4x+1, D is minimized when x = − 2 5, so the point on y = 2x+1 closest to (0, 0) is − 2 5, 1 5 31. x = ±y √ 10 34. y = √ 3 ± 16x + 9 2 32. x = ±(y − 2) 35. y = 2 ± x 33. x = 36. t = √ m ± m2 + 4 2 v0 ± v2 0 + 4gs0 2g 208 Linear and Quadratic Functions 2.4 Inequalities
with Absolute Value and Quadratic Functions In this section, not only do we develop techniques for solving various classes of inequalities analytically, we also look at them graphically. The ﬁrst example motivates the core ideas. Example 2.4.1. Let f (x) = 2x − 1 and g(x) = 5. 1. Solve f (x) = g(x). 2. Solve f (x) < g(x). 3. Solve f (x) > g(x). 4. Graph y = f (x) and y = g(x) on the same set of axes and interpret your solutions to parts 1 through 3 above. Solution. 1. To solve f (x) = g(x), we replace f (x) with 2x − 1 and g(x) with 5 to get 2x − 1 = 5. Solving for x, we get x = 3. 2. The inequality f (x) < g(x) is equivalent to 2x − 1 < 5. Solving gives x < 3 or (−∞, 3). 3. To ﬁnd where f (x) > g(x), we solve 2x − 1 > 5. We get x > 3, or (3, ∞). 4. To graph y = f (x), we graph y = 2x − 1, which is a line with a y-intercept of (0, −1) and a slope of 2. The graph of y = g(x) is y = 5 which is a horizontal line through (0, 5). y y = g(x) y = f (x1 To see the connection between the graph and the Algebra, we recall the Fundamental Graphing Principle for Functions in Section 1.6: the point (a, b) is on the graph of f if and only if f (a) = b. In other words, a generic point on the graph of y = f (x) is (x, f (x)), and a generic 2.4 Inequalities with Absolute Value and Quadratic Functions 209 point on the graph of y = g(x) is (x, g(x)). When we seek solutions to f (x) = g(x), we are looking for x values whose y values on the graphs of f and g are the same. In part 1, we found x = 3
is the solution to f (x) = g(x). Sure enough, f (3) = 5 and g(3) = 5 so that the point (3, 5) is on both graphs. In other words, the graphs of f and g intersect at (3, 5). In part 2, we set f (x) < g(x) and solved to ﬁnd x < 3. For x < 3, the point (x, f (x)) is below (x, g(x)) since the y values on the graph of f are less than the y values on the graph of g there. Analogously, in part 3, we solved f (x) > g(x) and found x > 3. For x > 3, note that the graph of f is above the graph of g, since the y values on the graph of f are greater than the y values on the graph of g for those values of x. y y = g(x) y = f (xx) y = g(x1 1 2 3 4 x −1 f (x) < g(x) on (−∞, 3) f (x) > g(x) on (3, ∞) The preceding example demonstrates the following, which is a consequence of the Fundamental Graphing Principle for Functions. Graphical Interpretation of Equations and Inequalities Suppose f and g are functions. The solutions to f (x) = g(x) are the x values where the graphs of y = f (x) and y = g(x) intersect. The solution to f (x) < g(x) is the set of x values where the graph of y = f (x) is below the graph of y = g(x). The solution to f (x) > g(x) is the set of x values where the graph of y = f (x) above the graph of y = g(x). The next example turns the tables and furnishes the graphs of two functions and asks for solutions to equations and inequalities. 210 Linear and Quadratic Functions Example 2.4.2. The graphs of f and g are below. (The graph of y = g(x) is bolded.) Use these graphs to answer the following questions. y 4 3 2 1 (−1, 2) y = g(x) (1, 2) −2 −1 1 2 x −1 y = f (
x) 1. Solve f (x) = g(x). 2. Solve f (x) < g(x). 3. Solve f (x) ≥ g(x). Solution. 1. To solve f (x) = g(x), we look for where the graphs of f and g intersect. These appear to be at the points (−1, 2) and (1, 2), so our solutions to f (x) = g(x) are x = −1 and x = 1. 2. To solve f (x) < g(x), we look for where the graph of f is below the graph of g. This appears to happen for the x values less than −1 and greater than 1. Our solution is (−∞, −1)∪(1, ∞). 3. To solve f (x) ≥ g(x), we look for solutions to f (x) = g(x) as well as f (x) > g(x). We solved the former equation and found x = ±1. To solve f (x) > g(x), we look for where the graph of f is above the graph of g. This appears to happen between x = −1 and x = 1, on the interval (−1, 1). Hence, our solution to f (x) ≥ g(x) is [−1, 1]. y 4 3 2 1 (−1, 2) y = g(x) (1, 2) (−1, 2) y 4 3 2 1 y = g(x) (1, 2) −2 −1 1 2 x −2 −1 1 2 x −1 y = f (x) f (x) < g(x) −1 y = f (x) f (x) ≥ g(x) 2.4 Inequalities with Absolute Value and Quadratic Functions 211 We now turn our attention to solving inequalities involving the absolute value. We have the following theorem from Intermediate Algebra to help us. Theorem 2.4. Inequalities Involving the Absolute Value: Let c be a real number. For c > 0, \|x\| < c is equivalent to −c < x < c. For c > 0, \|x\| ≤ c is equivalent to −c ≤ x ≤ c. For c ≤ 0, \|x\| < c has no solution, and for c < 0, \|x\| ≤ c has
no solution. For c ≥ 0, \|x\| > c is equivalent to x < −c or x > c. For c ≥ 0, \|x\| ≥ c is equivalent to x ≤ −c or x ≥ c. For c < 0, \|x\| > c and \|x\| ≥ c are true for all real numbers. As with Theorem 2.1 in Section 2.2, we could argue Theorem 2.4 using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if c > 0, the graph of y = c is a horizontal line which lies above the x-axis through (0, c). To solve \|x\| < c, we are looking for the x values where the graph of y = \|x\| is below the graph of y = c. We know that the graphs intersect when \|x\| = c, which, from Section 2.2, we know happens when x = c or x = −c. Graphing, we get y (−c, c) (c, c) −c c x We see that the graph of y = \|x\| is below y = c for x between −c and c, and hence we get \|x\| < c is equivalent to −c < x < c. The other properties in Theorem 2.4 can be shown similarly. Example 2.4.3. Solve the following inequalities analytically; check your answers graphically. 1. \|x − 1\| ≥ 3 3. 2 < \|x − 1\| ≤ 5 Solution. 2. 4 − 3\|2x + 1\| > −2 4. \|x + 1\| ≥ x + 4 2 1. From Theorem 2.4, \|x − 1\| ≥ 3 is equivalent to x − 1 ≤ −3 or x − 1 ≥ 3. Solving, we get x ≤ −2 or x ≥ 4, which, in interval notation is (−∞, −2] ∪ [4, ∞). Graphically, we have 212 Linear and Quadratic Functions x − 1\| −4 −3 −2 −1 1 2 3 4 5 x We see that the graph of y = \|x − 1\| is above the horizontal line y = 3 for x < −2 and x > 4 hence this is where \|x − 1\| > 3. The two graphs intersect when x = −2 and x
= 4, so we have graphical conﬁrmation of our analytic solution. 2. To solve 4 − 3\|2x + 1\| > −2 analytically, we ﬁrst isolate the absolute value before applying Theorem 2.4. To that end, we get −3\|2x + 1\| > −6 or \|2x + 1\| < 2. Rewriting, we now have. Graphically we −2 < 2x + 1 < 2 so that − 3 2 and 1 see that the graph of y = 4 − 3\|2x + 1\| is above y = −2 for x values between − 3 2. 2. In interval notation, we write − \|2x + 1\| −2 −1 y 4 3 2 1 −1 −2 −3 −4 1 2 x y = −2 3. Rewriting the compound inequality 2 < \|x − 1\| ≤ 5 as ‘2 < \|x − 1\| and \|x − 1\| ≤ 5’ allows us to solve each piece using Theorem 2.4. The ﬁrst inequality, 2 < \|x − 1\| can be re-written as \|x − 1\| > 2 so x − 1 < −2 or x − 1 > 2. We get x < −1 or x > 3. Our solution to the ﬁrst inequality is then (−∞, −1) ∪ (3, ∞). For \|x − 1\| ≤ 5, we combine results in Theorems 2.1 and 2.4 to get −5 ≤ x − 1 ≤ 5 so that −4 ≤ x ≤ 6, or [−4, 6]. Our solution to 2 < \|x − 1\| ≤ 5 is comprised of values of x which satisfy both parts of the inequality, so we take the intersection1 of (−∞, −1) ∪ (3, ∞) and [−4, 6] to get [−4, −1) ∪ (3, 6]. Graphically, we see that the graph of y = \|x − 1\| is ‘between’ the horizontal lines y = 2 and y = 5 for x values between −4 and −1 as well as those between 3 and 6. Including the x values where y = \|x − 1\| and y = 5 intersect, we get 1See Deﬁnition 1.2 in
Section 1.1.1. 2.4 Inequalities with Absolute Value and Quadratic Functions 213 x − 1\| y = 5 y = 2 −8 −7 −6 −5 −4 −3 −2 −. We need to exercise some special caution when solving \|x + 1\| ≥ x+4 2. As we saw in Example 2.2.1 in Section 2.2, when variables are both inside and outside of the absolute value, it’s usually best to refer to the deﬁnition of absolute value, Deﬁnition 2.4, to remove the absolute values and proceed from there. To that end, we have \|x + 1\| = −(x + 1) if x < −1 and \|x + 1\| = x + 1 if x ≥ −1. We break the inequality into cases, the ﬁrst case being when x < −1. For these values of x, our inequality becomes −(x + 1) ≥ x+4 2. Solving, we get −2x − 2 ≥ x + 4, so that −3x ≥ 6, which means x ≤ −2. Since all of these solutions fall into the category x < −1, we keep them all. For the second case, we assume x ≥ −1. Our inequality becomes x + 1 ≥ x+4 2, which gives 2x + 2 ≥ x + 4 or x ≥ 2. Since all of these values of x are greater than or equal to −1, we accept all of these solutions as well. Our ﬁnal answer is (−∞, −2] ∪ [2, ∞). y = \|x + 1\| y 4 3 2 y = x+4 2 −4 −3 −2 −1 1 2 3 4 x We now turn our attention to quadratic inequalities. In the last example of Section 2.3, we needed to determine the solution to x2 − x − 6 < 0. We will now re-visit this problem using some of the techniques developed in this section not only to reinforce our solution in Section 2.3, but to also help formulate a general analytic procedure for solving all quadratic inequalities. If we consider f (x) = x2 − x − 6 and g(x) = 0, then solving x2 − x − 6 < 0 corresponds graphically to ﬁnding 214 Linear and
Quadratic Functions the values of x for which the graph of y = f (x) = x2 − x − 6 (the parabola) is below the graph of y = g(x) = 0 (the x-axis). We’ve provided the graph again for reference. y 6 5 4 3 2 1 −3−2−1 −1 1 2 3 x −2 −3 −4 −5 −6 y = x2 − x − 6 We can see that the graph of f does dip below the x-axis between its two x-intercepts. The zeros of f are x = −2 and x = 3 in this case and they divide the domain (the x-axis) into three intervals: (−∞, −2), (−2, 3) and (3, ∞). For every number in (−∞, −2), the graph of f is above the x-axis; in other words, f (x) > 0 for all x in (−∞, −2). Similarly, f (x) < 0 for all x in (−2, 3), and f (x) > 0 for all x in (3, ∞). We can schematically represent this with the sign diagram below. (+) 0 (−) 0 (+) −2 3 Here, the (+) above a portion of the number line indicates f (x) > 0 for those values of x; the (−) indicates f (x) < 0 there. The numbers labeled on the number line are the zeros of f, so we place 0 above them. We see at once that the solution to f (x) < 0 is (−2, 3). Our next goal is to establish a procedure by which we can generate the sign diagram without graphing the function. An important property2 of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphically, this means that a parabola can’t be above the x-axis at one point and below the x-axis at another point without crossing the x-axis. This allows us to determine the sign of all of the function values on a given interval by testing the function at just one value in the interval. This gives us the following. 2We will give this property a name in Chapter 3 and revisit this concept then. 2.4 Inequalities with Absolute Value and Quad
ratic Functions 215 Steps for Solving a Quadratic Inequality 1. Rewrite the inequality, if necessary, as a quadratic function f (x) on one side of the in- equality and 0 on the other. 2. Find the zeros of f and place them on the number line with the number 0 above them. 3. Choose a real number, called a test value, in each of the intervals determined in step 2. 4. Determine the sign of f (x) for each test value in step 3, and write that sign above the corresponding interval. 5. Choose the intervals which correspond to the correct sign to solve the inequality. Example 2.4.4. Solve the following inequalities analytically using sign diagrams. Verify your answer graphically. 1. 2x2 ≤ 3 − x 3. x2 + 1 ≤ 2x Solution. 2. x2 − 2x > 1 4. 2x − x2 ≥ \|x − 1\| − 1 1. To solve 2x2 ≤ 3 − x, we ﬁrst get 0 on one side of the inequality which yields 2x2 + x − 3 ≤ 0. We ﬁnd the zeros of f (x) = 2x2 + x − 3 by solving 2x2 + x − 3 = 0 for x. Factoring gives (2x + 3)(x − 1) = 0, so x = − 3 2 or x = 1. We place these values on the number line with 0 2, 1 and (1, ∞). For the above them and choose test values in the intervals −∞, − 3 2 2, 1, we pick x = 0; and for (1, ∞), x = 2. interval −∞, − 3 2 Evaluating the function at the three test values gives us f (−2) = 3 > 0, so we place (+) above −∞, − 3 2, 1; and, f (2) = 7, 2 which means (+) is placed above (1, ∞). Since we are solving 2x2 + x − 3 ≤ 0, we look for solutions to 2x2 + x − 3 < 0 as well as solutions for 2x2 + x − 3 = 0. For 2x2 + x − 3 < 0, we need the intervals which we have a (−). Checking the sign diagram, we see this is − 3 2,
1. We know 2x2 + x − 3 = 0 when x = − 3 ; f (0) = −3 < 0, so (−) goes above the interval − 3, we choose3 x = −2; for − 3 2 and x = 1, so our ﬁnal answer is − 3, − 3 2, 1. To verify our solution graphically, we refer to the original inequality, 2x2 ≤ 3 − x. We let g(x) = 2x2 and h(x) = 3 − x. We are looking for the x values where the graph of g is below that of h (the solution to g(x) < h(x)) as well as the points of intersection (the solutions to g(x) = h(x)). The graphs of g and h are given on the right with the sign chart on the left. 3We have to choose something in each interval. If you don’t like our choices, please feel free to choose diﬀerent numbers. You’ll get the same sign chart. 216 Linear and Quadratic Functions (+) 0 (−) 0 (+) − 3 2 − = 2x2 y = 3 − x −2 −1 1 2 x √ 2. Once again, we re-write x2 − 2x > 1 as x2 − 2x − 1 > 0 and we identify f (x) = x2 − 2x − 1. When we go to ﬁnd the zeros of f, we ﬁnd, to our chagrin, that the quadratic x2 − 2x − 1 doesn’t factor nicely. Hence, we resort to the quadratic formula to solve x2 − 2x − 1 = 0, and arrive at x = 1 ± 2. As before, these zeros divide the number line into three pieces. To help us decide on test values, we approximate 1 − 2 ≈ 2.4. We choose x = −1, x = 0 and x = 3 as our test values and ﬁnd f (−1) = 2, which is (+); f (0) = −1 which is (−); and f (3) = 2 which is (+) again. Our solution to x2 − 2x − 1 > 0 is where 2, ∞. To check the inequality we have (+), so, in interval notation −∞, 1 −
x2 − 2x > 1 graphically, we set g(x) = x2 − 2x and h(x) = 1. We are looking for the x values where the graph of g is above the graph of h. As before we present the graphs on the right and the sign chart on the left. 2 ≈ −0.4 and 1 + 2 ∪ 1 + √ √ √ √ (+) −1 0 √ 1 − (−) 2 0 0 √ 1 + (+) 3 −2 −1 1 2 3 x y = x2 − 2x 3. To solve x2 + 1 ≤ 2x, as before, we solve x2 − 2x + 1 ≤ 0. Setting f (x) = x2 − 2x + 1 = 0, we ﬁnd the only one zero of f, x = 1. This one x value divides the number line into two intervals, from which we choose x = 0 and x = 2 as test values. We ﬁnd f (0) = 1 > 0 and f (2) = 1 > 0. Since we are looking for solutions to x2 − 2x + 1 ≤ 0, we are looking for x values where x2 − 2x + 1 < 0 as well as where x2 − 2x + 1 = 0. Looking at our sign diagram, there are no places where x2 − 2x + 1 < 0 (there are no (−)), so our solution is only x = 1 (where x2 − 2x + 1 = 0). We write this as {1}. Graphically, we solve x2 + 1 ≤ 2x by graphing g(x) = x2 + 1 and h(x) = 2x. We are looking for the x values where the graph of g is below the graph of h (for x2 +1 < 2x) and where the two graphs intersect (x2 +1 = 2x). Notice that the line and the parabola touch at (1, 2), but the parabola is always above the line otherwise.4 4In this case, we say the line y = 2x is tangent to y = x2 + 1 at (1, 2). Finding tangent lines to arbitrary functions is a fundamental problem solved, in general, with Calculus. 2.4 Inequalities with Absolute Value and Quadratic Functions 217 0 1 (+) 0 (+) 2
y 4 3 2 1 y = x2 + 1 y = 2x −1 1 x 4. To solve our last inequality, 2x − x2 ≥ \|x − 1\| − 1, we re-write the absolute value using cases. For x < 1, \|x − 1\| = −(x − 1) = 1 − x, so we get 2x − x2 ≥ 1 − x − 1, or x2 − 3x ≤ 0. Finding the zeros of f (x) = x2 − 3x, we get x = 0 and x = 3. However, we are only concerned with the portion of the number line where x < 1, so the only zero that we concern ourselves with is x = 0. This divides the interval x < 1 into two intervals: (−∞, 0) and (0, 1). We choose 2 as our test values. We ﬁnd f (−1) = 4 and f 1 x = −1 and x = 1 4. Hence, our solution to x2 − 3x ≤ 0 for x < 1 is [0, 1). Next, we turn our attention to the case x ≥ 1. Here, \|x − 1\| = x − 1, so our original inequality becomes 2x − x2 ≥ x − 1 − 1, or x2 − x − 2 ≤ 0. Setting g(x) = x2 − x − 2, we ﬁnd the zeros of g to be x = −1 and x = 2. Of these, only x = 2 lies in the region x ≥ 1, so we ignore x = −1. Our test intervals are now [1, 2) and (2, ∞). We choose x = 1 and x = 3 as our test values and ﬁnd g(1) = −2 and g(3) = 4. Hence, our solution to g(x) = x2 − x − 2 ≤ 0, in this region is [1, 2). = − 5 2 0 0 (+) −1 (−) 1 2 1 (−) 0 2 (+) 3 1 Combining these into one sign diagram, we have that our solution is [0, 2]. Graphically, to check 2x − x2 ≥ \|x − 1\| − 1, we set h(x) = 2x − x2 and i(x) = \|x − 1\| − 1 and look for the x values
where the graph of h is above the the graph of i (the solution of h(x) > i(x)) as well as the x-coordinates of the intersection points of both graphs (where h(x) = i(x)). The combined sign chart is given on the left and the graphs are on the right. (+) 0 (−) 0 (+) 0 2 −1 0 3 y y = 2x − x2 1 −1 1 2 3 x y = \|x − 1\| − 1 218 Linear and Quadratic Functions One of the classic applications of inequalities is the notion of tolerances.5 Recall that for real numbers x and c, the quantity \|x − c\| may be interpreted as the distance from x to c. Solving inequalities of the form \|x − c\| ≤ d for d ≥ 0 can then be interpreted as ﬁnding all numbers x which lie within d units of c. We can think of the number d as a ‘tolerance’ and our solutions x as being within an accepted tolerance of c. We use this principle in the next example. Example 2.4.5. The area A (in square inches) of a square piece of particle board which measures x inches on each side is A(x) = x2. Suppose a manufacturer needs to produce a 24 inch by 24 inch square piece of particle board as part of a home oﬃce desk kit. How close does the side of the piece of particle board need to be cut to 24 inches to guarantee that the area of the piece is within a tolerance of 0.25 square inches of the target area of 576 square inches? Solution. Mathematically, we express the desire for the area A(x) to be within 0.25 square inches of 576 as \|A − 576\| ≤ 0.25. Since A(x) = x2, we get \|x2 − 576\| ≤ 0.25, which is equivalent to −0.25 ≤ x2 − 576 ≤ 0.25. One way to proceed at this point is to solve the two inequalities −0.25 ≤ x2 − 576 and x2 − 576 ≤ 0.25 individually using sign diagrams and then taking the intersection of the solution sets. While this way will (eventually) lead to the correct answer, we take this opportunity to showcase the increasing property of the square root: if 0 ≤ a ≤ b, then √ �
