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tercept. We can approximate the slope of the line by extending it until we can estimate the rise _ run . Example 2 Finding a Line of Best Fit Find a linear function that fits the data in Table 1 by “eyeballing” a line that seems to fit. Solution On a graph, we could try sketching a line. Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of m = = 1.2 60 __ 50 and a y-intercept at 30. This gives an equation of T(c) = 1.2c + 30 where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure 3. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature 10 20 30 40 50 c, Number of Chirps Figure 3 Analysis This linear equation can then be used to approximate answers to various questions we might ask about the trend. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 4 CHAPTER 4 LINEAR FUNCTIONS Recognizing Interpolation or Extrapolation While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of the data. Figure 4 compares the two processes for the cricket-chirp data addressed in Example 2. We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44. There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called model breakdown. For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when x = 50, that is in 50 years, the model would not apply because we could not account for factors fifty years in the future. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature Extrapolation Interpolation 10 20 30 40 50 c, Number of Chirps Figure 4 Interpolation occurs within the domain and range of the provided data whereas extrapolation occurs outside. interpolation and extrapolation Different methods of making predictions are used to analyze data. The method of interpolation involves predicting a value inside the domain and/or range of the data. The method of extrapolation involves predicting a value outside the domain and/or range of the data. Model breakdown occurs at the point when the model no longer applies. Example 3 Understanding Interpolation and Extrapolation Use the cricket data from Table 1 to answer the following questions: a. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. b. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 FITTING LINEAR MODELS TO DATA 325 Solution a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model: T(30) = 30 + 1.2(30) = 66 degrees Based on the data we have, this value seems reasonable. b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model: 40 = 30 + 1.2c 10 = 1.2c c ≈ 8.33 We can compare the regions of interpolation and extrapolation using Figure 5. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature Interpolation Extrapolation 10 20 30 40 50 c, Number of Chirps Figure 5 Analysis Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees. Try It #1 According to the data from Table 1, what temperature can we predict it is if we counted 20 chirps in 15 seconds? Finding the Line of Best Fit Using a Graphing Utility While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values[11]. One such technique is called least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[12]. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression. 11 Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values. 12 For example, http://www.shodor.org/unchem/math/lls/leastsq.html Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 6 CHAPTER 4 LINEAR FUNCTIONS How To… Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression. 1. Enter the input in List 1 (L1). 2. Enter the output in List 2 (L2). 3. On a graphing utility, select Linear Regression (LinReg). Example 4 Finding a Least Squares Regression Line Find the least squares regression line using the cricket-chirp data in Table 2. Solution 1. Enter the input (chirps) in List 1 (L1). 2. Enter the output (temperature) in List 2 (L2). See Table 2. L1 L2 44 80.5 35 70.5 20.4 57 33 66 31 68 35 72 18.5 52 37 73.5 26 53 Table 2 3. On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation: T(c) = 30.281 + 1.143c Analysis Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to: T(30) = 30.281 + 1.143(30) = 64.571 ≈ 64.6 degrees The graph of the scatter plot with the least squares regression line is shown in Figure 90 80 70 60 50 40 30 Number of Cricket Chirps vs. Temperature 0 10 20 30 40 50 c, Number of Chirps Figure 6 Q & A… Will there ever be a case where two different lines will serve as the best fit for the data? No. There is only one best fit line. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 FITTING LINEAR MODELS TO DATA 327 Distinguishing Between Linear and Non-Linear Models As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a “diagnostic on” selection to find the correlation coefficient, which mathematicians label as r. The correlation coefficient provides an easy way to get an idea of how close to a line the data falls. We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of r and the graph of the data, Figure 7 shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output. 1.0 0.8 0.4 0.0 −0.4 −0.8 −1.0 1.0 1.0 1.0 −1.0 −1.0 −1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 Figure 7 Plotted data and related correlation coefﬁcients. (credit: “DenisBoigelot,” Wikimedia Commons) correlation coefficient The correlation coefficient is a value, r, between −1 and 1. • r > 0 suggests a positive (increasing) relationship • r < 0 suggests a negative (decreasing) relationship • The closer the value is to 0, the more scattered the data. • The closer the value is to 1 or −1, the less scattered the data is. Example 5 Finding a Correlation Coefﬁcient Calculate the correlation coefficient for cricket-chirp data in Table 1. Solution Because the data appear to follow a linear pattern, we can use technology to calculate r. Enter the inputs and corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation coefficient, r = 0.9509. This value is very close to 1, which suggests a strong increasing linear relationship. Note: For some calculators, the Diagnostics must be turned “on” in order to get the correlation coefficient when linear regression is performed: [2nd]> [0]> [alpha][x − 1], then scroll to DIAGNOSTICSON. Fitting a Regression Line to a Set of Data Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 8 CHAPTER 4 LINEAR FUNCTIONS Example 6 Using a Regression Line to Make Predictions Gasoline consumption in the United States has been steadily increasing. Consumption data from 19
94 to 2004 is shown in Table 3[13]. Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008. Year ‘94 ‘95 ‘96 ‘97 ‘98 ‘99 ‘00 ‘01 ‘02 ‘03 ‘04 Consumption (billions of gallons) 113 116 118 119 123 125 126 128 131 133 136 The scatter plot of the data, including the least squares regression line, is shown in Figure 8. Table ( 150 140 130 120 110 100 0 Gas Consumption vs. Year 0 1 2 3 4 6 5 7 Years After 1994 8 9 10 11 12 13 14 Figure 8 Solution We can introduce a new input variable, t, representing years since 1994. The least squares regression equation is: C(t) = 113.318 + 2.209t Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend. Using this to predict consumption in 2008 (t = 14), The model predicts 144.244 billion gallons of gasoline consumption in 2008. C(14) = 113.318 + 2.209(14) = 144.244 Try It #2 Use the model we created using technology in Example 6 to predict the gas consumption in 2011. Is this an interpolation or an extrapolation? Access these online resources for additional instruction and practice with fitting linear models to data. • Introduction to Regression Analysis (http://openstaxcollege.org/l/introregress) • Linear Regression (http://openstaxcollege.org/l/linearregress) 13 http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 SECTION EXERCISES 329 4.3 SECTION EXERCISES VERBAL 1. Describe what it means if there is a model breakdown when using a linear model. 2. What is interpolation when using a linear model? 3. What is extrapolation when using a linear model? 4. Explain the difference between a positive and a negative correlation coefficient. 5. Explain how to interpret the absolute value of a correlation coefficient. ALGEBRAIC 6. A regression was run to determine whether there is 7. A regression was run to determine whether there is a a relationship between hours of TV watched per day (x) and number of sit-ups a person can do (y). The results of the regression are given below. Use this to predict the number of situps a person who watches 11 hours of TV can do. y = ax + b a = −1.341 b = 32.234 r = −0.896 relationship between the diameter of a tree (x, in inches) and the tree’s age (y, in years). The results of the regression are given below. Use this to predict the age of a tree with diameter 10 inches. y = ax + b a = 6.301 b = −1.044 r = −0.970 For the following exercises, draw a scatter plot for the data provided. Does the data appear to be linearly related? 8. 10. 0 2 8 4 −22 −19 −15 −11 −6 6 10 −2 100 12 250 12.6 300 13.1 450 14 600 14.5 750 15.2 9. 11. 1 46 1 1 2 50 3 9 3 59 5 28 4 75 7 65 5 100 6 136 9 125 11 216 12. For the following data, draw a scatter plot. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Eyeball the line, and estimate the answer. 13. For the following data, draw a scatter plot. If we wanted to know when the temperature would reach 28°F, would the answer involve interpolation or extrapolation? Eyeball the line and estimate the answer. Year 2010 Population 11,500 12,100 12,700 13,000 13,750 2000 2005 1990 1995 Temperature, °F Time, seconds 16 46 18 50 20 54 25 55 30 62 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 0 CHAPTER 4 LINEAR FUNCTIONS GRAPHICAL For the following exercises, match each scatterplot with one of the four specified correlations in Figure 9 and Figure 10. (a) (b) Figure 9 14. r = 0.95 16. r = −0.26 (c) (d) Figure 10 15. r = −0.89 17. r = −0.39 For the following exercises, draw a best-fit line for the plotted data. 18. 12 10 8 6 4 2 0 19. 10 10 0 2 4 6 8 10 10 10 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 SECTION EXERCISES 331 20. 10 10 NUMERIC 21. 10 10 22. The U.S. Census tracks the percentage of persons 25 years or older who are college graduates. That data for several years is given in Table 4[14]. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the percentage exceed 35%? Year Percent Graduates 1990 21.3 1992 21.4 1994 22.2 1996 23.6 1998 24.4 2000 25.6 2002 26.7 2004 27.7 2006 28 2008 29.4 Table 4 23. The U.S. import of wine (in hectoliters) for several years is given in Table 5. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will imports exceed 12,000 hectoliters? Year Imports 1992 2665 1994 2688 1996 3565 1998 4129 2000 4584 2002 5655 2004 6549 2006 7950 2008 8487 2009 9462 Table 5 24. Table 6 shows the year and the number of people unemployed in a particular city for several years. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the number of unemployed reach 5 people? Year Number Unemployed 1990 750 1992 670 1994 650 1996 605 Table 6 1998 550 2000 510 2002 460 2004 420 2006 380 2008 320 TECHNOLOGY For the following exercises, use each set of data to calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to 3 decimal places of accuracy. 25. 26. x y x y 8 23 5 4 15 41 7 12 26 53 10 17 31 72 12 22 56 103 15 24 14 http://www.census.gov/hhes/socdemo/education/data/cps/historical/index.html. Accessed 5/1/2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 2 CHAPTER 4 LINEAR FUNCTIONS 3 21.9 10 18.54 4 22.22 11 15.76 5 22.74 12 13.68 6 22.26 13 14.1 7 20.78 14 14.02 8 17.6 15 11.94 9 16.52 16 12.76 4 44.8 5 43.1 6 38.8 7 39 8 38 9 32.7 10 30.1 11 29.3 12 27 13 25.8 21 17 100 2000 900 70 25 11 80 1798 988 80 30 2 60 1589 1000 82 31 −1 55 1580 1010 84 40 −18 40 1390 1200 105 50 −40 20 1202 1205 108 27. 28. 29. 30. 31 EXTENSIONS 32. Graph f (x) = 0.5x + 10. Pick a set of 5 ordered 33. Graph f (x) = −2x − 10. Pick a set of 5 ordered pairs using inputs x = −2, 1, 5, 6, 9 and use linear regression to verify that the function is a good fit for the data. pairs using inputs x = −2, 1, 5, 6, 9 and use linear regression to verify the function. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs shows dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years: (46, 600), (48, 550), (50, 505), (52, 540), (54, 495). 34. Use linear regression to determine a function P 35. Find to the nearest tenth and interpret the where the profit in thousands of dollars depends on the number of units sold in hundreds. x-intercept. 36. Find to the nearest tenth and interpret the y-intercept. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 SECTION EXERCISES 333 REAL-WORLD APPLICATIONS For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population, year) for specific recorded years: (2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010) 37. Use linear regression to determine a function y, 38. Predict when the population will hit 8,000. where the year depends on the population. Round to three decimal places of accuracy. For the following exercises, consider this scenario: The profit of a company increased steadily over a ten-year span. The following ordered pairs show the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years: (46, 250), (48, 305), (50, 350), (52, 390), (54, 410). 39. Use linear regression to determine a function y, 40. Predict when the profit will exceed one million where the profit in thousands of dollars depends on the number of units sold in hundreds. dollars. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs show dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span (number of units sold, profit) for specific recorded years: (46, 250), (48, 225), (50, 205), (52, 180), (54, 165). 41. Use linear regression to determine a function y, 42. Predict when the profit will dip below the $25,000 where the profit in thousands of dollars depends on the number of units sold in hundreds. threshold. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 4 CHAPTER 4 LINEAR FUNCTIONS CHAPTER 4 REVIEW Key Terms correlation coefficient a value, r, between −1 and 1 that indicates the degree of linear correlation of variables, or how closely a regression line fits a data set. decreasing linear function a function with a negative slope: If f (x) = mx + b, then m < 0. extrapolation predicting a value outside the domain and range of the data horizontal line a line defined by f (x) = b, where b is a real number. The slope of a horizontal line is 0. increasing linear function a function with a positive slope: If f (x) = mx + b, then m > 0. interpolation predicting a value inside the domain and range of the data least squares regression a statistical technique for fitting a line to data in a way that minimizes the differences between the line and data values linear function a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line model breakdown when a model no longer applies after a certain point parallel lines two or more lines with the same slope perpendicular lines two lines that intersect at right angles and have slopes that are negative reciprocals of each other point-slope form the equation for
a line that represents a linear function of the form y − y1 = m(x − x1) slope the ratio of the change in output values to the change in input values; a measure of the steepness of a line slope-intercept form the equation for a line that represents a linear function in the form f (x) = mx + b vertical line a line defined by x = a, where a is a real number. The slope of a vertical line is undefined. Key Concepts 4.1 Linear Functions • Linear functions can be represented in words, function notation, tabular form, and graphical form. See Example 1. • An increasing linear function results in a graph that slants upward from left to right and has a positive slope. A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. A constant linear function results in a graph that is a horizontal line. See Example 2. • Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line. See Example 3 and Example 4. • An equation for a linear function can be written from a graph. See Example 5. • The equation for a linear function can be written if the slope m and initial value b are known. See Example 6 and Example 7. • A linear function can be used to solve real-world problems given information in different forms. See Example 8, Example 9, and Example 10. • Linear functions can be graphed by plotting points or by using the y-intercept and slope. See Example 11 and Example 12. • Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See Example 13. • The equation for a linear function can be written by interpreting the graph. See Example 14. • The x-intercept is the point at which the graph of a linear function crosses the x-axis. See Example 15. • Horizontal lines are written in the form, f (x) = b. See Example 16. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 REVIEW 335 • Vertical lines are written in the form, x = b. See Example 17. • Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See Example 18. • A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x- and y-values of the given point into the equation, f (x) = mx + b, and using the b that results. Similarly, the point-slope form of an equation can also be used. See Example 19. • A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See Example 20 and Example 21. 4.2 Modeling with Linear Functions • We can use the same problem strategies that we would use for any type of function. • When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept. See Example 1. • Draw a diagram, where appropriate. See Example 2 and Example 3. • Check for reasonableness of the answer. • Linear models may be built by identifying or calculating the slope and using the y-intercept. ◦ The x-intercept may be found by setting y = 0, which is setting the expression mx + b equal to 0. ◦ The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See Example 4. ◦ A graph of the system may be used to identify the points where one line falls below (or above) the other line. 4.3 Fitting Linear Models to Data • Scatter plots show the relationship between two sets of data. See Example 1. • Scatter plots may represent linear or non-linear models. • The line of best fit may be estimated or calculated, using a calculator or statistical software. See Example 2. • Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data. See Example 3. • The correlation coefficient, r, indicates the degree of linear relationship between data. See Example 4. • A regression line best fits the data. See Example 5. • The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables. See Example 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 6 CHAPTER 4 LINEAR FUNCTIONS CHAPTER 4 REVIEW EXERCISES LINEAR FUNCTIONS 1. Determine whether the algebraic equation is linear. 2. Determine whether the algebraic equation is linear. 2x + 3y = 7 6x 2 − y = 5 3. Determine whether the function is increasing or 4. Determine whether the function is increasing or decreasing. f (x) = 7x − 2 decreasing. g(x) = −x + 2 5. Given each set of information, find a linear equation that satisfies the given conditions, if possible. Passes through (7, 5) and (3, 17) 6. Given each set of information, find a linear equation that satisfies the given conditions, if possible. x-intercept at (6, 0) and y-intercept at (0, 10) 7. Find the slope of the line shown in the graph. 8. Find the slope of the line shown in the graph. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 6 x 9. Write an equation in slope-intercept form for the line shown. 10. Does the following table represent a linear function? If so, find the linear equation that models the data. x g(x) –4 18 0 –2 2 10 –12 –52 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x 11. Does the following table represent a linear function? If so, find the linear equation that models the data. x 6 8 12 26 g(x) –8 –12 –18 –46 12. On June 1st, a company has $4,000,000 profit. If the company then loses 150,000 dollars per day thereafter in the month of June, what is the company’s profit nth day after June 1st? For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 13. 2x − 6y = 12 −x + 3y = 1 14. 1 __ x − 2 y = 3 3x + y = − 9 For the following exercises, find the x- and y-intercepts of the given equation 15. 7x + 9y = −63 16. f (x) = 2x − 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 REVIEW 337 For the following exercises, use the descriptions of the pairs of lines to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 17. Line 1: Passes through (5, 11) and (10, 1) Line 2: Passes through (−1, 3) and (−5, 11) 18. Line 1: Passes through (8, −10) and (0, −26) Line 2: Passes through (2, 5) and (4, 4) 19. Write an equation for a line perpendicular to f (x) = 5x − 1 and passing through the point (5, 20). 20. Find the equation of a line with a y-intercept of (0, 2) and slope − #1 __ . 2 21. Sketch a graph of the linear function f (t) = 2t − 5. 22. Find the point of intersection for the 2 linear x = y + 6 2x − y = 13 functions: 23. A car rental company offers two plans for renting a car. Plan A: 25 dollars per day and 10 cents per mile Plan B: 50 dollars per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? MODELING WITH LINEAR FUNCTIONS 24. Find the area of a triangle bounded by the y-axis, the line f (x) = 10 − 2x, and the line perpendicular to f that passes through the origin. 25. A town’s population increases at a constant rate. In 2010 the population was 55,000. By 2012 the population had increased to 76,000. If this trend continues, predict the population in 2016. 26. The number of people afflicted with the common cold in the winter months dropped steadily by 50 each year since 2004 until 2010. In 2004, 875 people were inflicted. Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t. When will no one be afflicted? For the following exercises, use the graph in Figure 1 showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. y 12,000 10,000 8,000 6,000 4,000 2,000 0 5 10 15 20 25 30 x Figure 1 27. Find the linear function y, where y depends on x, the number of years since 1980. 28. Find and interpret the y-intercept. For the following exercise, consider this scenario: In 2004, a school population was 1,700. By 2012 the population had grown to 2,500. 29. Assume the population is changing linearly. a. How much did the population grow between the year 2004 and 2012? b. What is the average population growth per year? c. Find an equation for the population, P, of the school t years after 2004. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 8 CHAPTER 4 LINEAR FUNCTIONS For the following exercises, consider this scenario: In 2000, the moose population in a park was measured to be 6,500. By 2010, the population was measured to be 12,500. Assume the population continues to change linearly. 30. Find a formula for the moose population, P. 31. What does your model predict the moose population to be in 2020? For the following exercises, consider this scenario: The median home values in subdivisions Pima Central and East Valley (adjusted for inflation) are shown in Table 1. Assume that the house values are changing linearly. Year 1970 2010 Pima Central 32,000 85,000 Table 1 East Valley 120,250 150,000 32. In which subdivision have home values increased at a higher rate? 33. If these trends were to continue, what would be the median home value in Pima Central in 2015? FITTING LINEAR MODELS TO DATA 34. Draw a scatter plot for the data in Table 2. Then determine whether the data appears to be linearly related. 0 –105 2 –50 4 1 6 55 8 105 10 160 Table 2 35. Draw a scatter plot for the data in Table 3. If we
wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Year Population 1990 5,600 1995 5,950 2000 6,300 2005 6,600 2010 6,900 Table 3 36. Eight students were asked to estimate their score on a 10-point quiz. Their estimated and actual scores are given in Table 4. Plot the points, then sketch a line that fits the data. Predicted Actual 6 6 7 7 7 8 8 8 Table 4 7 9 9 10 10 10 10 9 37. Draw a best-fit line for the plotted data. y 120 100 80 60 40 20 0 0 2 4 6 8 10 12 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 REVIEW 339 For the following exercises, consider the data in Table 5, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year. Year Percent Graduates 2000 6.5 2002 7.0 Table 5 2005 7.4 2007 8.2 2010 9.0 38. Determine whether the trend appears to be linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. 39. In what year will the percentage exceed 12%? 40. Based on the set of data given in Table 6, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. 41. Based on the set of data given in Table 7, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. x y 17 15 20 25 23 31 26 37 29 40 x y 10 36 12 34 15 30 18 28 20 22 Table 6 Table 7 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the population and the year over the ten-year span (population, year) for specific recorded years: (3,600, 2000); (4,000, 2001); (4,700, 2003); (6,000, 2006) 42. Use linear regression to determine a function y, 43. Predict when the population will hit 12,000. where the year depends on the population, to three decimal places of accuracy. 44. What is the correlation coefficient for this model to 45. According to the model, what is the population three decimal places of accuracy? in 2014? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 0 CHAPTER 4 LINEAR FUNCTIONS CHAPTER 4 PRACTICE TEST 1. Determine whether the following algebraic equation can be written as a linear function. 2x + 3y = 7 2. Determine whether the following function is increasing or decreasing. f (x) = −2x + 5 3. Determine whether the following function is increasing or decreasing. f (x) = 7x + 9 4. Given the following set of information, find a linear equation satisfying the conditions, if possible. Passes through (5, 1) and (3, −9) 5. Given the following set of information, find a linear equation satisfying the conditions, if possible. x-intercept at (−4, 0) and y-intercept at (0, −6) 6. Find the slope of the line in Figure 1. 7. Write an equation for line in Figure 2. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 6 x Figure 1 Figure 2 8. Does Table 1 represent a linear function? If so, find 9. Does Table 2 represent a linear function? If so, find a linear equation that models the data. a linear equation that models the data. x −6 14 g(x) 0 32 2 38 4 44 Table 1 x g(x) 1 4 3 9 7 19 11 12 Table 2 10. At 6 am, an online company has sold 120 items that day. If the company sells an average of 30 items per hour for the remainder of the day, write an expression to represent the number of items that were sold n after 6 am. For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 11. 3 __ x − 9 y = 4 −4x − 3y = 8 13. Find the x- and y-intercepts of the equation 2x + 7y = −14. 12. −2x + y = 3 3 __ y = 5 3x + 2 14. Given below are descriptions of two lines. Find the slopes of Line 1 and Line 2. Is the pair of lines parallel, perpendicular, or neither? Line 1: Passes through (−2, −6) and (3, 14) Line 2: Passes through (2, 6) and (4, 14) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 PRACTICE TEST 341 15. Write an equation for a line perpendicular to f (x) = 4x + 3 and passing through the point (8, 10). 16. Sketch a line with a y-intercept of (0, 5) and slope − #5 __ . 2 17. Graph of the linear function f (x) = −x + 6 . 18. For the two linear functions, find the point of intersection: x = y + 2 2x − 3y = −1 19. A car rental company offers two plans for renting 20. Find the area of a triangle bounded by the y-axis, a car. Plan A: $25 per day and $0.10 per mile Plan B: $40 per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? 21. A town’s population increases at a constant rate. In 2010 the population was 65,000. By 2012 the population had increased to 90,000. Assuming this trend continues, predict the population in 2018. the line f (x) = 12 − 4x, and the line perpendicular to f that passes through the origin. 22. The number of people afflicted with the common cold in the winter months dropped steadily by 25 each year since 2002 until 2012. In 2002, 8,040 people were inflicted. Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t. When will less than 6,000 people be afflicted? For the following exercises, use the graph in Figure 3, showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. y 35,000 30,000 25,000 20,000 15,000 10,000 5,000 0 5 10 15 20 25 30 x Figure 3 23. Find the linear function y , where y depends on x, the number of years since 1980. 24. Find and interpret the y-intercept. 25. In 2004, a school population was 1250. By 2012 the population had dropped to 875. Assume the population is changing linearly. a. How much did the population drop between the year 2004 and 2012? b. What is the average population decline per year? c. Find an equation for the population, P, of the school t years after 2004. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 2 CHAPTER 4 LINEAR FUNCTIONS 26. Draw a scatter plot for the data provided in Table 3. Then determine whether the data appears to be linearly related. 0 2 −450 −200 4 10 6 265 8 500 10 755 Table 3 27. Draw a best-fit line for the plotted data. 35 30 25 20 15 10 5 0 y 0 2 4 6 8 10 12 x For the following exercises, use Table 4 which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year. Year Percent Graduates 2000 8.5 2002 8.0 Table 4 2005 7.2 2007 6.7 2010 6.4 28. Determine whether the trend appears linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. 29. In what year will the percentage drop below 4%? 30. Based on the set of data given in Table 5, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracy. x y 16 106 18 110 20 115 24 120 26 125 Table 5 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population (in hundreds) and the year over the ten-year span, (population, year) for specific recorded years: (4,500, 2000); (4,700, 2001); (5,200, 2003); (5,800, 2006) 31. Use linear regression to determine a function y, where the year depends on the population. Round to three decimal places of accuracy. 32. Predict when the population will hit 20,000. 33. What is the correlation coefficient for this model? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 5 Polynomial and Rational Functions Figure 1 35-mm ﬁlm, once the standard for capturing photographic images, has been made largely obsolete by digital photography. (credit “ﬁlm”: modiﬁcation of work by Horia Varlan; credit “memory cards”: modiﬁcation of work by Paul Hudson) CHAPTER OUTLINE 5.1 Quadratic Functions 5.2 Power Functions and Polynomial Functions 5.3 Graphs of Polynomial Functions 5.4 Dividing Polynomials 5.5 Zeros of Polynomial Functions 5.6 Rational Functions 5.7 Inverses and Radical Functions 5.8 Modeling Using Variation Introduction Digital photography has dramatically changed the nature of photography. No longer is an image etched in the emulsion on a roll of film. Instead, nearly every aspect of recording and manipulating images is now governed by mathematics. An image becomes a series of numbers, representing the characteristics of light striking an image sensor. When we open an image file, software on a camera or computer interprets the numbers and converts them to a visual image. Photo editing software uses complex polynomials to transform images, allowing us to manipulate the image in order to crop details, change the color palette, and add special effects. Inverse functions make it possible to convert from one file format to another. In this chapter, we will learn about these concepts and discover how mathematics can be used in such applications. 343 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Gie the characteristics of parabolas. • Graph a quadratic function. • etermine a quadratic functions minimum or maimum alue. • ole problems inoling a quadratic function. 5.1 QUADRATIC FUNCTIONS Figure 1 An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr) Curved antennas, such as the ones shown in Figure 1 are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite a