� a ≤ b. To use this property, we proceed as follows −0.25 ≤ x2 − 576 ≤ 0.25 x2 575.75 ≤ √ x2 575.75 ≤ \|x\| 575.75 ≤ ≤ 576.25 √ ≤ √ ≤ √ √ √ (add 576 across the inequalities.) (take square roots.) √ x2 = \|x\|) ( 575.75 ∪ √ √ √ 575.75 ≤ \|x\| ≤ 576.25 576.25 575.75 ≤ \|x\| to be −∞, − 576.25. To solve 576.25, √ √ √ 575.75] ∪ [ 575.75, √ √ √ By Theorem 2.4, we ﬁnd the solution to 576.25 to be − the solution to \|x\| ≤ intersect these two sets to get [− length, we discard the negative answers and get [ √ the piece of particle board must be cut between tolerance of (approximately) 0.005 inches of the target length of 24 inches. 575.75, ∞ and 576.25, we 576.25]. Since x represents a 576.25]. This means that the side of 576.25 ≈ 24.005 inches, a 575.75 ≈ 23.995 and 576.25, − 575.75, √ √ √ √ √ Our last example in the section demonstrates how inequalities can be used to describe regions in the plane, as we saw earlier in Section 1.2. Example 2.4.6. Sketch the following relations. 1. R = {(x, y) : y > \|x\|} 2. S = {(x, y) : y ≤ 2 − x2} 3. T = {(x, y) : \|x\| < y ≤ 2 − x2} 5The underlying concept of Calculus can be phrased in terms of tolerances, so this is well worth your attention. 2.4 Inequalities with Absolute Value and Quadratic Functions 219 Solution. 1. The relation R consists of all points (x, y) whose y-coordinate is greater than \|x\|. If we graph y = \|x\|, then we want all of the points in the plane above
the points on the graph. Dotting the graph of y = \|x\| as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region below on the left. 2. For a point to be in S, its y-coordinate must be less than or equal to the y-coordinate on the parabola y = 2 − x2. This is the set of all points below or on the parabola y = 2 − x2. y 2 1 y 2 1 −2 −1 1 2 x −2 −1 1 2 x −1 −1 The graph of R The graph of S 3. Finally, the relation T takes the points whose y-coordinates satisfy both the conditions given in R and those of S. Thus we shade the region between y = \|x\| and y = 2 − x2, keeping those points on the parabola, but not the points on y = \|x\|. To get an accurate graph, we need to ﬁnd where these two graphs intersect, so we set \|x\| = 2 − x2. Proceeding as before, breaking this equation into cases, we get x = −1, 1. Graphing yields y 2 1 −1 −2 −1 1 2 x The graph of T 220 Linear and Quadratic Functions 2.4.1 Exercises In Exercises 1 - 32, solve the inequality. Write your answer using interval notation. 1. \|3x − 5\| ≤ 4 3. \|2x + 1\| − 5 < 0 5. \|3x + 5\| + 2 < 1 7. 2 ≤ \|4 − x\| < 7 2. \|7x + 2\| > 10 4. \|2 − x\| − 4 ≥ −3 6. 2\|7 − x\| + 4 > 1 8. 1 < \|2x − 9\| ≤ 3 9. \|x + 3\| ≥ \|6x + 9\| 10. \|x − 3\| − \|2x + 1\| < 0 11. \|1 − 2x\| ≥ x + 5 13. x ≥ \|x + 1\| 15. x + \|2x − 3\| < 2 17. x2 + 2x − 3 ≥ 0 19. x2 + 9 < 6x 21. x2 + 4 ≤ 4x 23. 3x2 ≤ 11x + 4 25. 2x2 − 4x
− 1 > 0 27. 2 ≤ \|x2 − 9\| < 9 29. x2 + x + 1 ≥ 0 31. x\|x + 5\| ≥ −6 12. x + 5 < \|x + 5\| 14. \|2x + 1\| ≤ 6 − x 16. \|3 − x\| ≥ x − 5 18. 16x2 + 8x + 1 > 0 20. 9x2 + 16 ≥ 24x 22. x2 + 1 < 0 24. x > x2 26. 5x + 4 ≤ 3x2 28. x2 ≤ \|4x − 3\| 30. x2 ≥ \|x\| 32. x\|x − 3\| < 2 33. The proﬁt, in dollars, made by selling x bottles of 100% All-Natural Certiﬁed Free-Trade Organic Sasquatch Tonic is given by P (x) = −x2 + 25x − 100, for 0 ≤ x ≤ 35. How many bottles of tonic must be sold to make at least $50 in proﬁt? 34. Suppose C(x) = x2 − 10x + 27, x ≥ 0 represents the costs, in hundreds of dollars, to produce x thousand pens. Find the number of pens which can be produced for no more than $1100. 35. The temperature T, in degrees Fahrenheit, t hours after 6 AM is given by T (t) = − 1 2 t2+8t+32, for 0 ≤ t ≤ 12. When is it warmer than 42◦ Fahrenheit? 2.4 Inequalities with Absolute Value and Quadratic Functions 221 36. The height h in feet of a model rocket above the ground t seconds after lift-oﬀ is given by h(t) = −5t2 + 100t, for 0 ≤ t ≤ 20. When is the rocket at least 250 feet oﬀ the ground? Round your answer to two decimal places. 37. If a slingshot is used to shoot a marble straight up into the air from 2 meters above the ground with an initial velocity of 30 meters per second, for what values of time t will the marble be over 35 meters above the ground? (Refer to Exercise 25 in Section 2.3 for assistance if needed.) Round your answers to two decimal places. 38. What temperature values in degrees Celsius are equivalent to the temperature range 50◦F to 95�
�F? (Refer to Exercise 35 in Section 2.1 for assistance if needed.) In Exercises 39 - 42, write and solve an inequality involving absolute values for the given statement. 39. Find all real numbers x so that x is within 4 units of 2. 40. Find all real numbers x so that 3x is within 2 units of −1. 41. Find all real numbers x so that x2 is within 1 unit of 3. 42. Find all real numbers x so that x2 is at least 7 units away from 4. 43. The surface area S of a cube with edge length x is given by S(x) = 6x2 for x > 0. Suppose the cubes your company manufactures are supposed to have a surface area of exactly 42 square centimeters, but the machines you own are old and cannot always make a cube with the precise surface area desired. Write an inequality using absolute value that says the surface area of a given cube is no more than 3 square centimeters away (high or low) from the target of 42 square centimeters. Solve the inequality and write your answer using interval notation. 44. Suppose f is a function, L is a real number and ε is a positive number. Discuss with your classmates what the inequality \|f (x) − L\| < ε means algebraically and graphically.6 In Exercises 45 - 50, sketch the graph of the relation. 45. R = {(x, y) : y ≤ x − 1} 47. R = {(x, y) : −1 < y ≤ 2x + 1} 46. R = (x, y) : y > x2 + 1 48. R = (x, y) : x2 ≤ y < x + 2 49. R = {(x, y) : \|x\| − 4 < y < 2 − x} 50. R = (x, y) : x2 < y ≤ \|4x − 3\| 51. Prove the second, third and fourth parts of Theorem 2.4. 6Understanding this type of inequality is really important in Calculus. 222 Linear and Quadratic Functions 2.4.2 Answers 1. 1 3, 3 3. (−3, 2) 5. No solution 7. (−3, 2] ∪ [6, 11) 9. − 12 7, − 6 5 11. −∞, − 4 3 ∪ [6, ∞) 13. No
Solution. 15. 1, 5 3 17. (−∞, −3] ∪ [1, ∞) 19. No solution 21. {2} 23. − 1 3, 4 25. −∞,, ∞ 27. √ √ 2, − 11 ∪ √ − 7, 0 √ 7 0, ∪ ∪ √ √ 11, 3 2 −3 2. −∞, − 12 7 ∪ 8 7, ∞ 4. (−∞, 1] ∪ [3, ∞) 6. (−∞, ∞) 8. [3, 4) ∪ (5, 6] 10. (−∞, −4) ∪ 2 3, ∞ 12. (−∞, −5) 14. −7, 5 3 16. (−∞, ∞) 18. −∞, − 1 4 ∪ − 1 4, ∞ 20. (−∞, ∞) 22. No solution 24. (0, 1) −∞, 5− 73 √ 6 ∪ √ 5+ 73 6, ∞ 26. 28. −2 − √ 7, −2 + √ 7 ∪ [1, 3] 29. (−∞, ∞) 30. (−∞, −1] ∪ {0} ∪ [1, ∞) 31. [−6, −3] ∪ [−2, ∞) 32. (−∞, 1) ∪ 2, 3+ 17 √ 2 33. P (x) ≥ 50 on [10, 15]. This means anywhere between 10 and 15 bottles of tonic need to be sold to earn at least $50 in proﬁt. 34. C(x) ≤ 11 on [2, 8]. This means anywhere between 2000 and 8000 pens can be produced and the cost will not exceed $1100. √ √ 35. T (t) > 42 on (8 − 2 11, 8 + 2 11) ≈ (1.37, 14.63), which corresponds to between 7:22 AM (1.37 hours after 6 AM) to 8:38 PM (14.63 hours after 6 AM.) However, since the model is valid only for t, 0 ≤ t ≤ 12, we restrict our answer and ﬁnd it is warmer than 42◦ Fahrenheit from 7:
22 AM to 6 PM. 2.4 Inequalities with Absolute Value and Quadratic Functions 223 36. h(t) ≥ 250 on [10 − 5 √ √ 2, 10 + 5 2] ≈ [2.93, 17.07]. This means the rocket is at least 250 feet oﬀ the ground between 2.93 and 17.07 seconds after lift oﬀ. 37. s(t) = −4.9t2 + 30t + 2. s(t) > 35 on (approximately) (1.44, 4.68). This means between 1.44 and 4.68 seconds after it is launched into the air, the marble is more than 35 feet oﬀ the ground. 38. From our previous work C(F ) = 5 9 (F − 32) so 50 ≤ F ≤ 95 becomes 10 ≤ C ≤ 35. 39. \|x − 2\| ≤ 4, [−2, 6] 40. \|3x + 1\| ≤ 2, −1, 1 3 41. \|x2 − 3\| ≤ 1, [−2, 2] √ 42. \|x2 − 4\| ≥ 7, (−∞, − 11 ] ∪ [ 11, ∞) 43. Solving \|S(x) − 42\| ≤ 3, and disregarding the negative solutions yields [2.550, 2.739]. The edge length must be within 2.550 and 2.739 centimeters. 13 2, 15 2 ≈ 45. 47. y 3 2 1 −2 −1 1 2 3 x −1 −2 −3 y 5 4 3 2 1 −2 −1 1 2 x 46. y 4 3 2 1 −2 −1 1 2 x 48. y 4 3 2 1 −1 1 2 x 224 49. y 4 3 2 1 Linear and Quadratic Functions 50. y 20 15 10 5 −2 −1 1 2 3 x −4 −3 −2 −1 1 2 3 x −1 −2 −3 −4 2.5 Regression 2.5 Regression 225 We have seen examples already in the text where linear and quadratic functions are used to model a wide variety of real world phenomena ranging from production costs to the height of a projectile above the ground. In this section, we use some basic tools from statistical analysis to quantify linear and quadratic trends that we may see in real world data in
order to generate linear and quadratic models. Our goal is to give the reader an understanding of the basic processes involved, but we are quick to refer the reader to a more advanced course1 for a complete exposition of this material. Suppose we collected three data points: {(1, 2), (3, 1), (4, 3)}. By plotting these points, we can clearly see that they do not lie along the same line. If we pick any two of the points, we can ﬁnd a line containing both which completely misses the third, but our aim is to ﬁnd a line which is in some sense ‘close’ to all the points, even though it may go through none of them. The way we measure ‘closeness’ in this case is to ﬁnd the total squared error between the data points and the line. Consider our three data points and the line y = 1 2. For each of our data points, we ﬁnd the vertical distance between the point and the line. To accomplish this, we need to ﬁnd a point on the line directly above or below each data point - in other words, a point on the line with the same x-coordinate as our data point. For example, to ﬁnd the point on the line directly below (1, 2), we plug x = 1 into y = 1 2 and we get the point (1, 1). Similarly, we get (3, 1) to correspond to (3, 2) and 4, 5 2 for (4, 3). We ﬁnd the total squared error E by taking the sum of the squares of the diﬀerences of the ycoordinates of each data point and its corresponding point on the line. For the data and line above E = (2 − 1)2 + (1 − 2)2 + 3 − 5 4. Using advanced mathematical machinery,2 it is possible to 2 ﬁnd the line which results in the lowest value of E. This line is called the least squares regression line, or sometimes the ‘line of best ﬁt’. The formula for the line of best ﬁt requires notation we won’t present until Chapter 9.1, so we will revisit it then. The graphing calculator can come to our assistance here, since it has a built-in feature to compute the regression line. We enter the data
and perform the Linear Regression feature and we get = 9 1and authors with more expertise in this area, 2Like Calculus and Linear Algebra 226 Linear and Quadratic Functions The calculator tells us that the line of best ﬁt is y = ax + b where the slope is a ≈ 0.214 and the y-coordinate of the y-intercept is b ≈ 1.428. (We will stick to using three decimal places for our approximations.) Using this line, we compute the total squared error for our data to be E ≈ 1.786. The value r is the correlation coeﬃcient and is a measure of how close the data is to being on the same line. The closer \|r\| is to 1, the better the linear ﬁt. Since r ≈ 0.327, this tells us that the line of best ﬁt doesn’t ﬁt all that well - in other words, our data points aren’t close to being linear. The value r2 is called the coeﬃcient of determination and is also a measure of the goodness of ﬁt.3 Plotting the data with its regression line results in the picture below. Our ﬁrst example looks at energy consumption in the US over the past 50 years.4 Year Energy Usage, in Quads5 34.6 45.1 67.8 78.3 84.6 98.9 1950 1960 1970 1980 1990 2000 Example 2.5.1. Using the energy consumption data given above, 1. Plot the data using a graphing calculator. 3We refer the interested reader to a course in Statistics to explore the signiﬁcance of r and r2. 4See this Department of Energy activity 5The unit 1 Quad is 1 Quadrillion = 1015 BTUs, which is enough heat to raise Lake Erie roughly 1◦F 2.5 Regression 227 2. Find the least squares regression line and comment on the goodness of ﬁt. 3. Interpret the slope of the line of best ﬁt. 4. Use the regression line to predict the annual US energy consumption in the year 2013. 5. Use the regression line to predict when the annual consumption will reach 120 Quads. Solution. 1. Entering the data into the calculator gives The data certainly appears to be linear in nature. 2. Performing a linear regression produces We
can tell both from the correlation coeﬃcient as well as the graph that the regression line is a good ﬁt to the data. 3. The slope of the regression line is a ≈ 1.287. To interpret this, recall that the slope is the rate of change of the y-coordinates with respect to the x-coordinates. Since the y-coordinates represent the energy usage in Quads, and the x-coordinates represent years, a slope of positive 1.287 indicates an increase in annual energy usage at the rate of 1.287 Quads per year. 4. To predict the energy needs in 2013, we substitute x = 2013 into the equation of the line of best ﬁt to get y = 1.287(2013) − 2473.890 ≈ 116.841. The predicted annual energy usage of the US in 2013 is approximately 116.841 Quads. 228 Linear and Quadratic Functions 5. To predict when the annual US energy usage will reach 120 Quads, we substitute y = 120 into the equation of the line of best ﬁt to get 120 = 1.287x − 2473.908. Solving for x yields x ≈ 2015.454. Since the regression line is increasing, we interpret this result as saying the annual usage in 2015 won’t yet be 120 Quads, but that in 2016, the demand will be more than 120 Quads. Our next example gives us an opportunity to ﬁnd a nonlinear model to ﬁt the data. According to the National Weather Service, the predicted hourly temperatures for Painesville on March 3, 2009 were given as summarized below. Time Temperature, ◦F 10AM 11AM 12PM 1PM 2PM 3PM 4PM 17 19 21 23 24 24 23 To enter this data into the calculator, we need to adjust the x values, since just entering the numbers could cause confusion. (Do you see why?) We have a few options available to us. Perhaps the easiest is to convert the times into the 24 hour clock time so that 1 PM is 13, 2 PM is 14, etc.. If we enter these data into the graphing calculator and plot the points we get While the beginning of the data looks linear, the temperature begins to fall in the afternoon hours. This sort of behavior reminds us of parabolas, and, sure enough, it is possible to ﬁnd a parabola
of best ﬁt in the same way we found a line of best ﬁt. The process is called quadratic regression and its goal is to minimize the least square error of the data with their corresponding points on the parabola. The calculator has a built in feature for this as well which yields 2.5 Regression 229 The coeﬃcient of determination R2 seems reasonably close to 1, and the graph visually seems to be a decent ﬁt. We use this model in our next example. Example 2.5.2. Using the quadratic model for the temperature data above, predict the warmest temperature of the day. When will this occur? Solution. The maximum temperature will occur at the vertex of the parabola. Recalling the Vertex Formula, Equation 2.4, x = − b 2(−0.321) ≈ 14.741. This corresponds to roughly 2 : 45 PM. To ﬁnd the temperature, we substitute x = 14.741 into y = −0.321x2 + 9.464x − 45.857 to get y ≈ 23.899, or 23.899◦F. 2a ≈ − 9.464 The results of the last example should remind you that regression models are just that, models. Our predicted warmest temperature was found to be 23.899◦F, but our data says it will warm to 24◦F. It’s all well and good to observe trends and guess at a model, but a more thorough investigation into why certain data should be linear or quadratic in nature is usually in order - and that, most often, is the business of scientists. 230 Linear and Quadratic Functions 2.5.1 Exercises 1. According to this website6, the census data for Lake County, Ohio is: Year Population 1970 197200 1980 212801 1990 215499 2000 227511 (a) Find the least squares regression line for these data and comment on the goodness of ﬁt.7 Interpret the slope of the line of best ﬁt. (b) Use the regression line to predict the population of Lake County in 2010. (The recorded ﬁgure from the 2010 census is 230,041) (c) Use the regression line to predict when the population of Lake County will reach 250,000. 2. According to this website8, the census
data for Lorain County, Ohio is: Year Population 1970 256843 1980 274909 1990 271126 2000 284664 (a) Find the least squares regression line for these data and comment on the goodness of ﬁt. Interpret the slope of the line of best ﬁt. (b) Use the regression line to predict the population of Lorain County in 2010. (The recorded ﬁgure from the 2010 census is 301,356) (c) Use the regression line to predict when the population of Lake County will reach 325,000. 3. Using the energy production data given below Year Production (in Quads) 1950 1960 1970 1980 1990 2000 35.6 42.8 63.5 67.2 70.7 71.2 (a) Plot the data using a graphing calculator and explain why it does not appear to be linear. (b) Discuss with your classmates why ignoring the ﬁrst two data points may be justiﬁed from a historical perspective. (c) Find the least squares regression line for the last four data points and comment on the goodness of ﬁt. Interpret the slope of the line of best ﬁt. (d) Use the regression line to predict the annual US energy production in the year 2010. (e) Use the regression line to predict when the annual US energy production will reach 100 Quads. 6http://www.ohiobiz.com/census/Lake.pdf 7We’ll develop more sophisticated models for the growth of populations in Chapter 6. For the moment, we use a theorem from Calculus to approximate those functions with lines. 8http://www.ohiobiz.com/census/Lorain.pdf 2.5 Regression 231 4. The chart below contains a portion of the fuel consumption information for a 2002 Toyota Echo that I (Jeﬀ) used to own. The ﬁrst row is the cumulative number of gallons of gasoline that I had used and the second row is the odometer reading when I reﬁlled the gas tank. So, for example, the fourth entry is the point (28.25, 1051) which says that I had used a total of 28.25 gallons of gasoline when the odometer read 1051 miles. Gasoline Used (Gallons) Odometer (Miles) 0 9.26 19.03 28.25 36.45 44.
64 53.57 62.62 71.93 81.69 90.43 41 356 731 1051 1347 1631 1966 2310 2670 3030 3371 Find the least squares line for this data. Is it a good ﬁt? What does the slope of the line represent? Do you and your classmates believe this model would have held for ten years had I not crashed the car on the Turnpike a few years ago? (I’m keeping a fuel log for my 2006 Scion xA for future College Algebra books so I hope not to crash it, too.) 5. On New Year’s Day, I (Jeﬀ, again) started weighing myself every morning in order to have an interesting data set for this section of the book. (Discuss with your classmates if that makes me a nerd or a geek. Also, the professionals in the ﬁeld of weight management strongly discourage weighing yourself every day. When you focus on the number and not your overall health, you tend to lose sight of your objectives. I was making a noble sacriﬁce for science, but you should not try this at home.) The whole chart would be too big to put into the book neatly, so I’ve decided to give only a small portion of the data to you. This then becomes a Civics lesson in honesty, as you shall soon see. There are two charts given below. One has my weight for the ﬁrst eight Thursdays of the year (January 1, 2009 was a Thursday and we’ll count it as Day 1.) and the other has my weight for the ﬁrst 10 Saturdays of the year. Day # (Thursday) My weight in pounds Day # (Saturday) My weight in pounds 1 8 15 22 29 36 43 50 238.2 237.0 235.6 234.4 233.0 233.8 232.8 232.0 3 10 17 24 31 38 45 52 59 66 238.4 235.8 235.0 234.2 236.2 236.2 235.2 233.2 236.8 238.2 (a) Find the least squares line for the Thursday data and comment on its goodness of ﬁt. (b) Find the least squares line for the Saturday data and comment on its goodness of ﬁt. (c) Use Quadratic Regression to ﬁnd a parabola which models the Saturday data
and com- ment on its goodness of ﬁt. (d) Compare and contrast the predictions the three models make for my weight on January 1, 2010 (Day #366). Can any of these models be used to make a prediction of my weight 20 years from now? Explain your answer. 232 Linear and Quadratic Functions (e) Why is this a Civics lesson in honesty? Well, compare the two linear models you obtained above. One was a good ﬁt and the other was not, yet both came from careful selections of real data. In presenting the tables to you, I have not lied about my weight, nor have you used any bad math to falsify the predictions. The word we’re looking for here is ‘disingenuous’. Look it up and then discuss the implications this type of data manipulation could have in a larger, more complex, politically motivated setting. (Even Obi-Wan presented the truth to Luke only “from a certain point of view.”) 6. (Data that is neither linear nor quadratic.) We’ll close this exercise set with two data sets that, for reasons presented later in the book, cannot be modeled correctly by lines or parabolas. It is a good exercise, though, to see what happens when you attempt to use a linear or quadratic model when it’s not appropriate. (a) This ﬁrst data set came from a Summer 2003 publication of the Portage County Animal Protective League called “Tattle Tails”. They make the following statement and then have a chart of data that supports it. “It doesn’t take long for two cats to turn into 80 million. If two cats and their surviving oﬀspring reproduced for ten years, you’d end up with 80,399,780 cats.” We assume N (0) = 2. Year x Number of Cats N (x 10 12 66 382 2201 12680 73041 420715 2423316 13968290 80399780 Use Quadratic Regression to ﬁnd a parabola which models this data and comment on its goodness of ﬁt. (Spoiler Alert: Does anyone know what type of function we need here?) (b) This next data set comes from the U.S. Naval Observatory. That site has loads of awesome stuﬀ on it, but for
this exercise I used the sunrise/sunset times in Fairbanks, Alaska for 2009 to give you a chart of the number of hours of daylight they get on the 21st of each month. We’ll let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Use Quadratic Regression to ﬁnd a parabola which models this data and comment on its goodness of ﬁt. (Spoiler Alert: Does anyone know what type of function we need here?) 2.5 Regression 2.5.2 Answers 233 1. (a) y = 936.31x − 1645322.6 with r = 0.9696 which indicates a good ﬁt. The slope 936.31 indicates Lake County’s population is increasing at a rate of (approximately) 936 people per year. (b) According to the model, the population in 2010 will be 236,660. (c) According to the model, the population of Lake County will reach 250,000 sometime between 2024 and 2025. 2. (a) y = 796.8x − 1309762.5 with r = 0.8916 which indicates a reasonable ﬁt. The slope 796.8 indicates Lorain County’s population is increasing at a rate of (approximately) 797 people per year. (b) According to the model, the population in 2010 will be 291,805. (c) According to the model, the population of Lake County will reach 325,000 sometime between 2051 and 2052. 3. (c) y = 0.266x−459.86 with r = 0.9607 which indicates a good ﬁt. The slope 0.266 indicates the country’s energy production is increasing at a rate of 0.266 Quad per year. (d) According to the model, the production in 2010 will be 74.8 Quad. (e) According to the model, the production will reach 100 Quad in the year 2105. 4. The line is y = 36.8x + 16.39. We have r =.99987 and r2
=.9997 so this is an excellent ﬁt to the data. The slope 36.8 represents miles per gallon. 5. (a) The line for the Thursday data is y = −.12x + 237.69. We have r = −.9568 and r2 =.9155 so this is a really good ﬁt. (b) The line for the Saturday data is y = −0.000693x + 235.94. We have r = −0.008986 and r2 = 0.0000807 which is horrible. This data is not even close to linear. (c) The parabola for the Saturday data is y = 0.003x2−0.21x+238.30. We have R2 =.47497 which isn’t good. Thus the data isn’t modeled well by a quadratic function, either. (d) The Thursday linear model had my weight on January 1, 2010 at 193.77 pounds. The Saturday models give 235.69 and 563.31 pounds, respectively. The Thursday line has my weight going below 0 pounds in about ﬁve and a half years, so that’s no good. The quadratic has a positive leading coeﬃcient which would mean unbounded weight gain for the rest of my life. The Saturday line, which mathematically does not ﬁt the data at all, yields a plausible weight prediction in the end. I think this is why grown-ups talk about “Lies, Damned Lies and Statistics.” 6. (a) The quadratic model for the cats in Portage county is y = 1917803.54x2−16036408.29x+ 24094857.7. Although R2 =.70888 this is not a good model because it’s so far oﬀ for small values of x. Case in point, the model gives us 24,094,858 cats when x = 0 but we know N (0) = 2. 234 Linear and Quadratic Functions (b) The quadratic model for the hours of daylight in Fairbanks, Alaska is y =.51x2 +6.23x−.36. Even with R2 =.92295 we should be wary of making predictions beyond the data. Case in point, the model gives −4.84 hours of daylight
when x = 13. So January 21, 2010 will be “extra dark”? Obviously a parabola pointing down isn’t telling us the whole story. Chapter 3 Polynomial Functions 3.1 Graphs of Polynomials Three of the families of functions studied thus far – constant, linear and quadratic – belong to a much larger group of functions called polynomials. We begin our formal study of general polynomials with a deﬁnition and some examples. Deﬁnition 3.1. A polynomial function is a function of the form f (x) = anxn + an−1xn−1 +... + a2x2 + a1x + a0, where a0, a1,..., an are real numbers and n ≥ 1 is a natural number. The domain of a polynomial function is (−∞, ∞). There are several things about Deﬁnition 3.1 that may be oﬀ-putting or downright frightening. The best thing to do is look at an example. Consider f (x) = 4x5 − 3x2 + 2x − 5. Is this a polynomial function? We can re-write the formula for f as f (x) = 4x5 + 0x4 + 0x3 + (−3)x2 + 2x + (−5). Comparing this with Deﬁnition 3.1, we identify n = 5, a5 = 4, a4 = 0, a3 = 0, a2 = −3, a1 = 2 and a0 = −5. In other words, a5 is the coeﬃcient of x5, a4 is the coeﬃcient of x4, and so forth; the subscript on the a’s merely indicates to which power of x the coeﬃcient belongs. The business of restricting n to be a natural number lets us focus on well-behaved algebraic animals.1 Example 3.1.1. Determine if the following functions are polynomials. Explain your reasoning. 1. g(x) = 4 + x3 x √ 4. f (x) = 3 x 2. p(x) = 4x + x3 x 3. q(x) = 4x + x3 x
2 + 4 5. h(x) = \|x\| 6. z(x) = 0 1Enjoy this while it lasts. Before we’re through with the book, you’ll have been exposed to the most terrible of algebraic beasts. We will tame them all, in time. 236 Solution. Polynomial Functions 1. We note directly that the domain of g(x) = x3+4 x is x = 0. By deﬁnition, a polynomial has all real numbers as its domain. Hence, g can’t be a polynomial. 2. Even though p(x) = x3+4x simpliﬁes to p(x) = x2 + 4, which certainly looks like the form given in Deﬁnition 3.1, the domain of p, which, as you may recall, we determine before we simplify, excludes 0. Alas, p is not a polynomial function for the same reason g isn’t. x 3. After what happened with p in the previous part, you may be a little shy about simplifying q(x) = x3+4x x2+4 to q(x) = x, which certainly ﬁts Deﬁnition 3.1. If we look at the domain of q before we simpliﬁed, we see that it is, indeed, all real numbers. A function which can be written in the form of Deﬁnition 3.1 whose domain is all real numbers is, in fact, a polynomial. √ 4. We can rewrite f (x) = 3 x as f (x) = x polynomial. 1 3. Since 1 3 is not a natural number, f is not a 5. The function h(x) = \|x\| isn’t a polynomial, since it can’t be written as a combination of powers of x even though it can be written as a piecewise function involving polynomials. As we shall see in this section, graphs of polynomials possess a quality2 that the graph of h does not. 6. There’s nothing in Deﬁnition 3.1 which prevents all the coeﬃcients an, etc., from being 0. Hence, z(x) = 0, is an honest-to-
goodness polynomial. Deﬁnition 3.2. Suppose f is a polynomial function. Given f (x) = anxn + an−1xn−1 +... + a2x2 + a1x + a0 with an = 0, we say – The natural number n is called the degree of the polynomial f. – The term anxn is called the leading term of the polynomial f. – The real number an is called the leading coeﬃcient of the polynomial f. – The real number a0 is called the constant term of the polynomial f. If f (x) = a0, and a0 = 0, we say f has degree 0. If f (x) = 0, we say f has no degree.a aSome authors say f (x) = 0 has degree −∞ for reasons not even we will go into. The reader may well wonder why we have chosen to separate oﬀ constant functions from the other polynomials in Deﬁnition 3.2. Why not just lump them all together and, instead of forcing n to be a natural number, n = 1, 2,..., allow n to be a whole number, n = 0, 1, 2,.... We could unify all 2One which really relies on Calculus to verify. 3.1 Graphs of Polynomials 237 of the cases, since, after all, isn’t a0x0 = a0? The answer is ‘yes, as long as x = 0.’ The function f (x) = 3 and g(x) = 3x0 are diﬀerent, because their domains are diﬀerent. The number f (0) = 3 is deﬁned, whereas g(0) = 3(0)0 is not.3 Indeed, much of the theory we will develop in this chapter doesn’t include the constant functions, so we might as well treat them as outsiders from the start. One good thing that comes from Deﬁnition 3.2 is that we can now think of linear functions as degree 1 (or ‘ﬁrst degree’) polynomial functions and quadratic functions as degree 2 (or ‘second degree’) polynomial functions. Example 3
.1.2. Find the degree, leading term, leading coeﬃcient and constant term of the following polynomial functions. 1. f (x) = 4x5 − 3x2 + 2x − 5 2. g(x) = 12x + x3 3. h(x) = 4 − x 5 Solution. 4. p(x) = (2x − 1)3(x − 2)(3x + 2) 1. There are no surprises with f (x) = 4x5 − 3x2 + 2x − 5. It is written in the form of Deﬁnition 3.2, and we see that the degree is 5, the leading term is 4x5, the leading coeﬃcient is 4 and the constant term is −5. 2. The form given in Deﬁnition 3.2 has the highest power of x ﬁrst. To that end, we re-write g(x) = 12x + x3 = x3 + 12x, and see that the degree of g is 3, the leading term is x3, the leading coeﬃcient is 1 and the constant term is 0. 3. We need to rewrite the formula for h so that it resembles the form given in Deﬁnition 3.2: 5 x, the leading 5. The degree of h is 1, the leading term is − 1 5 = − 1 h(x) = 4−x 5 = 4 coeﬃcient is − 1 5 − x 5 x + 4 5 and the constant term is 4 5. 4. It may seem that we have some work ahead of us to get p in the form of Deﬁnition 3.2. However, it is possible to glean the information requested about p without multiplying out the entire expression (2x − 1)3(x − 2)(3x + 2). The leading term of p will be the term which has the highest power of x. The way to get this term is to multiply the terms with the highest power of x from each factor together - in other words, the leading term of p(x) is the product of the leading terms of the factors of p(x). Hence, the leading term of p is (2x)3(x)(3x) = 24x5. This means that the degree of p
is 5 and the leading coeﬃcient is 24. As for the constant term, we can perform a similar trick. The constant term is obtained by multiplying the constant terms from each of the factors (−1)3(−2)(2) = 4. Our next example shows how polynomials of higher degree arise ‘naturally’4 in even the most basic geometric applications. 3Technically, 00 is an indeterminant form, which is a special case of being undeﬁned. The authors realize this is beyond pedantry, but we wouldn’t mention it if we didn’t feel it was neccessary. 4this is a dangerous word... 238 Polynomial Functions Example 3.1.3. A box with no top is to be fashioned from a 10 inch × 12 inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. Let x denote the length of the side of the square which is removed from each corner. 12 in x x x x x x x x 10 in height depth width 1. Find the volume V of the box as a function of x. Include an appropriate applied domain. 2. Use a graphing calculator to graph y = V (x) on the domain you found in part 1 and approximate the dimensions of the box with maximum volume to two decimal places. What is the maximum volume? Solution. 1. From Geometry, we know that Volume = width × height × depth. The key is to ﬁnd each of these quantities in terms of x. From the ﬁgure, we see that the height of the box is x itself. The cardboard piece is initially 10 inches wide. Removing squares with a side length of x inches from each corner leaves 10 − 2x inches for the width.5 As for the depth, the cardboard is initially 12 inches long, so after cutting out x inches from each side, we would have 12 − 2x inches remaining. As a function6 of x, the volume is V (x) = x(10 − 2x)(12 − 2x) = 4x3 − 44x2 + 120x To ﬁnd a suitable applied domain, we note that to make a box at all we need x > 0. Also the shorter of the two dimensions of the cardboard is 10 inches, and since we are removing 2x inches from this dimension, we also require 10 − 2
x > 0 or x < 5. Hence, our applied domain is 0 < x < 5. 2. Using a graphing calculator, we see that the graph of y = V (x) has a relative maximum. For 0 < x < 5, this is also the absolute maximum. Using the ‘Maximum’ feature of the calculator, we get x ≈ 1.81, y ≈ 96.77. This yields a height of x ≈ 1.81 inches, a width of 10 − 2x ≈ 6.38 inches, and a depth of 12 − 2x ≈ 8.38 inches. The y-coordinate is the maximum volume, which is approximately 96.77 cubic inches (also written in3). 5There’s no harm in taking an extra step here and making sure this makes sense. If we chopped out a 1 inch square from each side, then the width would be 8 inches, so chopping out x inches would leave 10 − 2x inches. 6When we write V (x), it is in the context of function notation, not the volume V times the quantity x. 3.1 Graphs of Polynomials 239 In order to solve Example 3.1.3, we made good use of the graph of the polynomial y = V (x), so we ought to turn our attention to graphs of polynomials in general. Below are the graphs of y = x2, y = x4 and y = x6, side-by-side. We have omitted the axes to allow you to see that as the exponent increases, the ‘bottom’ becomes ‘ﬂatter’ and the ‘sides’ become ‘steeper.’ If you take the the time to graph these functions by hand,7 you will see why. y = x2 y = x4 y = x6 All of these functions are even, (Do you remember how to show this?) and it is exactly because the exponent is even.8 This symmetry is important, but we want to explore a diﬀerent yet equally important feature of these functions which we can be seen graphically – their end behavior. The end behavior of a function is a way to describe what is happening to the function values (the y-values) as the x-values approach the ‘ends’ of the x-axis.9 That is, what happens to y as x becomes small without bound10
(written x → −∞) and, on the ﬂip side, as x becomes large without bound11 (written x → ∞). For example, given f (x) = x2, as x → −∞, we imagine substituting x = −100, x = −1000, etc., into f to get f (−100) = 10000, f (−1000) = 1000000, and so on. Thus the function values are becoming larger and larger positive numbers (without bound). To describe this behavior, we write: as x → −∞, f (x) → ∞. If we study the behavior of f as x → ∞, we see that in this case, too, f (x) → ∞. (We told you that the symmetry was important!) The same can be said for any function of the form f (x) = xn where n is an even natural number. If we generalize just a bit to include vertical scalings and reﬂections across the x-axis,12 we have 7Make sure you choose some x-values between −1 and 1. 8Herein lies one of the possible origins of the term ‘even’ when applied to functions. 9Of course, there are no ends to the x-axis. 10We think of x as becoming a very large (in the sense of its absolute value) negative number far to the left of zero. 11We think of x as moving far to the right of zero and becoming a very large positive number. 12See Theorems 1.4 and 1.5 in Section 1.7. 240 Polynomial Functions End Behavior of functions f (x) = axn, n even. Suppose f (x) = axn where a = 0 is a real number and n is an even natural number. The end behavior of the graph of y = f (x) matches one of the following: for a > 0, as x → −∞, f (x) → ∞ and as x → ∞, f (x) → ∞ for a < 0, as x → −∞, f (x) → −∞ and as x → ∞, f (x) → −∞ Graphically: a > 0 a < 0 We now turn our attention to functions of the form f (x) = xn where n ≥ 3 is an odd natural number. (We ignore the