nd spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function. In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior. Recognizing Characteristics of Parabolas The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. These features are illustrated in Figure 2. y 6 4 2 Axis of symmetry x-intercepts –6 –4 –2 2 4 6 x –2 y–intercept 4 –6 Vertex Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 345 The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of x at which y = 0. Example 1 Identifying the Characteristics of a Parabola Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown in Figure 3. y 10 8 6 4 2 –4 –2 2 4 6 8 x Figure 3 Solution The vertex is the turning point of the graph. We can see that the vertex is at (3, 1). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. This parabola does not cross the x-axis, so it has no zeros. It crosses the y-axis at (0, 7) so this is the y-intercept. Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions The general form of a quadratic function presents the function in the form f(x) = ax 2 + bx + c where a, b, and c are real numbers and a ≠ 0. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry. b2 − 4ac −b ± √ The axis of symmetry is defined by x = − # b _______________ __ , to solve 2a 2a ax 2 + bx + c = 0 for the x-intercepts, or zeros, we find the value of x halfway between them is always x = − # b __ 2a equation for the axis of symmetry. . If we use the quadratic formula, x = , the — Figure 4 represents the graph of the quadratic function written in general form as y = x2 + 4x + 3. In this form, a = 1, b = 4, and c = 3. Because a > 0, the parabola opens upward. The axis of symmetry is x = − # 4 ____ = −2. This 2(1) also makes sense because we can see from the graph that the vertical line x = −2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, (−2, −1). The x-intercepts, those points where the parabola crosses the x-axis, occur at (−3, 0) and (−1, 0). y y = x2 + 4x + 3 8 6 4 2 –4 –6 Vertex Axis of symmetry –2 –4 Figure 4 x-intercepts x 2 4 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS The standard form of a quadratic function presents the function in the form f(x) = a(x − h)2 + k where (h, k) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function. As with the general form, if a > 0, the parabola opens upward and the vertex is a minimum. If a < 0, the parabola opens downward, and the vertex is a maximum. Figure 5 represents the graph of the quadratic function written in standard form as y = −3(x + 2)2 + 4. Since x − h = x + 2 in this example, h = −2. In this form, a = −3, h = −2, and k = 4. Because a < 0, the parabola opens downward. The vertex is at (−2, 4). Vertex y 4 2 y = −3(x + 2)2 + 4 –6 –4 –2 2 4 6 x –2 –4 –6 –8 Figure 5 The standard form is useful for determining how the graph is transformed from the graph of y = x 2. Figure 6 is the graph of this basic function. y 10 8 6 4 2 y = x2 –6 –4 –2 2 4 6 x –2 Figure 6 If k > 0, the graph shifts upward, whereas if k < 0, the graph shifts downward. In Figure 5, k > 0, so the graph is shifted 4 units upward. If h > 0, the graph shifts toward the right and if h < 0, the graph shifts to the left. In Figure 5, h < 0, so the graph is shifted 2 units to the left. The magnitude of a indicates the stretch of the graph. If .a. > 1, the point associated with a particular x-value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if .a. < 1, the point associated with a particular x-value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5, .a. > 1, so the graph becomes narrower. The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form. a(x − h)2 + k = ax 2 + bx + c ax 2 − 2ahx + (ah2 + k) = ax 2 + bx + c For the linear terms to be equal, the coefficients must be equal. −2ah = b, so h = − # b __ . 2a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 347 This is the axis of symmetry we defined earlier. Setting the constant terms equal: ah2 + k = c k = c − ah2 = c − a ( b2 __ 4a = c − 2 b ) __ 2a In practice, though, it is usually easier to remember that k is the output value of the function when the input is h, so f(h) = k. forms of quadratic functions A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is f(x) = ax 2 + bx + c where a, b, and c are real numbers and a ≠ 0. The standard form of a quadratic function is f(x) = a(x − h)2 + k where a ≠ 0. The vertex (h, k) is located at h = − #b __ 2a , k = f(h) = f ( −b ) . ___ 2a How To… Given a graph of a quadratic function, write the equation of the function in general form. 1. Identify the horizontal shift of the parabola; this value is h. Identify the vertical shift of the parabola; this value is k. 2. Substitute the values of the horizontal and vertical shift for h and k. in the function f(x) = a(x − h)2 + k. 3. Substitute the values of any point, other than the vertex, on the graph of the parabola for x and f (x). 4. Solve for the stretch factor, .a.. 5. If the parabola opens up, a > 0. If the parabola opens down, a < 0 since this means the graph was reflected about the x-axis. 6. Expand and simplify to write in general form. Example 2 Writing the Equation of a Quadratic Function from the Graph Write an equation for the quadratic function g in Figure 7 as a transformation of f(x) = x 2, and then expand the formula, and simplify terms to write the equation in general form. y 6 4 2 –6 –4 –2 2 4 x –2 –4 Figure 7 Solution We can see the graph of g is the graph of f(x) = x 2 shifted to the left 2 and down 3, giving a formula in the form g(x) = a(x −(−2))2 − 3 = a(x + 2)2 − 3. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Substituting the coordinates of a point on the curve, such as (0, −1), we can solve for the stretch factor. −1 = a(0 + 2)2 − 3 2 = 4a a =# #1 __ 2 1 __ (x + 2)2 − 3. In standard form, the algebraic model for this graph is g (x) = 2 To write this in general polynomial form, we can expand the formula and simplify terms. 1 __ (x + 2)2 −#3 g(x) = 2 1 __ = (x + 2)(x + 2) −#3 2 1 __ = (x 2 + 4x + 4) −#3 2 1 __ = x 2 + 2x + 2 −#3 2 1 __ = x 2 + 2x − 1 2 Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions. 1 __ (x + 2) 2 − 3. Next, Analysis We can check our work using the table feature on a graphing utility. First enter Y1 = 2 select TBLSET, then use TblStart = − 6 and ΔTbl = 2, and select TABLE. See Table 1. x y −6 5 −4 −1 −2 −3 0 −1 2 5 Table 1 The ordered pairs in the table correspond to points on the graph. Try It #1 A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 8 Find an equation for the path of the ball. Does the shooter make the basket? Figure 8 (credit: modiﬁcation of work by Dan Meyer) How To… Given a quadratic function in general form, find the vertex of the parabola. 1. Identify a, b, and c. 2. Find h, the x-coordinate of the vertex, by substituting a and b into h = − b ) . 3. Find k, the y-coordinate of the vertex, by evaluating k = f (h) = f ( − __ 2a b __ . 2a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 349 Example 3 Finding the Vertex of a Quadratic Function Find the vertex of the quadratic function f (x) = 2x 2 − 6x + 7. Rewrite the quadratic in standard form (vertex form). Solution The horizontal coordinate of the vertex will be at h = − # b __ 2a −6 ____ 2(2) = − 6 __ = 4 3 __ = 2 k = f(h) The vertical coordinate of the vertex will be at 3 ) = f ( __ 2 3 ) = 2 ( __ 2 5 __ = 2 Rewriting into standard form, the stretch factor will be the same as the a in the original quadratic. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the “a” from the general form. 3 ) + 7 −
6 ( __ 2 2 f (x) = ax2 + bx + c f (x) = 2x2 − 6x + 7 The standard form of a quadratic function prior to writing the function then becomes the following: 3 ) f (x) = 2 ( x − __ 2 2 5 __ + 2 Analysis One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, k, and where it occurs, x. Try It #2 Given the equation g (x) = 13 + x 2 − 6x, write the equation in general form and then in standard form. Finding the Domain and Range of a Quadratic Function Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down. domain and range of a quadratic function The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions. The range of a quadratic function written in general form f(x) = ax2 + bx + c with a positive a value is ) , ∞#) . ) , or [ f ( − #b f (x) ≥ f ( − #b __ __ 2a 2a ) ] . ) , or ( −∞, f ( − #b The range of a quadratic function written in general form with a negative a value is f (x) ≤ f ( − #b __ __ 2a 2a The range of a quadratic function written in standard form f (x) = a(x − h)2 + k with a positive a value is f (x) ≥ k; the range of a quadratic function written in standard form with a negative a value is f (x) ≤ k. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS How To… Given a quadratic function, find the domain and range. 1. Identify the domain of any quadratic function as all real numbers. 2. Determine whether a is positive or negative. If a is positive, the parabola has a minimum. If a is negative, the parabola has a maximum. 3. Determine the maximum or minimum value of the parabola, k. 4. If the parabola has a minimum, the range is given by f(x) ≥ k, or [k, ∞). If the parabola has a maximum, the range is given by f (x) ≤ k, or (−∞, k]. Example 4 Finding the Domain and Range of a Quadratic Function Find the domain and range of f (x) = −5x2 + 9x − 1. Solution As with any quadratic function, the domain is all real numbers. Because a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x-value of the vertex. h = − # b __ 2a = − # 9 _____ 2(−5) = 9 __ 10 The maximum value is given by f (h). 9 f ( __ 10 2 9 ) ) = −5 ( __ 10 61 ___ 20 = 9 + 9 ( __ 10 ) − 1 The range is f (x) ≤ 61 ___ 20 , or ( −∞, 61 ] . ___ 20 Try It #3 Find the domain and range of f (x) = 2 ( x −# #4 ) __ 7 2 + 8 __ . 11 Determining the Maximum and Minimum Values of Quadratic Functions The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in Figure 9. y f (x) = (x − 2)2 + 1 6 4 2 (2, 1) (–3, 4) y 6 4 2 g(x) = −(x + 3)2 + 4 –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –2 –4 –6 (a) Minimum value of 1 occurs at x = 2 Figure 9 Minimum value of 4 occurs at x = −3 –2 –4 –6 (b) There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 351 Example 5 Finding the Maximum Value of a Quadratic Function A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. a. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L. b. What dimensions should she make her garden to maximize the enclosed area? Solution Let’s use a diagram such as Figure 10 to record the given information. It is also helpful to introduce a temporary variable, W, to represent the width of the garden and the length of the fence section parallel to the backyard fence. Garden L W Backyard Figure 10 a. We know we have only 80 feet of fence available, and L + W + L = 80, or more simply, 2L + W = 80. This allows us to represent the width, W, in terms of L. W = 80 − 2L Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so A = LW = L(80 − 2L) A(L) = 80L − 2L2 This formula represents the area of the fence in terms of the variable length L. The function, written in general form, is A(L) = −2L2 + 80L. b. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a is the coefficient of the squared term, a = −2, b = 80, and c = 0. To find the vertex: h = − # b _ 2a h = − # 80 _ 2(−2) k = A(20) = 80(20) − 2(20)2 = 20 and = 800 The maximum value of the function is an area of 800 square feet, which occurs when L = 20 feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet. Analysis This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11. (20, 800) A y 1000 900 800 700 600 500 400 300 200 100 ) A ( a e r A 0 10 30 20 Length (L) Figure 11 40 x 50 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS How To… Given an application involving revenue, use a quadratic equation to find the maximum. 1. Write a quadratic equation for a revenue function. 2. Find the vertex of the quadratic equation. 3. Determine the y-value of the vertex. Example 6 Finding Maximum Revenue The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue? Solution Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p for price per subscription and Q for quantity, giving us the equation Revenue = pQ. Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently p = 30 and Q = 84,000. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, p = 32 and Q = 79,000. From this we can find a linear equation relating the two quantities. The slope will be m =# 79,000 − 84,000 _____________ 32 − 30 = −5,000 _______ 2 = −2,500 This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept. Q = −2,500p + b Substitute in the point Q = 84,000 and p = 30 84,000 = −2,500(30) + b Solve for b b = 159,000 This gives us the linear equation Q = −2,500p + 159,000 relating cost and subscribers. We now return to our revenue equation. Revenue = pQ Revenue = p(−2,500p + 159,000) Revenue = −2,500p2 + 159,000p We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex. h = − 159,000 _________ 2(−2,500) = 31.8 The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function. maximum revenue = −2,500(31.8)2 + 159,000(31.8) = 2,528,100 Analysis This could also be solved by graphing the quadratic as in Figure 12. We can see the maximum revenue on a graph of the quadratic function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 353 (31.80, 2,528.1) y 3,000 2,500 2,000 1,500 1,000 500 ) 10 20 30 40 50 60 70 80 x Price (p) Figure 12 Finding the x- and y-Intercepts of a Quadratic Function Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y-intercept of a quadratic by evaluating the function at an input of zero, and we find the x-intercepts at locations where the output is zero. Notice in Figure 13 that the number of x-intercepts can vary depending upon the location of the graph. y 5 4 3 2 1 –3 –2 –1 –1 –2 –3 y 5 4 3 2 1 –1 –1 –2 –3 21 3 4 5 x –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 21 3 4 5 x 21 3 4 5 x –3 –2 No x-intercept One x-intercept Two x-intercepts Figure 13 Number of x-intercepts of a parabola How To… Given a quadratic function f (x), find the y- and x-intercepts. 1. Evaluate f (0) to find the y-intercept. 2. Solve the quadratic equation f (x) = 0 to find the x-intercepts. Example 7 Finding the y- and x-Intercepts of a Parabola Find the y- and x-intercepts of the quadratic f(x
) = 3x2 + 5x − 2. Solution We find the y-intercept by evaluating f (0). So the y-intercept is at (0, −2). For the x-intercepts, we find all solutions of f(x) = 0. 0 = 3x2 + 5x − 2 f(0) = 3(0)2 + 5(0) − 2 = −2 In this case, the quadratic can be factored easily, providing the simplest method for solution. 1 __ , 0 ) and (−2, 0). So the x-intercepts are at ( 3 0 = (3x − 1)(x + 2) 0 = 3x − 1 1 __ x = 3 0 = x + 2 or x = −2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis By graphing the function, we can confirm that the graph crosses the y-axis at (0, −2). We can also confirm that the graph crosses the x-axis 1 __ , 0 ) and (−2, 0). See Figure 14. at ( 3 y (–2, 0) 1 f (x) = 3 x2 + 5 x − 2 , 0) ) 1 3 –3 –2 –1 1 2 3 x (0, –2) –1 –2 –3 –4 –5 Figure 14 Rewriting Quadratics in Standard Form In Example 7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form. How To… Given a quadratic function, find the x-intercepts by rewriting in standard form. 1. Substitute a and b into h = − b __ . 2a 2. Substitute x = h into the general form of the quadratic function to find k. 3. Rewrite the quadratic in standard form using h and k. 4. Solve for when the output of the function will be zero to find the x-intercepts. Example 8 Finding the x-Intercepts of a Parabola Find the x-intercepts of the quadratic function f (x) = 2x2 + 4x − 4. Solution We begin by solving for when the output will be zero. 0 = 2x2 + 4x − 4 Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form. We know that a = 2. Then we solve for h and k. f (x) = a(x − h)2 + k So now we can rewrite in standard form. h = − # b __ 2a = − # 4 ___ 2(2) = −1 k = f (−1) = 2(−1)2 + 4(−1) − 4 = −6 f(x) = 2(x + 1)2 − 6 We can now solve for when the output will be zero. 0 = 2(x + 1)2 − 6 6 = 2(x + 1)2 3 = (x + 1)2 x + 1 = ± √ — 3 The graph has x-intercepts at (−1 − √ 3 , 0) and (−1 + √ — — 3 x = −1 ± √ 3 , 0). — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 355 We can check our work by graphing the given function on a graphing utility and observing the x-intercepts. See Figure 15. y 6 4 2 (0.732, 0) (−2.732, 0) –6 –4 –2 2 4 6 x –2 –4 –6 Figure 15 Analysis We could have achieved the same results using the quadratic formula. Identify a = 2, b = 4, and c = −4. x = x = — b2 − 4ac −b ± √ __ 2a — 42 − 4(2)(−4) −4 ± √ ___ 2(2) — x = −4 ± √ 48 __ 4 x = — −4 ± √ 3(16) __ 4 x = −1 ± √ — 3 So the x-intercepts occur at (−1 − √ — 3 , 0) and (−1 + √ — 3 , 0). Try It #4 In a separate Try It, we found the standard and general form for the function g(x) = 13 + x2 − 6x. Now find the y- and x-intercepts (if any). Example 9 Applying the Vertex and x-Intercepts of a Parabola A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H(t) = −16t 2 + 80t + 40. a. When does the ball reach the maximum height? b. What is the maximum height of the ball? c. When does the ball hit the ground? Solution a. The ball reaches the maximum height at the vertex of the parabola. h = − # 80 ______ 2(−16) = 80 ___ 32 5 __ = 2 = 2.5 The ball reaches a maximum height after 2.5 seconds. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS b. To find the maximum height, find the y-coordinate of the vertex of the parabola. k = H ( − # b ) __ 2a = H(2.5) = −16(2.5)2 + 80(2.5) + 40 = 140 The ball reaches a maximum height of 140 feet. c. To find when the ball hits the ground, we need to determine when the height is zero, H(t) = 0. We use the quadratic formula. t = — −80 ± √ _______________________ 802 − 4(−16)(40) 2(−16) = — −80 ± √ 8960 _____________ −32 Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions. t = — 8960 −80 − √ _____________ −32 ≈ 5.458 or t = — 8960 −80 + √ _____________ −32 ≈ −0.458 The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16. H 150 125 100 75 50 25 (2.5, 140) H(t) =− 16 t2 + 80t + 40 1 2 3 4 Figure 16. 5 6 t Notice that the graph does not represent the physical path of the ball upward and downward. Keep quantities on each axis in mind while interpreting the graph. Try It #5 A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation H(t) = −16t 2 + 96t + 112. a. When does the rock reach the maximum height? b. What is the maximum height of the rock? c. When does the rock hit the ocean? Access these online resources for additional instruction and practice with quadratic equations. • Graphing Quadratic Functions in General Form (http://openstaxcollege.org/l/graphquadgen) • Graphing Quadratic Functions in Standard Form (http://openstaxcollege.org/l/graphquadstan) • Quadratic Function Review (http://openstaxcollege.org/l/quadfuncrev) • Characteristics of a Quadratic Function (http://openstaxcollege.org/l/characterquad) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 SECTION EXERCISES 357 5.1 SECTION EXERCISES VERBAL 1. Explain the advantage of writing a quadratic 2. How can the vertex of a parabola be used in solving function in standard form. real-world problems? 3. Explain why the condition of a ≠ 0 is imposed in the definition of the quadratic function. 4. What is another name for the standard form of a quadratic function? 5. What two algebraic methods can be used to find the horizontal intercepts of a quadratic function? ALGEBRAIC For the following exercises, rewrite the quadratic functions in standard form and give the vertex. 6. f (x) = x 2 − 12x + 32 7. g(x) = x 2 + 2x − 3 8. f (x) = x 2 − x 9. f (x) = x 2 + 5x − 2 10. h(x) = 2x 2 + 8x − 10 11. k(x) = 3x 2 − 6x − 9 12. f (x) = 2x 2 − 6x 13. f (x) = 3x 2 − 5x − 1 For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry. 14. y(x) = 2x 2 + 10x + 12 15. f(x) = 2x 2 − 10x + 4 17. f(x) = 4x 2 + x − 1 18. h(t) = −4t 2 + 6t − 1 16. f(x) = −x 2 + 4x + 3 1 __ x 2 + 3x + 1 19. f(x) = 2 20. f(x) = − #1 __ x 2 − 2x + 3 3 For the following exercises, determine the domain and range of the quadratic function. 21. f(x) = (x − 3)2 + 2 22. f(x) = −2(x + 3)2 − 6 23. f(x) = x 2 + 6x + 4 24. f(x) = 2x 2 − 4x + 2 25. k(x) = 3x 2 − 6x − 9 For the following exercises, use the vertex (h, k) and a point on the graph (x, y) to find the general form of the equation of the quadratic function. 26. (h, k) = (2, 0), (x, y) = (4, 4) 27. (h, k) = (−2, −1), (x, y) = (−4, 3) 28. (h, k) = (0, 1), (x, y) = (2, 5) 29. (h, k) = (2, 3), (x, y) = (5, 12) 30. (h, k) = (−5, 3), (x, y) = (2, 9) 31. (h, k) = (3, 2), (x, y) = (10, 1) 32. (h, k) = (0, 1), (x, y) = (1, 0) 33. (h, k) = (1, 0), (x, y) = (0, 1) GRAPHICAL For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. 34. f(x) = x 2 − 2x 35. f(x) = x 2 − 6x − 1 36. f(x) = x 2 − 5x − 6 37. f(x) = x 2 − 7x + 3 38. f(x) = −2x 2 + 5x − 8 39. f(x) = 4x 2 − 12x − 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS For the following exercises, write the equation for the graphed function. 42. y y 5 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4 –5 40. 436 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6 –7 NUMERIC 416 –5 –4 –3 –2 –1 –1 –2 1 2 3 4 x 44. y 45. 7 6 5 4 3 2 1 –3 –2 –1 –1 –2 –2 –1 –1 –7 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function. 46. x −2 −1 y 5 2 49. x −2 −1 y −8 − 47. x −2 −1 y 1 0 50. x −2 − TECHNOLOGY For the following exercises, use a calculator to find the answer. 48. x −2 −1 y −2 1 0 2 1 2 1 −2 51. Graph on the same set of axes the functions 1 __ f (x) = x2, f(x) = 2x 2, and f(x) = x 2. What appears 3 to be the effect of changing the coefficient? 53. Graph on the same set of axes f(x) = x 2, f(x) = (x − 2)2, f(x − 3)2, and f(x) = (x + 4)2. What appears to be the effect of adding or subtracting those numbers? 52. Graph on the same set of axes f(x) = x 2, f(x) = x 2 + 2 and f(x) = x 2, f(x) = x 2 + 5 and f(x) = x 2 − 3. What appears to be the effect of adding a constant? 54. The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given −32 ____ by the function h(x) = (80)2 x 2 + x where x is the horizontal distance traveled and h(x) is the height in feet. Use the [TRACE] feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally. 55. A suspension bridge can be modeled by the quadratic function h(x) = 0.0001x 2 with −2000 ≤ x ≤ 2000 where ∣ x ∣ is the number of feet from the center and h(x) is height in feet. Use the [TRACE] feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 SECTION EXERCISES 359 EXTENSIONS For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function. 56. Vertex (1, −2), opens up. 58. Vertex (−5, 11), opens down. 57. Vertex (−1, 2) opens down. 59. Vertex (−100, 100), opens up. For the following exercises, write the equation
of the quadratic function that contains the given point and has the same shape as the given function. 60. Contains (1, 1) and has shape of f(x) = 2x 2. 61. Contains (−1, 4) and has the shape of f(x) = 2x 2. Vertex is on the y-axis. Vertex is on the y-axis. 62. Contains (2, 3) and has the shape of f(x) = 3x 2. 63. Contains (1, −3) and has the shape of f(x) = −x 2. Vertex is on the y-axis. Vertex is on the y-axis. 64. Contains (4, 3) and has the shape of f(x) = 5x 2. 65. Contains (1, −6) has the shape of f(x) = 3x 2. Vertex is on the y-axis. Vertex has x-coordinate of −1. REAL-WORLD APPLICATIONS 66. Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing. 67. Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing. 68. Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing. 69. Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. What is the product? 70. Among all of the pairs of numbers whose difference is 12, find the pair with the smallest product. What is the product? 71. Suppose that the price per unit in dollars of a cell phone production is modeled by p = $45 − 0.0125x, where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x ċ p. Find the production level that will maximize revenue. 72. A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by h(t) = −4.9t 2 + 229t + 234. Find the maximum height the rocket attains. 73. A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by h(t) = −4.9t2 + 24t + 8. How long does it take to reach maximum height? 74. A soccer stadium holds 62,000 spectators. With a 75. A farmer finds that if she plants 75 trees per acre, ticket price of $11, the average attendance has been 26,000. When the price dropped to $9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue? each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS G C F . and . In this section, you will: • Identify the degree and leading coefficient of polynomial functions. • dentify the end behaior of polynomial functions. • Gien a graph utilie the relationship between turning points and degree of a polynomial to determine the least possible degree. • Use factoring to find eros of a polynomial. • dentify eros and their multiplicity. • Graph polynomial functions using eros multiplicity and end behaior • rite a polynomial function that satisfies certain criteria. 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in Table 1. Figure 1 (credit: Jason Bay, Flickr) Year Bird Population 2009 800 2010 897 Table 1 2011 992 2012 1,083 2013 1,169 The population can be estimated using the function P(t) = −0.3t3 + 97t + 800, where P(t) represents the bird population on the island t years after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we will examine functions that we can use to estimate and predict these types of changes. Identifying Power Functions Before we can understand the bird problem, it will be helpful to understand a different type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number. (A number that multiplies a variable raised to an exponent is known as a coefficient.) As an example, consider functions for area or volume. The function for the area of a circle with radius r is A(r) = πr 2 and the function for the volume of a sphere with radius r is V(r) =# #4 __ πr 3 3 Both of these are examples of power functions because they consist of a coefficient, π or 4 _ 3 π, multiplied by a variable r raised to a power. power function A power function is a function that can be represented in the form f (x) = kx p where k and p are real numbers, and k is known as the coefficient. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 361 Q & A… Is f (x) = 2x a power function? No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function. Example 1 Identifying Power Functions Which of the following functions are power functions? f(x) = 1 f (x) = x f (x) = x 2 f (x) = x 3 f (x) =# #1 __ x 1 __ x2 — f (x) = √ x — x f (x) = f (x) = 3 √ Constant function Identify function Quadratic function Cubic function Reciprocal function Reciprocal squared function Square root function Cube root function Solution All of the listed functions are power functions. The constant and identity functions are power functions because they can be written as f (x) = x0 and f (x) = x1 respectively. The quadratic and cubic functions are power functions with whole number powers f (x) = x 2 and f (x) = x 3. The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as f (x) = x −1 and f (x) = x −2. The square and cube root functions are power functions with fractional powers because they can be written as f (x) = x1/2 or f (x) = x 1/3. Try It #1 Which functions are power functions? f (x) = 2x 2 ċ 4x 3 g(x) = −x 5 + 5x 3 h(x) = 2x5 − 1_ 3x2 + 4 Identifying End Behavior of Power Functions Figure 2 shows the graphs of f (x) = x 2, g(x) = x 4 and h(x) = x 6, which are all power functions with even, whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin. f (x) = x6 y g(x) = x4 h(x) = x2 4 3 2 1 –2 –1 1 2 x Figure 2 Even-power functions Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS To describe the behavior as numbers become larger and larger, we use the idea of in fi nity. We use the symbol ∞ for positive infinity and −∞ for negative infinity. When we say that “x approaches infinity,” which can be symbolically written as x → ∞, we are describing a behavior; we are saying that x is increasing without bound. With the positive even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as x approaches positive or negative infinity, the f (x) values increase without bound. In symbolic form, we could write Figure 3 shows the graphs of f (x) = x 3, g(x) = x 5, and h(x) = x 7, which are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin. as x → ±∞, f (x) → ∞ g(x) = x5 f (x) = x3 h(x) = x7 1 2 x y 4 2 –2 –4 –2 –1 Figure 3 Odd-power functions These examples illustrate that functions of the form f (x) = xn reveal symmetry of one kind or another. First, in Figure 2 we see that even functions of the form f (x) = xn, n even, are symmetric about the y-axis. In Figure 3 we see that odd functions of the form f (x) = xn, n odd, are symmetric about the origin. For these odd power functions, as x approaches negative infinity, f (x) decreases without bound. As x approaches positive infinity, f (x) increases without bound. In symbolic form we write as x → −∞, f (x) → −∞ as x → ∞, f (x) → ∞ The behavior of the graph of a function as the input values get very small (x → −∞) and get very large (x → ∞) is referred to as the end behavior of the function. We can use words or symbols to describe end behavior. Figure 4 shows the end behavior of power functions in the form f (x) = kxn where n is a non-negative integer depending on the power and the constant. Even power y Odd power y Positive constant k > 0 x x x → −∞, f (x) → ∞ and x → ∞, f (x) → ∞ y x → −∞, f (x) → ∞ and x → ∞, f (x) → −∞ y Negative constant k < 0 x x x → −∞, f (x) → ∞ and x → ∞, f (x) → −∞ x → −∞, f (x) → −∞ and x → ∞, f (x) → −∞ Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 363 How To… Given a power function f (x) = kx n where n is a non-negative integer, identify the end behavior. 1. Determine whether the power is even or odd. 2. Determine whether the constant is positive or negative. 3. Use Figure 4 to identify the end behavior. Example 2 Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f (x) = x8. Solution The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As x approaches infinity, the output (value of f (x)) increases without bound. We write as x → ∞, f (x) → ∞. As x approaches negative infinity, the output increases without bound. In symbolic form, as x → −∞, f (x) → ∞. We can graphically represent the function as shown in Figure 5. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 21 3 4 5 x Example 3 Id
entifying the End Behavior of a Power Function Describe the end behavior of the graph of f (x) = −x 9. Figure 5 Solution The exponent of the power function is 9 (an odd number). Because the coefficient is −1 (negative), the graph is the reflection about the x-axis of the graph of f (x) = x 9. Figure 6 shows that as x approaches infinity, the output decreases without bound. As x approaches negative infinity, the output increases without bound. In symbolic form, we would write as x → −∞, f (x) → ∞ as x → ∞, f (x) → −∞ 1 –1 –2 –3 –4 –5 –6 –7 –8 –5 –4 –3 –2 21 3 4 5 x Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis We can check our work by using the table feature on a graphing utility. x −10 −5 0 5 10 f (x) 1,000,000,000 1,953,125 0 −1,953,125 −1,000,000,000 Table 2 We can see from Table 2 that, when we substitute very small values for x, the output is very large, and when we substitute very large values for x, the output is very small (meaning that it is a very large negative value). Try It #2 Describe in words and symbols the end behavior of f (x) = −5x 4. Identifying Polynomial Functions An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil slick by combining two functions. The radius r of the spill depends on the number of weeks w that have passed. This relationship is linear. We can combine this with the formula for the area A of a circle. A(r) = πr 2 Composing these functions gives a formula for the area in terms of weeks. r(w) = 24 + 8w Multiplying gives the formula. A(w) = A(r(w)) = A(24 + 8w) = π(24 + 8w)2 A(w) = 576π + 384πw + 64πw 2 This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. polynomial functions Let n be a non-negative integer. A polynomial function is a function that can be written in the form f (x) = an x n + ... + a2 x 2 + a1 x + a0 This is called the general form of a polynomial function. Each ai is a coefficient and can be any real number other than zero. Each expression ai x i is a term of a polynomial function. Example 4 Identifying Polynomial Functions Which of the following are polynomial functions? f (x) = 2x 3 ċ 3x + 4 g(x) = −x(x2 − 4) h(x) = 5 √ — x + 2 Solution The first two functions are examples of polynomial functions because they can be written in the form f (x) = an xn + ... + a2 x2 + a1 x + a0, where the powers are non-negative integers and the coefficients are real numbers. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 365 • f (x) can be written as f (x) = 6x4 + 4. • g(x) can be written as g(x) = −x3 + 4x. • h(x) cannot be written in this form and is therefore not a polynomial function. Identifying the Degree and Leading Coefﬁcient of a Polynomial Function Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term. terminology of polynomial functions We often rearrange polynomials so that the powers are descending. Leading coefficient Degree f (x) = an x n + … + a2 x 2 + a1x + a0 Leading term When a polynomial is written in this way, we say that it is in general form. How To… Given a polynomial function, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree function. 2. Identify the term containing the highest power of x to find the leading term. 3. Identify the coefficient of the leading term. Example 5 Identifying the Degree and Leading Coefﬁcient of a Polynomial Function Identify the degree, leading term, and leading coefficient of the following polynomial functions. f (x) = 3 + 2x 2 − 4x 3 g(t) = 5t 5 − 2t 3 + 7t h(p) = 6p − p 3 − 2 Solution For the function f (x), the highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4x3. The leading coefficient is the coefficient of that term, −4. For the function g(t), the highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t5. The leading coefficient is the coefficient of that term, 5. For the function h(p), the highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, −p3; the leading coefficient is the coefficient of that term, −1. Try It #3 Identify the degree, leading term, and leading coefficient of the polynomial f (x) = 4x 2 − x 6 + 2x − 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Identifying End Behavior of Polynomial Functions Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the power function consisting of the leading term. See Table 3. Polynomial Function Leading Term Graph of Polynomial Function f (x) = 5x4 + 2x3 − x − 4 5x4 –5 –4 –3 –2 f (x) = −2x6 − x5 + 3x4 + x3 −2x6 –5 –4 –3 –2 f (x) = 3x5 − 4x4 + 2x2 + 1 3x5 –5 –4 –3 –2 f (x) = −6x3 + 7x2 + 3x + 1 −6x3 –5 –4 –3 –2 Table 3 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 x 21 3 4 5 x 21 3 4 5 x 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 367 Example 6 Identifying End Behavior and Degree of a Polynomial Function Describe the end behavior and determine a possible degree of the polynomial function in Figure 7. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 7 Solution As the input values x get very large, the output values f (x) increase without bound. As the input values x get very small, the output values f (x) decrease without bound. We can describe the end behavior symbolically by writing as x → −∞, as x → ∞, f (x) → −∞ f (x) → ∞ In words, we could say that as x values approach infinity, the function values approach infinity, and as x values approach negative infinity, the function values approach negative infinity. We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive. Try It #4 Describe the end behavior, and determine a possible degree of the polynomial function in Figure 8. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 8 Example 7 Identifying End Behavior and Degree of a Polynomial Function Given the function f (x) = −3x2(x − 1)(x + 4), express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function. Solution Obtain the general form by expanding the given expression for f (x). f (x) = −3x2(x − 1)(x + 4) = −3x2(x2 + 3x − 4) = −3x4 − 9x3 + 12x2 The general form is f (x) = −3x4 − 9x3 + 12x2. The leading term is −3x4; therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (−3), so the end behavior is as x → −∞, as x → ∞, f (x) → −∞ f (x) → −∞ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Try It #5 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. Identifying Local Behavior of Polynomial Functions In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing. We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-intercept (0, a0). The x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept. See Figure 9. y Turning point x-intercepts x Turning point ←y-intercept Figure 9 intercepts and turning points of polynomial functions A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero.