case when n = 1, since the graph of f (x) = x is a line and doesn’t ﬁt the general pattern of higher-degree odd polynomials.) Below we have graphed y = x3, y = x5, and y = x7. The ‘ﬂattening’ and ‘steepening’ that we saw with the even powers presents itself here as well, and, it should come as no surprise that all of these functions are odd.13 The end behavior of these functions is all the same, with f (x) → −∞ as x → −∞ and f (x) → ∞ as x → ∞. y = x3 y = x5 y = x7 As with the even degreed functions we studied earlier, we can generalize their end behavior. End Behavior of functions f (x) = axn, n odd. Suppose f (x) = axn where a = 0 is a real number and n ≥ 3 is an odd natural number. The end behavior of the graph of y = f (x) matches one of the following: for a > 0, as x → −∞, f (x) → −∞ and as x → ∞, f (x) → ∞ for a < 0, as x → −∞, f (x) → ∞ and as x → ∞, f (x) → −∞ Graphically: a > 0 a < 0 13And are, perhaps, the inspiration for the moniker ‘odd function’. 3.1 Graphs of Polynomials 241 Despite having diﬀerent end behavior, all functions of the form f (x) = axn for natural numbers n share two properties which help distinguish them from other animals in the algebra zoo: they are continuous and smooth. While these concepts are formally deﬁned using Calculus,14 informally, graphs of continuous functions have no ‘breaks’ or ‘holes’ in them, and the graphs of smooth functions have no ‘sharp turns’. It turns out that these traits are preserved when functions are added together, so general polynomial functions inherit these qualities. Below we ﬁnd the graph of a function which is neither smooth nor continuous, and to its right we have a graph of a polynomial, for comparison. The function whose graph appears on
the left fails to be continuous where it has a ‘break’ or ‘hole’ in the graph; everywhere else, the function is continuous. The function is continuous at the ‘corner’ and the ‘cusp’, but we consider these ‘sharp turns’, so these are places where the function fails to be smooth. Apart from these four places, the function is smooth and continuous. Polynomial functions are smooth and continuous everywhere, as exhibited in the graph on the right. ‘hole’ ‘corner’ ‘cusp’ ‘break’ Pathologies not found on graphs of polynomials The graph of a polynomial The notion of smoothness is what tells us graphically that, for example, f (x) = \|x\|, whose graph is the characteristic ‘∨’ shape, cannot be a polynomial. The notion of continuity is what allowed us to construct the sign diagram for quadratic inequalities as we did in Section 2.4. This last result is formalized in the following theorem. Theorem 3.1. The Intermediate Value Theorem (Zero Version): Suppose f is a continuous function on an interval containing x = a and x = b with a < b. If f (a) and f (b) have diﬀerent signs, then f has at least one zero between x = a and x = b; that is, for at least one real number c such that a < c < b, we have f (c) = 0. The Intermediate Value Theorem is extremely profound; it gets to the heart of what it means to be a real number, and is one of the most often used and under appreciated theorems in Mathematics. With that being said, most students see the result as common sense since it says, geometrically, that the graph of a polynomial function cannot be above the x-axis at one point and below the x-axis at another point without crossing the x-axis somewhere in between. The following example uses the Intermediate Value Theorem to establish a fact that that most students take for granted. Many students, and sadly some instructors, will ﬁnd it silly. 14In fact, if you take Calculus, you’ll ﬁnd that smooth functions are automatically continuous, so that saying ‘polynomials are continuous and smooth’ is redundant. 242
Polynomial Functions Example 3.1.4. Use the Intermediate Value Theorem to establish that Solution. Consider the polynomial function f (x) = x2 − 2. Then f (1) = −1 and f (3) = 7. Since f (1) and f (3) have diﬀerent signs, the Intermediate Value Theorem guarantees us a real number c between 1 and 3 with f (c) = 0. If c2 − 2 = 0 then c = ± 2. Since c is between 1 and 3, c is positive, so c = 2 is a real number. √ √ 2. √ Our primary use of the Intermediate Value Theorem is in the construction of sign diagrams, as in Section 2.4, since it guarantees us that polynomial functions are always positive (+) or always negative (−) on intervals which do not contain any of its zeros. The general algorithm for polynomials is given below. Steps for Constructing a Sign Diagram for a Polynomial Function Suppose f is a polynomial function. 1. Find the zeros of f and place them on the number line with the number 0 above them. 2. Choose a real number, called a test value, in each of the intervals determined in step 1. 3. Determine the sign of f (x) for each test value in step 2, and write that sign above the corresponding interval. Example 3.1.5. Construct a sign diagram for f (x) = x3(x − 3)2(x + 2) x2 + 1. Use it to give a rough sketch of the graph of y = f (x). Solution. First, we ﬁnd the zeros of f by solving x3(x − 3)2(x + 2) x2 + 1 = 0. We get x = 0, x = 3 and x = −2. (The equation x2 + 1 = 0 produces no real solutions.) These three points divide the real number line into four intervals: (−∞, −2), (−2, 0), (0, 3) and (3, ∞). We select the test values x = −3, x = −1, x = 1 and x = 4. We ﬁnd f (−3) is (+), f (−1) is (−) and f (1) is (+) as is f (4). Wherever f
is (+), its graph is above the x-axis; wherever f is (−), its graph is below the x-axis. The x-intercepts of the graph of f are (−2, 0), (0, 0) and (3, 0). Knowing f is smooth and continuous allows us to sketch its graph. y (+) 0 (−) −2 0 (+) 0 (+) 0 3 −3 −1 1 4 x A sketch of y = f (x) A couple of notes about the Example 3.1.5 are in order. First, note that we purposefully did not label the y-axis in the sketch of the graph of y = f (x). This is because the sign diagram gives us the zeros and the relative position of the graph - it doesn’t give us any information as to how high or low the graph strays from the x-axis. Furthermore, as we have mentioned earlier in the text, without Calculus, the values of the relative maximum and minimum can only be found approximately using a calculator. If we took the time to ﬁnd the leading term of f, we would ﬁnd it to be x8. Looking 3.1 Graphs of Polynomials 243 at the end behavior of f, we notice that it matches the end behavior of y = x8. This is no accident, as we ﬁnd out in the next theorem. Theorem 3.2. End Behavior for Polynomial Functions: The end behavior of a polynomial f (x) = anxn +an−1xn−1 +...+a2x2 +a1x+a0 with an = 0 matches the end behavior of y = anxn. To see why Theorem 3.2 is true, let’s ﬁrst look at a speciﬁc example. Consider f (x) = 4x3 − x + 5. If we wish to examine end behavior, we look to see the behavior of f as x → ±∞. Since we’re concerned with x’s far down the x-axis, we are far away from x = 0 so can rewrite f (x) for these values of x as f (x) = 4x3 1 − 1 4x2 + 5 4x3 4x2 and 5 1 As x becomes unbounded (in either direction), the terms 0, as the table below
indicates. 4x3 become closer and closer to x −1000 −100 −10 10 100 1000 1 4x2 5 4x3 0.00000025 −0.00000000125 −0.00000125 −0.00125 0.00125 0.00000125 0.00000000125 0.000025 0.0025 0.0025 0.000025 0.00000025 In other words, as x → ±∞, f (x) ≈ 4x3 (1 − 0 + 0) = 4x3, which is the leading term of f. The formal proof of Theorem 3.2 works in much the same way. Factoring out the leading term leaves f (x) = anxn 1 + an−1 anx +... + a2 anxn−2 + a1 anxn−1 + a0 anxn As x → ±∞, any term with an x in the denominator becomes closer and closer to 0, and we have f (x) ≈ anxn. Geometrically, Theorem 3.2 says that if we graph y = f (x) using a graphing calculator, and continue to ‘zoom out’, the graph of it and its leading term become indistinguishable. Below are the graphs of y = 4x3 − x + 5 (the thicker line) and y = 4x3 (the thinner line) in two diﬀerent windows. A view ‘close’ to the origin. A ‘zoomed out’ view. 244 Polynomial Functions Let’s return to the function in Example 3.1.5, f (x) = x3(x−3)2(x+2) x2 + 1, whose sign diagram and graph are reproduced below for reference. Theorem 3.2 tells us that the end behavior is the same as that of its leading term x8. This tells us that the graph of y = f (x) starts and ends above the x-axis. In other words, f (x) is (+) as x → ±∞, and as a result, we no longer need to evaluate f at the test values x = −3 and x = 4. Is there a way to eliminate the need to evaluate f at the other test values? What we would really need to know is how the function behaves near its zeros does it cross through the x-axis at these
points, as it does at x = −2 and x = 0, or does it simply touch and rebound like it does at x = 3. From the sign diagram, the graph of f will cross the x-axis whenever the signs on either side of the zero switch (like they do at x = −2 and x = 0); it will touch when the signs are the same on either side of the zero (as is the case with x = 3). What we need to determine is the reason behind whether or not the sign change occurs. y (+) 0 (−) −2 0 (+) 0 (+) 0 3 −3 −1 1 4 x A sketch of y = f (x) Fortunately, f was given to us in factored form: f (x) = x3(x − 3)2(x + 2). When we attempt to determine the sign of f (−4), we are attempting to ﬁnd the sign of the number (−4)3(−7)2(−2), which works out to be (−)(+)(−) which is (+). If we move to the other side of x = −2, and ﬁnd the sign of f (−1), we are determining the sign of (−1)3(−4)2(+1), which is (−)(+)(+) which gives us the (−). Notice that signs of the ﬁrst two factors in both expressions are the same in f (−4) and f (−1). The only factor which switches sign is the third factor, (x + 2), precisely the factor which gave us the zero x = −2. If we move to the other side of 0 and look closely at f (1), we get the sign pattern (+1)3(−2)2(+3) or (+)(+)(+) and we note that, once again, going from f (−1) to f (1), the only factor which changed sign was the ﬁrst factor, x3, which corresponds to the zero x = 0. Finally, to ﬁnd f (4), we substitute to get (+4)3(+2)2(+5) which is (+)(+)(+) or (+). The sign didn’t change for the middle factor (x − 3)2. Even though this is the factor which corresponds to the zero x = 3, the fact that the quantity is squared kept the sign of the middle
factor the same on either side of 3. If we look back at the exponents on the factors (x + 2) and x3, we see that they are both odd, so as we substitute values to the left and right of the corresponding zeros, the signs of the corresponding factors change which results in the sign of the function value changing. This is the key to the behavior of the function near the zeros. We need a deﬁnition and then a theorem. Deﬁnition 3.3. Suppose f is a polynomial function and m is a natural number. If (x − c)m is a factor of f (x) but (x − c)m+1 is not, then we say x = c is a zero of multiplicity m. Hence, rewriting f (x) = x3(x − 3)2(x + 2) as f (x) = (x − 0)3(x − 3)2(x − (−2))1, we see that x = 0 is a zero of multiplicity 3, x = 3 is a zero of multiplicity 2 and x = −2 is a zero of multiplicity 1. 3.1 Graphs of Polynomials 245 Theorem 3.3. The Role of Multiplicity: Suppose f is a polynomial function and x = c is a zero of multiplicity m. If m is even, the graph of y = f (x) touches and rebounds from the x-axis at (c, 0). If m is odd, the graph of y = f (x) crosses through the x-axis at (c, 0). Our last example shows how end behavior and multiplicity allow us to sketch a decent graph without appealing to a sign diagram. Example 3.1.6. Sketch the graph of f (x) = −3(2x − 1)(x + 1)2 using end behavior and the multiplicity of its zeros. 2 and x = −1 as zeros. To ﬁnd the multiplicity of x = 1 Solution. The end behavior of the graph of f will match that of its leading term. To ﬁnd the leading term, we multiply by the leading terms of each factor to get (−3)(2x)(x)2 = −6x3. This tells us that the graph will start above the x-axis, in Quadrant II, and ﬁn
ish below the x-axis, in Quadrant IV. Next, we ﬁnd the zeros of f. Fortunately for us, f is factored.15 Setting each factor equal to zero gives is x = 1 2 we note that it corresponds to the factor (2x − 1). This isn’t strictly in the form required in Deﬁnition 3.3. If we factor out the 2, however, we get (2x − 1) = 2 x − 1, and we see that the multiplicity of x = 1 2 2 is 1. Since 1 is an odd number, we know from Theorem 3.3 that the graph of f will cross through the x-axis at 1 2, 0. Since the zero x = −1 corresponds to the factor (x + 1)2 = (x − (−1))2, we ﬁnd its multiplicity to be 2 which is an even number. As such, the graph of f will touch and rebound from the x-axis at (−1, 0). Though we’re not asked to, we can ﬁnd the y-intercept by ﬁnding f (0) = −3(2(0) − 1)(0 + 1)2 = 3. Thus (0, 3) is an additional point on the graph. Putting this together gives us the graph below. y x 15Obtaining the factored form of a polynomial is the main focus of the next few sections. 246 3.1.1 Exercises Polynomial Functions In Exercises 1 - 10, ﬁnd the degree, the leading term, the leading coeﬃcient, the constant term and the end behavior of the given polynomial. 1. f (x) = 4 − x − 3x2 3. q(r) = 1 − 16r4 √ 5. f (x) = 3x17 + 22.5x10 − πx7 + 1 3 2. g(x) = 3x5 − 2x2 + x + 1 4. Z(b) = 42b − b3 6. s(t) = −4.9t2 + v0t + s0 7. P (x) = (x − 1)(x − 2)(x − 3)(x − 4) 8. p(t) = −t2(3 − 5t
)(t2 + t + 4) 9. f (x) = −2x3(x + 1)(x + 2)2 10. G(t) = 4(t − 2)2 t + 1 2 In Exercises 11 - 20, ﬁnd the real zeros of the given polynomial and their corresponding multiplicities. Use this information along with a sign chart to provide a rough sketch of the graph of the polynomial. Compare your answer with the result from a graphing utility. 11. a(x) = x(x + 2)2 12. g(x) = x(x + 2)3 13. f (x) = −2(x − 2)2(x + 1) 14. g(x) = (2x + 1)2(x − 3) 15. F (x) = x3(x + 2)2 16. P (x) = (x − 1)(x − 2)(x − 3)(x − 4) 17. Q(x) = (x + 5)2(x − 3)4 18. h(x) = x2(x − 2)2(x + 2)2 19. H(t) = (3 − t)(t2 + 1) 20. Z(b) = b(42 − b2) In Exercises 21 - 26, given the pair of functions f and g, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Track at least three points of your choice through the transformations. State the domain and range of g. 21. f (x) = x3, g(x) = (x + 2)3 + 1 22. f (x) = x4, g(x) = (x + 2)4 + 1 23. f (x) = x4, g(x) = 2 − 3(x − 1)4 24. f (x) = x5, g(x) = −x5 − 3 25. f (x) = x5, g(x) = (x + 1)5 + 10 26. f (x) = x6, g(x) = 8 − x6 27. Use the Intermediate Value Theorem to prove that f (x) = x3 − 9x + 5 has a real zero in each
of the following intervals: [−4, −3], [0, 1] and [2, 3]. 28. Rework Example 3.1.3 assuming the box is to be made from an 8.5 inch by 11 inch sheet of paper. Using scissors and tape, construct the box. Are you surprised?16 16Consider decorating the box and presenting it to your instructor. If done well enough, maybe your instructor will issue you some bonus points. Or maybe not. 3.1 Graphs of Polynomials 247 In Exercises 29 - 31, suppose the revenue R, in thousands of dollars, from producing and selling x hundred LCD TVs is given by R(x) = −5x3 + 35x2 + 155x for 0 ≤ x ≤ 10.07. 29. Use a graphing utility to graph y = R(x) and determine the number of TVs which should be sold to maximize revenue. What is the maximum revenue? 30. Assume that the cost, in thousands of dollars, to produce x hundred LCD TVs is given by C(x) = 200x + 25 for x ≥ 0. Find and simplify an expression for the proﬁt function P (x). (Remember: Proﬁt = Revenue - Cost.) 31. Use a graphing utility to graph y = P (x) and determine the number of TVs which should be sold to maximize proﬁt. What is the maximum proﬁt? 32. While developing their newest game, Sasquatch Attack!, the makers of the PortaBoy (from Example 2.1.5) revised their cost function and now use C(x) =.03x3 − 4.5x2 + 225x + 250, for x ≥ 0. As before, C(x) is the cost to make x PortaBoy Game Systems. Market research indicates that the demand function p(x) = −1.5x + 250 remains unchanged. Use a graphing utility to ﬁnd the production level x that maximizes the proﬁt made by producing and selling x PortaBoy game systems. 33. According to US Postal regulations, a rectangular shipping box must satisfy the inequality “Length + Girth ≤ 130 inches” for Parcel Post and “Length + Girth ≤ 108 inches” for other services. Let’s assume we have a closed rectangular box with a square face of side length
x as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, 4x. (a) Assuming that we’ll be mailing a box via Parcel Post where Length + Girth = 130 inches, express the length of the box in terms of x and then express the volume V of the box in terms of x. (b) Find the dimensions of the box of maximum volume that can be shipped via Parcel Post. (c) Repeat parts 33a and 33b if the box is shipped using “other services”. x x length 248 Polynomial Functions 34. We now revisit the data set from Exercise 6b in Section 2.5. In that exercise, you were given a chart of the number of hours of daylight they get on the 21st of each month in Fairbanks, Alaska based on the 2009 sunrise and sunset data found on the U.S. Naval Observatory website. We let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so on. The chart is given again for reference. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Find cubic (third degree) and quartic (fourth degree) polynomials which model this data and comment on the goodness of ﬁt for each. What can we say about using either model to make predictions about the year 2020? (Hint: Think about the end behavior of polynomials.) Use the models to see how many hours of daylight they got on your birthday and then check the website to see how accurate the models are. Knowing that Sasquatch are largely nocturnal, what days of the year according to your models are going to allow for at least 14 hours of darkness for ﬁeld research on the elusive creatures? 35. An electric circuit is built with a variable resistor installed. For each of the following resistance values (measured in kilo-ohms, kΩ), the corresponding power to the load (measured in milliwatts, mW ) is given in the table below. 17 Resistance: (kΩ) Power: (m