How To… Given a polynomial function, determine the intercepts. 1. Determine the y-intercept by setting x = 0 and finding the corresponding output value. 2. Determine the x-intercepts by solving for the input values that yield an output value of zero. Example 8 Determining the Intercepts of a Polynomial Function Given the polynomial function f (x) = (x − 2)(x + 1)(x − 4), written in factored form for your convenience, determine the y- and x-intercepts. Solution The y-intercept occurs when the input is zero so substitute 0 for x. The y-intercept is (0, 8). f (0) = (0 − 2)(0 + 1)(0 − 4) = (−2)(1)(− 4) = 8 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 369 The x-intercepts occur when the output is zero. 0 = (x − 2)(x + 1)(x − 4) The x-intercepts are (2, 0), (–1, 0), and (4, 0). x = 2 or x − 2 = 0 or x + 1 = 0 x = −1 or or x − 4 = 0 x = 4 We can see these intercepts on the graph of the function shown in Figure 10. y-intercept (0, 85 –4 –3 –2 21 3 –1 –1 –2 –3 –4 Figure 10 x 4 5 x-intercepts (−1, 0), (2, 0), and (4, 0) Example 9 Determining the Intercepts of a Polynomial Function with Factoring Given the polynomial function f (x) = x4 − 4x2 − 45, determine the y- and x-intercepts. Solution The y-intercept occurs when the input is zero. The y-intercept is (0, −45). f (0) = (0)4 − 4(0)2 − 45 = −45 The x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial. f (x) = x4 − 4x2 − 45 = (x2 − 9)(x2 + 5) = (x − 3)(x + 3)(x2 + 5) 0 = (x − 3)(x + 3)(x2 + 5) x − 3 = 0 or x + 3 = 0 or x2 + 5 = 0 x = 3 or x = −3 or (no real solution) The x-intercepts are (3, 0) and (−3, 0). We can see these intercepts on the graph of the function shown in Figure 11. We can see that the function is even because f (x) = f (−x). y 120 100 80 60 40 20 –2 –1 –20 –40 –60 –80 –100 –120 –5 –4 –3 x-intercepts (−3, 0) and (3, 0) x 21 3 4 5 y-intercept (0, −45) Try It #6 Given the polynomial function f (x) = 2x3 − 6x2 − 20x, determine the y- and x-intercepts. Figure 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Comparing Smooth and Continuous Graphs The degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function of nth degree is the product of n factors, so it will have at most n roots or zeros, or x-intercepts. The graph of the polynomial function of degree n must have at most n − 1 turning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors. A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth. intercepts and turning points of polynomials A polynomial of degree n will have, at most, n x-intercepts and n − 1 turning points. Example 10 Determining the Number of Intercepts and Turning Points of a Polynomial Without graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for f (x) = −3x 10 + 4x 7 − x 4 + 2x 3. Solution The polynomial has a degree of 10, so there are at most 10 x-intercepts and at most 9 turning points. Try It #7 Without graphing the function, determine the maximum number of x-intercepts and turning points for f (x) = 108 − 13x9 − 8x4 + 14x12 + 2x3 Example 11 Drawing Conclusions about a Polynomial Function from the Graph What can we conclude about the polynomial represented by the graph shown in Figure 12 based on its intercepts and turning points? y 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 12 Solution The end behavior of the graph tells us this is the graph of an even-degree polynomial. See Figure 13. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 x-intercepts 21 3 4 5 x Turning points Figure 13 The graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 371 Try It #8 What can we conclude about the polynomial represented by the graph shown in Figure 14 based on its intercepts and turning points? y –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x Figure 14 Example 12 Drawing Conclusions about a Polynomial Function from the Factors Given the function f (x) = −4x(x + 3)(x − 4), determine the local behavior. Solution The y-intercept is found by evaluating f (0). f (0) = −4(0)(0 + 3)(0 − 4) = 0 The y-intercept is (0, 0). The x-intercepts are found by determining the zeros of the function. 0 = −4x(x + 3)(x − 4) x = 0 or x + 3 = 0 or x − 4 = 0 x = 0 or x = −3 or x = 4 The x-intercepts are (0, 0), (−3, 0), and (4, 0). The degree is 3 so the graph has at most 2 turning points. Try It #9 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), determine the local behavior. Access these online resources for additional instruction and practice with power and polynomial functions. • Find Key Information About a Given Polynomial Function (http://openstaxcollege.org/l/keyinfopoly) • End Behavior of a Polynomial Function (http://openstaxcollege.org/l/endbehavior) • Turning Points and x-intercepts of Polynomial Functions (http://openstaxcollege.org/l/turningpoints) • Least Possible Degree of a Polynomial Function (http://openstaxcollege.org/l/leastposdegree) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.2 SECTION EXERCISES VERBAL 1. Explain the difference between the coefficient 2. If a polynomial function is in factored form, what of a power function and its degree. 3. In general, explain the end behavior of a power function with odd degree if the leading coefficient is positive. 5. What can we conclude if, in general, the graph of a polynomial function exhibits the following end behavior? As x → −∞, f (x) → −∞ and as x → ∞, f (x) → −∞. ALGEBRAIC would be a good first step in order to determine the degree of the function? 4. What is the relationship between the degree of a polynomial function and the maximum number of turning points in its graph? For the following exercises, identify the function as a power function, a polynomial function, or neither. 6. f (x) = x5 9. f (x) = x2 _____ x2 − 1 7. f (x) = (x2)3 8. f (x) = x − x4 10. f (x) = 2x(x + 2)(x − 1)2 11. f (x) = 3x + 1 For the following exercises, find the degree and leading coefficient for the given polynomial. 12. −3x 15. x(4 − x2)(2x + 1) 13. 7 − 2x2 16. x 2 (2x − 3)2 14. −2x2 − 3x5 + x − 6 For the following exercises, determine the end behavior of the functions. 17. f (x) = x4 20. f (x) = −x9 18. f (x) = x3 19. f (x) = −x4 21. f (x) = −2x4 − 3x2 + x − 1 22. f (x) = 3x2 + x − 2 23. f (x) = x2(2x3 − x + 1) 24. f (x) = (2 − x)7 For the following exercises, find the intercepts of the functions. 25. f (t) = 2(t − 1)(t + 2)(t − 3) 26. g(n) = −2(3n − 1)(2n + 1) 27. f (x) = x4 − 16 28. f (x) = x3 + 27 29. f (x) = x(x2 − 2x − 8) 30. f (x) = (x + 3)(4x2 − 1) GRAPHICAL For the following exercises, determine the least possible degree of the polynomial function shown. 31. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 32. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 33. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 SECTION EXERCISES 373 34. 37. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 –5 –4 –3 –2 35. 21 3 4 5 x –5 –4 –3 –2 38. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 36. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x 21 3 4 5 x For the following exercises, determine whether the graph of the function provided is a graph of a polynomial function. If so, determine the number of turning points and the least possible degree for the function. 42. y x 39. 43. y y y y 40. 44. x x 41. 45. x x y y x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS NUMERIC For the following exercises, make a table to confirm the end behavior of the function. 46. f (x) = −x3 49. f (x) = (x − 1)(x − 2)(3 − x) 47. f (x) = x4 − 5x2 50. f (x) = x5 __ 10 − x4 48. f (x) = x2(1 − x)2 TECHNOLOGY For the following exercises, graph the polynomial functions using a calculator. Based on the graph, determine the intercepts and the end behavior. 51. f (x) = x3(x − 2) 52. f (x) = x(x − 3)(x + 3) 53. f (x) = x(14 − 2x)(10 − 2x) 54. f (x) = x(14 − 2x)(10 − 2x)2 55. f (x) = x3 − 16x 56. f (x) = x3 − 27 58. f (x) = −x3 + x2 + 2x 59. f (x) = x3 − 2x2 − 15x 57. f (x) = x4 − 81 60. f (x) = x3 − 0.01x EXTENSIONS For the following exercises, use the information about the graph of a polynomial function to determine the function. Assume the leading coefficient is 1 or −1. There may be more than one correct answer. 61. The y-intercept is (0, −4). The x-intercepts are (−2, 0), (2, 0). Degree is 2. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → ∞. 62. The y-intercept is (0, 9). The x-intercepts are (−3, 0), (3, 0). Degree is 2. End behavior: as x → −∞, f (x) → −∞, as x → ∞, f (x) → −∞. 63. The y-intercept is (0, 0). The x-intercepts are (0, 0), (2, 0). Degree is 3. End behavior: as x → −∞, f (x) → −∞, as x → ∞, f (x) → ∞. 64. The y-intercept is (0, 1). The x-intercept is (1, 0). Degree is 3. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → −∞. 65. The y-intercept is (0, 1). There is
no x-intercept. Degree is 4. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → ∞. REAL-WORLD APPLICATIONS For the following exercises, use the written statements to construct a polynomial function that represents the required information. 66. An oil slick is expanding as a circle. The radius of the circle is increasing at the rate of 20 meters per day. Express the area of the circle as a function of d, the number of days elapsed. 67. A cube has an edge of 3 feet. The edge is increasing at the rate of 2 feet per minute. Express the volume of the cube as a function of m, the number of minutes elapsed. 68. A rectangle has a length of 10 inches and a width of 6 inches. If the length is increased by x inches and the width increased by twice that amount, express the area of the rectangle as a function of x. 69. An open box is to be constructed by cutting out square corners of x-inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of x. 70. A rectangle is twice as long as it is wide. Squares of side 2 feet are cut out from each corner. Then the sides are folded up to make an open box. Express the volume of the box as a function of the width (x). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 375 G C F . . In this section, you will: • dentify the degree and leading coefficient of polynomial functions. • dentify the end behaior of polynomial functions. • Gien a graph utilie the relationship between turning points and degree of a polynomial to determine the least possible degree. • Use factoring to find eros of a polynomial. • dentify eros and their multiplicity. • Graph polynomial functions using eros multiplicity and end behaior • rite a polynomial function that satisfies certain criteria. 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 1. Year 2006 2007 2008 2009 2010 2011 2012 2013 Revenues 52.4 52.8 51.2 49.5 48.6 48.6 48.7 47.1 The revenue can be modeled by the polynomial function Table 1 R(t) = −0.037t4 + 1.414t3 − 19.777t2 + 118.696t − 205.332 where R represents the revenue in millions of dollars and t represents the year, with t = 6 corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general. Recognizing Characteristics of Graphs of Polynomial Functions Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial. y y f x Figure 1 x f Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 1 Recognizing Polynomial Functions Which of the graphs in Figure 2 represents a polynomial function? y y f h y x x Figure 2 y g k x x Solution The graphs of f and h are graphs of polynomial functions. They are smooth and continuous. The graphs of g and k are graphs of functions that are not polynomials. The graph of function g has a sharp corner. The graph of function k is not continuous. Q & A… Do all polynomial functions have as their domain all real numbers? Yes. Any real number is a valid input for a polynomial function. Using Factoring to Find Zeros of Polynomial Functions Recall that if f is a polynomial function, the values of x for which f (x) = 0 are called zeros of f. If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros. We can use this method to find x-intercepts because at the x-intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases: 1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring. 2. The polynomial is given in factored form. 3. Technology is used to determine the intercepts. How To… Given a polynomial function f, find the x-intercepts by factoring. 1. Set f (x) = 0. 2. If the polynomial function is not given in factored form: a. Factor out any common monomial factors. b. Factor any factorable binomials or trinomials. 3. Set each factor equal to zero and solve to find the x-intercepts. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 377 Example 2 Finding the x-Intercepts of a Polynomial Function by Factoring Find the x-intercepts of f (x) = x 6 − 3x 4 + 2x2. Solution We can attempt to factor this polynomial to find solutions for f (x) = 0. x 6 − 3x 4 + 2x 2 = 0 Factor out the greatest common factor. x 2(x 4 − 3x2 + 2) = 0 Factor the trinomial. x 2(x 2 − 1)(x 2 − 2) = 0 Set each factor equal to zero. (x 2 − 1) = 0 (x 2 − 2) = 0 x 2 = 0 or x 2 = 1 or This gives us five x-intercepts: (0, 0), (1, 0), (−1, 0), ( √ even function because it is symmetric about the y-axis. — 2 , 0 ) , and ( − √ — 2 , 0 ) . See Figure 3. We can see that this is an (– , 0) √2 –2 f (x) = x6 – 3 x4 + 2x2 y 2 1 ( , 0)√2 x 2 (–1, 0) –1 (0, 0) (1, 0) –2 Figure 3 Example 3 Finding the x-Intercepts of a Polynomial Function by Factoring Find the x-intercepts of f (x) = x 3 − 5x 2 − x + 5. Solution Find solutions for f (x) = 0 by factoring. x 3 − 5x 2 − x + 5 = 0 Factor by grouping. x 2(x − 5) − (x − 5) = 0 Factor out the common factor. (x 2 − 1)(x − 5) = 0 Factor the difference of squares. (x + 1)(x − 1)(x − 5) = 0 Set each factor equal to zero. x + 1 = 0 or x − 1 = 0 or x − 5 = 0 x = −1 x = 1 x = 5 There are three x-intercepts: (−1, 0), (1, 0), and (5, 0). See Figure 4. y 18 12 6 f (x) = x3 − 5 x2 − x + 5 –6 –4 –2 2 4 6 x –6 –12 –18 Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 4 Finding the y- and x-Intercepts of a Polynomial in Factored Form Find the y- and x-intercepts of g(x) = (x − 2)2(2x + 3). Solution The y-intercept can be found by evaluating g(0). g(0) = (0 − 2)2(2(0) + 3) = 12 So the y-intercept is (0, 12). The x-intercepts can be found by solving g(x) = 0. (x − 2)2(2x + 3) = 0 (x − 2)2 = 0 (2x + 3) = 0 x − 2 = 0 or x = 2 x = − #3 __ 2 So the x-intercepts are (2, 0) and ( − #3 , 0 ) . __ 2 Analysis We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5. y 15 (0, 12) 12 9 6 3 (–1.5, 0) g(x) = (x − 2)2(2x + 3 ) (2, 0) –4 –3 –2 –1 1 2 3 4 x –3 Figure 5 Example 5 Finding the x-Intercepts of a Polynomial Function Using a Graph Find the x-intercepts of h(x) = x3 + 4x2 + x − 6. Solution This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities. Looking at the graph of this function, as shown in Figure 6, it appears that there are x-intercepts at x = −3, −2, and 1. h(x) = x3 + 4 x2 + x − 6 y 2 –4 –2 2 4 x –2 –4 –6 –8 Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 379 We can check whether these are correct by substituting these values for x and verifying that Since h(x) = x 3 + 4x 2 + x − 6, we have: h(−3) = h(−2) = h(1) = 0. h(−3) = (−3)3 + 4(−3)2 + (−3) − 6 = −27 + 36 − 3 − 6 = 0 h(−2) = (−2)3 + 4(−2)2 + (−2) − 6 = −8 + 16 − 2 − 6 = 0 h(1) = (1)3 + 4(1)2 + (1 Each x-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form. h(x) = x3 + 4x2 + x − 6 = (x + 3)(x + 2)(x − 1) Try It #1 Find the y- and x-intercepts of the function f (x) = x 4 − 19x 2 + 30x. Identifying Zeros and Their Multiplicities Graphs behave differently at various x-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and "bounce" off. Suppose, for example, we graph the function shown. f (x) = (x + 3)(x − 2)2(x + 1)3. Notice in Figure 7 that the behavior of the function at each of the x-intercepts is different. f (x) = (x + 3)(x − 2)2(x + 1)3 2 4 x y 40 30 20 10 –10 –20 –30 –40 –4 –2 Figure 7 Identifying the behavior of the graph at an x-intercept by examining the multiplicity of the zero. The x-intercept x = −3 is the solution of equation (x + 3) = 0. The graph passes directly through the x-intercept at x = −3. The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function. The x-intercept x = 2 is the repeated solution of equation (x − 2)2 = 0. The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. (x − 2)2 = (x − 2)(x
− 2) The factor is repeated, that is, the factor (x − 2) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, x = 2, has multiplicity 2 because the factor (x − 2) occurs twice. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS The x-intercept x = −1 is the repeated solution of factor (x + 1)3 = 0. The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic— with the same S-shape near the intercept as the toolkit function f (x) = x 3. We call this a triple zero, or a zero with multiplicity 3. For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3. y y y p = Single zero Zero with multiplicity 2 Zero with multiplicity 3 Figure 8 For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis. For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis. graphical behavior of polynomials at x-intercepts If a polynomial contains a factor of the form (x − h)p, the behavior near the x-intercept h is determined by the power p. We say that x = h is a zero of multiplicity p. The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities. The sum of the multiplicities is the degree of the polynomial function. How To… Given a graph of a polynomial function of degree n, identify the zeros and their multiplicities. 1. If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. 2. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. 3. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. 4. The sum of the multiplicities is n. Example 6 Identifying Zeros and Their Multiplicities Use the graph of the function of degree 6 in Figure 9 to identify the zeros of the function and their possible multiplicities. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 381 y 240 160 80 –6 –4 –2 2 4 6 x –80 –160 –240 Figure 9 Solution The polynomial function is of degree 6. The sum of the multiplicities must be 6. Starting from the left, the first zero occurs at x = −3. The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of −3 most likely has multiplicity 2. The next zero occurs at x = −1. The graph looks almost linear at this point. This is a single zero of multiplicity 1. The last zero occurs at x = 4. The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6. Try It #2 Use the graph of the function of degree 5 in Figure 10 to identify the zeros of the function and their multiplicities. y 60 40 20 –6 –4 –2 2 4 6 x –20 –40 –60 Figure 10 Determining End Behavior As we have already learned, the behavior of a graph of a polynomial function of the form f (x) = an x n + an − 1 x n − 1 + ... + a1 x + a0 will either ultimately rise or fall as x increases without bound and will either rise or fall as x decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say −100 or −1,000. Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, anxn, is an even power function, as x increases or decreases without bound, f (x) increases without bound. When the leading term is an odd power function, as x decreases without bound, f (x) also decreases without bound; as x increases without bound, f (x) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 11 summarizes all four cases. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Even Degree Odd Degree Positive Leading Coefficient, an > 0 y Positive Leading Coefficient, an > 0 y x x End Behavior: x → ∞, f (x) → ∞ x → −∞, f (x) → ∞ End Behavior: x → ∞, f (x) → ∞ x → −∞, f (x) → ∞ Negative Leading Coefficient, an < 0 y Negative Leading Coefficient, an < 0 y x x End Behavior: x → ∞, f (x) → −∞ x → −∞, f (x) → −∞ End Behavior: x → ∞, f (x) → −∞ x → −∞, f (x) → ∞ Figure 11 Understanding the Relationship Between Degree and Turning Points In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function f (x) = x4 − x3 − 4x2 + 4x in Figure 12. The graph has three turning points. y Increasing Decreasing Increasing Decreasing x Turning points Figure 12 This function f is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 383 interpreting turning points A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree n will have at most n − 1 turning points. Example 7 Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function Find the maximum number of turning points of each polynomial function. b. f (x) = −(x − 1)2(1 + 2x2) a. f (x) = −x 3 + 4x 5 − 3x 2 + 1 Solution a. f (x) = −x 3 + 4x 5 − 3x 2 + 1 First, rewrite the polynomial function in descending order: f (x) = 4x 5 − x 3 − 3x 2 + 1 Identify the degree of the polynomial function. This polynomial function is of degree 5. The maximum number of turning points is 5 − 1 = 4. b. f (x) = −(x − 1)2(1 + 2x2) First, identify the leading term of the polynomial function if the function were expanded. f (x) = −(x − 1)2(1 + 2x2) an = −(x 2)(2x 2) − 2x 4 Then, identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is 4 − 1 = 3. Graphing Polynomial Functions We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions. How To… Given a polynomial function, sketch the graph. 1. Find the intercepts. 2. Check for symmetry. If the function is an even function, its graph is symmetrical about the y-axis, that is, f (−x) = f (x). If a function is an odd function, its graph is symmetrical about the origin, that is, f (−x) = −f (x). 3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts. 4. Determine the end behavior by examining the leading term. 5. Use the end behavior and the behavior at the intercepts to sketch a graph. 6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial. 7. Optionally, use technology to check the graph. Example 8 Sketching the Graph of a Polynomial Function Sketch a graph of f (x) = −2(x + 3)2(x − 5). Solution This graph has two x-intercepts. At x = −3, the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-intercept. At x = 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept. The y-intercept is found by evaluating f (0). The y-intercept is (0, 90). f (0) = −2(0 + 3)2(0 − 5) = −2 ċ 9 ċ (−5) = 90 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Additionally, we can see the leading term, if this polynomial were multiplied out, would be −2x3, so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 13. y Figure 13 x To sketch this, we consider that: • As x → −∞ the function f (x) → ∞, so we know the graph starts in the second quadrant and is decreasing toward the x-axis. • Since f (−x) = −2(−x + 3)2 (−x − 5) is not equal to f (x), the graph does not display symmetry. • At (−3, 0), the graph bounces off of the x-axis, so the function must start increasing. At (0, 90), the graph crosses the y-axis at the y-intercept. See Figure 14. y (0, 90) (−3, 0) x Figure 14 Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at (5, 0). See Figure 15. y (0, 90) (−3, 0) (5, 0) x Figure 15 As x → ∞ the function f (x) → −∞, so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 385 Using technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like
our sketch in Figure 15. f (x) = −2(x + 3 )2(x − 5 ) y 180 120 60 –6 –4 –2 2 4 6 x –60 –120 –180 Figure 16 The complete graph of the polynomial function f (x ) = −2(x + 3)2(x − 5) Try It #3 1 __ x(x − 1)4(x + 3)3. Sketch a graph of f (x) = 4 Using the Intermediate Value Theorem In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial function f whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers a and b in the domain of f, if a < b and f (a) ≠ f (b), then the function f takes on every value between f (a) and f (b). (While the theorem is intuitive, the proof is actually quite complicated and require higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function f at x = a lies above the x-axis and another point at x = b lies below the x-axis, there must exist a third point between x = a and x = b where the graph crosses the x-axis. Call this point (c, f (c)). This means that we are assured there is a solution c where f (c) = 0. In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis. Figure 17 shows that there is a zero between a and b. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 f (b) is positive f (c) = 0 x 5 21 3 4 f(a) is negative Figure 17 Using the Intermediate Value Theorem to show there exists a zero Intermediate Value Theorem Let f be a polynomial function. The Intermediate Value Theorem states that if f (a) and f (b) have opposite signs, then there exists at least one value c between a and b for which f (c) = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 9 Using the Intermediate Value Theorem Show that the function f (x) = x3 − 5x2 + 3x + 6 has at least two real zeros between x = 1 and x = 4. Solution As a start, evaluate f (x) at the integer values x = 1, 2, 3, and 4. See Table 2. x f (x) 1 5 2 0 3 −3 4 2 Table 2 We see that one zero occurs at x = 2. Also, since f (3) is negative and f (4) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4. We have shown that there are at least two real zeros between x = 1 and x = 4. Analysis We can also see on the graph of the function in Figure 18 that there are two real zeros between x = 1 and x = 4. –5 –4 –3 –2 y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 f (1) = 5 is positive f (4) = 2 is positive 21 3 4 5 f (3) = −3 is negative Figure 18 Try It #4 Show that the function f (x) = 7x5 − 9x4 − x2 has at least one real zero between x = 1 and x = 2. Writing Formulas for Polynomial Functions Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors. factored form of polynomials If a polynomial of lowest degree p has horizontal intercepts at x = x1, x2, …, xn, then the polynomial can be written (x − x2)p2 … (x − xn)pn where the powers pi on each factor can be determined in the factored form: f (x) = a(x − x1)p1 by the behavior of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the x-intercept. How To… Given a graph of a polynomial function, write a formula for the function. 1. Identify the x-intercepts of the graph to find the factors of the polynomial. 2. Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor. 3. Find the polynomial of least degree containing all the factors found in the previous step. 4. Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 387 Example 10 Writing a Formula for a Polynomial Function from the Graph Write a formula for the polynomial function shown in Figure 19. y 6 4 2 –6 –4 –2 2 4 6 x –2 –4 –6 Figure 19 Solution This graph has three x-intercepts: x = −3, 2, and 5. The y-intercept is located at (0, −2). At x = −3 and x = 5, the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At x = 2, the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us f (x) = a(x + 3)(x − 2)2(x − 5) To determine the stretch factor, we utilize another point on the graph. We will use the y-intercept (0, −2), to solve for a. f (0) = a(0 + 3)(0 − 2)2(0 − 5) −2 = a(0 + 3)(0 − 2)2(0 − 5) −2 = −60a The graphed polynomial appears to represent the function f (x) = (x + 3)(x − 2)2 (x − 5). a = 1 __ 30 1 __ 30 Try It #5 Given the graph shown in Figure 20, write a formula for the function shown. y 12 8 4 –6 –4 –2 2 4 6 x –4 –8 –12 Figure 20 Using Local and Global Extrema With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph. Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS local and global extrema A local maximum or local minimum at x = a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around x = a. If a function has a local maximum at a, then f (a) ≥ f (x) for all x in an open interval around x = a. If a function has a local minimum at a, then f (a) ≤ f (x) for all x in an open interval around x = a. A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at a, then f (a) ≥ f (x) for all x. If a function has a global minimum at a, then f (a) ≤ f (x) for all x. We can see the difference between local and global extrema in Figure 21. –6 –5 –4 –3 –2 y Global maximum Local maximum 21 3 4 5 6 x Local minimum 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 Figure 21 Q & A… Do all polynomial functions have a global minimum or maximum? No. Only polynomial functions of even degree have a global minimum or maximum. For example, f (x) = x has neither a global maximum nor a global minimum. Example 11 Using Local Extrema to Solve Applications An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box. Solution We will start this problem by drawing a picture like that in Figure 22, labeling the width of the cut-out squares with a variable, w. w w w w Figure 22 w w w w Notice that after a square is cut out from each end, it leaves a (14 − 2w) cm by (20 − 2w) cm rectangle for the base of the box, and the box will be w cm tall. This gives the volume V(w) = (20 − 2w)(14 − 2w)w = 280w − 68w2 + 4w3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 389 Notice, since the factors are w, 20 − 2w and 14 − 2w, the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values w may take on are greater than zero or less than 7. This means we will restrict the domain of this function to 0 < w < 7. Using technology to sketch the graph of V(w) on this reasonable domain, we get a graph like that in Figure 23. We can use this graph to estimate the maximum value for the volume, restricted to values for w that are reasonable for this problem—values from 0 to 7. V(w) V(w) = 280w − 68w2 + 4w3 400 300 200 100 –2 2 –100 –200 4 6 8 10 12 w Figure 23 From this graph, we turn our focus to only the portion on the reasonable domain, [0, 7]. We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 24. V(w) 340 339 338 337 336 335 334 333 332 331 330 2.4 2.6 2.8 3 Figure 24 w From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side. Try It #6 Use technology to find the maximum and minimum values on the interval [−1, 4] of the function f (x) = −0.2(x − 2)3(x + 1)2(x − 4). Access the following online resource for additional instruction and practice with graphing polynomial functions. • Intermediate Value Theorem (http://openstaxcollege.org/l/ivt) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.3 SECTION EXERCISES VERBAL 1. What is the difference between an x-intercept and a zero of a polynomial function f ? 2. If
a polynomial function of degree n has n distinct zeros, what do you know about the graph of the function? 3. Explain how the Intermediate Value Theorem can 4. Explain how the factored form of the polynomial assist us in finding a zero of a function. helps us in graphing it. 5. If the graph of a polynomial just touches the x-axis and then changes direction, what can we conclude about the factored form of the polynomial? ALGEBRAIC For the following exercises, find the x- or t-intercepts of the polynomial functions. 6. C(t) = 2(t − 4)(t + 1)(t − 6) 7. C(t) = 3(t + 2)(t − 3)(t + 5) 8. C(t) = 4t(t − 2)2(t + 1) 9. C(t) = 2t(t − 3)(t + 1)2 10. C(t) = 2t4 − 8t3 + 6t2 11. C(t) = 4t4 + 12t3 − 40t2 12. f (x) = x4 − x2 13. f (x) = x3 + x2 − 20x 14. f (x) = x3 + 6x2 − 7x 15. f (x) = x3 + x2 − 4x − 4 16. f (x) = x3 + 2x2 − 9x − 18 17. f (x) = 2x3 − x2 − 8x + 4 18. f (x) = x6 − 7x3 − 8 19. f (x) = 2x4 + 6x2 − 8 20. f (x) = x3 − 3x2 − x + 3 21. f (x) = x6 − 2x4 − 3x2 22. f (x) = x6 − 3x4 − 4x2 23. f (x) = x5 − 5x3 + 4x For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval. 24. f (x) = x3 − 9x, between x = −4 and x = −2. 25. f (x) = x3 − 9x, between x = 2 and x = 4. 26. f (x) = x5 − 2x, between x = 1 and x = 2. 27. f (x) = −x4 + 4, between x = 1 and x = 3. 28. f (x) = −2x3 − x, between x = −1 and x = 1. 29. f (x) = x3 − 100x + 2, between x = 0.01 and x = 0.1 For the following exercises, find the zeros and give the multiplicity of each. 30. f (x) = (x + 2)3(x − 3)2 32. f (x) = x3 (x − 1)3(x + 2) 31. f (x) = x2(2x + 3)5(x − 4)2 33. f (x) = x2(x2 + 4x + 4) 34. f (x) = (2x + 1)3(9x2 − 6x + 1) 35. f (x) = (3x + 2)5(x2 − 10x + 25) 36. f (x) = x(4x2 − 12x + 9)(x2 + 8x + 16) 37. f (x) = x6 − x5 − 2x4 38. f (x) = 3x4 + 6x3 + 3x2 40. f (x) = 2x4(x3 − 4x2 + 4x) 39. f (x) = 4x5 − 12x4 + 9x3 41. f (x) = 4x4(9x4 − 12x3 + 4x2) GRAPHICAL For the following exercises, graph the polynomial functions. Note x- and y-intercepts, multiplicity, and end behavior. 42. f (x) = (x + 3)2(x − 2) 43. g(x) = (x + 4)(x − 1)2 44. h(x) = (x − 1)3(x + 3)2 45. k(x) = (x − 3)3(x − 2)2 46. m(x) = −2x(x − 1)(x + 3) 47. n(x) = −3x(x + 2)(x − 4) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 SECTION EXERCISES 391 For the following exercises, use the graphs to write the formula for a polynomial function of least degree. 48. f(x) 49. f(x) 50. f(x) 5 4 3 2 1 –1–1 –2 –3 –4 –5 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 51. f(x) 52. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, use the graph to identify zeros and multiplicity. 54. 21 3 4 5 x –5 –4 –3 –2 55. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x 53. 56. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 –5 –4 –3 –2 For the following exercises, use the given information about the polynomial graph to write the equation. 57. Degree 3. Zeros at x = −2, x = 1, and x = 3. 58. Degree 3. Zeros at x = −5, x = −2, and x = 1. y-intercept at (0, −4). y-intercept at (0, 6) 59. Degree 5. Roots of multiplicity 2 at x = 3 and x = 1, and a root of multiplicity 1 at x = −3. y-intercept at (0, 9) 60. Degree 4. Root of multiplicity 2 at x = 4, and roots of multiplicity 1 at x = 1 and x = −2. y-intercept at (0, −3). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 61. Degree 5. Double zero at x = 1, and triple zero at 62. Degree 3. Zeros at x = 4, x = 3, and x = 2. x = 3. Passes through the point (2, 15). y-intercept at (0, −24). 63. Degree 3. Zeros at x = −3, x = −2 and x = 1. y-intercept at (0, 12). 1 _ 65. Degree 4. Roots of multiplicity 2 at x = and roots 2 of multiplicity 1 at x = 6 and x = −2. y-intercept at (0,18). 64. Degree 5. Roots of multiplicity 2 at x = −3 and x = 2 and a root of multiplicity 1 at x = −2. y-intercept at (0, 4). 66. Double zero at x = −3 and triple zero at x = 0. Passes through the point (1, 32). TECHNOLOGY For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum. 67. f (x) = x3 − x − 1 68. f (x) = 2x3 − 3x − 1 69. f (x) = x4 + x 70. f (x) = −x4 + 3x − 2 71. f (x) = x4 − x3 + 1 EXTENSIONS For the following exercises, use the graphs to write a polynomial function of least degree. 72. 74. (0, 50,000,000) 100 200 300 400 500 600 700 x 6·107 5·107 4·107 3·107 2·107 1·107 –100 0 –1·107 –2·107 –3·107 –4·107 –5·107 –6·107 –7·107 f (x) 24 16 73. f (x) , 0 2 3 (0, 8) 4 3 , 0 –6 –4 –2 2 4 6 x –300 –200 –8 –16 –24 f(x) 2·105 1·105 (100, 0) (–300, 0) x 100 200 (0, –90,000) –400 –300 –200 –100 –1·105 –2·105 –3·105 –4·105 REAL-WORLD APPLICATIONS For the following exercises, write the polynomial function that models the given situation. 75. A rectangle has a length of 10 units and a width of 8 units. Squares of x by x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a polynomial function in terms of x. 77. A square has sides of 12 units. Squares x + 1 by x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of x. 79. A right circular cone has a radius of 3x + 6 and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is 1 _ V = πr 2h for radius r and height h. 3 76. Consider the same rectangle of the preceding problem. Squares of 2x by 2x units are cut out of each corner. Express the volume of the box as a polynomial in terms of x. 78. A cylinder has a radius of x + 2 units and a height of 3 units greater. Express the volume of the cylinder as a polynomial function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 393 LEARNING OBJECTIVES In this section, you will: • Use synthetic diision to diide polynomials. • Use synthetic diision to determine whether is a factor of a polynomial x + 5.4 DIVIDING POLYNOMIALS Figure 1 Lincoln Memorial, Washington, D.C. (credit: Ron Cogswell, Flickr) The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters ( m ), width 40 m, and height 30 m.[15] We can easily find the volume using elementary geometry. V = l ċ w ċ h = 61.5 ċ 40 ċ 30 = 73,800 So the volume is 73,800 cubic meters (m3). Suppose we knew the volume, length, and width. We could divide to find the height. h = V ____ l ċ w = 73,800 _______ 61.5 ċ 40 = 30 As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any or all of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x; the width is given by x − 2. To find the height of the solid, we can use polynomial division, which is the focus of this section. Using Long Division to Divide Polynomials We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division. Long Division 5 × 3 = 15 and 17 − 15 = 2 Step 1: Step 2: Bring down the 8 9 × 3 = 27 and 28 − 27 = 1 Step 3: 1 __ Answer: 59 R 1 or 59 3 59 3)178 −15 28 −27 1 15. National Park Service. “Lincoln Memorial Building Statistics.” http://www.nps.gov/linc/historyculture/lincoln-memorial-building-statistics.htm. Accessed 4/3/2014/ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic. dividend = (divisor ċ quotient) + remainder 178 = (3 ċ 59) + 1 = 177 + 1 = 178 We call this the Division Algorithm and will discuss it more formally after looking at an example. Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide 2x3 − 3x2 + 4x + 5 by x + 2 using the long division algorithm, it would look like this: Set up the division problem. 2x3 divided by x is 2x2. Multiply x + 2 by 2x2. Subtract. Bring down the next term. −7x 2 divided by x is −7x. Multiply x + 2 by −7x. Subtract. Bring down the next term. x + 2)2x3 − 3x2 + 4x + 5 2x2 x + 2)2x3 − 3x2 + 4x + 5 2x2 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x 2x2 − 7x x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 + 14x) 18x + 5 2x2 − 7x + 18 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 + 14x) 18x + 5 −18x + 36 −31 18x divided by x is 18. Multiply x + 2 by 18. Subtract. We have found or 2x3 − 3x2 + 4x + 5 __ x + 2 = 2x 2 − 7x + 18 − 31 _ x + 2 2x3 − 3x2 + 4x + 5 __ x + 2 = (x + 2)(2x2 − 7x + 18) − 31 We can identify the dividend, the divisor, the quotient, and the remainder. 2x3 – 3x2 + 4 x + 5 = (x + 2) (2x2 – 7 x + 18) + (–31) Dividend Divisor Quotient Remainder Writing the result in this manner illustrates the Division Algorithm. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 395 the Division Algorithm Th
e Division Algorithm states that, given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x), there exist unique polynomials q(x) and r(x) such that f (x) = d(x)q(x) + r(x) q(x) is the quotient and r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). If r(x) = 0, then d(x) divides evenly into f (x). This means that, in this case, both d(x) and q(x) are factors of f (x). How To… Given a polynomial and a binomial, use long division to divide the polynomial by the binomial. 1. Set up the division problem. 2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor. 3. Multiply the answer by the divisor and write it below the like terms of the dividend. 4. Subtract the bottom binomial from the top binomial. 5. Bring down the next term of the dividend. 6. Repeat steps 2–5 until reaching the last term of the dividend. 7. If the remainder is non-zero, express as a fraction using the divisor as the denominator. Example 1 Using Long Division to Divide a Second-Degree Polynomial Divide 5x2 + 3x − 2 by x + 1. Solution x + 1)5x2 + 3x − 2 5x Set up division problem. x + 1)5x2 + 3x − 2 5x2 divided by x is 5x. 5x x + 1)5x2 + 3x − 2 −(5x2 + 5x) −2x − 2 5x − 2 x + 1)5x2 + 3x − 2 −(5x2 + 5x) −2x − 2 −(−2x − 2) 0 Multiply x + 1 by 5x. Subtract. Bring down the next term. −2x divided by x is −2. Multiply x + 1 by −2. Subtract. The quotient is 5x − 2. The remainder is 0. We write the result as or 5x2 + 3x − 2 __________ x + 1 = 5x − 2 5x2 + 3x − 2 = (x + 1)(5x − 2) Analysis This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 2 Using Long Division to Divide a Third-Degree Polynomial Divide 6x3 + 11x2 − 31x + 15 by 3x − 2. Solution 2x2 + 5x − 7 3x − 2)6x3 + 11x2 − 31x + 1 −(6x3 − 4x2) 15x2 − 31x −(15x2 + 10x) −21x + 15 −(−21x + 14) 1 6x3 divided by 3x is 2x2. Multiply 3x − 2 by 2x2. Subtract. Bring down the next term. 15x2 divided by 3x is 5x. Multiply 3x − 2 by 5x. Subtract. Bring down the next term. −21x divided by 3x is −7. Multiply 3x − 2 by −7. Subtract. The remainder is 1. There is a remainder of 1. We can express the result as: 6x3 + 11x2 − 31x + 15 __________________ 3x − 2 = 2x 2 + 5x − 7 + 1 _____ 3x − 2 Analysis We can check our work by using the Division Algorithm to rewrite the solution. Then multiply. (3x − 2)(2x 2 + 5x − 7) + 1 = 6x 3 + 11x 2 − 31x + 15 Notice, as we write our result, • the dividend is 6x3 + 11x2 − 31x + 15 • the divisor is 3x − 2 • the quotient is 2x2 + 5x − 7 • the remainder is 1 Try It #1 Divide 16x 3 − 12x 2 + 20x − 3 by 4x + 5. Using Synthetic Division to Divide Polynomials As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1. To illustrate the process, recall the example at the beginning of the section. Divide 2x 3 − 3x 2 + 4x + 5 by x + 2 using the long division algorithm. The final form of the process looked like this: 2x2 + x + 18 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 − 14x) 18x + 5 −(18x + 36) −31 There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 397 2)2 −3 4 5 −2 −4 −7 14 18 −36 −31 Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to −2, multiply and add. The process starts by bringing down the leading coefficient. −2 2 2 −3 −4 −7 4 5 14 −36 18 −31 We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is 2x2 − 7x + 18 and the remainder is −31. The process will be made more clear in Example 3. synthetic division Synthetic division is a shortcut that can be used when the divisor is a binomial in the form x − k where k is a real number. In synthetic division, only the coefficients are used in the division process. How To… Given two polynomials, use synthetic division to divide. 1. Write k for the divisor. 2. Write the coefficients of the dividend. 3. Bring the lead coefficient down. 4. Multiply the lead coefficient by k. Write the product in the next column. 5. Add the terms of the second column. 6. Multiply the result by k. Write the product in the next column. 7. Repeat steps 5 and 6 for the remaining columns. 8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on. Example 3 Using Synthetic Division to Divide a Second-Degree Polynomial Use synthetic division to divide 5x2 − 3x − 36 by x − 3. Solution Begin by setting up the synthetic division. Write k and the coefficients. 3 5 −3 −36 Bring down the lead coefficient. Multiply the lead coefficient by k. 3 5 −3 −36 15 5 Continue by adding the numbers in the second column. Multiply the resulting number by k. Write the result in the next column. Then add the numbers in the third column. The result is 5x + 12. The remainder is 0. So x − 3 is a factor of the original polynomial. 3 5 −3 −36 36 15 12 0 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis remainder. Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the (x − 3)(5x + 12) + 0 = 5x2 − 3x − 36 Example 4 Using Synthetic Division to Divide a Third-Degree Polynomial Use synthetic division to divide 4x 3 + 10x 2 − 6x − 20 by x + 2. Solution The binomial divisor is x + 2 so k = −2. Add each column, multiply the result by −2, and repeat until the last column is reached. −2 4 4 −6 −20 10 −8 −4 20 2 −10 0 The result is 4x 2 + 2x − 10. The remainder is 0. Thus, x + 2 is a factor of 4x3 + 10x2 − 6x − 20. Analysis The graph of the polynomial function f (x) = 4x3 + 10x2 − 6x − 20 in Figure 2 shows a zero at x = k = −2. This confirms that x + 2 is a factor of 4x 3 + 10x2 − 6x − 20. −2 −1.8 –5 –4 –3 –2 y 14 12 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –12 –14 –16 –18 –20 –22 321 4 5 x Figure 2 Example 5 Using Synthetic Division to Divide a Fourth-Degree Polynomial Use synthetic division to divide −9x4 + 10x3 + 7x2 − 6 by x − 1. Solution Notice there is no x-term. We will use a zero as the coefficient for that term. 1 −9 −9 10 −9 1 7 1 8 0 −6 8 8 2 8 The result is −9x3 + x2 + 8x + 8 + 2 _ . x − 1 Try It #2 Use synthetic division to divide 3x4 + 18x3 − 3x + 40 by x + 7. Using Polynomial Division to Solve Application Problems Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 399 Example 6 Using Polynomial Division in an Application Problem The volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x and the width is given by x − 2. Find the height, t, of the solid. Solution There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in Figure 3. Height Width x − 2 Length 3 x Figure 3 We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid. V = l ċ w ċ h 3x4 − 3x3 − 33x2 + 54x = 3x ċ (x − 2) ċ h To solve for h, first divide both sides by 3x. 3x ċ (x − 2) ċ h ____________ 3x = 3x4 − 3x3 − 33x2 + 54x ___________________ 3x Now solve for h using synthetic division. (x − 2)h = x3 − x2 − 11x + 18 h = x3 − x2 − 11x + 18 ________________ x − 2 2 1 −1 −11 2 1 18 2 −18 −9 0 1 The quotient is x2 + x − 9 and the remainder is 0. The height of the solid is x2 + x − 9. Try It #3 The area of a rectangle is given by 3x3 + 14x2 − 23x + 6. The width of the rectangle is given by x + 6. Find an expression for the length of the rectangle. Access these online resources for additional instruction and practice with polynomial division. • Dividing a Trinomial by a Binomial Using Long Division (http://openstaxcollege.org/l/dividetribild) • Dividing a Polynomial by a Binomial Using Long Division (http://openstaxcollege.org/l/dividepolybild) • Ex 2: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd2) • Ex 4: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd4) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.4 SECTION EXERCISES VERBAL 1. If division of a polynomial by a binomial results in a remainder of zero, what can be conclude? 2. If a polynomial of degree n is divided by a binomial of degree 1, what is the degree of the quotient? ALGEBRAIC
For the following exercises, use long division to divide. Specify the quotient and the remainder. 3. (x2 + 5x − 1) ÷ (x − 1) 6. (4x2 − 10x + 6) ÷ (4x + 2) 9. (2x2 − 3x + 2) ÷ (x + 2) 12. (x3 − 3x2 + 5x − 6) ÷ (x − 2) 4. (2x2 − 9x − 5) ÷ (x − 5) 7. (6x2 − 25x − 25) ÷ (6x + 5) 10. (x3 − 126) ÷ (x − 5) 13. (2x3 + 3x2 − 4x + 15) ÷ (x + 3) 5. (3x2 + 23x + 14) ÷ (x + 7) 8. (−x2 − 1) ÷ (x + 1) 11. (3x2 − 5x + 4) ÷ (3x + 1) For the following exercises, use synthetic division to find the quotient. 14. (3x3 − 2x2 + x − 4) ÷ (x + 3) 16. (6x3 − 10x2 − 7x − 15) ÷ (x + 1) 18. (9x3 − 9x2 + 18x + 5) ÷ (3x − 1) 20. (−6x3 + x2 − 4) ÷ (2x − 3) 22. (3x3 − 5x2 + 2x + 3) ÷ (x + 2) 24. (x3 − 3x + 2) ÷ (x + 2) 26. (x3 − 15x2 + 75x − 125) ÷ (x − 5) 28. (6x3 − x2 + 5x + 2) ÷ (3x + 1) 30. (x4 − 3x2 + 1) ÷ (x − 1) 32. (x4 − 10x3 + 37x2 − 60x + 36) ÷ (x − 2) 34. (x4 + 5x3 − 3x2 − 13x + 10) ÷ (x + 5) 36. (4x4 − 2x3 − 4x + 2) ÷ (2x − 1) 15. (2x3 − 6x2 − 7x + 6) ÷ (x − 4) 17. (4x3 − 12x2 − 5x − 1) ÷ (2x + 1) 19. (3x3 − 2x2 + x − 4) ÷ (x + 3) 21. (2x3 + 7x2 − 13x − 3) ÷ (2x − 3) 23. (4x3 − 5x2 + 13) ÷ (x + 4) 25. (x3 − 21x2 + 147x − 343) ÷ (x − 7) 27. (9x3 − x + 2) ÷ (3x − 1) 29. (x4 + x3 − 3x2 − 2x + 1) ÷ (x + 1) 31. (x4 + 2x3 − 3x2 + 2x + 6) ÷ (x + 3) 33. (x4 − 8x3 + 24x2 − 32x + 16) ÷ (x − 2) 35. (x4 − 12x3 + 54x2 − 108x + 81) ÷ (x − 3) 37. (4x4 + 2x3 − 4x2 + 2x + 2) ÷ (2x + 1) For the following exercises, use synthetic division to determine whether the first expression is a factor of the second. If it is, indicate the factorization. 38. x − 2, 4x3 − 3x2 − 8x + 4 41. x − 2, 4x4 − 15x2 − 4 39. x − 2, 3x4 − 6x3 − 5x + 10 42. x − 1 __ , 2x4 − x3 + 2x − 1 2 40. x + 3, −4x3 + 5x2 + 8 43. x + 1 __ 3 , 3x4 + x3 − 3x + 1 GRAPHICAL For the following exercises, use the graph of the third-degree polynomial and one factor to write the factored form of the polynomial suggested by the graph. The leading coefficient is one. 44. Factor is x2 − x + 3 46. Factor is x2 + 2x + 5 45. Factor is x2 + 2x + 4 y 18 12 6 y 18 12 6 y 30 20 10 –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –6 –12 –18 –6 –12 –18 –10 –20 –30 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 SECTION EXERCISES 401 47. Factor is x2 + x + 1 48. Factor is x2 + 2x + 2 y 60 40 20 y 18 12 6 –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –20 –40 –60 – 6 –12 –18 For the following exercises, use synthetic division to find the quotient and remainder. 49. 4x3 − 33 _______ x − 2 51. 3x3 + 2x − 5 __________ x − 1 50. 2x3 + 25 _______ x + 3 53. x4 − 22 ______ x + 2 52. #−4x3 − x2 − 12 ____________ x + 4 TECHNOLOGY 54. Consider xk − 1 ______ x − 1 with k = 1, 2, 3. What do you For the following exercises, use a calculator with CAS to answer the questions. xk + 1 ______ x + 1 the result to be if k = 7? xk _____ x + 1 the result to be if k = 4? expect the result to be if k = 4? expect the result to be if k = 4? for k = 1, 2, 3. What do you x4 − k4 ______ x − k 55. Consider 56. Consider 57. Consider for k = 1, 3, 5. What do you expect with k = 1, 2, 3. What do you expect 58. Consider xk _____ x − 1 the result to be if k = 4? with k = 1, 2, 3. What do you expect EXTENSIONS For the following exercises, use synthetic division to determine the quotient involving a complex number. 59. x + 1 _____ x − i 62. x2 + 1 _____ x + i 60. x2 + 1 _____ x − i 63. x3 + 1 _____ x − i 61. x + 1 _____ x + i REAL-WORLD APPLICATIONS For the following exercises, use the given length and area of a rectangle to express the width algebraically. 64. Length is x + 5, area is 2x2 + 9x − 5. 66. Length is 3x − 4, area is 6x4 − 8x3 + 9x2 − 9x − 4 65. Length is 2x + 5, area is 4x3 + 10x2 + 6x + 15 For the following exercises, use the given volume of a box and its length and width to express the height of the box algebraically. 67. Volume is 12x3 + 20x2 − 21x − 36, length is 2x + 3, width is 3x − 4. 68. Volume is 18x3 − 21x2 − 40x + 48, length is 3x − 4, width is 3x − 4. 69. Volume is 10x3 + 27x2 + 2x − 24, length is 5x − 4, 70. Volume is 10x3 + 30x2 − 8x − 24, length is 2, width is 2x + 3. width is x + 3. For the following exercises, use the given volume and radius of a cylinder to express the height of the cylinder algebraically. 71. Volume is π(25x3 − 65x2 − 29x − 3), radius is 5x + 1. 73. Volume is π(3x4 + 24x3 + 46x2 − 16x − 32), radius is x + 4. 72. Volume is π(4x3 + 12x2 − 15x − 50), radius is 2x + 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: f • Use synthetic diision the emainder heorem to ealuate . • Use the Factor heorem to find the real eros of a polynomial equation. • Use the ational ero heorem to list all possible rational eros. • Find eros of a polynomial function. • Use the inear Factoriation heorem to find polynomials with gien eros. 5.5 ZEROS OF POLYNOMIAL FUNCTIONS ffTh ThTh Th Evaluating a Polynomial Using the Remainder Theorem Remainder Theoremx −k kf k f xdx dxf xqxrx dxx −k f x=dxqx+rx fx=x−kqx+r x −krx =k f k=k−kqk+r =ċqk+r =r f kf xx −k the Remainder Theorem f xx −kf k How To… ff xx =kTh 1. x −k 2. Thf k Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 403 Example 1 Using the Remainder Theorem to Evaluate a Polynomial Use the Remainder Theorem to evaluate f (x) = 6x4 − x3 − 15x2 + 2x − 7 at x = 2. Solution To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by x − 2. 2 6 −1 −15 2 −7 12 11 22 7 14 16 32 25 6 The remainder is 25. Therefore, f (2) = 25. Analysis We can check our answer by evaluating f (2). f (x) = 6x4 − x3 −#15x2 + 2x − 7 f (2) = 6(2)4 − (2)3 − 15(2)2 + 2(2) − 7 = 25 Try It #1 Use the Remainder Theorem to evaluate f (x) = 2x5 − 3x4 − 9x3 + 8x2 + 2 at x = −3. Using the Factor Theorem to Solve a Polynomial Equation The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm. f (x) = (x − k)q(x) + r If k is a zero, then the remainder r is f (k) = 0 and f (x) = (x − k)q(x) + 0 or f (x) = (x − k)q(x). Notice, written in this form, x − k is a factor of f (x). We can conclude if k is a zero of f (x), then x − k is a factor of f (x). Similarly, if x − k is a factor of f (x), then the remainder of the Division Algorithm f (x) = (x − k)q(x) + r is 0. This tells us that k is a zero. This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree n in the complex number system will have n zeros. We can use the Factor Theorem to completely factor a polynomial into the product of n factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial. the Factor Theorem According to the Factor Theorem, k is a zero of f (x) if and only if (x − k) is a factor of f (x). How To… Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial. 1. Use synthetic division to divide the polynomial by (x − k). 2. Confirm that the remainder is 0. 3. Write the polynomial as the product of (x − k) and the quadratic quotient. 4. If possible, factor the quadratic. 5. Write the polynomial as the product of factors. Example 2 Using the Factor Theorem to Solve a Polynomial Equation Show that (x + 2) is a factor of x3 − 6x2 − x + 30. Find the remaining factors. Use the factors to determine the zeros of the polynomial. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Solution We can use synthetic division to show that (x + 2) is a factor of the polynomial. −2 1 −6 −1 30 −2 16 −30 1 −8 15 0 The remainder is zero, so (x + 2) is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient: (x + 2)(x2 − 8x + 15) We can factor the quadratic factor to write the polynomial as By the Factor Theorem, the zeros of x3 − 6x2 − x + 30 are −2, 3, and 5. (x + 2)(x − 3)(x − 5) Try It #2 Use the Factor Theorem to find the zeros of f (x) = x3 + 4x2 − 4x − 16 given that (x − 2) is a factor of the polynomial. Using the Rational Zero Theorem to Find Rational Zeros Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial. 3 2 __ __ Consider a quadratic function with two zeros, x = and x = . By the Factor Theorem, these zeros have factors 4 5 associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor. 3 2 __ __ x − = 0 = 0 or x − 4 5 Set each factor equal to 0. 5x − 2 = 0 or 4x − 3 = 0 Multiply both sides of the equation to eliminate fractions. f (x) = (5x − 2)(4x − 3) Create the quadratic function, multiplying the factors. f (x) = 20x2 − 23x + 6 Expand the polynomial. f (x) = (5 ċ 4)x2 − 23x + (2 ċ 3) Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5 and 4. We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros. the Rational Zero Theorem The Rational Zero Theorem states that, if the polynomial f (x) = anxn + an − 1 xn − 1 + ... + a1 x + a0 has integer p _ q where p is a factor of the constant term a0 and q is a coefficients, t
hen every rational zero of f (x) has the form factor of the leading coefficient an. When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 405 How To… Given a polynomial function f (x), use the Rational Zero Theorem to find rational zeros. 1. Determine all factors of the constant term and all factors of the leading coefficient. p _ 2. Determine all possible values of q , where p is a factor of the constant term and q is a factor of the leading coefficient. Be sure to include both positive and negative candidates. p q ) . 3. Determine which possible zeros are actual zeros by evaluating each case of f ( _ Example 3 Listing All Possible Rational Zeros List all possible rational zeros of f (x) = 2x4 − 5x3 + x2 − 4. Solution The only possible rational zeros of f (x) are the quotients of the factors of the last term, −4, and the factors of the leading coefficient, 2. The constant term is −4; the factors of −4 are p = ±1, ±2, ±4. The leading coefficient is 2; the factors of 2 are q = ±1, ±2. If any of the four real zeros are rational zeros, then they will be of one of the following factors of −4 divided by one of the factors of 2. p 1 1 __ __ __ __ __ __ __ __ = 2, which have already been listed. So we can shorten our list. = 1 and Note that 2 2 p _ q = Factors of the last __ Factors of the first = ±1, ±2, ±4, ± #1 __ 2 Example 4 Using the Rational Zero Theorem to Find Rational Zeros Use the Rational Zero Theorem to find the rational zeros of f (x) = 2x3 + x2 − 4x + 1. p _ Solution The Rational Zero Theorem tells us that if q is a zero of f (x), then p is a factor of 1 and q is a factor of 2. p _ q = factor of constant term ___ factor of leading coefficient = factor of 1_ factor of 2 p q are ±1 and ± #1 _ 2 . These are the _ The factors of 1 are ±1 and the factors of 2 are ±1 and ±2. The possible values for possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for x in f (x). f (−1) = 2(−1)3 + (−1)2 − 4(−1) + 1 = 4 3 f (1) = 2(1)3 + (1)2 − 4(11 + ( − #1 ) = 2 ( − #1 f ( − #1 __ __ __ __ ) + 1_ __ __ __ __ − Of those, −1, −#1__ 2 , and #1 __ 2 are not zeros of f (x). 1 is the only rational zero of f (x). Try It #3 Use the Rational Zero Theorem to find the rational zeros of f (x) = x3 − 5x2 + 2x + 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Finding the Zeros of Polynomial Functions The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function. How To… Given a polynomial function f, use synthetic division to find its zeros. 1. Use the Rational Zero Theorem to list all possible rational zeros of the function. 2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate. 3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic. 4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula. Example 5 Finding the Zeros of a Polynomial Function with Repeated Real Zeros Find the zeros of f (x) = 4x3 − 3x − 1. p _ q is a zero of f (x), then p is a factor of −1 and q is a factor of 4. Solution The Rational Zero Theorem tells us that if p _ q = factor of constant term ___ factor of leading coefficient p , and ± #1 q are ±1, ± #1 _ __ __ The factors of −1 are ±1 and the factors of 4 are ±1, ±2, and ±4. The possible values for . These 4 2 are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1. = factor of −1__ factor of 4 1 4 4 0 −3 −1 1 4 4 0 1 4 Dividing by (x − 1) gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as The quadratic is a perfect square. f (x) can be written as (x − 1)(2x + 1)2. (x − 1)(4x2 + 4x + 1). We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0. 2x + 1 = 0 x = − #1 __ 2 The zeros of the function are 1 and − #1 __ with multiplicity 2. 2 Analysis Look at the graph of the function f in Figure 1. Notice, at x = −0.5, the graph bounces off the x-axis, indicating the even multiplicity (2, 4, 6…) for the zero −0.5. At x = 1, the graph crosses the x-axis, indicating the odd multiplicity (1, 3, 5…) for the zero x = 1. y Bounce 1.5 1 0.5 –2.5 –2 –1.5 0.5 1 1.5 2 2.5 x Cross –1 –0.5 –0.5 –1 –1.5 –2 –2.5 Figure 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 407 Using the Fundamental Theorem of Algebra Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations. Suppose f is a polynomial function of degree four, and f (x) = 0. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it c1. By the Factor Theorem, we can write f (x) as a product of x − c1 and a polynomial quotient. Since x − c1 is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it c2. So we can write the polynomial quotient as a product of x − c2 and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of f (x). the Fundamental Theorem of Algebra The Fundamental Theorem of Algebra states that, if f (x) is a polynomial of degree n > 0, then f (x) has at least one complex zero. We can use this theorem to argue that, if f (x) is a polynomial of degree n > 0, and a is a non-zero real number, then f (x) has exactly n linear factors where c1, c2, ..., cn are complex numbers. Therefore, f (x) has n roots if we allow for multiplicities. f (x) = a(x − c1)(x − c2)...(x − cn) Q & A… Does every polynomial have at least one imaginary zero? No. Real numbers are a subset of complex numbers, but not the other way around. A complex number is not necessarily imaginary. Real numbers are also complex numbers. Example 6 Finding the Zeros of a Polynomial Function with Complex Zeros Find the zeros of f (x) = 3x3 + 9x2 + x + 3. p _ Solution The Rational Zero Theorem tells us that if q is a zero of f (x), then p is a factor of 3 and q is a factor of 3. factor of constant term ___ factor of leading coefficient factor of 3_ factor of 3 p _ q = = p _ The factors of 3 are ±1 and ±3. The possible values for q , and therefore the possible rational zeros for the function, are ±3, ±1, and ± #1 __ . We will use synthetic division to evaluate each possible zero until we find one that gives a remainder 3 of 0. Let’s begin with −3. −3 3 3 9 −9 0 1 3 0 −3 0 1 Dividing by (x + 3) gives a remainder of 0, so −3 is a zero of the function. The polynomial can be written as We can then set the quadratic equal to 0 and solve to find the other zeros of the function. 3x2 + 1 = 0 (x + 3)(3x2 + 1) 3 The zeros of f (x) are −3 and ± #i √ _ . 3 — x2 = − #1 __ 3 x = ± √ ____ − #1 __ = ± 3 — 3 i √ _ 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis Look at the graph of the function f in Figure 2. Notice that, at x = −3, the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero x = −3. Also note the presence of the two turning points. This means that, since there is a 3rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of x = −3 is 1 and there are two complex solutions, which is what we found, or the multiplicity at x = −3 is three. Either way, our result is correct. y 18 12 6 Cross –6 –4 –2 2 4 6 x –6 –12 –18 Figure 2 Try It #4 Find the zeros of f (x) = 2x3 + 5x2 − 11x + 4. Using the Linear Factorization Theorem to Find Polynomials with Given Zeros A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree n will have n zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into n factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (x − c), where c is a complex number. Let f be a polynomial function with real coefficients, and suppose a + bi, b ≠ 0, is a zero of f (x). Then, by the Factor Theorem, x − (a + bi) is a factor of f (x). For f to have real coefficients, x − (a − bi) must also be a factor of f (x). This is true because any factor other than x − (a − bi), when multiplied by x − (a + bi), will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function f with real coefficients has a complex ze
ro a + bi, then the complex conjugate a − bi must also be a zero of f (x). This is called the Complex Conjugate Theorem. complex conjugate theorem According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form (x − c), where c is a complex number. If the polynomial function f has real coefficients and a complex zero in the form a + bi, then the complex conjugate of the zero, a − bi, is also a zero. How To… Given the zeros of a polynomial function f and a point (c, f (c)) on the graph of f, use the Linear Factorization Theorem to find the polynomial function. 1. Use the zeros to construct the linear factors of the polynomial. 2. Multiply the linear factors to expand the polynomial. 3. Substitute (c, f (c)) into the function to determine the leading coefficient. 4. Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 409 Example 7 Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros Find a fourth degree polynomial with real coefficients that has zeros of −3, 2, i, such that f (−2) = 100. Solution Because x = i is a zero, by the Complex Conjugate Theorem x = −i is also a zero. The polynomial must have factors of (x + 3), (x − 2), (x − i), and (x + i). Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors. f (x) = a(x + 3)(x − 2)(x − i)(x + i) f (x) = a(x2 + x − 6)(x2 + 1) f (x) = a(x4 + x3 − 5x2 + x − 6) We need to find a to ensure f (−2) = 100. Substitute x = −2 and f (2) = 100 into f (x). 100 = a((−2)4 + (−2)3 − 5(−2)2 + (−2) − 6) So the polynomial function is or 100 = a(−20) −5 = a f (x) = −5(x4 + x3 − 5x2 + x − 6) f (x) = −5x4 − 5x3 + 25x2 − 5x + 30 Analysis We found that both i and −i were zeros, but only one of these zeros needed to be given. If i is a zero of a polynomial with real coefficients, then −i must also be a zero of the polynomial because −i is the complex conjugate of i. Q & A… If 2 + 3i were given as a zero of a polynomial with real coefficients, would 2 − 3i also need to be a zero? Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial. Try It #5 Find a third degree polynomial with real coefficients that has zeros of 5 and −2i such that f (1) = 10. Using Descartes’ Rule of Signs There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in f (x) and the number of positive real zeros. For example, the polynomial function below has one sign change. f (x This tells us that the function must have 1 positive real zero. There is a similar relationship between the number of sign changes in f (−x) and the number of negative real zeros. f (−x) = (−x)4 + (−x)3 + (−x)2 + (−x) − 1 f (−x) =+x 4 −#x 3 + x 2 −#x − 1 In this case, f (−x) has 3 sign changes. This tells us that f (x) could have 3 or 1 negative real zeros. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Descartes’ Rule of Signs According to Descartes’ Rule of Signs, if we let f (x) = anxn + an − 1 xn − 1 + ... + a1 x + a0 be a polynomial function with real coefficients: • The number of positive real zeros is either equal to the number of sign changes of f (x) or is less than the number of sign changes by an even integer. • The number of negative real zeros is either equal to the number of sign changes of f (−x) or is less than the number of sign changes by an even integer. Example 8 Using Descartes’ Rule of Signs Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for f (x) = −x4 − 3x3 + 6x2 − 4x − 12. Solution Begin by determining the number of sign changes. f (x) = −x 4 − 3x 3 + 6x 2 − 4x − 12 Figure 3 There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine f (−x) to determine the number of negative real roots. f (−x) = −(−x)4 − 3(−x)3 + 6(−x)2 − 4(−x) − 12 f (−x) = −x 4 + 3x 3 + 6x 2 + 4x − 12 f (−x) = −x 4 +#3x 3 + 6x 2 + 4x − 12 Figure 4 Again, there are two sign changes, so there are either 2 or 0 negative real roots. There are four possibilities, as we can see in Table 1. Positive Real Zeros 2 2 0 0 Negative Real Zeros 2 0 2 0 Table 1 Complex Zeros 0 2 2 4 Total Zeros 4 4 4 4 Analysis We can confirm the numbers of positive and negative real roots by examining a graph of the function. See Figure 5. We can see from the graph that the function has 0 positive real roots and 2 negative real roots. y 60 50 40 30 20 10 0 –1 –10 –20 –30 x = −4.42 –5 –4 –3 –2 f (x) = − x4 − 3x3 + 6x2 − 4x − 12 x = −1 321 4 5 x Try It #6 Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for f (x) = 2x4 − 10x3 + 11x2 − 15x + 12. Use a graph to verify the numbers of positive and negative real zeros for the function. Figure 5 Solving Real-World Applications We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 411 Example 9 Solving Polynomial Equations A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be? Solution Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by V = lwh. We were given that the length must be four inches longer than the width, so we can express the length of the cake as l = w + 4. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as h = #1 __ w. Let’s write the volume of the cake in terms of width of the cake. 3 1 __ w ) V = (w + 4)(w) ( 3 1 4 __ __ V = w 3 + w 2 3 3 Substitute the given volume into this equation. 1 4 __ __ 351 = w 3 + w 2 3 3 1053 = w 3 + 4w 2 Substitute 351 for V. Multiply both sides by 3. 0 = w 3 + 4w 2 − 1053 Subtract 1053 from both sides. Descartes’ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are ±3, ±9, ±13, ±27, ±39, ±81, ±117, ±351, and ±1053. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check x = 1. Since 1 is not a solution, we will check x = 3. 1 3 Since 3 is not a solution either, we will test x = 91053 5 5 5 −1048 0 −1053 21 63 21 −990 4 9 13 0 −1053 1053 117 0 117 Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan. The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches = 13 and h =# #1 __ __ (9) = 3 w = 3 3 Try It #7 A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be? Access these online resources for additional instruction and practice with zeros of polynomial functions. • Real Zeros, Factors, and Graphs of Polynomial Functions (http://openstaxcollege.org/l/realzeros) • Complex Factorization Theorem (http://openstaxcollege.org/l/factortheorem) • Find the Zeros of a Polynomial Function (http://openstaxcollege.org/l/ﬁndthezeros) • Find the Zeros of a Polynomial Function 2 (http://openstaxcollege.org/l/ﬁndthezeros2) • Find the Zeros of a Polynomial Function 3 (http://openstaxcollege.org/l/ﬁndthezeros3) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.5 SECTION EXERCISES VERBAL 1. der Th 2. o Th 3. e diff 4. 5. ALGEBRAIC 6. = + + + a. b. c. +p 7. ()= + ,: a. b. c. p =+ + =+ 8. 10. = + 9. 11. = + 12. 14. 16. = ++ = + =++++ 13. 15. 17. =+ =++++ = + ++ conjugate theorem +p = p 18. 19. + 20. 21. p 22. +p 23. p 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 SECTION EXERCISES 413 24. 27. =+ + = + 25. 28. = + = ++ 26. = + 29. = + 32. 35. = = ++ + 30. 33. 36. = = + + = + 31. 34. = + = + GRAPHICAL Thfi 37. fx= x− 38. fx= x−x− 39. fx= x−x−x+ 40. fx= x−x+ x− 41. fx= x+ x−x+ x− 42. fx= x+ x+ x+ 43. fx= x−x−x+ 44. fx= x−x−x+ x+ fx= x−x−x+ x+ 45. 46. fx= x−x+ TECHNOLOGY fi 47. fx= x−x+ 48. fx= x−x−x− 49. fx= x−x−x+ 50. fx= x+ x+ x−x+ 51. fx= x−x+ x−x+ EXTENSIONS 52. −f= 54. 56. − −f−= − −−−f−= − 53. 55. − f= − " −f−= − REAL-WORLD APPLICATIONS For the following exercises, find the dimensions of the box described. 57. 59. 61. The length is twice as long as the width. The height is 2 inches greater than the width. The volume is 192 cubic inches. The length is one inch more than the width, which is one inch more than the height. The volume is 86.625 cubic inc
hes. The length is 3 inches more than the width. The width is 2 inches more than the height. The volume is 120 cubic inches. 58. The length, width, and height are consecutive whole numbers. The volume is 120 cubic inches. 60. The length is three times the height and the height is one inch less than the width. The volume is 108 cubic inches. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Find the domains of rational functions. • dentify ertical asymptotes and horiontal asymptotes. • Graph rational functions using asymptotes intercepts and appropriate points. • Find xand yintercepts. • rite a rational function that meets specific criteria. 5.6 RATIONAL FUNCTIONS Suppose we know that the cost of making a product is dependent on the number of items, x, produced. This is given by the equation C(x) = 15,000x − 0.1x2 + 1000. If we want to know the average cost for producing x items, we would divide the cost function by the number of items, x. The average cost function, which yields the average cost per item for x items produced, is f (x) = 15,000x − 0.1x2 + 1000 __________________ x Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power. In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator. Using Arrow Notation We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in Figure 1, and notice some of their features. Graphs of Toolkit Functions y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 f (x) = 1 x 321 4 5 x –5 –4 –3 –2 Figure 1 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 f (x) = 1 x2 321 4 5 x 1 _ Several things are apparent if we examine the graph of f (x) = x . 1. On the left branch of the graph, the curve approaches the x-axis (y = 0) as x → −∞. 2. As the graph approaches x = 0 from the left, the curve drops, but as we approach zero from the right, the curve rises. 3. Finally, on the right branch of the graph, the curves approaches the x-axis (y = 0) as x → ∞. To summarize, we use arrow notation to show that x or f (x) is approaching a particular value. See Table 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 415 Symbol x → a− x → a+ x → ∞ x → −∞ f (x) → ∞ f (x) → −∞ f (x) → a Meaning x approaches a from the left (x < a but close to a) x approaches a from the right (x > a but close to a) x approaches infinity (x increases without bound) x approaches negative infinity (x decreases without bound) The output approaches infinity (the output increases without bound) The output approaches negative infinity (the output decreases without bound) The output approaches a Table 1 Arrow Notation 1 __ Local Behavior of f (x ) = x Let’s begin by looking at the reciprocal function, f (x) =# #1 _ x . We cannot divide by zero, which means the function is undefined at x = 0; so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in Table 2. x f (x) = 1 __ x −0.1 −0.01 −0.001 −0.0001 −10 −100 −1000 −10,000 We write in arrow notation Table 2 as x → 0−, f (x) → −∞ As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in Table 3. x f (x) = 1 __ x 0.1 10 0.01 100 Table 3 0.001 0.0001 1000 10,000 We write in arrow notation See Figure 2. As x → 0+, f (x) → ∞. As x → 0 + f (x) → ∞ y As x → −∞ f (x) → 0 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 321 4 5 As x → ∞ f (x) → 0 x As x → 0 – f (x) → −∞ Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line x = 0 as the input becomes close to zero. See Figure 35 –4 –3 –2 321 x = 0 –1 –1 –2 –3 –4 –5 Figure 3 vertical asymptote A vertical asymptote of a graph is a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a. We write As x → a, f (x) → ∞, or as x → a, f (x) → −∞. 1 _ x End Behavior of f(x ) = As the values of x approach infinity, the function values approach 0. As the values of x approach negative infinity, the function values approach 0. See Figure 4. Symbolically, using arrow notation As x → ∞, f (x) → 0, and as x → −∞, f (x) → 0. y –2 –3 –4 –5 As x → −∞ f (x) → 0 As x → 0 + f (x) → ∞ As x → ∞ f (x) → 0 321 4 5 x As x → 0 – f (x) → −∞ 5 4 3 2 1 –1 –1 –2 –3 –4 –5 Figure 4 Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line y = 0. See Figure 5. y = 0 –2 –3 –4 –5 y 5 4 3 2 1 321 4 5 x x = 0 –1 –1 –2 –3 –4 –5 Figure 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 417 horizontal asymptote A horizontal asymptote of a graph is a horizontal line y = b where the graph approaches the line as the inputs increase or decrease without bound. We write As x → ∞ or x → −∞, f (x) → b. Example 1 Using Arrow Notation Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 6. y 12 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –12 –6 –5 –4 –3 –2 321 4 5 6 x Figure 6 Solution Notice that the graph is showing a vertical asymptote at x = 2, which tells us that the function is undefined at x = 2. As x → 2−, f (x) → −∞, and as x → 2+, f (x) → ∞. And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at y = 4. As the inputs increase without bound, the graph levels off at 4. As x → ∞, f (x) → 4 and as x → −∞, f (x) → 4. Try It #1 Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function. Example 2 Using Transformations to Graph a Rational Function Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any. Solution Shifting the graph left 2 and up 3 would result in the function 1 _____ x + 2 or equivalently, by giving the terms a common denominator, f (x) = + 3 The graph of the shifted function is displayed in Figure 7. f (x) = 3x + 7 ______ x + 2 x = −7 –6 –5 –4 –3 –2 –1 –1 –2 –3 y = 3 321 4 5 6 7 x Figure 7 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Notice that this function is undefined at x = −2, and the graph also is showing a vertical asymptote at x = −2. As x → −2−, f (x) → −∞, and as x → −2+, f (x) → ∞. As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at y = 3. As x → ±∞, f (x) → 3. Analysis Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function. Try It #2 Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units. Solving Applied Problems Involving Rational Functions In Example 2, we shifted a toolkit function in a way that resulted in the function f (x) = 3x + 7 ______ x + 2 . This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions. rational function A rational function is a function that can be written as the quotient of two polynomial functions P(x) and Q(x). f (x) = P(x) ____ Q(x) = ap x p + ap − 1 x p − 1 + ... + a1 x + a0 ___ bq x q + bq − 1 x q − 1 + ... + b1 x + b0 , Q(x) ≠ 0 Example 3 Solving an Applied Problem Involving a Rational Function A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning? Solution Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each: water: W(t) = 100 + 10t in gallons sugar: S(t) = 5 + 1t in pounds The concentration, C, will be the ratio of pounds of sugar to gallons of water The concentration after 12 minutes is given by evaluating C(t) at t = 12. C(t) = 5 + t ________ 100 + 10t This means the concentration is 17 pounds of sugar to 220 gallons of water. C(12) = 5 + 12 __________ 100 + 10(12) 17 ___ 220 = At the beginning, the concentration is C(0) = 5 + 0 _________ 100 + 1