W ) 1.012 1.063 2.199 1.496 3.275 1.610 4.676 1.613 6.805 1.505 9.975 1.314 (a) Make a scatter diagram of the data using the Resistance as the independent variable and Power as the dependent variable. (b) Use your calculator to ﬁnd quadratic (2nd degree), cubic (3rd degree) and quartic (4th degree) regression models for the data and judge the reasonableness of each. (c) For each of the models found above, ﬁnd the predicted maximum power that can be delivered to the load. What is the corresponding resistance value? (d) Discuss with your classmates the limitations of these models - in particular, discuss the end behavior of each. 36. Show that the end behavior of a linear function f (x) = mx + b is as it should be according to the results we’ve established in the section for polynomials of odd degree.18 (That is, show that the graph of a linear function is “up on one side and down on the other” just like the graph of y = anxn for odd numbers n.) 17The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set. 18Remember, to be a linear function, m = 0. 3.1 Graphs of Polynomials 249 37. There is one subtlety about the role of multiplicity that we need to discuss further; speciﬁcally we need to see ‘how’ the graph crosses the x-axis at a zero of odd multiplicity. In the section, we deliberately excluded the function f (x) = x from the discussion of the end behavior of f (x) = xn for odd numbers n and we said at the time that it was due to the fact that f (x) = x didn’t ﬁt the pattern we were trying to establish. You just showed in the previous exercise that the end behavior of a linear function behaves like every other polynomial of odd degree, so what doesn’t f (x) = x do that g(x) = x3 does? It’s the ‘ﬂattening’ for values of x near zero. It is this local behavior that will distinguish between a zero of multiplicity
1 and one of higher odd multiplicity. Look again closely at the graphs of a(x) = x(x + 2)2 and F (x) = x3(x + 2)2 from Exercise 3.1.1. Discuss with your classmates how the graphs are fundamentally diﬀerent at the origin. It might help to use a graphing calculator to zoom in on the origin to see the diﬀerent crossing behavior. Also compare the behavior of a(x) = x(x + 2)2 to that of g(x) = x(x + 2)3 near the point (−2, 0). What do you predict will happen at the zeros of f (x) = (x − 1)(x − 2)2(x − 3)3(x − 4)4(x − 5)5? 38. Here are a few other questions for you to discuss with your classmates. (a) How many local extrema could a polynomial of degree n have? How few local extrema can it have? (b) Could a polynomial have two local maxima but no local minima? (c) If a polynomial has two local maxima and two local minima, can it be of odd degree? Can it be of even degree? (d) Can a polynomial have local extrema without having any real zeros? (e) Why must every polynomial of odd degree have at least one real zero? (f) Can a polynomial have two distinct real zeros and no local extrema? (g) Can an x-intercept yield a local extrema? Can it yield an absolute extrema? (h) If the y-intercept yields an absolute minimum, what can we say about the degree of the polynomial and the sign of the leading coeﬃcient? 250 3.1.2 Answers 1. f (x) = 4 − x − 3x2 Degree 2 Leading term −3x2 Leading coeﬃcient −3 Constant term 4 As x → −∞, f (x) → −∞ As x → ∞, f (x) → −∞ 3. q(r) = 1 − 16r4 Degree 4 Leading term −16r4 Leading coeﬃcient −16 Constant term 1 As r → −∞, q(
r) → −∞ As r → ∞, q(r) → −∞ √ 5. f (x) = 3x17 + 22.5x10 − πx7 + 1 3 √ 3x17 √ Degree 17 Leading term Leading coeﬃcient Constant term 1 3 As x → −∞, f (x) → −∞ As x → ∞, f (x) → ∞ 3 Polynomial Functions 2. g(x) = 3x5 − 2x2 + x + 1 Degree 5 Leading term 3x5 Leading coeﬃcient 3 Constant term 1 As x → −∞, g(x) → −∞ As x → ∞, g(x) → ∞ 4. Z(b) = 42b − b3 Degree 3 Leading term −b3 Leading coeﬃcient −1 Constant term 0 As b → −∞, Z(b) → ∞ As b → ∞, Z(b) → −∞ 6. s(t) = −4.9t2 + v0t + s0 Degree 2 Leading term −4.9t2 Leading coeﬃcient −4.9 Constant term s0 As t → −∞, s(t) → −∞ As t → ∞, s(t) → −∞ 7. P (x) = (x − 1)(x − 2)(x − 3)(x − 4) 8. p(t) = −t2(3 − 5t)(t2 + t + 4) Degree 4 Leading term x4 Leading coeﬃcient 1 Constant term 24 As x → −∞, P (x) → ∞ As x → ∞, P (x) → ∞ Degree 5 Leading term 5t5 Leading coeﬃcient 5 Constant term 0 As t → −∞, p(t) → −∞ As t → ∞, p(t) → ∞ 3.1 Graphs of Polynomials 251 9. f (x) = −2x3(x + 1)(x + 2)2 Degree 6 Leading term −2x6 Leading coeﬃcient −2 Constant term 0 As x → −∞, f (x) → −∞ As x → ∞, f (x)
→ −∞ 10. G(t) = 4(t − 2)2 t + 1 2 Degree 3 Leading term 4t3 Leading coeﬃcient 4 Constant term 8 As t → −∞, G(t) → −∞ As t → ∞, G(t) → ∞ 11. a(x) = x(x + 2)2 x = 0 multiplicity 1 x = −2 multiplicity 2 12. g(x) = x(x + 2)3 x = 0 multiplicity 1 x = −2 multiplicity 3 y y −2 −1 x −2 −1 x 13. f (x) = −2(x − 2)2(x + 1) x = 2 multiplicity 2 x = −1 multiplicity 1 14. g(x) = (2x + 1)2(x − 3) x = − 1 x = 3 multiplicity 1 2 multiplicity 2 y y −2 −1 1 2 x −1 1 2 3 x 252 Polynomial Functions 15. F (x) = x3(x + 2)2 x = 0 multiplicity 3 x = −2 multiplicity 2 y −2 −1 x 17. Q(x) = (x + 5)2(x − 3)4 x = −5 multiplicity 2 x = 3 multiplicity 4 y 16. P (x) = (x − 1)(x − 2)(x − 3)(x − 4) x = 1 multiplicity 1 x = 2 multiplicity 1 x = 3 multiplicity 1 x = 4 multiplicity 1 y 1 2 3 4 x 18. f (x) = x2(x − 2)2(x + 2)2 x = −2 multiplicity 2 x = 0 multiplicity 2 x = 2 multiplicity 2 y −5−4−3−2−1 1 2 3 4 5 x −2 −1 1 2 x 19. H(t) = (3 − t) t2 + 1 x = 3 multiplicity 1 y 1 2 3 t 20. Z(b) = b(42 − b2) √ 42 multiplicity 1 b = − b = 0 multiplicity 1 b = 42 multiplicity 1 √ y −6−5−4−3−2−1 1 2 3 4 5 6 b 3.1 Graphs of Polynomials 253 21. g(x) = (x + 2)
3 + 1 domain: (−∞, ∞) range: (−∞, ∞) 22. g(x) = (x + 2)4 + 1 domain: (−∞, ∞) range: [1, ∞) y x 12 11 10 1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −4 −3 −2 −1 23. g(x) = 2 − 3(x − 1)4 domain: (−∞, ∞) range: (−∞, 2] y 1 2 x 2 1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 y 21 20 19 18 17 16 15 14 13 12 11 10 4 −3 −2 −1 x 24. g(x) = −x5 − 3 domain: (−∞, ∞) range: (−∞, ∞) y 10 1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −1 1 x 254 Polynomial Functions 25. g(x) = (x + 1)5 + 10 domain: (−∞, ∞) range: (−∞, ∞) 26. g(x) = 8 − x6 domain: (−∞, ∞) range: (−∞, 8] y 21 20 19 18 17 16 15 14 13 12 11 10 4 −3 −2 −1 x y 10 1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −1 1 x 27. We have f (−4) = −23, f (−3) = 5, f (0) = 5, f (1) = −3, f (2) = −5 and f (3) = 5 so the Intermediate Value Theorem tells us that f (x) = x3 − 9x + 5 has real zeros in the intervals [−4, −3], [0, 1] and [2, 3]. 28. V (x) = x(8.5 − 2x)(11 − 2x) = 4x3 − 39x2 + 93.5x, 0 < x < 4.25. Volume is maximized when x ≈ 1.58, so the dimensions of the box with maximum volume are: height ≈ 1.58 inches, width ≈ 5.
34 inches, and depth ≈ 7.84 inches. The maximum volume is ≈ 66.15 cubic inches. 29. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x ≈ 6.305 and y ≈ 1115.417. Since x represents the number of TVs sold in hundreds, x = 6.305 corresponds to 630.5 TVs. Since we can’t sell half of a TV, we compare R(6.30) ≈ 1115.415 and R(6.31) ≈ 1115.416, so selling 631 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $1,115,416. 30. P (x) = R(x) − C(x) = −5x3 + 35x2 − 45x − 25, 0 ≤ x ≤ 10.07. 31. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x ≈ 3.897 and y ≈ 35.255. Since x represents the number of TVs sold in hundreds, x = 3.897 corresponds to 389.7 TVs. Since we can’t sell 0.7 of a TV, we compare P (3.89) ≈ 35.254 and P (3.90) ≈ 35.255, so selling 390 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $35,255. 32. Making and selling 71 PortaBoys yields a maximized proﬁt of $5910.67. 3.1 Graphs of Polynomials 255 33. (a) Our ultimate goal is to maximize the volume, so we’ll start with the maximum Length + Girth of 130. This means the length is 130 − 4x. The volume of a rectangular box is always length × width × height so we get V (x) = x2(130 − 4x) = −4x3 + 130x2. (b) Graphing y = V (x) on [0, 33] × [0, 21000] shows a maximum at (21.67, 20342.59) so the dimensions of the box with maximum volume are 21.67in. × 21.67in. × 43.32in. for a
volume of 20342.59in.3. (c) If we start with Length + Girth = 108 then the length is 108 − 4x and the volume is V (x) = −4x3 + 108x2. Graphing y = V (x) on [0, 27] × [0, 11700] shows a maximum at (18.00, 11664.00) so the dimensions of the box with maximum volume are 18.00in. × 18.00in. × 36in. for a volume of 11664.00in.3. (Calculus will conﬁrm that the measurements which maximize the volume are exactly 18in. by 18in. by 36in., however, as I’m sure you are aware by now, we treat all calculator results as approximations and list them as such.) 34. The cubic regression model is p3(x) = 0.0226x3 − 0.9508x2 + 8.615x − 3.446. It has R2 = 0.93765 which isn’t bad. The graph of y = p3(x) in the viewing window [−1, 13] × [0, 24] along with the scatter plot is shown below on the left. Notice that p3 hits the x-axis at about x = 12.45 making this a bad model for future predictions. To use the model to approximate the number of hours of sunlight on your birthday, you’ll have to ﬁgure out what decimal value of x is close enough to your birthday and then plug it into the model. My (Jeﬀ’s) birthday is July 31 which is 10 days after July 21 (x = 7). Assuming 30 days in a month, I think x = 7.33 should work for my birthday and p3(7.33) ≈ 17.5. The website says there will be about 18.25 hours of daylight that day. To have 14 hours of darkness we need 10 hours of daylight. We see that p3(1.96) ≈ 10 and p3(10.05) ≈ 10 so it seems reasonable to say that we’ll have at least 14 hours of darkness from December 21, 2008 (x = 0) to February 21, 2009 (x = 2) and then again from October 21,2009 (x = 10) to December 21, 2009
(x = 12). The quartic regression model is p4(x) = 0.0144x4 −0.3507x3 +2.259x2 −1.571x+5.513. It has R2 = 0.98594 which is good. The graph of y = p4(x) in the viewing window [−1, 15] × [0, 35] along with the scatter plot is shown below on the right. Notice that p4(15) is above 24 making this a bad model as well for future predictions. However, p4(7.33) ≈ 18.71 making it much better at predicting the hours of daylight on July 31 (my birthday). This model says we’ll have at least 14 hours of darkness from December 21, 2008 (x = 0) to about March 1, 2009 (x = 2.30) and then again from October 10, 2009 (x = 9.667) to December 21, 2009 (x = 12). y = p3(x) y = p4(x) 256 Polynomial Functions 35. (a) The scatter plot is shown below with each of the three regression models. (b) The quadratic model is P2(x) = −0.02x2 + 0.241x + 0.956 with R2 = 0.77708. The cubic model is P3(x) = 0.005x3 − 0.103x2 + 0.602x + 0.573 with R2 = 0.98153. The quartic model is P4(x) = −0.000969x4 + 0.0253x3 − 0.240x2 + 0.944x + 0.330 with R2 = 0.99929. (c) The maximums predicted by the three models are P2(5.737) ≈ 1.648, P3(4.232) ≈ 1.657 and P4(3.784) ≈ 1.630, respectively. y = P2(x) y = P3(x) y = P4(x) 3.2 The Factor Theorem and the Remainder Theorem 257 3.2 The Factor Theorem and the Remainder Theorem Suppose we wish to ﬁnd the zeros of f (x) = x3 + 4x2 − 5x − 14. Setting f (
x) = 0 results in the polynomial equation x3 + 4x2 − 5x − 14 = 0. Despite all of the factoring techniques we learned1 in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing calculator, we get The graph suggests that the function has three zeros, one of which is x = 2. It’s easy to show that f (2) = 0, but the other two zeros seem to be less friendly. Even though we could use the ‘Zero’ command to ﬁnd decimal approximations for these, we seek a method to ﬁnd the remaining zeros exactly. Based on our experience, if x = 2 is a zero, it seems that there should be a factor of (x − 2) lurking around in the factorization of f (x). In other words, we should expect that x3 + 4x2 − 5x − 14 = (x − 2) q(x), where q(x) is some other polynomial. How could we ﬁnd such a q(x), if it even exists? The answer comes from our old friend, polynomial division. Dividing x3 + 4x2 − 5x − 14 by x − 2 gives x2 + 6x + 7 x−2 x3 + 4x2 − 5x − 14 −x3 − 2x2 6x2 − 5x −6x2 − 12x) 7x − 14 − (7x − 14) 0 As you may recall, this means x3 + 4x2 − 5x − 14 = (x − 2) x2 + 6x + 7, so to ﬁnd the zeros of f, we now solve (x − 2) x2 + 6x + 7 = 0. We get x − 2 = 0 (which gives us our known zero, x = 2) as well as x2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic Formula to get x = −3 ± 2. The point of this section is to generalize the technique applied here. First up is a friendly reminder of what we can expect when we divide polynomials. √ 1and probably forgot 2pun intended 258 Polynomial Functions Theorem 3.4. Polynomial Division: Suppose d(
x) and p(x) are nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q(x) and r(x), such that p(x) = d(x) q(x) + r(x), where either r(x) = 0 or the degree of r is strictly less than the degree of d. As you may recall, all of the polynomials in Theorem 3.4 have special names. The polynomial p is called the dividend; d is the divisor; q is the quotient; r is the remainder. If r(x) = 0 then d is called a factor of p. The proof of Theorem 3.4 is usually relegated to a course in Abstract Algebra,3 but we can still use the result to establish two important facts which are the basis of the rest of the chapter. Theorem 3.5. The Remainder Theorem: Suppose p is a polynomial of degree at least 1 and c is a real number. When p(x) is divided by x − c the remainder is p(c). The proof of Theorem 3.5 is a direct consequence of Theorem 3.4. When a polynomial is divided by x − c, the remainder is either 0 or has degree less than the degree of x − c. Since x − c is degree 1, the degree of the remainder must be 0, which means the remainder is a constant. Hence, in either case, p(x) = (x − c) q(x) + r, where r, the remainder, is a real number, possibly 0. It follows that p(c) = (c − c) q(c) + r = 0 · q(c) + r = r, so we get r = p(c) as required. There is one last ‘low hanging fruit’4 to collect which we present below. Theorem 3.6. The Factor Theorem: Suppose p is a nonzero polynomial. The real number c is a zero of p if and only if (x − c) is a factor of p(x). The proof of The Factor Theorem is a consequence of what we already know. If (x − c) is a factor of p(x), this means p(x) = (x − c) q(
x) for some polynomial q. Hence, p(c) = (c − c) q(c) = 0, so c is a zero of p. Conversely, if c is a zero of p, then p(c) = 0. In this case, The Remainder Theorem tells us the remainder when p(x) is divided by (x − c), namely p(c), is 0, which means (x − c) is a factor of p. What we have established is the fundamental connection between zeros of polynomials and factors of polynomials. Of the things The Factor Theorem tells us, the most pragmatic is that we had better ﬁnd a more eﬃcient way to divide polynomials by quantities of the form x − c. Fortunately, people like Ruﬃni and Horner have already blazed this trail. Let’s take a closer look at the long division we performed at the beginning of the section and try to streamline it. First oﬀ, let’s change all of the subtractions into additions by distributing through the −1s. 3Yes, Virginia, there are Algebra courses more abstract than this one. 4Jeﬀ hates this expression and Carl included it just to annoy him. 3.2 The Factor Theorem and the Remainder Theorem 259 x2 + 6x + 7 x−2 x3 + 4x2 − 5x −14 −x3+ 2x2 6x2 − 5x −6x2+ 12x 7x −14 −7x+14 0 Next, observe that the terms −x3, −6x2 and −7x are the exact opposite of the terms above them. The algorithm we use ensures this is always the case, so we can omit them without losing any information. Also note that the terms we ‘bring down’ (namely the −5x and −14) aren’t really necessary to recopy, so we omit them, too. x2 + 6x + 7 x−2 x3+4x2− 5x −14 2x2 6x2 12x 7x 14 0 Now, let’s move things up a bit and, for reasons which will become clear in a moment, copy the x3 into the last row. x2 + 6x + 7 x−2 x3+4x2−
5x −14 2x2 12x 14 6x2 0 7x Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of the ﬁrst three terms in the last row by x and adding the results. If you take the time to work back through the original division problem, you will ﬁnd that this is exactly the way we determined the quotient polynomial. This means that we no longer need to write the quotient polynomial down, nor the x in the divisor, to determine our answer. x3 −2 x3+4x2− 5x −14 2x2 12x 14 6x2 0 7x x3 260 Polynomial Functions We’ve streamlined things quite a bit so far, but we can still do more. Let’s take a moment to remind ourselves where the 2x2, 12x and 14 came from in the second row. Each of these terms was obtained by multiplying the terms in the quotient, x2, 6x and 7, respectively, by the −2 in x − 2, then by −1 when we changed the subtraction to addition. Multiplying by −2 then by −1 is the same as multiplying by 2, so we replace the −2 in the divisor by 2. Furthermore, the coeﬃcients of the quotient polynomial match the coeﬃcients of the ﬁrst three terms in the last row, so we now take the plunge and write only the coeﬃcients of the terms to get 2 1 4 −5 −14 14 12 0 7 2 1 6 We have constructed a synthetic division tableau for this polynomial division problem. Let’s rework our division problem using this tableau to see how it greatly streamlines the division process. To divide x3 + 4x2 − 5x − 14 by x − 2, we write 2 in the place of the divisor and the coeﬃcients of x3 + 4x2 − 5x − 14 in for the dividend. Then ‘bring down’ the ﬁrst coeﬃcient of the dividend. 2 1 4 −5 −14 2 1 4 −5 −14 ↓ 1 Next
, take the 2 from the divisor and multiply by the 1 that was ‘brought down’ to get 2. Write this underneath the 4, then add to get 6. 2 1 4 −5 −14 ↓ 2 1 2 1 4 −5 −14 ↓ 2 1 6 Now take the 2 from the divisor times the 6 to get 12, and add it to the −5 to get 7. 2 1 4 −5 −14 ↓ 2 12 1 6 2 1 4 −5 −14 ↓ 2 12 7 1 6 Finally, take the 2 in the divisor times the 7 to get 14, and add it to the −14 to get 0. 2 1 4 −5 −14 ↓ 2 12 14 7 1 6 2 1 4 −5 −14 ↓ 2 14 12 0 7 1 6 3.2 The Factor Theorem and the Remainder Theorem 261 The ﬁrst three numbers in the last row of our tableau are the coeﬃcients of the quotient polynomial. Remember, we started with a third degree polynomial and divided by a ﬁrst degree polynomial, so the quotient is a second degree polynomial. Hence the quotient is x2 + 6x + 7. The number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials by divisors of the form x − c. It is important to note that it works only for these kinds of divisors.5 Also take note that when a polynomial (of degree at least 1) is divided by x − c, the result will be a polynomial of exactly one less degree. Finally, it is worth the time to trace each step in synthetic division back to its corresponding step in long division. While the authors have done their best to indicate where the algorithm comes from, there is no substitute for working through it yourself. Example 3.2.1. Use synthetic division to perform the following polynomial divisions. Find the quotient and the remainder polynomials, then write the dividend, quotient and remainder in the form given in Theorem 3.4. 1. 5x3 − 2x2 + 1 ÷ (x − 3) 2. x3 + 8 ÷ (x + 2) 3. 4 − 8x − 12x2 2x − 3 Solution. 1. When setting up