0(0) 1 __ 20 = Since ≈ 0.08 > = 0.05, the concentration is greater after 12 minutes than at the beginning. 17 ___ 220 1 __ 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 419 Try It #3 There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m. Finding the Domains of Rational Functions A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero. domain of a rational function The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. How To… Given a rational function, find the domain. 1. Set the denominator equal to zero. 2. Solve to find the x-values that cause the denominator to equal zero. 3. The domain is all real numbers except those found in Step 2. Example 4 Finding the Domain of a Rational Function Find the domain of f (x) = x + 3 _ x2 − 9 . Solution Begin by setting the denominator equal to zero and solving. x2 − 9 = 0 x2 = 9 x = ±3 The denominator is equal to zero when x = ±3. The domain of the function is all real numbers except x = ±3. Analysis A graph of this function, as shown in Figure 8, confirms that the function is not defined when x = ±3. y = 0 –2 –3 –4 –6 –5 y 4 3 2 1 –1 –1 –2 –3 –4 321 4 5 6 x x = 3 Figure 8 There is a vertical asymptote at x = 3 and a hole in the graph at x = −3. We will discuss these types of holes in greater detail later in this section. Try It #4 Find the domain of f (x) = 4x _____________ . 5(x − 1)(x − 5) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Identifying Vertical Asymptotes of Rational Functions By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location. Vertical Asymptotes The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors. How To… Given a rational function, identify any vertical asymptotes of its graph. 1. Factor the numerator and denominator. 2. Note any restrictions in the domain of the function. 3. Reduce the expression by canceling common factors in the numerator and the denominator. 4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur. 5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities or "holes." Example 5 Identifying Vertical Asymptotes Find the vertical asymptotes of the graph of k(x) = 5 + 2x2 _________ 2 − x − x2 . Solution First, factor the numerator and denominator. k(x) = 5 + 2x2 ________ 2 − x − x2 = 5 + 2x2 ___________ (2 + x)(1 − x) To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero: Neither x = −2 nor x = 1 are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in Figure 9 confirms the location of the two vertical asymptotes. (2 + x)(1 − x) = 0 x = −2, 1 x = − 21 3 4 5 6 x –5 –4 –3 –2 –6 y = −2 –1 –1 –2 –3 –4 –5 –6 –7 Figure 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 421 Removable Discontinuities Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity. For example, the function f (x) = may be re-written by factoring the numerator and the denominator. x2 − 1 __________ x2 − 2x − 3 f (x) = #(x + 1)(x − 1) ____________ (x + 1)(x − 3) Notice that x + 1 is a common factor to the numerator and the denominator. The zero of this factor, x = −1, is the location of the removable discontinuity. Notice also that x − 3 is not a factor in both the numerator and denominator. The zero of this factor, x = 3, is the vertical asymptote. See Figure 10. [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.] Removable discontinuity at x = −1 –3 y 6 4 2 –1 –2 –4 –6 1 3 5 7 9 x Vertical asymptote at x = 3 Figure 10 removable discontinuities of rational functions A removable discontinuity occurs in the graph of a rational function at x = a if a is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value. Example 6 Identifying Vertical Asymptotes and Removable Discontinuities for a Graph Find the vertical asymptotes and removable discontinuities of the graph of k(x) = x − 2 _____ . x2 − 4 Solution Factor the numerator and the denominator. k(x) = x − 2 ___________ (x − 2)(x + 2) Notice that there is a common factor in the numerator and the denominator, x − 2. The zero for this factor is x = 2. This is the location of the removable discontinuity. Notice that there is a factor in the denominator that is not in the numerator, x + 2. The zero for this factor is x = −2. The vertical asymptote is x = −2. See Figure 11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS y x = −2 6 4 2 –8 –6 –4 –2 2 4 x –2 –4 –6 Figure 11 The graph of this function will have the vertical asymptote at x = −2, but at x = 2 the graph will have a hole. Try It #5 Find the vertical asymptotes and removable discontinuities of the graph of f (x) = x2 − 25 ___________ x3 − 6x2 + 5x . Identifying Horizontal Asymptotes of Rational Functions While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading term. There are three distinct outcomes when checking for horizontal asymptotes: Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at y = 0. Example: f (x) = 4x + 2 _________ x2 + 4x − 5 4x 4 _ _ In this case, the end behavior is f (x) ≈ x2 = x . This tells us that, as the inputs increase or decrease without bound, this 4 _ function will behave similarly to the function g(x) = x , and the outputs will approach zero, resulting in a horizontal asymptote at y = 0. See Figure 12. Note that this graph crosses the horizontal asymptote. y 6 4 2 y = 0 –8 –6 –4 –2 2 4 x –2 –4 –6 x = 1 x = −5 Figure 12 Horizontal asymptote y = 0 when f(x ) = , q(x ) ≠ 0 where degree of p < degree of q. p(x ) ____ q(x ) Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote. Example: f (x) = 3x2 − 2x + 1 __________ x − 1 3x2 _ x = 3x. This tells us that as the inputs increase or decrease without bound, this In this case, the end behavior is f (x) ≈ function will behave similarly to the function g(x) = 3x. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of g(x) = 3x looks like a diagonal line, and since f will behave similarly to g, it will approach a line close to y = 3x. This line is a slant asymptote. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 423 To find the equation of the slant asymptote, divide slant asymptote is the graph of the line g(x) = 3x + 1. See Figure 13. 3x2 − 2x + 1 ___________ x − 1 . The quotient is 3x + 1, and the remainder is 2. The y = 3x + 1 2 4 6 8 x y 14 12 10 8 6 4 2 –2 –4 –6 –4 –2 Figure 13 Slant asymptote when f( x ) = x = 1 , q( x ) ≠ 0 where degree of p > degree of q by 1. p( x ) ____ q( x ) an _ Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at y = , where an and bn bn are the leading coefficients of p(x) and q(x) for f (x) = , q(x) ≠ 0. p(x) ____ q(x) Example: f (x) = 3x2 + 2 _________ x2 + 4x − 5 3x2 _ x2 = 3. This tells us that as the inputs grow large, this function will behave like In this case, the end behavior is f (x) ≈ the function g(x) = 3, which is a horizontal line. As x → ±∞, f (x) → 3, resulting in a horizontal asymptote at y = 3. See Figure 14. Note that this graph crosses the horizontal asymptote. y –15 –12 –9 –6 –3 12 9 6 3 –3 –6 –5 Figure 14 Horizontal asymptote when f ( x ) =# # p( x ) _ q( x ) , q( x ) ≠ 0 where degree of p = degree of q. Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote. It should be noted that, if the degree of the numerator is larger than the de
gree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function with end behavior f (x) = f (x) ≈ 3x5 − x2 _______ x + 3 3x5 ___ x = 3x4, the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient. x → ±∞, f (x) → ∞ horizontal asymptotes of rational functions The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator. • Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0. • Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. • Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 7 Identifying Horizontal and Slant Asymptotes For the functions listed, identify the horizontal or slant asymptote. a. g(x) = 6x3 − 10x ________ 2x3 + 5x2 b. h(x) = c. k(x) = x2 + 4x ______ x3 − 8 Solution For these solutions, we will use f (x) = , q(x) ≠ 0. x2 − 4x + 1 _________ x + 2 p(x) ____ q(x) a. g(x) = 6x3 − 10x _ 2x3 + 5x2 : The degree of p = degree of q = 3, so we can find the hori zontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at y = 6 _ 2 or y = 3. : The degree of p = 2 and degree of q = 1. Since p > q by 1, there is a slant asymptote found b. h(x) = x2 − 4x + 1 _ x + 2 at x2 − 4x + 1_ x + 2 . 2 1 −4 −2 1 −6 1 12 13 The quotient is x − 2 and the remainder is 13. There is a slant asymptote at y = x − 2. c. k(x) = : The degree of p = 2 < degree of q = 3, so there is a horizontal asymptote y = 0. x2 + 4x _ x3 − 8 Example 8 Identifying Horizontal Asymptotes In the sugar concentration problem earlier, we created the equation C(t) = 5 + t ________ . 100 + 10t Find the horizontal asymptote and interpret it in context of the problem. Solution Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t, with coefficient 1. In the denominator, the leading term is 10t, with coefficient 10. The horizontal asymptote will be at the ratio of these values: This function will have a horizontal asymptote at y = 1 __ . 10 t → ∞, C(t) → 1 __ 10 This tells us that as the values of t increase, the values of C will approach . In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or pounds per gallon. 1 __ 10 1 __ 10 Example 9 Identifying Horizontal and Vertical Asymptotes Find the horizontal and vertical asymptotes of the function f (x) = (x − 2)(x + 3) _________________ (x − 1)(x + 2)(x − 5) Solution First, note that this function has no common factors, so there are no potential removable discontinuities. The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at x = 1, −2, and 5, indicating vertical asymptotes at these values. The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as x → ±∞, f (x) → 0. This function will have a horizontal asymptote at y = 0. See Figure 15. y 6 4 2 y = 0 –6 –4 –2 2 4 6 8 x –2 –4 –6 x = −2 x = 1 x = 5 Figure 15 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 425 Try It #6 Find the vertical and horizontal asymptotes of the function: f (x) = (2x − 1)(2x + 1) _____________ (x − 2)(x + 3) intercepts of rational functions A rational function will have a y-intercept when the input is zero, if the function is defined at zero. A rational function will not have a y-intercept if the function is not defined at zero. Likewise, a rational function will have x-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function is equal to zero. Example 10 Finding the Intercepts of a Rational Function Find the intercepts of f (x) = (x − 2)(x + 3) __________________ (x − 1)(x + 2)(x − 5) . Solution We can find the y-intercept by evaluating the function at zero f (0) = = (0 − 2)(0 + 3) _________________ (0 − 1)(0 + 2)(0 − 5) −6 ___ 10 = − #3 __ 5 The x-intercepts will occur when the function is equal to zero: = −0.6 0 = (x − 2)(x + 3) _________________ (x − 1)(x + 2)(x − 5) This is zero when the numerator is zero. The y-intercept is (0, −0.6), the x-intercepts are (2, 0) and (−3, 0). See Figure 16. 0 = (x − 2)(x + 3) x = 2, −3 y 6 5 4 3 2 1 (−3, 0) –4 –3 –5 –6 (2, 0) 321 4 5 6 7 8 (0, –0.61 –2 0 –1 –2 –3 –4 –5 –6 x = −2 Figure 16 Try It #7 Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x- and y-intercepts and the horizontal and vertical asymptotes. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Graphing Rational Functions In Example 9, we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials. The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See Figure 17. y –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 1 y = x 321 4 5 x x = 0 Figure 17 When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure 18. –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y = 1 x2 321 4 5 x x = 0 Figure 18 For example, the graph of f (x) = is shown in Figure 19. (x + 1)2(x − 3) _____________ (x + 3)2(x − 2) y f (x) = (x + 1)2 (x − 3) (x + 3)2 (x − 23, 0) –8 –6 (−1, 0) –4 –2 2 –2 –4 –6 x = −3 x = 2 Figure 19 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 427 • At the x-intercept x = −1 corresponding to the (x + 1)2 factor of the numerator, the graph "bounces," consistent with the quadratic nature of the factor. • At the x-intercept x = 3 corresponding to the (x − 3) factor of the numerator, the graph passes through the axis as we would expect from a linear factor. • At the vertical asymptote x = −3 corresponding to the (x + 3)2 factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function f (x) = • At the vertical asymptote x = 2, corresponding to the (x − 2) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the 1 __ behavior of the function f (x) = . x 1 _ x 2 . How To… Given a rational function, sketch a graph. 1. Evaluate the function at 0 to find the y-intercept. 2. Factor the numerator and denominator. 3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x-intercepts. 4. Find the multiplicities of the x-intercepts to determine the behavior of the graph at those points. 5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve. 6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve. 7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes. 8. Sketch the graph. Example 11 Graphing a Rational Function Sketch a graph of f (x) = (x + 2)(x − 3) ____________ . (x + 1)2(x − 2) Solution We can start by noting that the function is already factored, saving us a step. Next, we will find the intercepts. Evaluating the function at zero gives the y-intercept: f (0) = (0 + 2)(0 − 3)__ (0 + 1)2(0 − 2) = 3 To fi nd the x-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find x-intercepts at x = −2 and x = 3. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept. We have a y-intercept at (0, 3) and x-intercepts at (−2, 0) and (3, 0). To fi nd the vertical asymptotes, we determine when the denominator is equal to zero. Th is occurs when x + 1 = 0 and when x − 2 = 0, giving us vertical asymptotes at x = −1 and x = 2. There are no common factors in the numerator and denominator. This means there are no removable discontinuities. Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at y = 0. To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x-intercepts between the vertical asymptote
s, and the y-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in Figure 20. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 –2 321 4 5 6 x Figure 20 The factor associated with the vertical asymptote at x = −1 was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well. For the vertical asymptote at x = 2, the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See Figure 21. After passing through the x-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote. y = 0 –6 –5 –4 y 6 5 4 3 2 1 –3 –2 –1 –1 –2 –3 –4 x = – Figure 21 Try It #8 Given the function f (x) = (x + 2)2(x − 2) ______________ 2(x − 1)2 (x − 3) , use the characteristics of polynomials and rational functions to describe its behavior and sketch the function. Writing Rational Functions Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors. writing rational functions from intercepts and asymptotes If a rational function has x-intercepts at x = x1, x2, ... , xn, vertical asymptotes at x = v1, v2, … , vm, and no xi = any vj, then the function can be written in the form: f (x) = a (x − x1) p ___ (x − v1) q 2 … (x − xn) p 2 … (x − vm) q 1(x − x2) p 1(x − v2) q n n where the powers pi or qi on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor a can be determined given a value of the function other than the x-intercept or by the horizontal asymptote if it is nonzero. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 429 How To… Given a graph of a rational function, write the function. 1. Determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.) 2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers. 3. Use any clear point on the graph to find the stretch factor. Example 12 Writing a Rational Function from Intercepts and Asymptotes Write an equation for the rational function shown in Figure 22. –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –7 321 4 5 6 x Figure 22 Solution The graph appears to have x-intercepts at x = −2 and x = 3. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at x = −1 seems to exhibit the basic behavior 1 __ similar to , with the graph heading toward positive infinity on one side and heading toward negative infinity on the x 1 _ other. The asymptote at x = 2 is exhibiting a behavior similar to x 2 , with the graph heading toward negative infinity on both sides of the asymptote. See Figure 23. y 6 4 2 Vertical asymptotes –6 –4 –2 2 4 6 x x-intercepts –2 –4 –6 Figure 23 We can use this information to write a function of the form f (x) = a (x + 2)(x − 3) __ (x + 1)(x − 2)2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS To find the stretch factor, we can use another clear point on the graph, such as the y-intercept (0, −2). −2 = a −2 = a a = (0 + 2)(0 − 3)__ (0 + 1)(0 − 2)2 −6_ 4 −8 _ −6 4 _ = 3 This gives us a final function of f (x) = 4(x + 2)(x − 3) ______________ 3(x + 1)(x − 2)2 . Access these online resources for additional instruction and practice with rational functions. • Graphing Rational Functions (http://openstaxcollege.org/l/graphrational) • Find the Equation of a Rational Function (http://openstaxcollege.org/l/equatrational) • Determining Vertical and Horizontal Asymptotes (http://openstaxcollege.org/l/asymptote) • Find the Intercepts, Asymptotes, and Hole of a Rational Function (http://openstaxcollege.org/l/interasymptote) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 SECTION EXERCISES 431 5.6 SECTION EXERCISES VERBAL 1. What is the fundamental difference in the algebraic representation of a polynomial function and a rational function? 3. If the graph of a rational function has a removable discontinuity, what must be true of the functional rule? 5. Can a graph of a rational function have no x-intercepts? If so, how? 2. What is the fundamental difference in the graphs of polynomial functions and rational functions? 4. Can a graph of a rational function have no vertical asymptote? If so, how? ALGEBRAIC For the following exercises, find the domain of the rational functions. 6. f (x) = x − 1 _____ x + 2 9. f (x) = x2 + 4x − 3 _________ x4 − 5x2 + 4 7. f (x) = x + 1 _____ x2 − 1 8. f (x) = x2 + 4 _________ x2 − 2x − 8 For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. 10. f (x) = 4 ____ x − 1 13. f (x) = 16. f (x) = x __________ x2 + 5x − 36 x2 − 1 ___________ x3 + 9x2 + 14x 11. f (x) = 2 _____ 5x + 2 14. f (x) = 3 + x ______ x3 − 27 17. f (x) = x + 5 ______ x2 − 25 12. f (x) = x _____ x2 − 9 15. f (x) = 3x − 4 _______ x3 − 16x 18. f (x) = x − 4 _____ x − 6 19. f (x) = 4 − 2x ______ 3x − 1 For the following exercises, find the x- and y-intercepts for the functions. 20. f (x) = x + 5 _____ x2 + 4 21. f (x) = x _____ x2 − x 22. f (x) = x2 + 8x + 7 ___________ x2 + 11x + 30 23. f (x) = x2 + x + 6 ___________ x2 − 10x + 24 24. f (x) = 94 − 2x2 _______ 3x2 − 12 For the following exercises, describe the local and end behavior of the functions. 25. f (x) = x _____ 2x + 1 26. f (x) = 2x _____ x − 6 27. f (x) = −2x _____ x − 6 28. f (x) = x2 − 4x + 3 _________ x2 − 4x − 5 29. f (x) = 2x2 − 32 ___________ 6x2 + 13x − 5 For the following exercises, find the slant asymptote of the functions. 30. f (x) = 24x2 + 6x ________ 31. f (x) = 4x2 − 10 _______ 2x − 4 2x + 1 32. f (x) = 81x2 − 18 ________ 3x − 2 33. f (x) = 6x3 − 5x _______ 3x2 + 4 34. f (x) = x2 + 5x + 4 _________ x − 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS GRAPHICAL For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes. 35. The reciprocal function shifted up two units. 36. The reciprocal function shifted down one unit and left three units. 37. The reciprocal squared function shifted to the right 38. The reciprocal squared function shifted down 2 units 2 units. and right 1 unit. For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph. 39. p(x) = 2x − 3 ______ x + 4 40. q(x) = x − 5 _____ 3x − 1 4 ______ (x − 2)2 41. s(x) = 42. r(x) = 5 ______ (x + 1)2 43. f (x) = 3x 2 − 14x − 5 __ 3x 2 + 8x − 16 44. g(x) = 2x 2 + 7x − 15 __ 3x 2 − 14 + 15 45. a(x) = x 2 + 2x − 3 _ x 2 − 1 46. b(x 47. h(x) = 2x 2 + x − 1 _ x − 4 48. k(x) = 2x 2 − 3x − 20 __ x − 5 49. w(x) = (x − 1)(x + 3)(x − 5) __ (x + 2)2(x − 4) 50. z(x) = (x + 2)2(x − 5) __ (x − 3)(x + 1)(x + 4) For the following exercises, write an equation for a rational function with the given characteristics. 51. Vertical asymptotes at x = 5 and x = −5, x-intercepts at (2, 0) and (−1, 0), y-intercept at (0, 4) 52. Vertical asymptotes at x = −4 and x = −1, x-intercepts at (1, 0) and (5, 0), y-intercept at (0, 7) 53. Vertical asymptotes at x = −4 and x = −5, x-intercepts at (4, 0) and (−6, 0), horizontal asymptote at y = 7 54. Vertical asymptotes at x = −3 55. Vertical asymptote at x = −1, 56. Vertical asymptote at x = 3, and x = 6, x-intercepts at (−2, 0) and (1, 0), horizontal asymptote at y = −2 double zero at x = 2, y-intercept at (0, 2) double zero at x = 1, y-intercept at (0, 4) For the following exercises, use the graphs to write an equation for the function. 57. y 5 4 3 2 1 –2 0 –1 –2 –3 –4 –5 –10 –8 –6 –4 58. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 59. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 642 8 10 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 SECTION EXERCISES 433 62. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 642 8 10 x 642 8 10 x 61. 642 8 10 x –10 –8 –6 –4 64. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 60. 63. y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 –10 –8 –6 –4 –10 –8 –6 –4 NUMERIC For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote 65. f (x) = 1 _____ x − 2 66. f (x) = x _____ x − 3 67. f (x) = 2x _____ x + 4 68. f (x) = 2x ______ (x − 3)2 69. f (x) = x2 _________ x2 + 2x + 1 TECHNOLOGY
For the following exercises, use a calculator to graph f (x). Use the graph to solve f (x) > 0. 70. f (x) = 2 _____ x + 1 4 _____ 2x − 3 71. f (x) = 72. f (x) = 2 ___________ (x − 1)(x + 2) 73. f (x) = x + 2 ___________ (x − 1)(x − 4) 74. f (x) = (x + 3)2 ____________ (x − 1)2(x + 1) EXTENSIONS For the following exercises, identify the removable discontinuity. 75. f (x) = x2 − 4 _____ x − 2 76. f (x) = x3 + 1 _____ x + 1 78. f (x) = 2x2 + 5x − 3 __________ x + 3 79. f (x) = x3 + x2 ______ x + 1 77. f (x) = x2 + x − 6 ________ x − 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS REAL-WORLD APPLICATIONS For the following exercises, express a rational function that describes the situation. 80. A large mixing tank currently contains 200 gallons of water, into which 10 pounds of sugar have been mixed. A tap will open, pouring 10 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 3 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t minutes. 81. A large mixing tank currently contains 300 gallons of water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t minutes. For the following exercises, use the given rational function to answer the question. 82. The concentration C of a drug in a patient’s 83. The concentration C of a drug in a patient’s bloodstream t hours after injection in given by C(t) = 2t _ 3 + t2 . What happens to the concentration of the drug as t increases? bloodstream t hours after injection is given by C(t) = . Use a calculator to approximate the 100t _ 2t2 + 75 time when the concentration is highest. For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question. 84. An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let x = length of the side of the base. 85. A rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let x = length of the side of the base. 86. A right circular cylinder has volume of 100 cubic inches. Find the radius and height that will yield minimum surface area. Let x = radius. 87. A right circular cylinder with no top has a volume of 50 cubic meters. Find the radius that will yield minimum surface area. Let x = radius. 88. A right circular cylinder is to have a volume of 40 cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost. Let x = radius. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 435 LEARNING OBJECTIVES In this section, you will: • • Find the inverse of an invertible polynomial function. Restrict the domain to ﬁnd the inverse of a polynomial function. 5.7 INVERSES AND RADICAL FUNCTIONS A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume is found using a formula from elementary geometry. Figure 1 V = 1 __ πr 2 h 3 = 1 __ 3 = 2 __ 3 We have written the volume V in terms of the radius r. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula πr 2(2r) πr 3 3 r =# ____ 3V ___ 2π √ This function is the inverse of the formula for V in terms of r. In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process. Finding the Inverse of a Polynomial Function Two functions f and g are inverse functions if for every coordinate pair in f, (a, b), there exists a corresponding coordinate pair in the inverse function, g, (b, a). In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test. For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in Figure 2. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water. 12 in 18 in 3 ft Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x measured horizontally and y measured vertically, with the origin at the vertex of the parabola. See Figure 3. y 18 16 14 12 10 8 6 4 2 –2 –2 –4 –6 –10 –8 –6 –4 642 8 10 x Figure 3 From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form y(x) = ax2. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a. Our parabolic cross section has the equation 18 = a62 a = 18 __ 36 = 1 __ 2 y(x) = 1 __ x2 2 We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y the width will be given by 2x, so we need to solve the equation above for x and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative. To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x values. On this domain, we can find an inverse by solving for the input variable: 1 __ y = x2 2 2y = x2 x = ± √ — 2y This is not a function as written. We are limiting ourselves to positive x values, so we eliminate the negative solution, giving us the inverse function we’re looking for. y = x2 __ , x > 0 2 Because x is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x. The trough is 3 feet (36 inches) long, so the surface area will then be: Area = l ċ w = 36 ċ 2x = 72x = 72 √ — 2y This example illustrates two important points: 1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one. 2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 437 Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation f −1(x). Warning: f −1(x) is not the same as the reciprocal of the function f (x). This use of “−1” is reserved to denote inverse functions. To denote the reciprocal of a function f (x), we would need to write ( f (x))−1 = 1 _ . f (x) An important relationship between inverse functions is that they “undo” each other. If f −1 is the inverse of a function f, then f is the inverse of the function f −1. In other words, whatever the function f does to x, f −1 undoes it—and viceversa. More formally, we write f −1 ( f (x)) = x, for all x in the domain of f and f ( f −1 (x)) = x, for all x in the domain of f −1 Note that the inverse switches the domain and range of th original function. verifying two functions are inverses of one another Two functions, f and g, are inverses of one another if for all x in the domain of f and g. g( f (x)) = f ( g(x)) = x How To… Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one. 1. Replace f (x) with y. 2. Interchange x and y. 3. Solve for y, and rename the function f −1(x). Example 1 Show that f (x) = Verifying Inverse Functions 1 _ x + 1 1 and f −1(x) = __ − 1 are inverses, for x ≠ 0, −1. x Solution We must show that f −1( f (x)) = x and f ( f −1(x)) = x. 1 _____ ) x + 1 f −1(f (x)) = f −1 ( 1 _ 1 _____ x + 1 = (x + 1) − 1 − 1 = = x 1 __ f (f −1(x)) = f ( − 1 ) x 1 __ = 1 __ − __ x = x Therefore, f (x) = 1 _____ x + 1 1 and f −1 (x) = __ − 1 are inverses. x Try It #1 Show that f (x) = x + 5 _____ 3 and f −1(x) = 3x − 5 are inverses. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 2 Finding the Inverse of a Cubic Function Find the inverse of the function f (x) = 5x3 + 1. Solution This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x. y = 5x3 + 1 x = 5y3 + 1 x − 1 = 5y3 x − 1 _____ 5 = y3 f −1(x) = # √ Analysis Look at the graph of f and f −1. Notice that one graph is the reflection of the other about the line y = x. This is always the case when graphing a function and its inverse function. 3 — x − 1 _____ 5 Also, since the method involved interchanging x and y, notice corresponding poin
ts. If (a, b) is on the graph of f , then (b, a) is on the graph of f −1. Since (0, 1) is on the graph of f, then (1, 0) is on the graph of f −1. Similarly, since (1, 6) is on the graph of f, then (6, 1) is on the graph of f −1. See Figure 4. f (x) = 5x3 + 1 y (1, 6) y = x (6, 1) 6 4 2 (0, 1) –6 –4 –2 2 4 6 x (1, 0) –2 –4 –6 Figure 4 Try It #2 Find the inverse function of f (x) = # 3 √ — x + 4 Restricting the Domain to Find the Inverse of a Polynomial Function So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses. restricting the domain If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 439 How To… Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse. 1. Restrict the domain by determining a domain on which the original function is one-to-one. 2. Replace f (x) with y. 3. Interchange x and y. 4. Solve for y, and rename the function or pair of functions f −1(x). 5. Revise the formula for f −1(x) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function. Example 3 Restricting the Domain to Find the Inverse of a Polynomial Function Find the inverse function of f : a. f (x) = (x − 4)2, x ≥ 4 b. f (x) = (x − 4)2, x ≤ 4 Solution The original function f (x) = (x − 4)2 is not one-to-one, but the function is restricted to a domain of x ≥ 4 or x ≤ 4 on which it is one-to-one. See Figure 5. f (x) = (x − 4)2, x ≥ 4 y 10 8 6 4 2 f (x) = (x − 4)2, x ≤ 4 y 10 8 6 4 2 –10 –8 –6 –4 –2 –2 642 8 10 x –10 –8 –6 –4 –2 –2 642 8 10 x To find the inverse, start by replacing f (x) with the simple variable y. y = (x − 4)2 Interchange x and y. Figure 5 — x = (y − 4)2 Take the square root. x = y − 4 x = y Add 4 to both sides. — ± √ 4 ± √ This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of x and y for the original f (x), we looked at the domain: the values x could assume. When we reversed the roles of x and y, this gave us the values y could assume. For this function, x ≥ 4, so for the inverse, we should have y ≥ 4, which is what our inverse function gives. a. The domain of the original function was restricted to x ≥ 4, so the outputs of the inverse need to be the same, f (x) ≥ 4, and we must use the + case: f −1(x) = 4 + √ — x b. The domain of the original function was restricted to x ≤ 4, so the outputs of the inverse need to be the same, f (x) ≤ 4, and we must use the − case: f −1(x) = 4 − √ — x Analysis On the graphs in Figure 6, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line y = x. The coordinate pair (4, 0) is on the graph of f and the coordinate pair (0, 4) is on the graph of f −1. For any coordinate pair, if (a, b) is on the graph of f, then (b, a) is on the graph of f −1. Finally, observe that the graph of f intersects the graph of f −1 on the line y = x. Points of intersection for the graphs of f and f −1 will always lie on the line y = x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS y 10 8 6 2 (0, 4) f (x) y = x f –1(x) (0, 4) 10 8 6 2 y f (x) y = x –10 –8 –6 –4 –2 –2 2 6 8 10 x –10 –8 –6 –4 –2 –2 2 6 (4, 0) (4, 0) Figure 6 f –1(x) x 8 10 Example 4 Finding the Inverse of a Quadratic Function When the Restriction Is Not Speciﬁed Restrict the domain and then find the inverse of f (x) = (x − 2)2 − 3. Solution We can see this is a parabola with vertex at (2, −3) that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to x ≥ 2. To fi nd the inverse, we will use the vertex form of the quadratic. We start by replacing f (x) with a simple variable, y, then solve for x. y = (x − 2)2 − 3 Interchange x and y. x = (y − 2)2 − 3 Add 3 to both sides. Take the square rooty − 2)1(x Add 2 to both sides. Rename the function. Now we need to determine which case to use. Because we restricted our original function to a domain of x ≥ 2, the outputs of the inverse should be the same, telling us to utilize the + case f −1(x) = 2 + √ — x + 3 If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain. Analysis Notice that we arbitrarily decided to restrict the domain on x ≥ 2. We could just have easily opted to restrict the domain on x ≤ 2, in which case f −1(x) = 2 − √ x + 3 . Observe the original function graphed on the same set of axes as its inverse function in Figure 7. Notice that both graphs show symmetry about the line y = x. The coordinate pair (2, −3) is on the graph of f and the coordinate pair (−3, 2) is on the graph of f−1. Observe from the graph of both functions on the same set of axes that — and domain of f = range of f −1 = [2, ∞) domain of f −1 = range of f = [−3, ∞) Finally, observe that the graph of f intersects the graph of f −1 along the line y = x. (−3, 2) –6 –8 –4 –10 y y = x f (x) f –1(x) x 8 10 642 (2, −3) 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 7 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 441 Try It #3 Find the inverse of the function f (x) = x2 + 1, on the domain x ≥ 0. Solving Applications of Radical Functions Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the inverse of a radical function, we will need to restrict the domain of the answer because the range of the original function is limited. How To… Given a radical function, find the inverse. 1. Determine the range of the original function. 2. Replace f (x) with y, then solve for x. 3. If necessary, restrict the domain of the inverse function to the range of the original function. Example 5 Finding the Inverse of a Radical Function Restrict the domain of the function f (x) = √ — x − 4 and then find the inverse. Solution Note that the original function has range f (x) ≥ 0. Replace f (x) with y, then solve for x Replace f (x) with y. Interchange x and y. Square each side. x2 = y − 4 x2 + 4 = y f −1(x) = x2 + 4 Add 4. Rename the function f−1(x). Recall that the domain of this function must be limited to the range of the original function. f −1(x) = x2 + 4, x ≥ 0 Analysis Notice in Figure 8 that the inverse is a reflection of the original function over the line y = x. Because the original function has only positive outputs, the inverse function has only positive inputs. 12 10 8 6 4 2 y f –1(x) y = x f (x) x 2 4 6 8 10 12 Figure 8 Try It #4 Restrict the domain and then find the inverse of the function f (x) = √ — 2x + 3 . Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Solving Applications of Radical Functions Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section. Example 6 Solving an Application with a Cubic Function A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by V = 2 __ 3 πr 3 2 __ Find the inverse of the function V = πr 3 that determines the volume V of a cone and is a function of the radius r. 3 Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use π = 3.14. Solution Start with the given function for V. Notice that the meaningful domain for the function is r > 0 since negative radii would not make sense in this context nor would a radius of 0. Also note the range of the function (hence, the domain of the inverse function) is V ≥ 0. Solve for r in terms of V, using the method outlined previously. Note that in real-world applications, we do not swap the variables when finding inverses. Instead, we change which variable is considered to be the independent variable. V = 2 __ πr 3 3 r 3 = 3V ___ 2π ____ √ 3 3V ___ 2π r =# Solve for r 3. Solve for r. This is the result stated in the section opener. Now evaluate this for V = 100 and π = 3.14. 3 r = ____ 3V ___ 2π _______ 3 ċ 100 ______ 2 ċ 3.14 √ √ 3 = Therefore, the radius is about 3.63 ft. ≈ 3 √ — 47.7707 ≈ 3.63 Determining the Domain of a Radical Function Composed with Other Functions When radical functions are composed with other functions, determining domain can become more complicated. Example 7 Finding the Domain of a Radical Function Composed with a Rational Function Find the domain of the function f (x) = √ ___________ (x + 2)(x − 3) ____________ # # (x − 1) . Solution Because a square root is only defined when the quantity under the radical is non-negative, we need to determine ≥ 0. The output of a rational function can change signs (change from positive to negative or vice where versa) at x-intercepts and at vertical asymptotes. For this equation, the graph could change signs at x = −2, 1, and 3. (x + 2)(x − 3) __________
__ (x − 1) To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as shown in Figure 9. y x = 1 Outputs are non-negative (−2, 0) –3 –2 –4 –7 –6 –5 10 8 6 4 2 –1 –2 –4 –6 –8 –10 Outputs are non-negative (3, 0) 321 4 5 6 7 x Figure 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 443 This function has two x-intercepts, both of which exhibit linear behavior near the x-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a y-intercept at (0, 6). From the y-intercept and x-intercept at x = −2, we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph. From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function f (x) will be defined. f (x) has domain −2 ≤ x < 1 or x ≥ 3, or in interval notation, [−2, 1) ∪ [3, ∞). Finding Inverses of Rational Functions As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as in concentration applications. Example 8 Finding the Inverse of a Rational Function The function C = represents the concentration C of an acid solution after n mL of 40% solution has been 20 + 0.4n ________ 100 + n added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for n in terms of C. Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution. Solution We first want the inverse of the function in order to determine how many mL we need for a given concentration. We will solve for n in terms of C. C = 20 + 0.4n ________ 100 + n C(100 + n) = 20 + 0.4n 100C + Cn = 20 + 0.4n 100C − 20 = 0.4n − Cn 100C − 20 = (0.4 − C)n Now evaluate this function at 35%, which is C = 0.35 . n = 100C − 20 _________ 0.4 − C n = 100(0.35) − 20 ____________ 0.4 − 0.35 = 15 ___ 0.05 = 300 We can conclude that 300 mL of the 40% solution should be added. Try It #5 Find the inverse of the function f (x) = x + 3 _____ . x − 2 Access these online resources for additional instruction and practice with inverses and radical functions. • Graphing the Basic Square Root Function (http://openstaxcollege.org/l/graphsquareroot) • Find the Inverse of a Square Root Function (http://openstaxcollege.org/l/inversesquare) • Find the Inverse of a Rational Function (http://openstaxcollege.org/l/inverserational) • Find the Inverse of a Rational Function and an Inverse Function Value (http://openstaxcollege.org/l/rationalinverse) • Inverse Functions (http://openstaxcollege.org/l/inversefunction) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.7 SECTION EXERCISES VERBAL 1. Explain why we cannot find inverse functions for all 2. Why must we restrict the domain of a quadratic polynomial functions. function when finding its inverse? 3. When finding the inverse of a radical function, what 4. The inverse of a quadratic function will always take restriction will we need to make? what form? ALGEBRAIC For the following exercises, find the inverse of the function on the given domain. 5. f (x) = (x − 4)2, [4, ∞) 6. f (x) = (x + 2)2, [−2, ∞) 7. f (x) = (x + 1)2 − 3, [−1, ∞) 8. f (x) = 3x 2 + 5, (−∞, 0] 9. f (x) = 12 − x 2, [0, ∞) 10. f (x) = 9 − x 2, [0, ∞) 11. f (x) = 2x 2 + 4, [0, ∞) For the following exercises, find the inverse of the functions. 12. f (x) = x 3 + 5 15. f (x) = 4 − 2x 3 13. f (x) = 3x 3 + 1 14. f (x) = 4 − x 3 For the following exercises, find the inverse of the functions. 16. f (x) = √ — 2x + 1 19. f (x) = √ — 6x − 8 + 5 22. f (x) = 2 _____ x + 8 25. f (x) = x − 2 _____ x + 7 17. f (x) = √ — 3 − 4x 20. f (x) = 9 + 2 3 √ — x 23. f (x) = 3 _____ x − 4 26. f (x) = 3x + 4 ______ 5 − 4x 18. f (x) = 9 + √ — 4x − 4 21. f (x) = 3 − 3 √ — x 24. f (x) = x + 3 _____ x + 7 27. f (x) = 5x + 1 ______ 2 − 5x 28. f (x) = x 2 + 2x, [−1, ∞) 29. f (x) = x 2 + 4x + 1, [−2, ∞) 30. f (x) = x 2 − 6x + 3, [3, ∞) GRAPHICAL For the following exercises, find the inverse of the function and graph both the function and its inverse. 31. f (x) = x 2 + 2, x ≥ 0 32. f (x) = 4 − x 2, x ≥ 0 33. f (x) = (x + 3)2, x ≥ −3 34. f (x) = (x − 4)2, x ≥ 4 35. f (x) = x 3 + 3 37. f (x) = x 2 + 4x, x ≥ −2 38. f (x) = x 2 − 6x + 1, x ≥ 3 36. f (x) = 1 − x 3 39. f (x) =# #2 __ x 40. f (x) = 1 __ x2 , x ≥ 0 For the following exercises, use a graph to help determine the domain of the functions. 41. f (x) = √ 44. f (x) = √ _____________ (x + 1)(x − 1) __ x ___________ x2 − x − 20 _ x − 2 _____________ (x + 2)(x − 3) __ x − 1 42. f (x) = √ 45. f (x) = √ ______ 9 − x2 _ x + 4 43. f (x) = √ ________ x(x + 3) _ x − 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 SECTION EXERCISES 445 TECHNOLOGY For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with y-coordinates given. 46. f (x) = x3 − x − 2, y = 1, 2, 3 47. f (x) = x3 + x − 2, y = 0, 1, 2 48. f (x) = x3 + 3x − 4, y = 0, 1, 2 49. f (x) = x3 + 8x − 4, y = −1, 0, 1 50. f (x) = x4 + 5x + 1, y = −1, 0, 1 EXTENSIONS For the following exercises, find the inverse of the functions with a, b, c positive real numbers. 51. f (x) = ax3 + b 54. f (x) = 3 √ — ax + b 52. f (x) = x2 + bx 55. f (x) = ax + b ______ x + c REAL-WORLD APPLICATIONS 53. f (x) = √ — ax2 + b For the following exercises, determine the function described and then use it to answer the question. 56. An object dropped from a height of 200 meters has a height, h(t), in meters after t seconds have lapsed, such that h(t) = 200 − 4.9t 2. Express t as a function of height, h, and find the time to reach a height of 50 meters. 57. An object dropped from a height of 600 feet has a height, h(t), in feet after t seconds have elapsed, such that h(t) = 600 − 16t 2. Express t as a function of height h, and find the time to reach a height of 400 feet. 58. The volume, V, of a sphere in terms of its radius, r, 4 __ is given by V(r) = πr 3. Express r as a function of 3 V, and find the radius of a sphere with volume of 200 cubic feet. 59. The surface area, A, of a sphere in terms of its radius, r, is given by A(r) = 4πr 2. Express r as a function of V, and find the radius of a sphere with a surface area of 1000 square inches. 60. A container holds 100 ml of a solution that is 25 ml acid. If n ml of a solution that is 60% acid is added, the function C(n) = gives the 25 + 0.6n ________ 100 + n concentration, C, as a function of the number of ml added, n. Express n as a function of C and determine the number of ml that need to be added to have a solution that is 50% acid. 61. The period T, in seconds, of a simple pendulum as a function of its length l, in feet, is given by T(l) = 2π √ ____ l ____ 32.2 . Express l as a function of T and determine the length of a pendulum with period of 2 seconds. 62. The volume of a cylinder, V, in terms of radius, r, 63. The surface area, A, of a cylinder in terms of its and height, h, is given by V = πr 2h. If a cylinder has a height of 6 meters, express the radius as a function of V and find the radius of a cylinder with volume of 300 cubic meters. radius, r, and height, h, is given by A = 2πr2 + 2πrh. If the height of the cylinder is 4 feet, express the radius as a function of V and find the radius if the surface area is 200 square feet. 64. The volume of a right circular cone, V, in terms of its 1 _ radius, r, and its height, h, is given by V = πr 2h. 3 Express r in terms of h if the height of the cone is 12 feet and find the radius of a cone with volume of 50 cubic inches. 65. Consider a cone with height of 30 feet. Express the radius, r, in terms of the volume, V, and find the radius of a cone with volume of 1000 cubic feet. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: • ole algebraic and numeric ariation problems. • ole application problems including direct inerse andor oint ariation. 5.8 MODELING USING VARIATION A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate. Solving Direct Variation Problems In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula e = 0.16s tells us her earnings, e, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 1. s, sales price $4,600 $9,200 $18,400 e = 0.16s e = 0.16(4,600) = 736 Interpretation A sale of a $4,600 vehicle results in $736 earnings. e = 0.16(9,200) = 1,472 A sale of a $9,200 vehicle results in $1472 earnings. e = 0.16(18,400) = 2,944 A sale of a $18,400 vehicle results in $2944 earnings. Table 1 Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $73
6 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation. Each variable in this type of relationship varies directly with the other. Figure 1 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula y = kxn is used for direct variation. The value k is a nonzero constant greater than zero and is called the constant of variation. In this case, k = 0.16 and n = 1. We saw functions like this one when we discussed power functions. 5,000 4,000 3,000 2,000 1,000 $ , 18,400, 2,944) (9,200, 1,472) (4,600, 736) 6,000 12,000 18,000 24,000 30,000 s, Sales Prices in Dollars Figure 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.8 MODELING USING VARIATION 447 direct variation If x and y are related by an equation of the form y = kx n then we say that the relationship is direct variation and y varies directly with, or is proportional to, the nth power of x. In direct variation relationships, there is a nonzero constant ratio k = constant of variation, which help defines the relationship between the variables. y _ xn , where k is called the How To… Given a description of a direct variation problem, solve for an unknown. 1. Identify the input, x, and the output, y. 2. Determine the constant of variation. You may need to divide y by the specified power of x to determine the constant of variation. 3. Use the constant of variation to write an equation for the relationship. 4. Substitute known values into the equation to find the unknown. Example 1 Solving a Direct Variation Problem The quantity y varies directly with the cube of x. If y = 25 when x = 2, find y when x is 6. Solution The general formula for direct variation with a cube is y = kx 3. The constant can be found by dividing y by the cube of x. k = y _ x3 = = 25 __ 23 25__ 8 Now use the constant to write an equation that represents this relationship. Substitute x = 6 and solve for y. y = 25 __ x3 8 (6)3 y = 25 __ 8 = 675 Analysis The graph of this equation is a simple cubic, as shown in Figure 2. y 800 600 400 200 (6, 675) (2, 25) 0 2 x 8 10 4 6 Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Q & A… Do the graphs of all direct variation equations look like Example 1? No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0,0). Try It #1 The quantity y varies directly with the square of x. If y = 24 when x = 3, find y when x is 4. Solving Inverse Variation Problems 14,000 ______ d Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula T = gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F. If we create Table 2, we observe that, as the depth increases, the water temperature decreases. d, depth 500 ft 1,000 ft 2,000 ft T = 14,000_ d 14,000 _ 500 = 28 14,000 _ 1,000 = 14 14,000 _ 2,000 = 7 Interpretation At a depth of 500 ft, the water temperature is 28° F. At a depth of 1,000 ft, the water temperature is 14° F. At a depth of 2,000 ft, the water temperature is 7° F. Table 2 We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations. For our example, Figure 3 depicts the inverse variation. We say the water temperature varies inversely with the depth k _ of the water because, as the depth increases, the temperature decreases. The formula y = x for inverse variation in this case uses k = 14,000 ° ( 42 35 28 21 14 7 0 (500, 28) (1000, 14) (2000, 7) 1,000 2,000 3,000 4,000 inverse variation If x and y are related by an equation of the form Depth, d (ft) Figure 3 y = k __ xn where k is a nonzero constant, then we say that y varies inversely with the nth power of x. In inversely proportional relationships, or inverse variations, there is a constant multiple k = xny. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.8 MODELING USING VARIATION 449 Example 2 Writing a Formula for an Inversely Proportional Relationship A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives. Solution Recall that multiplying speed by time gives distance. If we let t represent the drive time in hours, and v represent the velocity (speed or rate) at which the tourist drives, then vt = distance. Because the distance is fixed at 100 miles, vt = 100 so t = . Because time is a function of velocity, we can write t(v). 100 _ v t(v) = 100___ v = 100v −1 We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. We say that the time varies inversely with velocity. How To… Given a description of an indirect variation problem, solve for an unknown. 1. Identify the input, x, and the output, y. 2. Determine the constant of variation. You may need to multiply y by the specified power of x to determine the constant of variation. 3. Use the constant of variation to write an equation for the relationship. 4. Substitute known values into the equation to find the unknown. Example 3 Solving an Inverse Variation Problem A quantity y varies inversely with the cube of x. If y = 25 when x = 2, find y when x is 6. Solution The general formula for inverse variation with a cube is y = y by the cube of x. k = x3 y k __ x3 . The constant can be found by multiplying Now we use the constant to write an equation that represents this relationship. = 23 ċ 25 = 200 Substitute x = 6 and solve for y. y = k __ x3 , k = 200 200___ x3 200___ 63 25 __ 27 y = y = = Analysis The graph of this equation is a rational function, as shown in Figure 4. y 30 25 20 15 10 5 0 (2, 25) 6, 25 27 x 2 4 6 8 10 Figure 4 Try It #2 A quantity y varies inversely with the square of x. If y = 8 when x = 3, find y when x is 4. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Solving Problems Involving Joint Variation Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c, cost, varies jointly with the number of students, n, and the distance, d. joint variation Joint variation occurs when a variable varies directly or inversely with multiple variables. For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and inversely with z, we have x = . Notice that we only use one constant in a joint variation equation. ky __ z Example 4 Solving Problems Involving Joint Variation A quantity x varies directly with the square of y and inversely with the cube root of z. If x = 6 when y = 2 and z = 8, find x when y = 1 and z = 27. Solution Begin by writing an equation to show the relationship between the variables. Substitute x = 6, y = 2, and z = 8 to find the value of the constant k. x = ky2 _ — 3 z √ 6 = k22 _ — 3 8 √ 6 = 4k__ 2 3 = k Now we can substitute the value of the constant into the equation for the relationship. To find x when y = 1 and z = 27, we will substitute values for y and z into our equation. x = 3y2 _ — 3 z √ x = 3(1)2 _ — 3 27 √ = 1 Try It #3 A quantity x varies directly with the square of y and inversely with z. If x = 40 when y = 4 and z = 2, find x when y = 10 and z = 25. Access these online resources for additional instruction and practice with direct and inverse variation. • Direct Variation (http://openstaxcollege.org/l/directvariation) • Inverse Variation (http://openstaxcollege.org/l/inversevariatio) • Direct and Inverse Variation (http://openstaxcollege.org/l/directinverse) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.8 SECTION EXERCISES 451 5.8 SECTION EXERCISES VERBAL 1. 2. 3. ALGEBRAIC 4. yxx=y= 6. yx 5. yxx=y= 7. yxx=y= x=y= 8. yx 9. yxx= x=y= 10. yxx=y= 12. yx x=y= y= 11. yxx = 3, y= 13. yx x=y= 14. yx 15. yx x=y= x=y= 16. yxzx= 17. yxzwx=z= z=y= w=y= 18. yx zx=z=y= 20. yxz wx=z= w=y= 19. yxz x=z=y= 21. yxzwx= z=w=y= 22. yx zw x=z=w=y= 23. yxz wtx=z=w= t=y= NUMERIC o fi 24. yxx=y= 25. yxx= yx= y=yx= 26. yxx= y=yx= 28. yx x=y=yx= 30. yxx= y=yx= 27. yx x=y=yx= 29. yxx=y= yx= 31. yxx= y=yx= 32. yx 33. yx x=y=yx= x=y=yx= 34. yxzx=z= 35. yxzwx=z= y=yx=z= 36. yxz x=z=y=yx= z= w=y=yx=z=w= 37. yx zx=z=y=y x=z= 38. yxzw x=z=w=y=y x=z=w= 39. yxz wx= z=w=y=yx= z=w= Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 40. yxzwtx=z= w=t=y=yx=z=w=t= TECHNOLOGY 41. yxx= 42. yxx=y= y= 43. yx 44. yxx=y= x=y= 45. yx x=y= EXTENSIONS T a 46. Ta 48. Ta 47. 49. 50. REAL-WORLD APPLICATIONS 51. 53. 55. 52. 54. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 453 CHAPTER 5 REVIEW Key Terms arrow notation a way to represent the local and end behavior of a function by using arrows to indicate that an input or output approaches a value axis of symmetry a vertical line drawn through the vertex of a parabola, that
opens up or down, around which the parabola is symmetric; it is defined by x = − # b __ . 2a coefficient a nonzero real number multiplied by a variable raised to an exponent constant of variation the non-zero value k that helps define the relationship between variables in direct or inverse variation continuous function a function whose graph can be drawn without lifting the pen from the paper because there are no breaks in the graph degree the highest power of the variable that occurs in a polynomial Descartes’ Rule of Signs a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of f (x) and f (−x) direct variation the relationship between two variables that are a constant multiple of each other; as one quantity increases, so does the other Division Algorithm given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x), there exist unique polynomials q(x) and r(x) such that f (x) = d(x) q(x) + r(x) where q(x) is the quotient and r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). end behavior the behavior of the graph of a function as the input decreases without bound and increases without bound Factor Theorem k is a zero of polynomial function f (x) if and only if (x − k) is a factor of f (x) Fundamental Theorem of Algebra a polynomial function with degree greater than 0 has at least one complex zero general form of a quadratic function the function that describes a parabola, written in the form f (x) = ax 2 + bx + c, where a, b, and c are real numbers and a ≠ 0. global maximum highest turning point on a graph; f (a) where f (a) ≥ f (x) for all x. global minimum lowest turning point on a graph; f (a) where f (a) ≤ f (x) for all x. horizontal asymptote a horizontal line y = b where the graph approaches the line as the inputs increase or decrease without bound. Intermediate Value Theorem for two numbers a and b in the domain of f, if a < b and f (a) ≠ f (b), then the function f takes on every value between f (a) and f (b); specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis inverse variation the relationship between two variables in which the product of the variables is a constant inversely proportional a relationship where one quantity is a constant divided by the other quantity; as one quantity increases, the other decreases invertible function any function that has an inverse function −1 imaginary number a number in the form bi where i = √ — joint variation a relationship where a variable varies directly or inversely with multiple variables leading coefficient the coefficient of the leading term leading term the term containing the highest power of the variable Linear Factorization Theorem allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form (x − c), where c is a complex number multiplicity the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form (x − h)p, x = h is a zero of multiplicity p. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS polynomial function a function that consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. power function a function that can be represented in the form f (x) = kxp where k is a constant, the base is a variable, and the exponent, p, is a constant rational function a function that can be written as the ratio of two polynomials p _ Rational Zero Theorem the possible rational zeros of a polynomial function have the form q where p is a factor of the constant term and q is a factor of the leading coefficient. Remainder Theorem if a polynomial f (x) is divided by x − k, then the remainder is equal to the value f (k) removable discontinuity a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function roots in a given function, the values of x at which y = 0, also called zeros smooth curve a graph with no sharp corners standard form of a quadratic function the function that describes a parabola, written in the form f (x) = a(x − h)2 + k, where (h, k) is the vertex. synthetic division a shortcut method that can be used to divide a polynomial by a binomial of the form x − k term of a polynomial function any aixi of a polynomial function in the form f (x) = anxn + ... + a2x2 + a1x + a0 turning point the location at which the graph of a function changes direction varies directly a relationship where one quantity is a constant multiplied by the other quantity varies inversely a relationship where one quantity is a constant divided by the other quantity vertex the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function vertex form of a quadratic function another name for the standard form of a quadratic function vertical asymptote a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a zeros in a given function, the values of x at which y = 0, also called roots Key Equations general form of a quadratic function f (x) = ax 2 + bx + c standard form of a quadratic function f (x) = a(x − h)2 + k general form of a polynomial function f (x) = an xn + ... + a2 x2 + a1x + a0 Division Algorithm f (x) = d(x)q(x) + r(x) where q(x) ≠ 0 Rational Function Direct variation Inverse variation f (x) = P(x) ____ Q(x) =# # ap x ___ bq x p−1 + ... + a1x + a0 q−1 + ... + b1 x + b0 p + ap − 1x q + bq − 1 x , Q(x) ≠ 0 y = kx n, k is a nonzero constant. y = k _ xn , k is a nonzero constant. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 455 Key Concepts 5.1 Quadratic Functions • A polynomial function of degree two is called a quadratic function. • The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down. • The axis of symmetry is the vertical line passing through the vertex. The zeros, or x-intercepts, are the points at which the parabola crosses the x-axis. The y-intercept is the point at which the parabola crosses the y-axis. See Example 1, Example 7, and Example 8. • Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph. See Example 2. • The vertex can be found from an equation representing a quadratic function. See Example 3. • The domain of a quadratic function is all real numbers. The range varies with the function. See Example 4. • A quadratic function’s minimum or maximum value is given by the y-value of the vertex. • The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. See Example 5 and Example 6. • The vertex and the intercepts can be identified and interpreted to solve real-world problems. See Example 9. 5.2 Power Functions and Polynomial Functions • A power function is a variable base raised to a number power. See Example 1. • The behavior of a graph as the input decreases beyond bound and increases beyond bound is called the end behavior. • The end behavior depends on whether the power is even or odd. See Example 2 and Example 3. • A polynomial function is the sum of terms, each of which consists of a transformed power function with positive whole number power. See Example 4. • The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term containing the highest power of the variable is called the leading term. The coefficient of the leading term is called the leading coefficient. See Example 5. • The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. See Example 6 and Example 7. • A polynomial of degree n will have at most n x-intercepts and at most n −#1 turning points. See Example 8, Example 9, Example 10, Example 11, and Example 12. 5.3 Graphs of Polynomial Functions • Polynomial functions of degree 2 or more are smooth, continuous functions. See Example 1. • To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero. See Example 2, Example 3, and Example 4. • Another way to find the x-intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the x-axis. See Example 5. • The multiplicity of a zero determines how the graph behaves at the x-intercepts. See Example 6. • The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity. • The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity. • The end behavior of a polynomial function depends on the leading term. • The graph of a polynomial function changes direction at its turning points. • A polynomial function of degree n has at most n − 1 turning points. See Example 7. • To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most n − 1 turning points. See Example 8 and Example 10. • Graphing a polynomial function helps to estimate local and global extremas. See Example 11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS • The Intermediate Value Theorem t
ells us that if f (a) and f (b) have opposite signs, then there exists at least one value c between a and b for which f (c) = 0. See Example 9. 5.4 Dividing Polynomials • Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See Example 1 and Example 2. • The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder. • Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form x − k. See Example 3, Example 4, and Example 5. • Polynomial division can be used to solve application problems, including area and volume. See Example 6. 5.5 Zeros of Polynomial Functions • To fi nd f (k), determine the remainder of the polynomial f (x) when it is divided by x − k. this is known as the Remainder Theorem. See Example 1. • According to the Factor Theorem, k is a zero of f (x) if and only if (x − k) is a factor of f (x). See Example 2. • According to the Rational Zero Theorem, each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient. See Example 3 and Example 4. • When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. • Synthetic division can be used to find the zeros of a polynomial function. See Example 5. • According to the Fundamental Theorem, every polynomial function has at least one complex zero. See Example 6. • Every polynomial function with degree greater than 0 has at least one complex zero. • Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form (x − c), where c is a complex number. See Example 7. • The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer. • The number of negative real zeros of a polynomial function is either the number of sign changes of f (−x) or less than the number of sign changes by an even integer. See Example 8. • Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division. See Example 9. 5.6 Rational Functions 1 • We can use arrow notation to describe local behavior and end behavior of the toolkit functions f (x) =# #1 _ _ x and f (x) = x2 . See Example 1. • A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote. See Example 2. • Application problems involving rates and concentrations often involve rational functions. See Example 3. • The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. See Example 4. • The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. See Example 5. • A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. See Example 6. • A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. See Example 7, Example 8, Example 9, and Example 10. • Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. See Example 11. • If a rational function has x-intercepts at x = x1, x2, …, xn, vertical asymptotes at x = v1, v2, ..., vm, and no xi = any vj, then the function can be written in the form Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 457 f (x) =#a (x − x1)p ___ (x − v1)q 2...(x − xn)p n 2...(x − vm)q 1(x − x2)p 1(x − v2)q n See Example 12. 5.7 Inverses and Radical Functions • The inverse of a quadratic function is a square root function. • If f −l is the inverse of a function f, then f is the inverse of the function f −l. See Example 1. • While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See Example 2. • To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one. See Example 3 and Example 4. • When finding the inverse of a radical function, we need a restriction on the domain of the answer. See Example 5 and Example 7. • Inverse and radical and functions can be used to solve application problems. See Example 6 and Example 8. 5.8 Modeling Using Variation • A relationship where one quantity is a constant multiplied by another quantity is called direct variation. See Example 1. • Two variables that are directly proportional to one another will have a constant ratio. • A relationship where one quantity is a constant divided by another quantity is called inverse variation. See Example 2. • Two variables that are inversely proportional to one another will have a constant multiple. See Example 3. • In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation. See Example 4. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS CHAPTER 5 REVIEW EXERCISES QUADRATIC FUNCTIONS For the following exercises, write the quadratic function in standard form. Then, give the vertex and axes intercepts. Finally, graph the function. 1. f (x) = x2 − 4x − 5 2. f (x) = −2x2 − 4x For the following problems, find the equation of the quadratic function using the given information. 3. The vertex is (−2, 3) and a point on the graph is (3, 6). 4. The vertex is (−3, 6.5) and a point on the graph is (2, 6). For the following exercises, complete the task. 5. A rectangular plot of land is to be enclosed by fencing. One side is along a river and so needs no fence. If the total fencing available is 600 meters, find the dimensions of the plot to have maximum area. 6. An object projected from the ground at a 45 degree angle with initial velocity of 120 feet per second has height, h, in terms of horizontal distance traveled, −32 _____ (120)2 x2 + x. Find the maximum x, given by h(x) = height the object attains. POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS For the following exercises, determine if the function is a polynomial function and, if so, give the degree and leading coefficient. 7. f (x) = 4x5 − 3x3 + 2x − 1 8. f (x) = 5x + 1 − x2 9. f (x) = x2(3 − 6x + x2) For the following exercises, determine end behavior of the polynomial function. 10. f (x) = 2x4 + 3x3 − 5x2 + 7 11. f (x) = 4x3 − 6x2 + 2 12. f (x) = 2x2(1 + 3x − x2) GRAPHS OF POLYNOMIAL FUNCTIONS For the following exercises, find all zeros of the polynomial function, noting multiplicities. 13. f (x) = (x + 3)2(2x − 1)(x + 1)3 14. f (x) = x5 + 4x4 + 4x3 15. f (x) = x3 − 4x2 + x − 4 For the following exercises, based on the given graph, determine the zeros of the function and note multiplicity. 16. y 17. y –5 –4 –3 –2 20 16 12 8 4 –1 –4 –8 –12 –16 –20 321 4 5 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 642 8 10 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 459 18. Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the function f (x) = x3 − 5x + 1 DIVIDING POLYNOMIALS For the following exercises, use long division to find the quotient and remainder. 19. x3 − 2x2 + 4x + 4 ______________ x − 2 20. 3x4 − 4x2 + 4x + 8 _______________ x + 1 For the following exercises, use synthetic division to find the quotient. If the divisor is a factor, then write the factored form. 21. x3 − 2x2 + 5x − 1 x + 3 23. 2x3 + 6x2 − 11x − 12 _________________ x + 4 22. x3 + 4x + 10 __________ x − 3 24. 3x4 + 3x3 + 2x + 2 _______________ x + 1 ______________ ZEROS OF POLYNOMIAL FUNCTIONS For the following exercises, use the Rational Zero Theorem to help you solve the polynomial equation. 25. 2x3 − 3x2 − 18x − 8 = 0 26. 3x3 + 11x2 + 8x − 4 = 0 27. 2x4 − 17x3 + 46x2 − 43x + 12 = 0 28. 4x4 + 8x3 + 19x2 + 32x + 12 = 0 For the following exercises, use Descartes’ Rule of Signs to find the possible number of positive and negative solutions. 29. x3 − 3x2 − 2x + 4 = 0 30. 2x4 − x3 + 4x2 − 5x + 1 = 0 RATIONAL FUNCTIONS For the following exercises, find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph of the function. 31. f (x) = x + 2 _____ x − 5 33. f (x) = 3x2 − 27 32. f (x) = x2 + 1 _____ x2 − 4 34. f (x) = x + 2 _____ x2 − 9 ________ x2 + x − 2 For the following exercises, find the slant asymptote. 35. f (x) = x2 − 1 _____ x + 2 36. f (x) = 2x3 − x2 + 4 __________ x2 + 1 INVERSES AND RADICAL FUNCTIONS For the following exercises, find the inverse of the function with the domain given. 37. f (x) = (x − 2)2, x ≥ 2 38. f (x) = (x + 4)2 − 3, x ≥ −4 40. f (x) = 2x3 − 3 41. f (x) = √ — 4x + 5 − 3 39. f (x) = x2 + 6x − 2, x ≥ −3 42. f (x) = x − 3 _____ 2x + 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS MODELING USING VARIATION For the following exercises, find the unknown value. 43. y varies directly as the square of x. If when x = 3, y = 36, find y if x = 4. 45. y varies jointly as the cube of x and as z. If when x = 1 and z = 2, y = 6, find y if x = 2 and z = 3. 44. y varies inversely as the square root of x. If when x = 25, y = 2, find y if x = 4. 46. y varies jointly as x and the square of z and inversely as the cube of w. If when x = 3, z = 4, and w = 2, y = 48, find y if x = 4, z = 5, and w = 3. For the following exercises, solve the application problem. 47. The weight of an object above the surface of the earth varies inversely with the distance from the center of the earth.If a person weighs 150 pounds when he is on the surface of the earth (3,960 miles from center), find t