the synthetic division tableau, we need to enter 0 for the coeﬃcient of x in the dividend. Doing so gives 3 5 −2 ↓ 5 0 15 39 13 39 1 117 118 Since the dividend was a third degree polynomial, the quotient is a quadratic polynomial with coeﬃcients 5, 13 and 39. Our quotient is q(x) = 5x2 + 13x + 39 and the remainder is r(x) = 118. According to Theorem 3.4, we have 5x3 −2x2 +1 = (x−3) 5x2 + 13x + 39+118. 2. For this division, we rewrite x + 2 as x − (−2) and proceed as before −2 8 0 0 1 ↓ −2 4 −8 1 −2 4 0 We get the quotient q(x) = x2 − 2x + 4 and the remainder r(x) = 0. Relating the dividend, quotient and remainder gives x3 + 8 = (x + 2) x2 − 2x + 4. 3. To divide 4 − 8x − 12x2 by 2x − 3, two things must be done. First, we write the dividend in descending powers of x as −12x2 − 8x + 4. Second, since synthetic division works only. Our strategy is to ﬁrst divide for factors of the form x − c, we factor 2x − 3 as 2 x − 3 2. The tableau becomes −12x2 − 8x + 4 by 2, to get −6x2 − 4x + 2. Next, we divide by x − 3 2 5You’ll need to use good old-fashioned polynomial long division for divisors of degree larger than 1. 262 Polynomial Functions 3 2 −6 −4 2 ↓ −9 − 39 2 −6 −13 − 35 2 From this, we get −6x2 − 4x + 2 = x − 3 2. Multiplying both sides by 2 and 2 distributing gives −12x2 − 8x + 4 = (2x − 3) (−6x − 13) − 35. At this stage, we have written −12x2 − 8x + 4 in the form (2x − 3)q(x) + r(x), but how can we be sure the quotient
polynomial is −6x − 13 and the remainder is −35? The answer is the word ‘unique’ in Theorem 3.4. The theorem states that there is only one way to decompose −12x2 − 8x + 4 into a multiple of (2x − 3) plus a constant term. Since we have found such a way, we can be sure it is the only way. (−6x − 13) − 35 The next example pulls together all of the concepts discussed in this section. Example 3.2.2. Let p(x) = 2x3 − 5x + 3. 1. Find p(−2) using The Remainder Theorem. Check your answer by substitution. 2. Use the fact that x = 1 is a zero of p to factor p(x) and then ﬁnd all of the real zeros of p. Solution. 1. The Remainder Theorem states p(−2) is the remainder when p(x) is divided by x − (−2). We set up our synthetic division tableau below. We are careful to record the coeﬃcient of x2 as 0, and proceed as above. −2 0 −5 2 ↓ −4 2 −4 3 8 −6 3 −3 According to the Remainder Theorem, p(−2) = −3. We can check this by direct substitution into the formula for p(x): p(−2) = 2(−2)3 − 5(−2) + 3 = −16 + 10 + 3 = −3. 2. The Factor Theorem tells us that since x = 1 is a zero of p, x − 1 is a factor of p(x). To factor p(x), we divide 1 2 0 −5 ↓ 2 2 2 −3 3 2 −3 0 We get a remainder of 0 which veriﬁes that, indeed, p(1) = 0. Our quotient polynomial is a second degree polynomial with coeﬃcients 2, 2, and −3. So q(x) = 2x2 + 2x − 3. Theorem 3.4 tells us p(x) = (x − 1) 2x2 + 2x − 3. To ﬁnd the remaining real zeros of p, we need to solve 2x2 + 2x − 3 = 0 for x
. Since this doesn’t factor nicely, we use the quadratic formula to ﬁnd that the remaining zeros are x = −1±. 2 √ 7 3.2 The Factor Theorem and the Remainder Theorem 263 In Section 3.1, we discussed the notion of the multiplicity of a zero. Roughly speaking, a zero with multiplicity 2 can be divided twice into a polynomial; multiplicity 3, three times and so on. This is illustrated in the next example. Example 3.2.3. Let p(x) = 4x4 − 4x3 − 11x2 + 12x − 3. Given that x = 1 2, ﬁnd all of the real zeros of p. Solution. We set up for synthetic division. Since we are told the multiplicity of 1 continue our tableau and divide 1 2 into the quotient polynomial 2 is a zero of multiplicity 2 is two, we 1 2 1 2 12 −3 3 0 4 −4 −11 ↓ 4 −2 −12 ↓ 2 0 −12 4 2 −1 −6 6 0 −6 0 From the ﬁrst division, we get 4x4 − 4x3 − 11x2 + 12x − 3 = x − 1 2 second division tells us 4x3 − 2x2 − 12x + 6 = x − 1 2 have 4x4 − 4x3 − 11x2 + 12x − 3 = x − 1 2 4x2 − 12 = 0 and get x = ± 4x3 − 2x2 − 12x + 6. The 4x2 − 12. Combining these results, we 2 4x2 − 12. To ﬁnd the remaining zeros of p, we set √ 3. A couple of things about the last example are worth mentioning. First, the extension of the synthetic division tableau for repeated divisions will be a common site in the sections to come. Typically, we will start with a higher order polynomial and peel oﬀ one zero at a time until we are left with a quadratic, whose roots can always be found using the Quadratic Formula. Secondly, we 3 are found x = ± both factors of p. We can certainly put the Factor Theorem to the test and continue the synthetic division tableau from above to see what happens. 3 are zeros of p. The Factor Theorem
guarantees x − 3 and − 12 −3 3 0 −4 −11 −2 −12 2 −1 −6 6 0 −4 4 2 0 −12 12 3 0 3 3 0 This gives us 4x4 − 4x3 − 11x2 + 12x − 3 = x − 1 2 with the constant in front √ 4), or, when written p(x 264 Polynomial Functions We have shown that p is a product of its leading coeﬃcient times linear factors of the form (x − c) where c are zeros of p. It may surprise and delight the reader that, in theory, all polynomials can be reduced to this kind of factorization. We leave that discussion to Section 3.4, because the zeros may not be real numbers. Our ﬁnal theorem in the section gives us an upper bound on the number of real zeros. Theorem 3.7. Suppose f is a polynomial of degree n ≥ 1. Then f has at most n real zeros, counting multiplicities. Theorem 3.7 is a consequence of the Factor Theorem and polynomial multiplication. Every zero c of f gives us a factor of the form (x − c) for f (x). Since f has degree n, there can be at most n of these factors. The next section provides us some tools which not only help us determine where the real zeros are to be found, but which real numbers they may be. We close this section with a summary of several concepts previously presented. You should take the time to look back through the text to see where each concept was ﬁrst introduced and where each connection to the other concepts was made. Connections Between Zeros, Factors and Graphs of Polynomial Functions Suppose p is a polynomial function of degree n ≥ 1. The following statements are equivalent: The real number c is a zero of p p(c) = 0 x = c is a solution to the polynomial equation p(x) = 0 (x − c) is a factor of p(x) The point (c, 0) is an x-intercept of the graph of y = p(x) 3.2 The Factor Theorem and the Remainder Theorem 265 3.2.1 Exercises In Exercises 1 - 6, use polynomial long division to perform the indicated division. Write the polynomial in
the form p(x) = d(x)q(x) + r(x). 1. 4x2 + 3x − 1 ÷ (x − 3) 2. 2x3 − x + 1 ÷ x2 + x + 1 3. 5x4 − 3x3 + 2x2 − 1 ÷ x2 + 4 4. −x5 + 7x3 − x ÷ x3 − x2 + 1 5. 9x3 + 5 ÷ (2x − 3) 6. 4x2 − x − 23 ÷ x2 − 1 In Exercises 7 - 20 use synthetic division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x). 7. 3x2 − 2x + 1 ÷ (x − 1) 9. 3 − 4x − 2x2 ÷ (x + 1) 11. x3 + 8 ÷ (x + 2) 13. 18x2 − 15x − 25 ÷ x − 5 3 15. 2x3 + x2 + 2x + 1 ÷ x + 1 2 8. x2 − 5 ÷ (x − 5) 10. 4x2 − 5x + 3 ÷ (x + 3) 12. 4x3 + 2x − 3 ÷ (x − 3) 14. 4x2 − 1 ÷ x − 1 2 16. 3x3 − x + 4 ÷ x − 2 3 17. 2x3 − 3x + 1 ÷ x − 1 2 √ 19. x4 − 6x2 + 9 ÷ x − 3 18. 4x4 − 12x3 + 13x2 − 12x + 9 ÷ x − 3 2 20. x6 − 6x4 + 12x2 − 8 ÷ x + √ 2 In Exercises 21 - 30, determine p(c) using the Remainder Theorem for the given polynomial functions and value of c. If p(c) = 0, factor p(x) = (x − c)q(x). 21. p(x) = 2x2 − x + 1, c = 4 22. p(x) = 4x2 − 33x − 180, c = 12 23. p(x) = 2x3 − x + 6, c = −
3 24. p(x) = x3 + 2x2 + 3x + 4, c = −1 25. p(x) = 3x3 − 6x2 + 4x − 8, c = 2 26. p(x) = 8x3 + 12x2 + 6x + 1, c = − 1 2 27. p(x) = x4 − 2x2 + 4, c = 3 2 29. p(x) = x4 + x3 − 6x2 − 7x − 7, c = − √ 28. p(x) = 6x4 − x2 + 2, c = − 2 3 7 30. p(x) = x2 − 4x + 1, c = 2 − √ 3 266 Polynomial Functions In Exercises 31 - 40, you are given a polynomial and one of its zeros. Use the techniques in this section to ﬁnd the rest of the real zeros and factor the polynomial. 31. x3 − 6x2 + 11x − 6, c = 1 32. x3 − 24x2 + 192x − 512, c = 8 33. 3x3 + 4x2 − x − 2, c = 2 3 35. x3 + 2x2 − 3x − 6, c = −2 34. 2x3 − 3x2 − 11x + 6, c = 1 2 36. 2x3 − x2 − 10x + 5, c = 1 2 37. 4x4 − 28x3 + 61x2 − 42x + 9, c = 1 2 is a zero of multiplicity 2 38. x5 + 2x4 − 12x3 − 38x2 − 37x − 12, c = −1 is a zero of multiplicity 3 39. 125x5 − 275x4 − 2265x3 − 3213x2 − 1728x − 324, c = − 3 5 is a zero of multiplicity 3 40. x2 − 2x − 2, c = 1 − √ 3 In Exercises 41 - 45, create a polynomial p which has the desired characteristics. You may leave the polynomial in factored form. 41. 42. 43. 44. 45. The zeros of p are c = ±2 and c = ±1 The leading term of p(x) is 117x4. The
zeros of p are c = 1 and c = 3 c = 3 is a zero of multiplicity 2. The leading term of p(x) is −5x3 The solutions to p(x) = 0 are x = ±3 and x = 6 The leading term of p(x) is 7x4 The point (−3, 0) is a local minimum on the graph of y = p(x). The solutions to p(x) = 0 are x = ±3, x = −2, and x = 4. The leading term of p(x) is −x5. The point (−2, 0) is a local maximum on the graph of y = p(x). p is degree 4. as x → ∞, p(x) → −∞ p has exactly three x-intercepts: (−6, 0), (1, 0) and (117, 0) The graph of y = p(x) crosses through the x-axis at (1, 0). 46. Find a quadratic polynomial with integer coeﬃcients which has x = √ 29 5 3 5 ± as its real zeros. 3.2 The Factor Theorem and the Remainder Theorem 267 3.2.2 Answers 8 4 x + 81 + 283 8 2 x2 + 27 1. 4x2 + 3x − 1 = (x − 3)(4x + 15) + 44 2. 2x3 − x + 1 = x2 + x + 1 (2x − 2) + (−x + 3) 3. 5x4 − 3x3 + 2x2 − 1 = x2 + 4 5x2 − 3x − 18 + (12x + 71) 4. −x5 + 7x3 − x = x3 − x2 + 1 −x2 − x + 6 + 7x2 − 6 5. 9x3 + 5 = (2x − 3) 9 6. 4x2 − x − 23 = x2 − 1 (4) + (−x − 19) 7. 3x2 − 2x + 1 = (x − 1) (3x + 1) + 2 8. x2 − 5 = (x − 5) (x + 5) + 20 9. 3 − 4x − 2x2 = (x + 1) (−2x − 2) + 5 10. 4x2 − 5x
+ 3 = (x + 3) (4x − 17) + 54 11. x3 + 8 = (x + 2) x2 − 2x + 4 + 0 12. 4x3 + 2x − 3 = (x − 3) 4x2 + 12x + 38 + 111 13. 18x2 − 15x − 25 = x − 5 3 (4x + 2) + 0 14. 4x2 − 1 = x − 1 2 15. 2x3 + x2 + 2x + 1 = x + 1 2 16. 3x3 − x + 4 = x − 2 3 17. 2x3 − 3x + 1 = x − 1 2 18. 4x4 − 12x3 + 13x2 − 12x + 9 = x − 3 2 19. x4 − 6x2 + 9 = x − 20. x6 − 6x4 + 12x2 − 8 = x + 2x2 + 2 + 0 + 38 9 − 1 4 3x2 + 2x + 1 3 2x2 + x − 5 2 3 x2 − 3x − 3 √ (18x + 15) + 0 3 x3 + √ 2 x4 − 4x3 + 4 2 x5 − √ √ √ 4x3 − 6x2 + 4x − 6 + 0 3 + 0 √ 2 x2 + 4x − 4 √ 2 + 0 21. p(4) = 29 23. p(−3) = −45 22. p(12) = 0, p(x) = (x − 12)(4x + 15) 24. p(−1) = 2 25. p(2) = 0, p(x) = (x − 2) 3x2 + 4 26. p − 1 2 = 0, p(x) = x + 1 2 8x2 + 8x + 2 268 27. p 3 2 29. p(− = 73 16 √ 7) = 0, p(x) = (x + √ √ 7) x3 + (1 − √ 30. p(2 − 3) = 0, p(x) = (x − (2 − 3))(x − (2 + 3)) √ 28. p − 2 = 74 27 3 √ 7)x − 7)x2 + (1 − √ Polynomial Functions �
� 7 31. x3 − 6x2 + 11x − 6 = (x − 1)(x − 2)(x − 3) 32. x3 − 24x2 + 192x − 512 = (x − 8)3 33. 3x3 + 4x2 − x − 2 = 3 x − 2 3 34. 2x3 − 3x2 − 11x + 6 = 2 x − 1 2 (x + 1)2 (x + 2)(x − 3) √ √ 35. x3 + 2x2 − 3x − 6 = (x + 2)(x + 36. 2x3 − x2 − 10x + 5 = 2 x − 1 2 (x + 3)(x − √ 5)(x − 3) √ 5) 37. 4x4 − 28x3 + 61x2 − 42x + 9 = 4 x − 1 2 2 (x − 3)2 38. x5 + 2x4 − 12x3 − 38x2 − 37x − 12 = (x + 1)3(x + 3)(x − 4) 39. 125x5 − 275x4 − 2265x3 − 3213x2 − 1728x − 324 = 125 x + 3 5 3 (x + 2)(x − 6) 40. x2 − 2x − 2 = (x − (1 − √ 3))(x − (1 + √ 3)) 41. p(x) = 117(x + 2)(x − 2)(x + 1)(x − 1) 42. p(x) = −5(x − 1)(x − 3)2 43. p(x) = 7(x + 3)2(x − 3)(x − 6) 44. p(x) = −(x + 2)2(x − 3)(x + 3)(x − 4) 45. p(x) = a(x + 6)2(x − 1)(x − 117) or p(x) = a(x + 6)(x − 1)(x − 117)2 where a can be any negative real number 46. p(x) = 5x2 − 6x − 4 3.3 Real Zeros of Polynomials 269 3.3 Real Zeros of Polynomials In Section 3.2, we found that we can use synthetic division to determine if a given real number is a zero
of a polynomial function. This section presents results which will help us determine good candidates to test using synthetic division. There are two approaches to the topic of ﬁnding the real zeros of a polynomial. The ﬁrst approach (which is gaining popularity) is to use a little bit of Mathematics followed by a good use of technology like graphing calculators. The second approach (for purists) makes good use of mathematical machinery (theorems) only. For completeness, we include the two approaches but in separate subsections.1 Both approaches beneﬁt from the following two theorems, the ﬁrst of which is due to the famous mathematician Augustin Cauchy. It gives us an interval on which all of the real zeros of a polynomial can be found. Theorem 3.8. Cauchy’s Bound: Suppose f (x) = anxn + an−1xn−1 +... + a1x + a0 is a \|an\|,..., \|an−1\| polynomial of degree n with n ≥ 1. Let M be the largest of the numbers: \|an\|. Then all the real zeros of f lie in in the interval [−(M + 1), M + 1]. \|an\|, \|a1\| \|a0\| The proof of this fact is not easily explained within the conﬁnes of this text. This paper contains the result and gives references to its proof. Like many of the results in this section, Cauchy’s Bound is best understood with an example. Example 3.3.1. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Determine an interval which contains all of the real zeros of f. Solution. To ﬁnd the M stated in Cauchy’s Bound, we take the absolute value of the leading coeﬃcient, in this case \|2\| = 2 and divide it into the largest (in absolute value) of the remaining coeﬃcients, in this case \| − 6\| = 6. This yields M = 3 so it is guaranteed that all of the real zeros of f lie in the interval [−4, 4]. Whereas the previous result tells us where we can ﬁnd the real zeros of a polyn
omial, the next theorem gives us a list of possible real zeros. Theorem 3.9. Rational Zeros Theorem: Suppose f (x) = anxn + an−1xn−1 +... + a1x + a0 is a polynomial of degree n with n ≥ 1, and a0, a1,... an are integers. If r is a rational zero of f, then r is of the form ± p q, where p is a factor of the constant term a0, and q is a factor of the leading coeﬃcient an. The Rational Zeros Theorem gives us a list of numbers to try in our synthetic division and that is a lot nicer than simply guessing. If none of the numbers in the list are zeros, then either the polynomial has no real zeros at all, or all of the real zeros are irrational numbers. To see why the Rational Zeros Theorem works, suppose c is a zero of f and c = p q in lowest terms. This means p and q have no common factors. Since f (c) = 0, we have p q +... + a1 + a0 = 0. p q p q + an−1 n−1 n an 1Carl is the purist and is responsible for all of the theorems in this section. Jeﬀ, on the other hand, has spent too much time in school politics and has been polluted with notions of ‘compromise.’ You can blame the slow decline of civilization on him and those like him who mingle Mathematics with technology. 270 Polynomial Functions Multiplying both sides of this equation by qn, we clear the denominators to get anpn + an−1pn−1q +... + a1pqn−1 + a0qn = 0 Rearranging this equation, we get anpn = −an−1pn−1q −... − a1pqn−1 − a0qn Now, the left hand side is an integer multiple of p, and the right hand side is an integer multiple of q. (Can you see why?) This means anpn is both a multiple of p and a multiple of q. Since p and q have no common factors, an must be a multiple of q. If we rearrange the equation as anpn + an−1pn−1q
+... + a1pqn−1 + a0qn = 0 a0qn = −anpn − an−1pn−1q −... − a1pqn−1 we can play the same game and conclude a0 is a multiple of p, and we have the result. Example 3.3.2. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Use the Rational Zeros Theorem to list all of the possible rational zeros of f. Solution. To generate a complete list of rational zeros, we need to take each of the factors of constant term, a0 = −3, and divide them by each of the factors of the leading coeﬃcient a4 = 2. The factors of −3 are ± 1 and ± 3. Since the Rational Zeros Theorem tacks on a ± anyway, for the moment, we consider only the positive factors 1 and 3. The factors of 2 are 1 and 2, so the Rational Zeros Theorem gives the list ± 1 2, ± 3 2, ± 1, ± 3 or ± 1 2, ± 3 Our discussion now diverges between those who wish to use technology and those who do not. 3.3.1 For Those Wishing to use a Graphing Calculator At this stage, we know not only the interval in which all of the zeros of f (x) = 2x4 +4x3 −x2 −6x−3 are located, but we also know some potential candidates. We can now use our calculator to help us determine all of the real zeros of f, as illustrated in the next example. Example 3.3.3. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. 1. Graph y = f (x) on the calculator using the interval obtained in Example 3.3.1 as a guide. 2. Use the graph to shorten the list of possible rational zeros obtained in Example 3.3.2. 3. Use synthetic division to ﬁnd the real zeros of f, and state their multiplicities. Solution. 1. In Example 3.3.1, we determined all of the real zeros of f lie in the interval [−4, 4]. We set our window accordingly and get 3.3 Real Zeros of Polynomials 271 2
. In Example 3.3.2, we learned that any rational zero of f must be in the list ± 1 2, ± 3. From the graph, it looks as if we can rule out any of the positive rational zeros, since the graph seems to cross the x-axis at a value just a little greater than 1. On the negative side, −1 looks good, so we try that for our synthetic division. 2, ± 1, ± 3 −1 2 ↓ −2 −2 2 4 −1 −6 −3 3 0 3 2 −3 −3 We have a winner! Remembering that f was a fourth degree polynomial, we know that our quotient is a third degree polynomial. If we can do one more successful division, we will have knocked the quotient down to a quadratic, and, if all else fails, we can use the quadratic formula to ﬁnd the last two zeros. Since there seems to be no other rational zeros to try, we continue with −1. Also, the shape of the crossing at x = −1 leads us to wonder if the zero x = −1 has multiplicity 3. −1 −1 4 −1 −6 −3 3 0 2 ↓ −2 −2 2 ↓ −2 2 3 2 −3 −3 3 0 0 −3 0 2, which gives us x = ± Success! Our quotient polynomial is now 2x2 − 3. Setting this to zero gives 2x2 − 3 = 0, or √ 6 x2 = 3 2. Concerning multiplicities, based on our division, we have that −1 has a multiplicity of at least 2. The Factor Theorem tells us our remaining zeros, √ 6 ± 2, each have multiplicity at least 1. However, Theorem 3.7 tells us f can have at most 4 real zeros, counting multiplicity, and so we conclude that −1 is of multiplicity exactly 2 and √ 6 2 each has multiplicity 1. (Thus, we were wrong to think that −1 had multiplicity 3.) ± It is interesting to note that we could greatly improve on the graph of y = f (x) in the previous example given to us by the calculator. For instance, from our determination of the zeros of f and √ 6 their multiplicities, we know the graph crosses at x = − 2 ≈
−1.22 then turns back upwards to touch the x−axis at x = −1. This tells us that, despite what the calculator showed us the ﬁrst time, there is a relative maximum occurring at x = −1 and not a ‘ﬂattened crossing’ as we originally 272 Polynomial Functions believed. After resizing the window, we see not only the relative maximum but also a relative minimum2 just to the left of x = −1 which shows us, once again, that Mathematics enhances the technology, instead of vice-versa. Our next example shows how even a mild-mannered polynomial can cause problems. Example 3.3.4. Let f (x) = x4 + x2 − 12. 1. Use Cauchy’s Bound to determine an interval in which all of the real zeros of f lie. 2. Use the Rational Zeros Theorem to determine a list of possible rational zeros of f. 3. Graph y = f (x) using your graphing calculator. 4. Find all of the real zeros of f and their multiplicities. Solution. 1. Applying Cauchy’s Bound, we ﬁnd M = 12, so all of the real zeros lie in the interval [−13, 13]. 2. Applying the Rational Zeros Theorem with constant term a0 = −12 and leading coeﬃcient a4 = 1, we get the list {± 1, ± 2, ± 3, ± 4, ± 6, ± 12}. 3. Graphing y = f (x) on the interval [−13, 13] produces the graph below on the left. Zooming in a bit gives the graph below on the right. Based on the graph, none of our rational zeros will work. (Do you see why not?) 2This is an example of what is called ‘hidden behavior.’ 3.3 Real Zeros of Polynomials 273 4. From the graph, we know f has two real zeros, one positive, and one negative. Our only hope at this point is to try and ﬁnd the zeros of f by setting f (x) = x4 + x2 − 12 = 0 and solving. If we stare at this equation long enough, we may recognize it as a ‘quadratic in disguise’ or ‘quadr