he weight of the person if he is 20 miles above the surface. 48. The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 PRACTICE TEST 461 CHAPTER 5 PRACTICE TEST Give the degree and leading coefficient of the following polynomial function. 1. f (x) = x3(3 − 6x2 − 2x2) Determine the end behavior of the polynomial function. 2. f (x) = 8x3 − 3x2 + 2x − 4 3. f (x) = −2x2(4 − 3x − 5x2) Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function. 4. f (x) = x2 + 2x − 8 Given information about the graph of a quadratic function, find its equation. 5. Vertex (2, 0) and point on graph (4, 12). Solve the following application problem. 6. A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed. Find all zeros of the following polynomial functions, noting multiplicities. 7. f (x) = (x − 3)3(3x − 1)(x − 1)2 8. f (x) = 2x6 − 6x5 + 18x4 Based on the graph, determine the zeros of the function and multiplicities. 9. y 125 100 75 50 25 –5 –4 –3 –2 –1 –25 –50 –75 –100 –125 321 4 5 x Use long division to find the quotient. 10. 2x3 + 3x − 4 __________ x + 2 Use synthetic division to find the quotient. If the divisor is a factor, write the factored form. 11. x4 + 3x2 − 4 __________ x − 2 12. 2x3 + 5x2 − 7x − 12 ________________ x + 3 Use the Rational Zero Theorem to help you find the zeros of the polynomial functions. 13. f (x) = 2x3 + 5x2 − 6x − 9 14. f (x) = 4x4 + 8x3 + 21x2 + 17x + 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 15. f (x) = 4x4 + 16x3 + 13x2 − 15x − 18 16. f (x) = x5 + 6x4 + 13x3 + 14x2 + 12x + 8 Given the following information about a polynomial function, find the function. 17. It has a double zero at x = 3 and zeroes at x = 1 and x = −2. It’s y-intercept is (0, 12). 18. It has a zero of multiplicity 3 at x =# #1 _ 2 and another zero at x = −3. It contains the point (1, 8). Use Descartes’ Rule of Signs to determine the possible number of positive and negative solutions. 19. 8x3 − 21x2 + 6 = 0 For the following rational functions, find the intercepts and horizontal and vertical asymptotes, and sketch a graph. 20. f (x) = x + 4 _________ x2 − 2x − 3 21. f (x) = x2 + 2x − 3 _________ x2 − 4 Find the slant asymptote of the rational function. 22. f (x) = x2 + 3x − 3 _________ x − 1 Find the inverse of the function. — x − 2 + 4 23. f (x) = √ 25. f (x) = 2x + 3 ______ 3x − 1 24. f (x) = 3x3 − 4 Find the unknown value. 26. y varies inversely as the square of x and when x = 3, y = 2. Find y if x = 1. 27. y varies jointly with x and the cube root of z. If when x = 2 and z = 27, y = 12, find y if x = 5 and z = 8. Solve the following application problem. 28. The distance a body falls varies directly as the square of the time it falls. If an object falls 64 feet in 2 seconds, how long will it take to fall 256 feet? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Exponential and Logarithmic Functions 6 Figure 1 Electron micrograph of E. Coli bacteria (credit: “Mattosaurus,” Wikimedia Commons) CHAPTER OUTLINE 6.1 Exponential Functions 6.2 Graphs of Exponential Functions 6.3 Logarithmic Functions 6.4 Graphs of Logarithmic Functions 6.5 Logarithmic Properties 6.6 Exponential and Logarithmic Equations 6.7 Exponential and Logarithmic Models 6.8 Fitting Exponential Models to Data Introduction Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see hundreds of thousands of microscopic organisms. They are bacteria, and they are not only on your skin, but in your mouth, nose, and even your intestines. In fact, the bacterial cells in your body at any given moment outnumber your own cells. But that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to the body. Bacteria commonly reproduce through a process called binary fission, during which one bacterial cell splits into two. When conditions are right, bacteria can reproduce very quickly. Unlike humans and other complex organisms, the time required to form a new generation of bacteria is often a matter of minutes or hours, as opposed to days or years.[16] For simplicity’s sake, suppose we begin with a culture of one bacterial cell that can divide every hour. Table 1 shows the number of bacterial cells at the end of each subsequent hour. We see that the single bacterial cell leads to over one thousand bacterial cells in just ten hours! And if we were to extrapolate the table to twenty-four hours, we would have over 16 million! Hour Bacteria 0 1 1 2 2 4 3 8 5 32 4 16 Table 1 6 64 7 128 8 256 9 10 512 1024 In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data. 16 Todar, PhD, Kenneth. Todar’s Online Te xtbook of Bacteriology. http://te xtbookofbacteriology.net/growth_3.html. 463 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS G C F . . In this section, you will: • efine and eponential function and determine its domain. • Graph an eponential function including using transformations. • ntroduce . • rite a function that results from a gien transformation. 6.1 EXPONENTIAL FUNCTIONS India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year[17]. If this rate continues, the population of India will exceed China’s population by the year 2031. When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth. Identifying Exponential Functions When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input. For example, in the equation f (x) = 3x + 4, the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people. Deﬁning an Exponential Function A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2050.. What exactly does it mean to grow exponentially? What does the word double have in common with percent increase? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media. • Percent change refers to a change based on a percent of the original amount. • Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time. • Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time. For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth. We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See Table 1x) = 2x 1 2 4 8 16 32 64 Table 1 g(x) = 2x 0 2 4 6 8 10 12 17 http://www.worldometers.info/world-population/. Accessed February 24, 2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 465 From Table 2 we can infer that for these two functions, exponential growth dwarfs linear growth. • Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain. • Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain. Apparently, the difference between “the same percentage” and “the same amount” is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one. The general form of the exponential function is f (x) = ab x, where a is any nonzero number, b is a positive real number not equal to 1. • If b > 1, the function grows at a rate proportional to its size. • If
0 < b < 1, the function decays at a rate proportional to its size. Let’s look at the function f (x) = 2x from our example. We will create a table (Table 2) to determine the corresponding outputs over an interval in the domain from −3 to 3. x f (x) = 2x −3 −2 −1 0 1 2 3 2−3 = 1 _ 8 2−2 = 1 _ 4 2−1 = 1 _ 2 Table 2 20 = 1 21 = 2 22 = 4 23 = 8 Let us examine the graph of f by plotting the ordered pairs we observe on the table in Figure 1, and then make a few observations. y f(x) = 2x (3, 8) (2, 4) (1, 2) (0, 15 –4 –3 –2 –1 –1 –2 Figure 1 Let’s define the behavior of the graph of the exponential function f (x) = 2x and highlight some its key characteristics. • the domain is (−∞, ∞), • the range is (0, ∞), • as x → ∞, f (x) → ∞, • as x → −∞, f (x) → 0, • f (x) is always increasing, • the graph of f (x) will never touch the x-axis because base two raised to any exponent never has the result of zero. • y = 0 is the horizontal asymptote. • the y-intercept is 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS exponential function For any real number x, an exponential function is a function with the form where • a is the a non-zero real number called the initial value and f (x) = ab x • b is any positive real number such that b ≠ 1. • The domain of f is all real numbers. • The range of f is all positive real numbers if a > 0. • The range of f is all negative real numbers if a < 0. • The y-intercept is (0, a), and the horizontal asymptote is y = 0. Example 1 Identifying Exponential Functions Which of the following equations are not exponential functions? f (x) = 43(x − 2) g (x) = x3 j (x) = (−2)x x 1 ) h (x) = ( _ 3 Solution B y definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, g(x) = x3 does not represent an exponential function because the base is an independent variable. In fact, g(x) = x3 is a power function. Recall that the base b of an exponential function is always a positive constant, and b ≠ 1. Thus, j(x) = (−2)x does not represent an exponential function because the base, −2, is less than 0. Try It #1 Which of the following equations represent exponential functions? • f (x) = 2x2 − 3x + 1 • g(x) = 0.875x • h(x) = 1.75x + 2 • j(x) = 1095.6−2x Evaluating Exponential Functions Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive: 1 __ = √ 2 1 1 ) = (−9) . Then f (x) = f ( _ _ • Let b = −9 and x = 2 2 −9 , which is not a real number. — Why do we limit the base to positive values other than 1? Because base 1 results in the constant function. Observe what happens if the base is 1: • Let b = 1. Then f (x) = 1x = 1 for any value of x. To evaluate an exponential function with the form f (x) = b x, we simply substitute x with the given value, and calculate the resulting power. For example: Let f (x) = 2x. What is f (3)? f (x) = 2x f (3) = 23 = 8 Substitute x = 3. Evaluate the power. To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 467 For example: Let f (x) = 30(2)x. What is f (3)? f (x) = 30(2)x f (3) = 30(2)3 = 30(8) = 240 Substitute x = 3. Simplify the power first. Multiply. Note that if the order of operations were not followed, the result would be incorrect: f (3) = 30(2)3 ≠ 603 = 216,000 Example 2 Evaluating Exponential Functions Let f (x) = 5(3)x + 1. Evaluate f (2) without using a calculator. Solution Follow the order of operations. Be sure to pay attention to the parentheses. f (x) = 5(3)x + 1 f (2) = 5(3)2 + 1 = 5(3)3 = 5(27) = 135 Substitute x = 2. Add the exponents. Simplify the power. Multiply. Try It #2 Let f (x) = 8(1.2)x − 5. Evaluate f (3) using a calculator. Round to four decimal places. Deﬁning Exponential Growth Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth. exponential growth A function that models exponential growth grows by a rate proportional to the amount present. For any real number x and any positive real numbers a and b such that b ≠ 1, an exponential growth function has the form where f (x) = ab x • a is the initial or starting value of the function. • b is the growth factor or growth multiplier per unit x. In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function A(x) = 100 + 50x. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function B(x) = 100(1 + 0.5)x. A few years of growth for these companies are illustrated in Table 3. Year, x 0 1 2 3 x Stores, Company A 100 + 50(0) = 100 100 + 50(1) = 150 100 + 50(2) = 200 100 + 50(3) = 250 A(x) = 100 + 50x Table 3 Stores, Company B 100(1 + 0.5)0 = 100 100(1 + 0.5)1 = 150 100(1 + 0.5)2 = 225 100(1 + 0.5)3 = 337.5 B(x) = 100(1 + 0.5)x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 8 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS The graphs comparing the number of stores for each company over a five-year period are shown in Figure 2. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth. y B(x) = 100(1 + 0.5) 500 450 400 350 300 250 200 150 100 A(x) = 100 + 50x 0 1 2 3 Years x 4 5 Figure 2 The graph shows the numbers of stores Companies A and B opened over a ﬁve-year period. Notice that the domain for both functions is [0, ∞), and the range for both functions is [100, ∞). After year 1, Company B always has more stores than Company A. Now we will turn our attention to the function representing the number of stores for Company B, B(x) = 100(1 + 0.5)x. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and 1 + 0.5 = 1.5 represents the growth factor. Generalizing further, we can write this function as B(x) = 100(1.5)x, where 100 is the initial value, 1.5 is called the base, and x is called the exponent. Example 3 Evaluating a Real-World Exponential Model At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function P(t) = 1.25(1.012)t, where t is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031? Solution To estimate the population in 2031, we evaluate the models for t = 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth, There will be about 1.549 billion people in India in the year 2031. P(18) = 1.25(1.012)18 ≈ 1.549 Try It #3 The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function P(t) = 1.39(1.006)t, where t is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 3? Finding Equations of Exponential Functions In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a and b, and evaluate the function. How To… Given two data points, write an exponential model. 1. If one of the data points has the form (0, a), then a is the initial value. Using a, substitute the second point into the equation f (x) = a(b)x, and solve for b. 2. If neither of the data points have the form (0, a), substitute both points into two equations with the form f (x) = a(b)x. Solve the resulting system of two equations in two unknowns to find a and b. 3. Using the a and b found in the steps above, write the exponential function in the form f (x) = a(b)x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 469 Example 4 Writing an Exponential Model When the Initial Value Is Known In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population (N) of deer over time t. Solution We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation N(t) = 80bt to find b: N(t) = 80b t 180 = 80b 6 9 __ = b 6 4 9 __ ) b = ( 4 b ≈ 1.1447 1 __ 6 Substitute using point (6, 180). Divide and write in lowest terms. Isolate b using properties of exponents. Round to 4 decimal places. NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section. The exponential model for th
e population of deer is N(t) = 80(1.1447)t. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.) We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure 3 passes through the initial points given in the problem, (0, 80) and (6, 180). We can also see that the domain for the function is [0, ∞), and the range for the function is [80, ∞). N(t) 320 300 280 260 240 220 200 180 160 140 120 100 80 6, 180) (0, 80) 0 1 2 3 7 8 9 10 t 4 6 5 Years Figure 3 Graph showing the population of deer over time, N(t) = 80(1.1447)t, t years after 2006. Try It #4 A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013, the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N of wolves over time t. Example 5 Writing an Exponential Model When the Initial Value is Not Known Find an exponential function that passes through the points (−2, 6) and (2, 1). Solution Because we don’t have the initial value, we substitute both points into an equation of the form f (x) = ab x, and then solve the system for a and b. • Substituting (−2, 6) gives 6 = ab−2 • Substituting (2, 1) gives 1 = ab2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 0 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Use the first equation to solve for a in terms of b: 6 = ab−2 6 _ b−2 = a a = 6b 2 Substitute a in the second equation, and solve for b: Divide. Use properties of exponents to rewrite the denominator. 1 = ab 2 1 = 6b 2b 2 = 6b 4 1 1 __ .6389 Substitute a. Use properties of exponents to isolate b. Round 4 decimal places. Use the value of b in the first equation to solve for the value of a: a = 6b 2 ≈ 6(0.6389)2 ≈ 2.4492 Thus, the equation is f (x) = 2.4492(0.6389)x. We can graph our model to check our work. Notice that the graph in Figure 4 passes through the initial points given in the problem, (−2, 6) and (2, 1). The graph is an example of an exponential decay function. f (x) 10 1 –1 –2 (−2, 6) –5 –4 –3 –2 (2, 1) 1 2 3 4 5 x Figure 4 The graph of f (x) = 2.4492(0.6389)x models exponential decay. Try It #5 Given the two points (1, 3) and (2, 4.5), find the equation of the exponential function that passes through these two points. Q & A… Do two points always determine a unique exponential function? Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x, which in many real world cases involves time. How To… Given the graph of an exponential function, write its equation. 1. First, identify two points on the graph. Choose the y-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error. 2. If one of the data points is the y-intercept (0, a), then a is the initial value. Using a, substitute the second point into the equation f (x) = a(b)x, and solve for b. 3. If neither of the data points have the form (0, a), substitute both points into two equations with the form f (x) = a(b)x. Solve the resulting system of two equations in two unknowns to find a and b. 4. Write the exponential function, f (x) = a(b)x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 471 Example 6 Writing an Exponential Function Given Its Graph Find an equation for the exponential function graphed in Figure 5. f (x) 21 18 15 12 9 6 3 –3.5 –3 –2.5 –2 –1.5 –1 –0.5 –3 0.5 1 1.5 2 2.5 3 3.5 x Figure 5 Solution We can choose the y-intercept of the graph, (0, 3), as our first point. This gives us the initial value, a = 3. Ne xt, choose a point on the curve some distance away from (0, 3) that has integer coordinates. One such point is (2, 12). Substitute the initial value 3 for a. y = ab x Write the general form of an exponential equation. y = 3b x 12 = 3b2 4 = b2 b = ± 2 Substitute in 12 for y and 2 for x. Take the square root. Divide by 3. Because we restrict ourselves to positive values of b, we will use b = 2. Substitute a and b into the standard form to yield the equation f (x) = 3(2)x. Try It #6 Find an equation for the exponential function graphed in Figure 6. f(x) 5 4 3 2 1 (0, √2 ) –5 –4 –3 –2 –1–1 21 3 4 5 x Figure 6 How To… Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 1. Press [STAT]. 2. Clear any existing entries in columns L1 or L2. 3. In L1, enter the x-coordinates given. 4. In L2, enter the corresponding y-coordinates. 5. Press [STAT] again. Cursor right to CALC, scroll down to ExpReg (Exponential Regression), and press [ENTER]. 6. The screen displays the values of a and b in the exponential equation y = a ⋅ b x Example 7 Using a Graphing Calculator to Find an Exponential Function Use a graphing calculator to find the exponential equation that includes the points (2, 24.8) and (5, 198.4). Solution Follow the guidelines above. First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Ne xt, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4. Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values a = 6.2 and b = 2 will be displayed. The exponential equation is y = 6.2 ⋅ 2x. Try It #7 Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 2 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Applying the Compound-Interest Formula Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account. The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing. We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t, principal P, APR r, and number of compounding periods in a year n: For example, observe Table 4, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases. r _ A(t) = P ( 1 + ) n nt Frequency Annually Semiannually Quarterly Monthly Daily Value after 1 year $1100 $1102.50 $1103.81 $1104.71 $1105.16 Table 4 the compound interest formula Compound interest can be calculated using the formula A(t nt where • A(t) is the account value, • t is measured in years, • P is the starting amount of the account, often called the principal, or more generally present value, • r is the annual percentage rate (APR) expressed as a decimal, and • n is the number of compounding periods in one year. Example 8 Calculating Compound Interest If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years? Solution Because we are starting with $3,000, P = 3000. Our interest rate is 3%, so r = 0.03. Because we are compounding quarterly, we are compounding 4 times per year, so n = 4. We want to know the value of the account in 10 years, so we are looking for A(10), the value when t = 10. r _ A(t) = P ( 1 + ) n nt Use the compound interest formula. A(10) = 3000 ( 1 + ≈ $4,045.05 4 ⋅ 10 0.03 ____ ) 4 Substitute using given values. Round to two decimal places. The account will be worth about $4,045.05 in 10 years. Try It #8 An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 473 Example 9 Using the Compound Interest Formula to Solve for the Principal A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now? Solution The nominal interest rate is 6%, so r = 0.06. Interest is compounded twice a year, so n = 2. We want to find the initial investment, P, needed so that the value of the account will be worth $40,000 in 18 years. Substitute the given values into the compound interest formula, and solve for P. nt r _ A(t) = P ( 1 + ) n 0.06 ____ ) 40,000 = P ( 1 + 2 2(18) Use the compound interest formula. Substitute using given values A, r, n, and t. 40,000 = P(1.03)36 40,000 ______ (1.03)36 = P Simplify. Isolate P. Lily will need to invest $13,801 to have $40,000 in 18 years. P ≈ $13, 801 Divide and round to the nearest dollar. Try It #9 Refer to Example 9. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly? Evaluating Functions with Base e As we saw earlier, the amount earned on an account increases as the compounding frequency increa
ses. Table 5 shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue. Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table 5. Frequency Annually Semiannually Quarterly Monthly Daily Hourly Once per minute Once per second 1 _ n ) A(t + 12 1 ) _ 12 ( 1 + 365 1 ) _ 365 ( 1 + 1 ) _ 8766 8766 ( 1 + 1 ) _ 525960 525960 ( 1 + 1 ) ________ 31557600 31557600 Table 5 Value $2 $2.25 $2.441406 $2.613035 $2.714567 $2.718127 $2.718279 $2.718282 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS These values appear to be approaching a limit as n increases without bound. In fact, as n gets larger and larger, the 1 expression ( 1 + _ n ) approaches a number used so frequently in mathematics that it has its own name: the letter e. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below. n the number e The letter e represents the irrational number ( 1 + 1 __ ) n n , as n increases without bound The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, e ≈ 2.718282. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties. Example 10 Using a Calculator to Find Powers of e Calculate e3.14. Round to five decimal places. Solution On a calculator, press the button labeled [e x]. The window shows [e^(]. Type 3.14 and then close parenthesis, [)]. Press [ENTER]. Rounding to 5 decimal places, e 3.14 ≈ 23.10387. Caution: Many scientific calculators have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of e. Try It #10 Use a calculator to find e −0.5. Round to five decimal places. Investigating Continuous Growth So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use e as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics. the continuous growth/decay formula For all real numbers t, and all positive numbers a and r, continuous growth or decay is represented by the formula where • a is the initial value, • r is the continuous growth rate per unit time, A(t) = aert • and t is the elapsed time. If r > 0, then the formula represents continuous growth. If r < 0, then the formula represents continuous decay. For business applications, the continuous growth formula is called the continuous compounding formula and takes the form A(t) = Pert where • P is the principal or the initial invested, • r is the growth or interest rate per unit time, • and t is the period or term of the investment. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 475 How To… Given the initial value, rate of growth or decay, and time t, solve a continuous growth or decay function. 1. Use the information in the problem to determine a, the initial value of the function. 2. Use the information in the problem to determine the growth rate r. a. If the problem refers to continuous growth, then r > 0. b. If the problem refers to continuous decay, then r < 0. 3. Use the information in the problem to determine the time t. 4. Substitute the given information into the continuous growth formula and solve for A(t). Example 11 Calculating Continuous Growth A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year? Solution Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. The initial investment was $1,000, so P = 1000. We use the continuous compounding formula to find the value after t = 1 year: A(t) = Pert Use the continuous compounding formula. = 1000(e)0.1 Substitute known values for P, r, and t. ≈ 1105.17 Use a calculator to approximate. The account is worth $1,105.17 after one year. Try It #11 A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years? Example 12 Calculating Continuous Decay Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days? Solution Since the substance is decaying, the rate, 17.3%, is negative. So, r = −0.173. The initial amount of radon222 was 100 mg, so a = 100. We use the continuous decay formula to find the value after t = 3 days: A(t) = aert Use the continuous growth formula. = 100e−0.173(3) ≈ 59.5115 Substitute known values for a, r, and t. Use a calculator to approximate. So 59.5115 mg of radon-222 will remain. Try It #12 Using the data in Example 12, how much radon-222 will remain after one year? Access these online resources for additional instruction and practice with exponential functions. • Exponential Growth Function (http://openstaxcollege.org/l/expgrowth) • Compound Interest (http://openstaxcollege.org/l/compoundint) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.1 SECTION EXERCISES Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 SECTION EXERCISES 477 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 8 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 479 G C F . . In this section, you will: • efine an eponential function and determine its domain. • Graph an eponential function including using transformations. • ntroduce . • rite a function that results from a gien transformation. 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a realworld situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events. Graphing Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form f (x) = b x whose base is greater than one. We’ll use the function f (x) = 2x. Observe how the output values in Table 1 change as the input increases by 1. x f (x) = 2x −3 1 _ 8 −2 1 _ 4 0 1 −1 1_ 2 Table 1 1 2 2 4 3 8 Each output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio. In fact, for any exponential function with the form f (x) = ab x, b is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a. Notice from the table that • the output values are positive for all values of x; • as x increases, the output values increase without bound; and • as x decreases, the output values grow smaller, approaching zero. Figure 1 shows the exponential growth function f (x) = 2x. f (x) 10 1, 1 2 1 4 −2, −3, 1 8 –5 –4 –3 –2 – –1 1 (3, 8) f (x) = 2x (2, 4) (1, 2) (0 , 1) 21 3 4 5 x The x-axis is an asymptote. Figure 1 Notice that the graph gets close to the x-axis, but never touches it. The domain of f (x) = 2x is all real numbers, the range is (0, ∞), and the horizontal asymptote is y = 0. To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form f (x) = b x whose 1 ) base is between zero and one. We’ll use the function g(x) = ( _ . Observe how the output values in Table 2 change as 2 the input increases by 1. x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 0 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS x g(x) = ( 1 __ 2 ) x −3 8 −2 4 0 1 −1 2 Table 2 1 1_ 2 2 1 _ 4 3 1 _ 8 Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or 1 _ . constant ratio 2 Notice from the table that • the output values are positive for all values of x; • as x increases, the output values grow smaller, approaching zero; and • as x decreases, the output values grow without bound. x 1 ) Figure 2 shows the exponential decay function, g(x) = ( _ . 2 g(x) x g(x) = 1 2 (−3, 8) (–2, 4) (–1, 2) (0, 1) 10 21, 1 42, 1 83, –5 –4 –3 –2 –1 21 3 4 5 x The x-axis is an asymptote. Figure 2 1 ) The domain of g(x) = ( _ 2 x is all real numbers, the range is (0, ∞), and the horizontal asymptote is y = 0. characteristics of the graph of the parent function f (x) = b x An exponential function with the form f (x) = b x, b > 0, b ≠ 1, has these characteristics: • one-to-one function • horizontal asymptote: y = 0 • domain: (–∞, ∞) • range: (0, ∞) • x-intercept: none • y-intercept: (0, 1) • increasing if b > 1 • decreasing if b < 1 Figure 3 compares the graphs of exponential growth and decay functions. f(x) f(x) f (x) = bx b > 1 f (x) = bx 0 < b < 1 (1, b) (0, 1) x Figure 3 (0, 1) (1, b) x How To… Given an exponential function of the form f (x) = b x, graph the function. 1. Create a table of points. 2. Plot at least 3 point from the table, including the y-intercept (0, 1). 3. Draw a smooth curve through the points
. 4. State the domain, (−∞, ∞), the range, (0, ∞), and the horizontal asymptote, y = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 481 Example 1 Sketching the Graph of an Exponential Function of the Form f (x) = b x Sketch a graph of f (x) = 0.25x. State the domain, range, and asymptote. Solution Before graphing, identify the behavior and create a table of points for the graph. • Since b = 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = 0. • Create a table of points as in Table 3. x f (x) = 0.25x −3 64 −2 16 −1 4 Table 3 0 1 1 2 3 0.25 0.0625 0.015625 • Plot the y-intercept, (0, 1), along with two other points. We can use (−1, 4) and (1, 0.25). Draw a smooth curve connecting the points as in Figure 4. f(x) f(x) = 0.25x (−1, 4) 6 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 –6 (0, 1) (1, 0.25) 21 3 4 5 x The domain is (−∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. Figure 4 Try It #1 Sketch the graph of f (x) = 4x. State the domain, range, and asymptote. Graphing Transformations of Exponential Functions Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function f (x) = b x without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied. Graphing a Vertical Shift The first transformation occurs when we add a constant d to the parent function f (x) = b x, giving us a vertical shift d units in the same direction as the sign. For example, if we begin by graphing a parent function, f (x) = 2x, we can then graph two vertical shifts alongside it, using d = 3: the upward shift, g(x) = 2x + 3 and the downward shift, h(x) = 2x − 3. Both vertical shifts are shown in Figure 5. y 12 10 8 6 4 2 –1– 2 –4 –6 g(x) = 2 x + 3 f (x) = 2 x h(x) = 2 x − 3 –6 –5 –4 –3 –2 21 3 Figure 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 2 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Observe the results of shifting f (x) = 2x vertically: • The domain, (−∞, ∞) remains unchanged. • When the function is shifted up 3 units to g(x) = 2x + 3: ◦ The y-intercept shifts up 3 units to (0, 4). ◦ The asymptote shifts up 3 units to y = 3. ◦ The range becomes (3, ∞). • When the function is shifted down 3 units to h(x) = 2x − 3: ◦ The y-intercept shifts down 3 units to (0, −2). ◦ The asymptote also shifts down 3 units to y = −3. ◦ The range becomes (−3, ∞). Graphing a Horizontal Shift The next transformation occurs when we add a constant c to the input of the parent function f (x) = b x, giving us a horizontal shift c units in the opposite direction of the sign. For example, if we begin by graphing the parent function f (x) = 2x, we can then graph two horizontal shifts alongside it, using c = 3: the shift left, g(x) = 2x + 3, and the shift right, h (x) = 2x − 3. Both horizontal shifts are shown in Figure 6. y 10 8 6 4 2 g(x) = 2 x + 3 f (x) = 2 x h(x 10 –10 –8 –6 –4 –2–2 42 6 8 Figure 6 Observe the results of shifting f (x) = 2x horizontally: • The domain, (−∞, ∞), remains unchanged. • The asymptote, y = 0, remains unchanged. • The y-intercept shifts such that: ◦ When the function is shifted left 3 units to g(x) = 2x + 3, the y-intercept becomes (0, 8). This is because 2x + 3 = (8)2x, so the initial value of the function is 8. 1 1 ◦ When the function is shifted right 3 units to h(x) = 2x − 3, the y-intercept becomes ( 0, ) 2x, ) . Again, see that 2x − 3 = ( _ _ 8 8 1 _ . so the initial value of the function is 8 shifts of the parent function f (x) = b x For any constants c and d, the function f (x) = b x + c + d shifts the parent function f (x) = b x • vertically d units, in the same direction of the sign of d. • horizontally c units, in the opposite direction of the sign of c. • The y-intercept becomes (0, bc + d). • The horizontal asymptote becomes y = d. • The range becomes (d, ∞). • The domain, (−∞, ∞), remains unchanged. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 483 How To… Given an exponential function with the form f (x) = b x + c + d, graph the translation. 1. Draw the horizontal asymptote y = d. 2. Identify the shift as (−c, d). Shift the graph of f (x) = b x left c units if c is positive, and right c units if c is negative. 3. Shift the graph of f (x) = b x up d units if d is positive, and down d units if d is negative. 4. State the domain, (−∞, ∞), the range, (d, ∞), and the horizontal asymptote y = d. Example 2 Graphing a Shift of an Exponential Function Graph f (x) = 2x + 1 − 3. State the domain, range, and asymptote. Solution We have an exponential equation of the form f (x) = b x + c + d, with b = 2, c = 1, and d = −3. Draw the horizontal asymptote y = d, so draw y = −3. Identify the shift as (−c, d), so the shift is (−1, −3). Shift the graph of f (x) = b x left 1 units and down 3 units. f (x) f (x) = 2 x + 1 − 3 10 8 6 4 2 (0, −1) –6 –5 –4 –3 –2 (−1, −2) –1–2 –4 –6 –8 –10 (1, 1) 3 21 4 5 6 x y = −3 The domain is (−∞, ∞); the range is (−3, ∞); the horizontal asymptote is y = −3. Figure 7 Try It #2 Graph f (x) = 2x − 1 + 3. State domain, range, and asymptote. How To… Given an equation of the form f (x) = b x + c + d for x, use a graphing calculator to approximate the solution. 1. Press [Y=]. Enter the given exponential equation in the line headed “Y1=”. 2. Enter the given value for f (x) in the line headed “Y2=”. 3. Press [WINDOW]. Adjust the y-axis so that it includes the value entered for “Y2=”. 4. Press [GRAPH] to observe the graph of the exponential function along with the line for the specified value of f (x). 5. To fi nd the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x for the indicated value of the function. Example 3 Approximating the Solution of an Exponential Equation Solve 42 = 1.2(5)x + 2.8 graphically. Round to the nearest thousandth. Solution Press [Y=] and enter 1.2(5)x + 2.8 next to Y1=. Then enter 42 next to Y2=. For a window, use the values −3 to 3 for x and −5 to 55 for y. Press [GRAPH]. The graphs should intersect somewhere near x = 2. For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth, x ≈ 2.166. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #3 Solve 4 = 7.85(1.15)x − 2.27 graphically. Round to the nearest thousandth. Graphing a Stretch or Compression While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function f (x) = b x by a constant ∣ a ∣ > 0. For example, if we begin by graphing the parent function f (x) = 2x, we can then graph the stretch, using a = 3, to get g(x) = 3(2)x as shown on the 1 1 _ _ , to get h(x) = left in Figure 8, and the compression, using a = (2)x as shown on the right in Figure 8. 3 3 Vertical stretch y g(x) = 3(2)x f (x) = 2x Vertical compression y f (x) = 2x h(x) = (2)x 1 3 10 8 6 4 2 –1–2 –4 (a) –5 –4 –3 –2 21 3 4 5 y = 0 x –5 –4 –3 –2 10 8 6 4 2 –1–2 –4 (b) 21 3 4 5 x y = 0 Figure 8 (a) g(x) = 3(2)x stretches the graph of f (x) = 2x vertically by a factor of 3. 1 1 __ __ (2)x compresses the graph of f (x) = 2x vertically by a factor of . (b) h(x) = 3 3 stretches and compressions of the parent function f ( x ) = b x For any factor a > 0, the function f (x) = a(b)x • is stretched vertically by a factor of a if ∣ a ∣ > 1. • is compressed vertically by a factor of a if ∣ a ∣ < 1. • has a y-intercept of (0, a). • has a horizontal asymptote at y = 0, a range of (0, ∞), and a domain of (−∞, ∞), which are unchanged from the parent function. Example 4 Graphing the Stretch of an Exponential Function 1 ) Sketch a graph of f (x) = 4 ( _ 2 x . State the domain, range, and asymptote. Solution Before graphing, identify the behavior and key points on the graph. 1 _ • Since b = is between zero and one, the left tail of the graph will increase without bound as x decreases, and the 2 right tail will approach the x-axis as x increases. • Since a = 4, the graph of f (x) = ( 1 ) _ 2 • Create a table of points as shown in Table 4. x will be stretched by a factor of 4. x f (x) = 4 ( 1 __ 2 ) x −3 32 −2 16 −1 8 Table 4 0 4 1 2 2 1 3 0.5 • Plot the y-intercept, (0, 4), along with two other points. We can use (−1, 8) and (1, 2). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 485 Draw a smooth curve connecting the points, as shown in Figure 9. (–1, 8) f(x) 10 8 6 4 2 (0, 4) (1, 2) –6 –5 –4 –3 –2 –1–2 –4 321 4 5 6 Figure 9 y = 0 x f (x) = 4 1 2 x The domain is (−∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. Try It #4 1 _ Sketch the graph of f (x) = (4)x. State the domain, range, and asymptote. 2 Graphing Reﬂections In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x-axis or the y-axis. When we multiply the parent function f (x) = b x by −1, we get a reflection about the x-axis. When we multiply the input by −1, we get a reflection about the y-axis. For example, if we begin by graphing the parent function f (x) = 2x, we can then