atic in form’. In other words, we have three terms: x4, x2 and 12, and the exponent on the ﬁrst term, x4, is exactly twice that of the second term, x2. We may rewrite this as x22 + x2 − 12 = 0. To better see the forest for the trees, we momentarily replace x2 with the variable u. In terms of u, our equation becomes u2 + u − 12 = 0, which we can readily factor as (u + 4)(u − 3) = 0. In terms of x, this means x4 + x2 − 12 = x2 − 3 x2 + 4 = 0. 3, or x2 = −4, which admits no real solutions. Since We get x2 = 3, which gives us x = ± √ 3 ≈ 1.73, the two zeros match what we expected from the graph. In terms of multiplicity, 3 are factors of f (x). Since f (x) can the Factor Theorem guarantees x − 3 and x + be factored as f (x) = x2 − 3 x2 + 4, and x2 + 4 has no real zeros, the quantities x − 3 3 must both be factors of x2 − 3. According to Theorem 3.7, x2 − 3 can have at and x + √ most 2 zeros, counting multiplicity, hence each of ± 3 is a zero of f of multiplicity 1. √ √ √ √ √ The technique used to factor f (x) in Example 3.3.4 is called u-substitution. We shall see more of this technique in Section 5.3. In general, substitution can help us identify a ‘quadratic in disguise’ provided that there are exactly three terms and the exponent of the ﬁrst term is exactly twice that of the second. It is entirely possible that a polynomial has no real roots at all, or worse, it has real roots but none of the techniques discussed in this section can help us ﬁnd them exactly. In the latter case, we are forced to approximate, which in this subsection means we use the ‘Zero’ command on the graphing calculator. 3.3.2 For Those Wishing NOT to use a Graphing Calculator Suppose we wish to ﬁnd the zeros of f (x
) = 2x4 + 4x3 − x2 − 6x − 3 without using the calculator. In this subsection, we present some more advanced mathematical tools (theorems) to help us. Our ﬁrst result is due to Ren´e Descartes. Theorem 3.10. Descartes’ Rule of Signs: Suppose f (x) is the formula for a polynomial function written with descending powers of x. If P denotes the number of variations of sign in the formula for f (x), then the number of positive real zeros (counting multiplicity) is one of the numbers {P, P − 2, P − 4,... }. If N denotes the number of variations of sign in the formula for f (−x), then the number of negative real zeros (counting multiplicity) is one of the numbers {N, N − 2, N − 4,... }. A few remarks are in order. First, to use Descartes’ Rule of Signs, we need to understand what is meant by a ‘variation in sign’ of a polynomial function. Consider f (x) = 2x4 + 4x3 − x2 − 6x − 3. If we focus on only the signs of the coeﬃcients, we start with a (+), followed by another (+), then switch to (−), and stay (−) for the remaining two coeﬃcients. Since the signs of the coeﬃcients switched once as we read from left to right, we say that f (x) has one variation in sign. When 274 Polynomial Functions we speak of the variations in sign of a polynomial function f we assume the formula for f (x) is written with descending powers of x, as in Deﬁnition 3.1, and concern ourselves only with the nonzero coeﬃcients. Second, unlike the Rational Zeros Theorem, Descartes’ Rule of Signs gives us an estimate to the number of positive and negative real zeros, not the actual value of the zeros. Lastly, Descartes’ Rule of Signs counts multiplicities. This means that, for example, if one of the zeros has multiplicity 2, Descsartes’ Rule of Signs would count this as two zeros. Lastly, note that the number of positive or
negative real zeros always starts with the number of sign changes and decreases by an even number. For example, if f (x) has 7 sign changes, then, counting multplicities, f has either 7, 5, 3 or 1 positive real zero. This implies that the graph of y = f (x) crosses the positive x-axis at least once. If f (−x) results in 4 sign changes, then, counting multiplicities, f has 4, 2 or 0 negative real zeros; hence, the graph of y = f (x) may not cross the negative x-axis at all. The proof of Descartes’ Rule of Signs is a bit technical, and can be found here. Example 3.3.5. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Use Descartes’ Rule of Signs to determine the possible number and location of the real zeros of f. Solution. As noted above, the variations of sign of f (x) is 1. This means, counting multiplicities, f has exactly 1 positive real zero. Since f (−x) = 2(−x)4 + 4(−x)3 − (−x)2 − 6(−x) − 3 = 2x4 − 4x3 − x2 + 6x − 3 has 3 variations in sign, f has either 3 negative real zeros or 1 negative real zero, counting multiplicities. Cauchy’s Bound gives us a general bound on the zeros of a polynomial function. Our next result helps us determine bounds on the real zeros of a polynomial as we synthetically divide which are often sharper3 bounds than Cauchy’s Bound. Theorem 3.11. Upper and Lower Bounds: Suppose f is a polynomial of degree n ≥ 1. If c > 0 is synthetically divided into f and all of the numbers in the ﬁnal line of the division tableau have the same signs, then c is an upper bound for the real zeros of f. That is, there are no real zeros greater than c. If c < 0 is synthetically divided into f and the numbers in the ﬁnal line of the division tableau alternate signs, then c is a lower bound for the real zeros of f. That is, there are no real zeros less than c.
NOTE: If the number 0 occurs in the ﬁnal line of the division tableau in either of the above cases, it can be treated as (+) or (−) as needed. The Upper and Lower Bounds Theorem works because of Theorem 3.4. For the upper bound part of the theorem, suppose c > 0 is divided into f and the resulting line in the division tableau contains, for example, all nonnegative numbers. This means f (x) = (x − c)q(x) + r, where the coeﬃcients of the quotient polynomial and the remainder are nonnegative. (Note that the leading coeﬃcient of q is the same as f so q(x) is not the zero polynomial.) If b > c, then f (b) = (b − c)q(b) + r, where (b − c) and q(b) are both positive and r ≥ 0. Hence f (b) > 0 which shows b cannot be a zero of f. Thus no real number b > c can be a zero of f, as required. A similar argument proves 3That is, better, or more accurate. 3.3 Real Zeros of Polynomials 275 f (b) < 0 if all of the numbers in the ﬁnal line of the synthetic division tableau are non-positive. To prove the lower bound part of the theorem, we note that a lower bound for the negative real zeros of f (x) is an upper bound for the positive real zeros of f (−x). Applying the upper bound portion to f (−x) gives the result. (Do you see where the alternating signs come in?) With the additional mathematical machinery of Descartes’ Rule of Signs and the Upper and Lower Bounds Theorem, we can ﬁnd the real zeros of f (x) = 2x4 + 4x3 − x2 − 6x − 3 without the use of a graphing calculator. Example 3.3.6. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. 1. Find all of the real zeros of f and their multiplicities. 2. Sketch the graph of y = f (x). Solution. 2, ± 1, ± 3 1. We know from Cauchy’s Bound that all of the
real zeros lie in the interval [−4, 4] and that our possible rational zeros are ± 1 2 and ± 3. Descartes’ Rule of Signs guarantees us at least one negative real zero and exactly one positive real zero, counting multiplicity. We try our positive rational zeros, starting with the smallest, 1 2. Since the remainder isn’t zero, we know 1 2 isn’t a zero. Sadly, the ﬁnal line in the division tableau has both positive and negative numbers, so 1 2 is not an upper bound. The only information we get from this division is courtesy of the Remainder Theorem which tells us f 1 is 2 on the graph of f. We continue to our next possible zero, 1. As before, the only information we can glean from this is that (1, −4) is on the graph of f. When we try our next possible zero, 3 2, we get that it is not a zero, and we also see that it is an upper bound on the zeros of f, since all of the numbers in the ﬁnal line of the division tableau are positive. This means there is no point trying our last possible rational zero, 3. Descartes’ Rule of Signs guaranteed us a positive real zero, and at this point we have shown this zero is irrational. Furthermore, the Intermediate Value Theorem, Theorem 3.1, tells us the zero lies between 1 and 3 2, since f (1) < 0 and f 3 2 8 so the point 1 = − 45 2, − 45 > 0. 8 1 2 2 4 −1 −6 3 5 ↓ 1 4 2 −3 − 21 8 2 5 3 2 − 21 4 − 45 8 1 2 4 −1 −6 −3 5 −1 ↓ 2 6 2 6 5 −1 −4 3 2 2 4 −1 −6 −3 99 ↓ 3 8 57 4 21 2 2 7 19 2 33 4 75 8 We now turn our attention to negative real zeros. We try the largest possible zero, − 1 2. Synthetic division shows us it is not a zero, nor is it a lower bound (since the numbers in the ﬁnal line of the division tableau do not alternate), so we proceed to −1. This division shows −1 is a zero. Descartes’ Rule of Signs told us that we may have up to three
negative real zeros, counting multiplicity, so we try −1 again, and it works once more. At this point, we have taken f, a fourth degree polynomial, and performed two successful divisions. Our quotient polynomial is quadratic, so we look at it to ﬁnd the remaining zeros. 276 − 1 2 4 −1 −6 5 4 2 ↓ −1 − 3 2 3 − 5 2 − 19 2 4 − 5 8 −3 19 8 −1 −1 Polynomial Functions 4 −1 −6 −3 3 0 2 ↓ −2 −2 2 ↓ −2 2 3 2 −3 −3 0 3 0 −3 0 √ 6 Setting the quotient polynomial equal to zero yields 2x2 − 3 = 0, so that x2 = 3 2, or x = ± 2. √ 6 Descartes’ Rule of Signs tells us that the positive real zero we found, 2, has multiplicity 1. Descartes also tells us the total multiplicity of negative real zeros is 3, which forces −1 to be a zero of multiplicity 2 and − √ 6 2 to have multiplicity 1. 2. We know the end behavior of y = f (x) resembles that of its leading term y = 2x4. This √ 6 means that the graph enters the scene in Quadrant II and exits in Quadrant I. Since ± 2 are zeros of odd multiplicity, we have that the graph crosses through the x-axis at the points. Since −1 is a zero of multiplicity 2, the graph of y = f (x) touches and √ − 6 2, 0 √ 6 2, 0 and rebounds oﬀ the x-axis at (−1, 0). Putting this together, we get y x You can see why the ‘no calculator’ approach is not very popular these days. It requires more computation and more theorems than the alternative.4 In general, no matter how many theorems you throw at a polynomial, it may well be impossible5 to ﬁnd their zeros exactly. The polynomial f (x) = x5 − x − 1 is one such beast.6 According to Descartes’ Rule of Signs, f has exactly one positive real zero, and it could have two negative real zeros, or none at all. The Rational Zeros
4This is apparently a bad thing. 5We don’t use this word lightly; it can be proven that the zeros of some polynomials cannot be expressed using the usual algebraic symbols. 6See this page. 3.3 Real Zeros of Polynomials 277 Test gives us ±1 as rational zeros to try but neither of these work since f (1) = f (−1) = −1. If we try the substitution technique we used in Example 3.3.4, we ﬁnd f (x) has three terms, but the exponent on the x5 isn’t exactly twice the exponent on x. How could we go about approximating the positive zero without resorting to the ‘Zero’ command of a graphing calculator? We use the Bisection Method. The ﬁrst step in the Bisection Method is to ﬁnd an interval on which f changes sign. We know f (1) = −1 and we ﬁnd f (2) = 29. By the Intermediate Value Theorem, we know that the zero of f lies in the interval [1, 2]. Next, we ‘bisect’ this interval and ﬁnd the midpoint is 1.5. We have that f (1.5) ≈ 5.09. This means that our zero is between 1 and 1.5, since f changes sign on this interval. Now, we ‘bisect’ the interval [1, 1.5] and ﬁnd f (1.25) ≈ 0.80, so now we have the zero between 1 and 1.25. Bisecting [1, 1.25], we ﬁnd f (1.125) ≈ −0.32, which means the zero of f is between 1.125 and 1.25. We continue in this fashion until we have ‘sandwiched’ the zero between two numbers which diﬀer by no more than a desired accuracy. You can think of the Bisection Method as reversing the sign diagram process: instead of ﬁnding the zeros and checking the sign of f using test values, we are using test values to determine where the signs switch to ﬁnd the zeros. It is a slow and tedious, yet fool-proof, method for approximating a real zero. Our next example reminds us of the role �
��nding zeros plays in solving equations and inequalities. Example 3.3.7. 1. Find all of the real solutions to the equation 2x5 + 6x3 + 3 = 3x4 + 8x2. 2. Solve the inequality 2x5 + 6x3 + 3 ≤ 3x4 + 8x2. 3. Interpret your answer to part 2 graphically, and verify using a graphing calculator. Solution. 1. Finding the real solutions to 2x5 + 6x3 + 3 = 3x4 + 8x2 is the same as ﬁnding the real solutions to 2x5 − 3x4 + 6x3 − 8x2 + 3 = 0. In other words, we are looking for the real zeros of p(x) = 2x5 − 3x4 + 6x3 − 8x2 + 3. Using the techniques developed in this section, we get 1 1 − 1 2 2 −3 ↓ 2 −1 ↓ 2 2 1 ↓ −1 0 2 2 −1 6 −8 3 0 5 −3 −3 0 5 −3 −3 3 6 1 6 0 3 0 −3 0 6 The quotient polynomial is 2x2 + 6 which has no real zeros so we get x = − 1 2 and x = 1. 2. To solve this nonlinear inequality, we follow the same guidelines set forth in Section 2.4: we get 0 on one side of the inequality and construct a sign diagram. Our original inequality can be rewritten as 2x5−3x4+6x3−8x2+3 ≤ 0. We found the zeros of p(x) = 2x5−3x4+6x3−8x2+3 in part 1 to be x = − 1 2 and x = 1. We construct our sign diagram as before. 278 Polynomial Functions (−) 0 (+) 0 (+) − 1 2 −1 1 0 2, and we know p(x) = 0 at x = − 1 The solution to p(x) < 0 is −∞, − 1 2 ∪ {1}. the solution to p(x) ≤ 0 is −∞, − 1 2 2 and x = 1. Hence, 3. To interpret this solution graphically, we set f (x) = 2x5 + 6x3 + 3 and g(x) =
3x4 + 8x2. We recall that the solution to f (x) ≤ g(x) is the set of x values for which the graph of f is below the graph of g (where f (x) < g(x)) along with the x values where the two graphs intersect (f (x) = g(x)). Graphing f and g on the calculator produces the picture on the lower left. (The end behavior should tell you which is which.) We see that the graph of f is below the graph of g on −∞, − 1. However, it is diﬃcult to see what is happening near 2 x = 1. Zooming in (and making the graph of g thicker), we see that the graphs of f and g do intersect at x = 1, but the graph of g remains below the graph of f on either side of x = 1. Our last example revisits an application from page 247 in the Exercises of Section 3.1. Example 3.3.8. Suppose the proﬁt P, in thousands of dollars, from producing and selling x hundred LCD TVs is given by P (x) = −5x3 + 35x2 − 45x − 25, 0 ≤ x ≤ 10.07. How many TVs should be produced to make a proﬁt? Check your answer using a graphing utility. Solution. To ‘make a proﬁt’ means to solve P (x) = −5x3 + 35x2 − 45x − 25 > 0, which we do analytically using a sign diagram. To simplify things, we ﬁrst factor out the −5 common to all the coeﬃcients to get −5 x3 − 7x2 + 9x − 5 > 0, so we can just focus on ﬁnding the zeros of f (x) = x3 − 7x2 + 9x + 5. The possible rational zeros of f are ±1 and ±5, and going through the usual computations, we ﬁnd x = 5 is the only rational zero. Using this, we factor f (x) = x3 − 7x2 + 9x + 5 = (x − 5) x2 − 2x − 1, and we ﬁnd the remaining zeros by applying the Quadratic Formula to x2 − 2x −
1 = 0. We ﬁnd three real zeros, x = 1 − 2 = −0.414..., x = 1 + 2 = 2.414..., and x = 5, of which only the last two fall in the applied domain of [0, 10.07]. We choose x = 0, x = 3 and x = 10.07 as our test values and plug them into the function P (x) = −5x3 + 35x2 − 45x − 25 (not f (x) = x3 − 7x2 + 9x − 5) to get the sign diagram below. √ √ 3.3 Real Zeros of Polynomials 279 (−) 1 + 0 (+) 0 √ 5 2 (−) 0 3 10.07 √ √ 2, 5). Since x measures the number of TVs in hundreds, We see immediately that P (x) > 0 on (1+ 2 corresponds to 241.4... TVs. Since we can’t produce a fractional part of a TV, we need x = 1 + to choose between producing 241 and 242 TVs. From the sign diagram, we see that P (2.41) < 0 but P (2.42) > 0 so, in this case we take the next larger integer value and set the minimum production to 242 TVs. At the other end of the interval, we have x = 5 which corresponds to 500 TVs. Here, we take the next smaller integer value, 499 TVs to ensure that we make a proﬁt. Hence, in order to make a proﬁt, at least 242, but no more than 499 TVs need to be produced. To check our answer using a calculator, we graph y = P (x) and make use of the ‘Zero’ command. We see that the calculator approximations bear out our analysis.7 7Note that the y-coordinates of the points here aren’t registered as 0. They are expressed in Scientiﬁc Notation. For instance, 1E − 11 corresponds to 0.00000000001, which is pretty close in the calculator’s eyes8to 0. 8but not a Mathematician’s 280 3.3.3 Exercises Polynomial Functions In Exercises 1 - 10, for the given polynomial: Use Cauchy’s Bound to �
�nd an interval containing all of the real zeros. Use the Rational Zeros Theorem to make a list of possible rational zeros. Use Descartes’ Rule of Signs to list the possible number of positive and negative real zeros, counting multiplicities. 1. f (x) = x3 − 2x2 − 5x + 6 2. f (x) = x4 + 2x3 − 12x2 − 40x − 32 3. f (x) = x4 − 9x2 − 4x + 12 4. f (x) = x3 + 4x2 − 11x + 6 5. f (x) = x3 − 7x2 + x − 7 6. f (x) = −2x3 + 19x2 − 49x + 20 7. f (x) = −17x3 + 5x2 + 34x − 10 8. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 9. f (x) = 3x3 + 3x2 − 11x − 10 10. f (x) = 2x4 + x3 − 7x2 − 3x + 3 In Exercises 11 - 30, ﬁnd the real zeros of the polynomial using the techniques speciﬁed by your instructor. State the multiplicity of each real zero. 11. f (x) = x3 − 2x2 − 5x + 6 12. f (x) = x4 + 2x3 − 12x2 − 40x − 32 13. f (x) = x4 − 9x2 − 4x + 12 14. f (x) = x3 + 4x2 − 11x + 6 15. f (x) = x3 − 7x2 + x − 7 16. f (x) = −2x3 + 19x2 − 49x + 20 17. f (x) = −17x3 + 5x2 + 34x − 10 18. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 19. f (x) = 3x3 + 3x2 − 11x − 10 20. f (x) = 2x4 + x3 − 7x2 − 3x + 3 21. f (x) = 9x3 − 5x2 − x 22.
f (x) = 6x4 − 5x3 − 9x2 23. f (x) = x4 + 2x2 − 15 24. f (x) = x4 − 9x2 + 14 25. f (x) = 3x4 − 14x2 − 5 26. f (x) = 2x4 − 7x2 + 6 27. f (x) = x6 − 3x3 − 10 28. f (x) = 2x6 − 9x3 + 10 29. f (x) = x5 − 2x4 − 4x + 8 30. f (x) = 2x5 + 3x4 − 18x − 27 3.3 Real Zeros of Polynomials 281 In Exercises 31 - 33, use your calculator,9 to help you ﬁnd the real zeros of the polynomial. State the multiplicity of each real zero. 31. f (x) = x5 − 60x3 − 80x2 + 960x + 2304 32. f (x) = 25x5 − 105x4 + 174x3 − 142x2 + 57x − 9 33. f (x) = 90x4 − 399x3 + 622x2 − 399x + 90 34. Find the real zeros of f (x) = x3 − 1 72 x + 1 72 by ﬁrst ﬁnding a polynomial q(x) with integer coeﬃcients such that q(x) = N · f (x) for some integer N. (Recall that the Rational Zeros Theorem required the polynomial in question to have integer coeﬃcients.) Show that f and q have the same real zeros. 12 x2 − 7 In Exercises 35 - 44, ﬁnd the real solutions of the polynomial equation. (See Example 3.3.7.) 35. 9x3 = 5x2 + x 37. x3 + 6 = 2x2 + 5x 39. x3 − 7x2 = 7 − x 41. x3 + x2 = 11x + 10 3 43. 14x2 + 5 = 3x4 36. 9x2 + 5x3 = 6x4 38. x4 + 2x3 = 12x2 + 40x + 32 40. 2x3 = 19x2
− 49x + 20 42. x4 + 2x2 = 15 44. 2x5 + 3x4 = 18x + 27 In Exercises 45 - 54, solve the polynomial inequality and state your answer using interval notation. 45. −2x3 + 19x2 − 49x + 20 > 0 46. x4 − 9x2 ≤ 4x − 12 47. (x − 1)2 ≥ 4 49. x4 ≤ 16 + 4x − x3 48. 4x3 ≥ 3x + 1 50. 3x2 + 2x < x4 51. x3 + 2x2 2 53. 2x4 > 5x2 + 3 < x + 2 52. x3 + 20x 8 54. x6 + x3 ≥ 6 ≥ x2 + 2 55. In Example 3.1.3 in Section 3.1, a box with no top is constructed from a 10 inch × 12 inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. We determined the volume of that box (in cubic inches) is given by V (x) = 4x3 − 44x2 + 120x, where x denotes the length of the side of the square which is removed from each corner (in inches), 0 < x < 5. Solve the inequality V (x) ≥ 80 analytically and interpret your answer in the context of that example. 9You can do these without your calculator, but it may test your mettle! 282 Polynomial Functions 56. From Exercise 32 in Section 3.1, C(x) =.03x3 − 4.5x2 + 225x + 250, for x ≥ 0 models the cost, in dollars, to produce x PortaBoy game systems. If the production budget is $5000, ﬁnd the number of game systems which can be produced and still remain under budget. 57. Let f (x) = 5x7 − 33x6 + 3x5 − 71x4 − 597x3 + 2097x2 − 1971x + 567. With the help of your classmates, ﬁnd the x- and y- intercepts of the graph of f. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Sketch the graph of f, using more than one picture if necessary to show all of
the important features of the graph. 58. With the help of your classmates, create a list of ﬁve polynomials with diﬀerent degrees whose real zeros cannot be found using any of the techniques in this section. 3.3 Real Zeros of Polynomials 283 3.3.4 Answers 1. For f (x) = x3 − 2x2 − 5x + 6 All of the real zeros lie in the interval [−7, 7] Possible rational zeros are ±1, ±2, ±3, ±6 There are 2 or 0 positive real zeros; there is 1 negative real zero 2. For f (x) = x4 + 2x3 − 12x2 − 40x − 32 All of the real zeros lie in the interval [−41, 41] Possible rational zeros are ±1, ±2, ±4, ±8, ±16, ±32 There is 1 positive real zero; there are 3 or 1 negative real zeros 3. For f (x) = x4 − 9x2 − 4x + 12 All of the real zeros lie in the interval [−13, 13] Possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12 There are 2 or 0 positive real zeros; there are 2 or 0 negative real zeros 4. For f (x) = x3 + 4x2 − 11x + 6 All of the real zeros lie in the interval [−12, 12] Possible rational zeros are ±1, ±2, ±3, ±6 There are 2 or 0 positive real zeros; there is 1 negative real zero 5. For f (x) = x3 − 7x2 + x − 7 All of the real zeros lie in the interval [−8, 8] Possible rational zeros are ±1, ±7 There are 3 or 1 positive real zeros; there are no negative real zeros 6. For f (x) = −2x3 + 19x2 − 49x + 20 All of the real zeros lie in the interval − 51 2, ±1, ±2, ± 5 Possible rational zeros are ± 1 There are 3 or 1 positive real zeros; there are no negative real zeros 2, 51 2, ±4, ±5, ±10, ±20 2 7. For f (x) = −17x
3 + 5x2 + 34x − 10 All of the real zeros lie in the interval [−3, 3] Possible rational zeros are ± 1 17, ± 2 There are 2 or 0 positive real zeros; there is 1 negative real zero 17, ±1, ±2, ±5, ±10 17, ± 10 17, ± 5 284 Polynomial Functions 8. For f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 All of the real zeros lie in the interval − 4 Possible rational zeros are ± 1 18, ± 1 There are 2 or 0 positive real zeros; there are 2 or 0 negative real zeros 3, 4 12, ± 1 36, ± 1 3 9. For f (x) = 3x3 + 3x2 − 11x − 10 All of the real zeros lie in the interval − 14 3, 14 Possible rational zeros are ± 1 3, ± 10 3, ± 2 3, ±1, ±2, ±5, ±10 There is 1 positive real zero; there are 2 or 0 negative real zeros 3, ± 5 3 10. For f (x) = 2x4 + x3 − 7x2 − 3x + 3 All of the real zeros lie in the interval − 9 2, 9 Possible rational zeros are ± 1 2, ±1, ± 3 2, ±3 There are 2 or 0 positive real zeros; there are 2 or 0 negative real zeros 2 11. f (x) = x3 − 2x2 − 5x + 6 x = −2, x = 1, x = 3 (each has mult. 1) 12. f (x) = x4 + 2x3 − 12x2 − 40x − 32 x = −2 (mult. 3), x = 4 (mult. 1) 13. f (x) = x4 − 9x2 − 4x + 12 x = −2 (mult. 2), x = 1 (mult. 1), x = 3 (mult. 1) 14. f (x) = x3 + 4x2 − 11x + 6 x = −6 (mult. 1), x = 1 (mult. 2) 15. f (x) = x3 − 7x2 + x − 7 x = 7 (mult. 1) 16. f (x) = −2x3 + 19x2
− 49x + 20 x = 1 2, x = 4, x = 5 (each has mult. 1) 17. f (x) = −17x3 + 5x2 + 34x − 10 √ x = 5 17, x = ± 2 (each has mult. 1) 18. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 2 (mult. 2), x = − 1 3 (mult. 2) x = 1 19. f (x) = 3x3 + 3x2 − 11x − 10 x = −2, x = 3± 69 (each has mult. 1) √ 6 3.3 Real Zeros of Polynomials 285 20. f (x) = 2x4 + x3 − 7x2 − 3x + 3 √ x = −1, x = 1 2, x = ± 21. f (x) = 9x3 − 5x2 − x √ x = 0, x = 5± 18 61 (each has mult. 1) 3 (each mult. 1) 22. f (x) = 6x4 − 5x3 − 9x2 x = 0 (mult. 2), x = 5± √ 12 241 (each has mult. 1) 23. f (x) = x4 + 2x2 − 15 x = ± 3 (each has mult. 1) 24. f (x) = x4 − 9x2 + 14 √ x = ± 2, x = ± 7 (each has mult. 1) 25. f (x) = 3x4 − 14x2 − 5 x = ± 5 (each has mult. 1) √ √ √ 26. f (x) = 2x4 − 7x2 + each has mult. 1) 27. f (x) = x6 − 3x3 − 10 √ x = 3 √ −2 = − 3 √ 2, x = 3 5 (each has mult. 1) 28. f (x) = 2x6 − 9x3 + 10 √ 3√ 20 2, x = 3 29. f (x) = x5 − 2x4 − 4x + 8 x = √ 2 (each has mult. 1) x = 2, x = ± 2 (each has mult. 1) 30. f (x) = 2x5
+ 3x4 − 18x − 27 √ each has mult. 1) 31. f (x) = x5 − 60x3 − 80x2 + 960x + 2304 x = −4 (mult. 3), x = 6 (mult. 2) 32. f (x) = 25x5 − 105x4 + 174x3 − 142x2 + 57x − 9 x = 3 5 (mult. 2), x = 1 (mult. 3) 33. f (x) = 90x4 − 399x3 + 622x2 − 399x + 90 each has mult. 1) 34. We choose q(x) = 72x3 − 6x2 − 7x + 1 = 72 · f (x). Clearly f (x) = 0 if and only if q(x) = 0 4 are the real zeros so they have the same real zeros. In this case, x = − 1 of both f and q. 6 and x = 1 3, x = 1 286 √ 35. x = 0, 5± 18 61 37. x = −2, 1, 3 39. x = 7 41. x = −2, 3± √ 6 69 √ 43. x = ± 5 45. (−∞, 1 2 ) ∪ (4, 5) 47. (−∞, −1] ∪ [3, ∞) Polynomial Functions 36. x = 0, 5± √ 241 12 38. x = −2, 4 40. x = 1 2, 4, 5 √ 42. x = ± 3 44 46. {−2} ∪ [1, 3] 48. − 1 2 ∪ [1, ∞) 49. [−2, 2] 50. (−∞, −1) ∪ (−1, 0) ∪ (2, ∞) 51. (−∞, −2) ∪ − √ √ 2 52. {2} ∪ [4, ∞) 2, 53. (−∞, − 3) ∪ ( √ √ 3, ∞) √ √ √ 54. (−∞, − 3 √ 3 ) ∪ ( 3 √ 2, ∞) 55. V (x) ≥ 80 on [1, 5 − 5] ∪ [5 + 5, ∞). Only the portion [1
, 5 − 5] lies in the applied domain, however. In the context of the problem, this says for the volume of the box to be at least 80 cubic inches, the square removed from each corner needs to have a side length of at least 1 inch, but no more than 5 − 5 ≈ 2.76 inches. √ 56. C(x) ≤ 5000 on (approximately) (−∞, 82.18]. The portion of this which lies in the applied domain is (0, 82.18]. Since x represents the number of game systems, we check C(82) = 4983.04 and C(83) = 5078.11, so to remain within the production budget, anywhere between 1 and 82 game systems can be produced. 3.4 Complex Zeros and the Fundamental Theorem of Algebra 287 3.4 Complex Zeros and the Fundamental Theorem of Algebra In Section 3.3, we were focused on ﬁnding the real zeros of a polynomial function. In this section, we expand our horizons and look for the non-real zeros as well. Consider the polynomial p(x) = x2 + 1. The zeros of p are the solutions to x2 + 1 = 0, or x2 = −1. This equation has no real solutions, but you may recall from Intermediate Algebra that we can formally extract the square roots of both −1 is usually re-labeled i, the so-called imaginary unit.1 sides to get x = ± The number i, while not a real number, plays along well with real numbers, and acts very much like any other radical expression. For instance, 3(2i) = 6i, 7i − 3i = 4i, (2 − 7i) + (3 + 4i) = 5 − 3i, and so forth. The key properties which distinguish i from the real numbers are listed below. −1. The quantity √ √ Deﬁnition 3.4. The imaginary unit i satisﬁes the two following properties 1. i2 = −1 2. If c is a real number with c ≥ 0 then √ √ −c = i c Property 1 in Deﬁnition 3.4 establishes that i does act as a square root2 of −1, and property 2 establishes what we mean by the ‘principal square root
’ of a negative real number. In property 2, it is important to remember the restriction on c. For example, it is perfectly acceptable to say √ √ √ −4 = i 4 = i(2) = 2i. However, −(−4) = i −4, otherwise, we’d get √ 2 = 4 = −(−4) = i √ −4 = i(2i) = 2i2 = 2(−1) = −2, which is unacceptable.3 We are now in the position to deﬁne the complex numbers. Deﬁnition 3.5. A complex number is a number of the form a + bi, where a and b are real numbers and i is the imaginary unit. Complex numbers include things you’d normally expect, like 3 + 2i and 2 3. However, don’t forget that a or b could be zero, which means numbers like 3i and 6 are also complex numbers. In other words, don’t forget that the complex numbers include the real numbers, so 0 and π − 21 are both considered complex numbers.4 The arithmetic of complex numbers is as you would expect. The only things you need to remember are the two properties in Deﬁnition 3.4. The next example should help recall how these animals behave. 5 − i √ √ 1Some Technical Mathematics textbooks label it ‘j’. 2Note the use of the indeﬁnite article ‘a’. Whatever beast is chosen to be i, −i is the other square root of −1. 3We want to enlarge the number system so we can solve things like x2 = −1, but not at the cost of the established rules already set in place. For that reason, the general properties of radicals simply do not apply for even roots of negative quantities. 4See the remarks in Section 1.1.1. 288 Polynomial Functions Example 3.4.1. Perform the indicated operations. Write your answer in the form5 a + bi. 1. (1 − 2i) − (3 + 4i) 2. (1 − 2i)(3 + 4i) 3. 1 − 2i 3 − 4i √ √ −3 −12 4. Solution. (−3)(−12) 5. 6. (x − [1 + 2i
])(x − [1 − 2i]) 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We combine like terms to get (1 − 2i) − (3 + 4i) = 1 − 2i − 3 − 4i = −2 − 6i. 2. Using the distributive property, we get (1 − 2i)(3 + 4i) = (1)(3) + (1)(4i) − (2i)(3) − (2i)(4i) = 3 + 4i − 6i − 8i2. Since i2 = −1, we get 3 + 4i − 6i − 8i2 = 3 − 2i − (−8) = 11 − 2i. 3. How in the world are we supposed to simplify 1−2i 3−4i? Well, we deal with the denominator 3 − 4i as we would any other denominator containing a radical, and multiply both numerator and denominator by 3 + 4i (the conjugate of 3 − 4i).6 Doing so produces 1 − 2i 3 − 4i · 3 + 4i 3 + 4i = (1 − 2i)(3 + 4i) (3 − 4i)(3 + 4i) = 11 − 2i 25 = 11 25 − 2 25 i 4. We use property 2 of Deﬁnition 3.4 ﬁrst, then apply the rules of radicals applicable to real radicals to get √ √ −3 √ √ −12 = i 3 i 12 = i2 √ √ 3 · 12 = − 36 = −6. 5. We adhere to the order of operations here and perform the multiplication before the radical to get (−3)(−12) = 36 = 6. √ 6. We can brute force multiply using the distributive property and see that (x − [1 + 2i])(x − [1 − 2i]) = x2 − x[1 − 2i] − x[1 + 2i] + [1 − 2i][1 + 2i] = x2 − x + 2ix − x − 2ix + 1 − 2i + 2i − 4i2 = x2 − 2x + 5 A couple of remarks about the last example are in order. First, the conjugate of a complex number a + bi is the number a

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