graph the two reflections alongside it. The reflection about the x-axis, g(x) = −2x, is shown on the left side of Figure 10, and the reflection about the y-axis h(x) = 2−x, is shown on the right side of Figure 10. Reflection about the x-axis y Reflection about the y-axis y –5 –4 –3 –2 10 8 6 4 2 – –2 1 –4 –6 –8 –10 f (x) = 2 x h(x) = 2 −x y = 0 21 3 4 5 x –5 –4 –3 –2 g(x) = −2 x f (x) = 2 x y = 0 21 3 4 5 x 10 8 6 4 2 –1–2 –4 –6 –8 –10 Figure 10 (a) g(x) = −2x reﬂects the graph of f (x) = 2x about the x-axis. (b) g(x) = 2−x reﬂects the graph of f (x) = 2x about the y-axis. reflections of the parent function f (x) = b x The function f (x) = −b x • reflects the parent function f (x) = b x about the x-axis. • has a y-intercept of (0, −1). • has a range of (−∞, 0). • has a horizontal asymptote at y = 0 and domain of (−∞, ∞), which are unchanged from the parent function. The function f (x) = b−x • reflects the parent function f (x) = b x about the y-axis. • has a y-intercept of (0, 1), a horizontal asymptote at y = 0, a range of (0, ∞), and a domain of (−∞, ∞), which are unchanged from the parent function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 5 Writing and Graphing the Reﬂection of an Exponential Function 1 ) Find and graph the equation for a function, g (x), that reflects f (x) = ( _ 4 and asymptote. 1 Solution Since we want to reflect the parent function f (x) = ( ) _ 4 g (x) = − ( 1 ) _ 4 . Ne xt we create a table of points as in Table 5. x x x about the x-axis. State its domain, range, about the x-axis, we multiply f (x) by −1 to get, x −3 −2 1 ) g(x) = − ( __ 4 x −64 −16 −1 −4 Table 5 0 1 2 3 −1 −0.25 −0.0625 −0.0156 Plot the y-intercept, (0, −1), along with two other points. We can use (−1, −4) and (1, −0.25). Draw a smooth curve connecting the points: g(x) 10 8 6 4 2 –1–2 –4 –6 –8 –10 y = 0 –2 –5 –3 –4 (−1, −4) (1, −0.25) 5 3 21 4 (0, −1) x x g(x) = − 1 4 The domain is (−∞, ∞); the range is (−∞, 0); the horizontal asymptote is y = 0. Figure 11 Try It #5 Find and graph the equation for a function, g(x), that reflects f (x) = 1.25x about the y-axis. State its domain, range, and asymptote. Summarizing Translations of the Exponential Function Now that we have worked with each type of translation for the exponential function, we can summarize them in Table 6 to arrive at the general equation for translating exponential functions. Translations of the Parent Function f (x) = b x Translation Form Shift • Horizontally c units to the left • Vertically d units up Stretch and Compress • Stretch if \| a \| > 1 • Compression if 0 < \| a \| < 1 Reflect about the x-axis Reflect about the y-axis General equation for all translations Table 6 f (xx) = ab x f (x) = −b x 1 ) f (x) = b−x = ( _ b x f (x) = ab x + c + d Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 487 translations of exponential functions A translation of an exponential function has the form f (x) = ab x + c + d Where the parent function, y = b x, b > 1, is • shifted horizontally c units to the left. • stretched vertically by a factor of ∣ a ∣ if ∣ a ∣ > 0. • compressed vertically by a factor of ∣ a ∣ if 0 < ∣ a ∣ < 1. • shifted vertically d units. • reflected about the x-axis when a < 0. Note the order of the shifts, transformations, and reflections follow the order of operations. Example 6 Writing a Function from a Description Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range. • f (x) = e x is vertically stretched by a factor of 2, reflected across the y-axis, and then shifted up 4 units. Solution We want to find an equation of the general form f (x) = ab x + c + d. We use the description provided to find a, b, c, and d. • We are given the parent function f (x) = e x, so b = e. • The function is stretched by a factor of 2, so a = 2. • The function is reflected about the y-axis. We replace x with −x to get: e−x. • The graph is shifted vertically 4 units, so d = 4. Substituting in the general form we get, f (x) = ab x + c + d = 2e−x + 0 + 4 = 2e−x + 4 The domain is (−∞, ∞); the range is (4, ∞); the horizontal asymptote is y = 4. Try It #6 Write the equation for function described below. Give the horizontal asymptote, the domain, and the range. 1 _ • f (x) = e x is compressed vertically by a factor of , reflected across the x-axis and then shifted down 2 units. 3 Access this online resource for additional instruction and practice with graphing exponential functions. • Graph Exponential Functions (http://openstaxcollege.org/l/graphexpfunc) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 SECTION EXERCISES 489 Figure 13 fx=ab x y B C D A E F x Figure 13 19. b 21. a 20. b 22. a flx 23. fx= x 24. fx=x− 25. fx=−x+ f x=x 26. fx=−x 27. hx=x+ 28. fx=x− 29. fx=−x− ) 30. fx=( x − 31. fx=−x+ f x=xTh 32. fx 35. fx 33. fx 36. fxx 34. fxts left 37. fxy y =x 38. y –– – – – – – – – – 39. x – – – – y –– – – – – 40. x – – – – y –– – – – – x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 0 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS , fi 41. y 42. x – – – – – – – – –– – – – – NUMERIC y –– – – – – x x 43. gx= x−g 44. f x=x−−f 45. hx=− ) ( x +h− TECHNOLOGY f x=ab x +d 46. −=− ( ) 47. = ) ( ) 49. =( 48. =x+ 50. −=−x++ − x− −x x EXTENSIONS 51. fx=bx x . Th gx> b 52. fx=x gx=x−hx=( bn ) b xn b x( b> )x. Th 53. 54. REAL-WORLD APPLICATIONS 55. =+/) = 56. a. b. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 LOGARITHMIC FUNCTIONS 491 LEARNING OBJECTIVES In this section, you will: • efine a logarithmic function emphasiing it as an inerse of an eponential function. • Conert a logarithm to eponential form. • Conert from eponential to logarithmic form. • aluate logarithms. • aluate logarithms using natural and common logarithms. 6.3 LOGARITHMIC FUNCTIONS Figure 1 Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce) In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes[19]. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[20] like those shown in Figure 1. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[21] whereas the Japanese earthquake registered a 9.0.[22] The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is 108 − 4 = 104 = 10,000 times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends. Converting from Logarithmic to Exponential Form In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is 10 x = 500, where x represents the difference in magnitudes on the Richter Scale. How would we solve for x? We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve 10 x = 500. We know that 102 = 100 and 103 = 1000, so it is clear that x must be some value between 2 and 3, since y = 10x is increasing. We can examine a graph, as in Figure 2, to better estimate the solution. y 1,000 800 600 400 200 –3 –2 –1 –200 1 2 3 Figure 2 y = 10x y = 500 x 19 http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/#summary. Accessed 3/4/2013. 20 http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc001xgp/#summary. Accessed 3/4/2013. 21 http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/. Accessed 3/4/2013. 22 http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc001xgp/#details. Accessed 3/4/2013. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 2 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure 2 passes the horizontal line test. The exponential function y = b x is one-to-one, so its inverse, x = b y is also a function. As is the case with all inverse functions, we simply interchange x and y and solve for y to find the inverse function. To represent y as a function of x, we use a logarithmic function of the form y = logb(x). The base b logarithm of a number is the exponent by which we must raise b to get that number. We read a logarithmic expression as, “The logarithm with base b of x is equal to y,” or, simplified, “log base b of x is y.” We can also say, “b raised to the power of y is x,” because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since 25 = 32, we can write log2 32 = 5. We read this as “log base 2 of 32 is 5.” We can express the relationship between logarithmic form and its corresponding exponential form as follows: Note that the base b is always positive. logb(x) = y ⇔ b y = x, b > 0, b ≠ 1 = logb(x) = y to Think b to the y = x Because logarithm is a function, it is most correctly written as logb(x), using parentheses to denote function evaluation, just as we would with f (x). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without
parentheses, as logb x. Note that many calculators require parentheses around the x. We can illustrate the notation of logarithms as follows: = logb(c) = a means ba = c to Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means y = logb (x) and y = b x are inverse functions. definition of the logarithmic function A logarithm base b of a positive number x satisfies the following definition. For x > 0, b > 0, b ≠ 1, where, • we read logb (x) as, “the logarithm with base b of x” or the “log base b of x.” • the logarithm y is the exponent to which b must be raised to get x. y = logb(x) is equivalent to b y = x Also, since the logarithmic and exponential functions switch the x and y values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore, • the domain of the logarithm function with base b is (0, ∞). • the range of the logarithm function with base b is ( −∞, ∞). Q & A… Can we take the logarithm of a negative number? No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number. How To… Given an equation in logarithmic form logb(x) = y, convert it to exponential form. 1. Examine the equation y = logb(x) and identify b, y, and x. 2. Rewrite logb(x) = y as b y = x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 LOGARITHMIC FUNCTIONS 493 Example 1 Converting from Logarithmic Form to Exponential Form Write the following logarithmic equations in exponential form. a. log6. log3(9) = 2 — Solution First, identify the values of b, y, and x. Then, write the equation in the form b y = x. 6 . Therefore, the equation log6( √ a. log6( √ 1 _ Here, b = 6, y = , and x = √ 2 1 __ 6 = √ 6 . 2 Here, b = 3, y = 2, and x = 9. Therefore, the equation log3(9) = 2 is equivalent to 32 = 9. 1 6 ) = _ is equivalent to 2 b. log3(9) = 2 1 6 ) = _ 2 — — — Try It #1 Write the following logarithmic equations in exponential form. a. log10(1,000,000) = 6 b. log5(25) = 2 Converting from Exponential to Logarithmic Form To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base b, exponent x, and output y. Then we write x = logb(y). Example 2 Converting from Exponential Form to Logarithmic Form Write the following exponential equations in logarithmic form. 1 ______ 10,000 b. 52 = 25 c. 10−4 = a. 23 = 8 Solution First, identify the values of b, y, and x. Then, write the equation in the form x = logb(y). a. 23 = 8 Here, b = 2, x = 3, and y = 8. Therefore, the equation 23 = 8 is equivalent to log2(8) = 3. b. 52 = 25 Here, b = 5, x = 2, and y = 25. Therefore, the equation 52 = 25 is equivalent to log5(25) = 2. c. 10−4 = 1 ______ 10,000 Here, b = 10, x = −4, and y = ) = −4. log10 ( 1 _ 10,000 1 _ 10,000 . Therefore, the equation 10−4 = is equivalent to 1 _ 10,000 Try It #20 Write the following exponential equations in logarithmic form. a. 32 = 9 b. 53 = 125 1 __ c. 2−1 = 2 Evaluating Logarithms Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider log2(8). We ask, “To what exponent must 2 be raised in order to get 8?” Because we already know 23 = 8, it follows that log2(8) = 3. Now consider solving log7(49) and log3(27) mentally. • We ask, “To what exponent must 7 be raised in order to get 49?” We know 72 = 49. Therefore, log7(49) = 2 • We ask, “To what exponent must 3 be raised in order to get 27?” We know 33 = 27. Therefore, log3(27) = 3 Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate log 2 _ 4 ) mentally. 3 ( _ 9 2 2 4 ) ?” We know 22 = 4 and 32 = 9, so ( _ _ _ • We ask, “To what exponent must be raised in order to get 3 9 3 4 ) = 2. 3 ( _ 9 Therefore, log 2 _ 2 4 _ = . 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How To… Given a logarithm of the form y = logb(x), evaluate it mentally. 1. Rewrite the argument x as a power of b : b y = x. 2. Use previous knowledge of powers of b identify y by asking, “To what exponent should b be raised in order to get x?” Example 3 Solving Logarithms Mentally Solve y = log4(64) without using a calculator. Solution First we rewrite the logarithm in exponential form: 4y = 64. Ne xt, we ask, “To what exponent must 4 be raised in order to get 64?” We know 43 = 64 therefore, log4(64) = 3. Try It #3 Solve y = log121(11) without using a calculator. Example 4 Evaluating the Logarithm of a Reciprocal Evaluate y = log3 ( ) without using a calculator. 1 _ 27 y Solution First we rewrite the logarithm in exponential form: 3 raised in order to get ?” 1 _ 27 We know 33 = 27, but what must we do to get the reciprocal, We use this information to write 1 _ 27 = 1 _ 27 . Ne xt, we ask, “To what exponent must 3 be ? Recall from working with exponents that b−a = 1 _ ba . Therefore, log3 ( 1 _ 27 ) = −3. 3−3 = 1 __ 33 1 __ 27 = Try It #4 Evaluate y = log2 ( 1 _ 32 ) without using a calculator. Using Common Logarithms Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression log(x) means log10(x). We call a base-10 logarithm a common logarithm. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms. definition of the common logarithm A common logarithm is a logarithm with base 10. We write log10(x) simply as log(x). The common logarithm of a positive number x satisfies the following definition. For x > 0, y = log(x) is equivalent to 10 y = x We read log(x) as, “the logarithm with base 10 of x” or “log base 10 of x.” The logarithm y is the exponent to which 10 must be raised to get x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 LOGARITHMIC FUNCTIONS 495 How To… Given a common logarithm of the form y = log(x), evaluate it mentally. 1. Rewrite the argument x as a power of 10: 10 y = x. 2. Use previous knowledge of powers of 10 to identify y by asking, “To what exponent must 10 be raised in order to get x?” Example 5 Finding the Value of a Common Logarithm Mentally Evaluate y = log(1,000) without using a calculator. Solution First we rewrite the logarithm in exponential form: 10y = 1,000. Ne xt, we ask, “To what exponent must 10 be raised in order to get 1,000?” We know 103 = 1,000 therefore, log(1,000) = 3. Try It #5 Evaluate y = log(1,000,000). How To… Given a common logarithm with the form y = log(x), evaluate it using a calculator. 1. Press [LOG]. 2. Enter the value given for x, followed by [ ) ]. 3. Press [ENTER]. Example 6 Finding the Value of a Common Logarithm Using a Calculator Evaluate y = log(321) to four decimal places using a calculator. Solution • Press [LOG]. • Enter 321, followed by [ ) ]. • Press [ENTER]. Rounding to four decimal places, log(321) ≈ 2.5065. Analysis Note that 102 = 100 and that 103 = 1000. Since 321 is between 100 and 1000, we know that log(321) must be between log(100) and log(1000). This gives us the following: 100 < 321 < 1000 2 < 2.5065 < 3 Try It #6 Evaluate y = log(123) to four decimal places using a calculator. Example 7 Rewriting and Solving a Real-World Exponential Model The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation 10x = 500 represents this situation, where x is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes? Solution We begin by rewriting the exponential equation in logarithmic form. 10x = 500 log(500) = x Use the definition of the common log. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Ne xt we evaluate the logarithm using a calculator: • Press [LOG]. • Enter 500, followed by [ ) ]. • Press [ENTER]. • To the nearest thousandth, log(500) ≈ 2.699. The difference in magnitudes was about 2.699. Try It #7 The amount of energy released from one earthquake was 8,500 times greater than the amount of energy released from another. The equation 10x = 8500 represents this situation, where x is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes? Using Natural Logarithms The most frequently used base for logarithms is e. Base e logarithms are important in calculus and some scientific applications; they are called natural logarithms. The base e logarithm, loge(x), has its own notation, ln(x). Most values of ln(x) can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, ln1 = 0. For other natural logarithms, we can use the ln key that can be found on most scientific calculators. We can also find the natural logarithm of any power of e using the inverse property of logarithms. definition of the natural logarithm A natural logarithm is a logarithm with base e. We write loge(x) simply as ln(x). The natural logarithm of a positive number x satisfies the following definition. For x > 0, y = ln(x) is equivalent to e y = x We read ln(x) as, “the logarithm with base e of x” or “the natural logarithm of x.” The logarithm y is the exponent to which e must be raised to get x. Since the functions y = e and y = ln(x) are inverse functions, ln(e x) = x for all x and e = x for x > 0. How To… Given a natural logarithm with the form y = ln(x), evaluate it using a calculator. 1. Press [LN]. 2. Enter the value given
for x, followed by [ ) ]. 3. Press [ENTER]. Example 8 Evaluating a Natural Logarithm Using a Calculator Evaluate y = ln(500) to four decimal places using a calculator. Solution • Press [LN]. • Enter 500, followed by [ ) ]. • Press [ENTER]. Rounding to four decimal places, ln(500) ≈ 6.2146 Try It #8 Evaluate ln(−500). Access this online resource for additional instruction and practice with logarithms. • Introduction to Logarithms (http://openstaxcollege.org/l/intrologarithms) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 SECTION EXERCISES 497 6.3 SECTION EXERCISES VERBAL 1. b b y= x bx = y b > b ≠ 2. f x=bx gx=b x 3. b x = y x 5. b n diff 4. b n diff ALGEBRAIC 6. q=m 10. yx=− 14. v=t 7. b=c 11. a=b 15. w=n 8. y=x 9. x=y 12. y=x 13. =a 16. x=y 18. m−=n 22. ( ) 20. x − =y 21. n= 17. c d=k m =n 24. a=b 25. e k=h 19. x=y 23. y x= x 26. x= 28. x= 29. x= 32. x= 33. x=− 30. x= 34. x= 27. x=− 31. x= 35. x= fi 36. 40. e − NUMERIC 37. 41. e + 38. 39. e b 42. 43. 44. 46. 47. 48. 50. p 54. () 4p 51. ✓ ◆ 55. 52. ✓ ◆ 45. 49. 53. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 8 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 56. ) ( 57. — √ 58. )+ ( 59. 60. 61. 62. + 63. − 64. ( e ) 65. 66. e−− 67. ( e ) TECHNOLOGY 68. 69. 70. ) ( 71. √ — 72. √ — EXTENSIONS 73. x =f(x=x x = 74. f(x=fx=x x 75. x x= 76. =− ) ( 77. e = EI− REAL-WORLD APPLICATIONS 78. ThEI 79. f ) EI=( t ft 80. ThI I =M−M I M Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 499 LEARNING OBJECTIVES In this section, you will: • dentify the domain of a logarithmic function. • Graph logarithmic functions including using transformations. 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS In Graphs of Exponential Functions, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the cause for an effect. To illustrate, suppose we invest $2,500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year t can be found with the equation A = 2500e0.05t. But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1 shows this point on the logarithmic graph. Logarithmic Model Showing Years as a Function of the Balance in the Account s r a e Y 20 18 16 14 12 10 8 6 4 2 0 The balance reaches $5,000 near year 14 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 5,000 5,500 6,000 Account balance Figure 1 In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions. Finding the Domain of a Logarithmic Function Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined. Recall that the exponential function is defined as y = b x for any real number x and constant b > 0, b ≠ 1, where • The domain of y is (−∞, ∞). • The range of y is (0, ∞). In the last section we learned that the logarithmic function y = logb(x) is the inverse of the exponential function y = b x. So, as inverse functions: • The domain of y = logb(x) is the range of y = b x : (0, ∞). • The range of y = logb(x) is the domain of y = b x : (−∞, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 500 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Transformations of the parent function y = logb(x) behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, stretches, compressions, and reflections—to the parent function without loss of shape. In Graphs of Exponential Functions we saw that certain transformations can change the range of y = b x. Similarly, applying transformations to the parent function y = logb(x) can change the domain. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. That is, the argument of the logarithmic function must be greater than zero. For example, consider f (x) = log4(2x − 3). This function is defined for any values of x such that the argument, in this case 2x − 3, is greater than zero. To find the domain, we set up an inequality and solve for x : 2x − 3 > 0 Show the argument greater than zero. 2x > 3 Add 3. x > 1.5 Divide by 2. In interval notation, the domain of f (x) = log4(2x − 3) is (1.5, ∞). How To… Given a logarithmic function, identify the domain. 1. Set up an inequality showing the argument greater than zero. 2. Solve for x. 3. Write the domain in interval notation. Example 1 Identifying the Domain of a Logarithmic Shift What is the domain of f (x) = log2(x + 3)? Solution The logarithmic function is defined only when the input is positive, so this function is defined when x + 3 > 0. Solving this inequality, x + 3 > 0 The input must be positive. x > −3 The domain of f (x) = log2(x + 3) is (−3, ∞). Subtract 3. Try It #1 What is the domain of f (x) = log5(x − 2) + 1? Example 2 Identifying the Domain of a Logarithmic Shift and Reﬂection What is the domain of f (x) = log(5 − 2x)? Solution The logarithmic function is defined only when the input is positive, so this function is defined when 5 − 2x > 0. Solving this inequality, 5 − 2x > 0 The input must be positive. −2x > −5 Subtract 5. 5 __ x < 2 Divide by −2 and switch the inequality. 5 ) . The domain of f (x) = log(5 − 2x) is ( –∞, _ 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 501 Try It #2 What is the domain of f (x) = log(x − 5) + 2? Graphing Logarithmic Functions Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function y = logb(x) along with all its transformations: shifts, stretches, compressions, and reflections. We begin with the parent function y = logb(x). Because every logarithmic function of this form is the inverse of an exponential function with the form y = b x, their graphs will be reflections of each other across the line y = x. To illustrate this, we can observe the relationship between the input and output values of y = 2x and its equivalent x = log2(y) in Table 1. x 2x = y log2(y) = x −3 1 _ 8 −3 −2 1 _ 4 −2 0 1 0 −1 1 _ 2 −1 Table Using the inputs and outputs from Table 1, we can build another table to observe the relationship between points on the graphs of the inverse functions f (x) = 2x and g(x) = log2(x). See Table 2. f (x) = 2x g(x) = log2(x) 1 ) ( −3, _ 8 1 , −3 ) ( _ 8 1 ) ( −2, _ 4 1 , −2 ) ( _ 4 1 ) ( −1, _ 2 1 , −1 ) ( _ 2 (0, 1) (1, 2) (2, 4) (3, 8) (1, 0) (2, 1) (4, 2) (8, 3) Table 2 As we’d expect, the x- and y-coordinates are reversed for the inverse functions. Figure 2 shows the graph of f and g. f (x) = 2 x –5 –4 –3 –2 y = x g(x) = log2 (x) 321 1 –1 –2 –3 –4 –5 Figure 2 Notice that the graphs of f (x) = 2x and g(x) = log2(x) are reﬂections about the line y = x. Observe the following from the graph: • f (x) = 2x has a y-intercept at (0, 1) and g(x) = log2(x) has an x-intercept at (1, 0). • The domain of f (x) = 2x, (−∞, ∞), is the same as the range of g(x) = log2(x). • The range of f (x) = 2x, (0, ∞), is the same as the domain of g(x) = log2(x). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 502 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS characteristics of the graph of the parent function, f (x) = logb(x) For any real number x and constant b > 0, b ≠ 1, we can see the following characteristics in the graph of f (x) = logb(x): • one-to-one function • vertical asymptote: x = 0 • domain: (0, ∞) • range: (−∞, ∞) • x-intercept: (1, 0) f (x) = logb(x) 0 < b < 1 f(x) = logb(x) b > 1 (b, 1) (b, 1) (1, 0) f (x) f(x) x x (1, 0) x = 0 and key point (b, 1) • y-intercept: none • increasing if b > 1 • decreasing if 0 < b < 1 See Figure 3. Figure 4 shows how changing the base b in f (x) = logb(x) can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function ln(x) has base e ≈ 2.718.) x = 0 Figure 3 x = 0 642 8 10 12 log2(x) ln(x) log(x) x y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 –12 –10 –8 –6 –4 Figure 4 The graphs of three logarithmic functions with different bases, all greater than 1. How To… Given a logarithmic function with the form f (x) = logb(x), graph the function. 1. Draw and label the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 3. Plot the key point (b, 1). 4. Draw a smooth curve through the points. 5. State the domain, (0, ∞), the range, (−∞, ∞), and the vertical asymptote, x = 0. Example 3 Graphing a Logarithmic Function with the Form f ( x) = logb( x). Graph f (x) = log5(x). State the domain, range, and asymptote. Solution Before graphing, identify the behavior and key points for the graph. • Since b = 5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote x = 0, and the right tail will increase slowly without bound. • The x-intercept is (1, 0). • The key point (5, 1) is on the graph. • We d
raw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see Figure 5). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 503 –10 –8 –6 –4 f (x) x = 0 (5, 1) 642 8 10 (1, 0) 5 4 3 2 1 –2 –1 –2 –3 –4 –5 Figure 5 f (x) = log5(x) x The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. Try It #3 Graph f (x) = log 1 _ 5 (x). State the domain, range, and asymptote. Graphing Transformations of Logarithmic Functions As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function y = logb(x) without loss of shape. Graphing a Horizontal Shift of f (x ) = logb(x ) When a constant c is added to the input of the parent function f (x) = logb(x), the result is a horizontal shift c units in the opposite direction of the sign on c. To visualize horizontal shifts, we can observe the general graph of the parent function f (x) = logb(x) and for c > 0 alongside the shift left, g(x ) = logb(x + c), and the shift right, h(x) = logb(x − c). See Figure 6. Shift left g (x) = logb(x + c) x = –c y x = 0 g(x) = logb(x + c) y x = 0 Shift right h(x) = logb(x − c) x = c f (x) = logb(x) (b − c, 1) (1 − c, 0) f (x) = logb(x) x (b, 1) (1, 0) (b, 1) (b + c, 1) (1, 0) (1 + c, 0) x h(x) = logb(x − c) • The asymptote changes to x = −c. • The domain changes to (−c, ∞). • The range remains (−∞, ∞). • The asymptote changes to x = c. • The domain changes to (c, ∞). • The range remains (−∞, ∞). Figure 6 horizontal shifts of the parent function y = logb(x) For any constant c, the function f (x) = logb (x + c) • shifts the parent function y = logb(x) left c units if c > 0. • shifts the parent function y = logb(x) right c units if c < 0. • has the vertical asymptote x = −c. • has domain (−c, ∞). • has range (−∞, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 504 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How To… Given a logarithmic function with the form f (x) = logb(x + c), graph the translation. 1. Identify the horizontal shift: a. If c > 0, shift the graph of f (x) = logb(x) left c units. b. If c < 0, shift the graph of f (x) = logb(x) right c units. 2. Draw the vertical asymptote x = −c. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting c from the x coordinate. 4. Label the three points. 5. The domain is (−c, ∞), the range is (−∞, ∞), and the vertical asymptote is x = −c. Example 4 Graphing a Horizontal Shift of the Parent Function y = logb( x) Sketch the horizontal shift f (x) = log3(x − 2) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = log3(x − 2), we notice x + (−2) = x − 2. Thus c = −2, so c < 0. This means we will shift the function f (x) = log3(x) right 2 units. The vertical asymptote is x = −(−2) or x = 2. 1 , −1 ) , (1, 0), and (3, 1). Consider the three key points from the parent function, ( _ 3 The new coordinates are found by adding 2 to the x coordinates. 7 , −1 ) , (3, 0), and (5, 1). Label the points ( _ 3 The domain is (2, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 2. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 (1, 0) y = log3(x) f (x) = log3(x − 2) x (3, 1) (5, 1) 321 4 5 6 7 8 9 (3, 0) x = 2 x = 0 Figure 7 Try It #4 Sketch a graph of f (x) = log3(x + 4) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote. Graphing a Vertical Shift of y = logb(x ) When a constant d is added to the parent function f (x) = logb(x), the result is a vertical shift d units in the direction of the sign on d. To visualize vertical shifts, we can observe the general graph of the parent function f (x) = logb(x) alongside the shift up, g (x) = logb(x) + d and the shift down, h(x) = logb(x) − d. See Figure 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 505 Shift up g (x) = logb(x) + d Shift down h(x) = logb(x) − d y x = 0 (b1 − d, 1) −d, 0) (b, 1) (b (1, 0) g(x) = logb(x) + d f (x) = logb(x) x y x = 0 f (x) = logb(x) (b, 1) (b1+d, 1) h(x) = logb(x) − d (1, 0) (bd, 0) x • The asymptote remains x = 0. • The domain remains to (0, ∞). • The range remains (−∞, ∞). • The asymptote remains x = 0. • The domain remains to (0, ∞). • The range remains (−∞, ∞). Figure 8 vertical shifts of the parent function y = logb(x) For any constant d, the function f (x) = logb(x) + d • shifts the parent function y = logb(x) up d units if d > 0. • shifts the parent function y = logb(x) down d units if d < 0. • has the vertical asymptote x = 0. • has domain (0, ∞). • has range (−∞, ∞). How To… Given a logarithmic function with the form f (x) = logb(x) + d, graph the translation. 1. Identify the vertical shift: a. If d > 0, shift the graph of f (x) = logb(x) up d units. b. If d < 0, shift the graph of f (x) = logb(x) down d units. 2. Draw the vertical asymptote x = 0. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by adding d to the y coordinate. 4. Label the three points. 5. The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. Example 5 Graphing a Vertical Shift of the Parent Function y = logb(x) Sketch a graph of f (x) = log3(x) − 2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = log3(x) − 2, we will notice d = −2. Thus d < 0. This means we will shift the function f (x) = log3(x) down 2 units. The vertical asymptote is x = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 506 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 1 , −1 ) , (1, 0), and (3, 1). Consider the three key points from the parent function, ( _ 3 The new coordinates are found by subtracting 2 from the y coordinates. 1 Label the points ( , −3 ) , (1, −2), and (3, −1). _ 3 The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. y 5 4 3 2 1 (1, 0) –1–1 –2 –3 –4 –5 (3, 1) 321 4 5 6 7 8 9 (3, −1) (1, −2) y = log3(x) x f (x) = log3(x − 2) x = 0 Figure 9 The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. Try It #5 Sketch a graph of f (x) = log2(x) + 2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Graphing Stretches and Compressions of y = logb(x ) When the parent function f (x) = logb(x) is multiplied by a constant a > 0, the result is a vertical stretch or compression of the original graph. To visualize stretches and compressions, we set a > 1 and observe the general graph of the parent 1 _ a logb(x). function f (x) = logb(x) alongside the vertical stretch, g (x) = alogb(x) and the vertical compression, h(x) = See Figure 10. Vertical Stretch g (x) = alogb(x), a > 1 y x = 0 g(x) = alogb(x) 1/a (b , 1) f(x) = logb(x) (b, 1) (1, 0) x Vertical Compression 1 _ a logb(x), a > 1 h(x) = y x = 0 (b, 1) (1, 0) f(x) = logb(x) 1 a h(x) = logb(x) (ba, 1) x • The asymptote remains x = 0. • The x-intercept remains (1, 0). • The domain remains (0, ∞). • The range remains (−∞, ∞). • The asymptote remains x = 0. • The x-intercept remains (1, 0). • The domain remains (0, ∞). • The range remains (−∞, ∞). Figure 10 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 507 vertical stretches and compressions of the parent function y = logb(x) For any constant a > 1, the function f (x) = alogb(x) • stretches the parent function y = logb(x) vertically by a factor of a if a > 1. • compresses the parent function y = logb(x) vertically by a factor of a if 0 < a < 1. • has the vertical asymptote x = 0. • has the x-intercept (1, 0). • has domain (0, ∞). • has range (−∞, ∞). How To… Given a logarithmic function with the form f (x) = alogb(x), a > 0, graph the translation. 1. Identify the vertical stretch or compressions: a. If ∣ a ∣ > 1, the graph of f (x) = logb(x) is stretched by a factor of a units. b. If ∣ a ∣ < 1, the graph of f (x) = logb(x) is compressed by a factor of a units. 2. Draw the vertical asymptote x = 0. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the y coordinates by a. 4. Label the three points. 5. The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. Example 6 Graphing a Stretch or Compression of the Parent Function y = logb( x ) Sketch a graph of f (x) = 2log4(x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = 2log4(x), we will notice a = 2. This means we will stretch the function f (x) = log4(x) by a factor of 2. The vertical asymptote is x = 0. 1 , −1 ) , (1, 0), and (4, 1). Consider the three key points from the parent function, ( _ 4 The new coordinates are found by multiplying the y coordinates by 2. 1 Label the points ( , −2 ) , (1, 0) , and (4, 2). _ 4 The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. See Figure 11. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 (4, 2) (2, 1) (4, 1) 4 5 6 7 8 9 321 (1, 0) f (x) = 2log4(x) y = log4(x) x x = 0 Figure 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 508 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #6 1 _ Sketch a graph of f (x) = log4(x) alongside its parent function. Include the key points and asymptote on the graph. State 2 the domain, range, and asymptote. Example 7 Combining a Shift and a Stretch Sketch a graph of f (x) = 5log(x + 2). State the do
main, range, and asymptote. Solution Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5, as in Figure 12. The vertical asymptote will be shifted to x = −2. The x-intercept will be (−1, 0). The domain will be (−2, ∞). Two points will help give the shape of the graph: (−1, 0) and (8, 5). We chose x = 8 as the x-coordinate of one point to graph because when x = 8, x + 2 = 10, the base of the common logarithm. y y = 5 log(x + 2) y = log(x + 2) 321 4 5 x y = log(x) 5 4 3 2 1 –5 –4 –2 –3 –2 –1 –2 –3 –4 –5 x = −2 The domain is (−2, ∞), the range is (−∞, ∞), and the vertical asymptote is x = −2. Figure 12 Try It #7 Sketch a graph of the function f (x) = 3log(x − 2) + 1. State the domain, range, and asymptote. Graphing Reﬂections of f (x ) = logb(x ) When the parent function f (x) = logb(x) is multiplied by −1, the result is a reflection about the x-axis. When the input is multiplied by −1, the result is a reflection about the y-axis. To visualize reflections, we restrict b > 1, and observe the general graph of the parent function f (x) = logb(x) alongside the reflection about the x-axis, g(x) = −logb(x) and the reflection about the y-axis, h(x) = logb(−x). Reflection about the x-axis g (x) = logb(x), b > 1 Reflection about the y-axis h(x) = logb(−x), x) = logb(x) −1, 1) (b (b, 1) (1, 0) g(x) = −logb(x) h(x) = logb(−x) f (x) = logb(x) x (−b, 1) (−1, 0) (b, 1) (1, 0) x • The reflected function is decreasing as x moves from zero to infinity. • The asymptote remains x = 0. • The x-intercept remains (1, 0). • The key point changes to (b − 1, 1). • The domain remains (0, ∞). • The range remains (−∞, ∞). • The reflected function is decreasing as x moves from infinity to zero. • The asymptote remains x = 0. • The x-intercept remains (−1, 0). • The key point changes to (−b, 1). • The domain changes to (−∞, 0). • The range remains (−∞, ∞). Figure 13 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 509 reflections of the parent function y = logb(x) The function f (x) = −logb(x) • reflects the parent function y = logb(x) about the x-axis. • has domain, (0, ∞), range, (−∞, ∞), and vertical asymptote, x = 0, which are unchanged from the parent function. The function f (x) = logb(−x) • reflects the parent function y = logb(x) about the y-axis. • has domain (−∞, 0). • has range, (−∞, ∞), and vertical asymptote, x = 0, which are unchanged from the parent function. How To… Given a logarithmic function with the parent function f (x) = logb(x), graph a translation. If f (x) = −logb(x) 1. Draw the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 3. Reflect the graph of the parent function f (x) = logb(x) If f (x) = logb(−x) 1. Draw the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 3. Reflect the graph of the parent function f (x) = logb(x) about the x-axis. about the y-axis. 4. Draw a smooth curve through the points. 5. State the domain, (0, ∞), the range, (−∞, ∞), and 4. Draw a smooth curve through the points. 5. State the domain, (−∞, 0), the range, (−∞, ∞), and the vertical asymptote x = 0. the vertical asymptote x = 0. Table 3 Example 8 Graphing a Reﬂection of a Logarithmic Function Sketch a graph of f (x) = log(−x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Before graphing f (x) = log(−x), identify the behavior and key points for the graph. • Since b = 10 is greater than one, we know that the parent function is increasing. Since the input value is multiplied by −1, f is a reflection of the parent graph about the y-axis. Thus, f (x) = log(−x) will be decreasing as x moves from negative infinity to zero, and the right tail of the graph will approach the vertical asymptote x = 0. • The x-intercept is (−1, 0). • We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points. y x = 0 (–10, 0) f (x) = log(−x) y = log(x) (10, 0) (–1, 0) (1, 0) x The domain is (−∞, 0), the range is (−∞, ∞), and the vertical asymptote is x = 0. Figure 14 Try It #8 Graph f (x) = −log(−x). State the domain, range, and asymptote. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 510 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How To… Given a logarithmic equation, use a graphing calculator to approximate solutions. Press [Y=]. Enter the given logarithm equation or equations as Y1= and, if needed, Y2= . 1. 2. Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection. 3. To fi nd the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x, for the point(s) of intersection. Example 9 Approximating the Solution of a Logarithmic Equation Solve 4ln(x) + 1 = −2ln(x − 1) graphically. Round to the nearest thousandth. Solution Press [Y=] and enter 4ln(x) + 1 next to Y1= . Then enter −2ln(x − 1) next to Y2= . For a window, use the values 0 to 5 for x and –10 to 10 for y. Press [GRAPH]. The graphs should intersect somewhere a little to right of x = 1. For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for Guess?) So, to the nearest thousandth, x ≈ 1.339. Try It #9 Solve 5log(x + 2) = 4 − log(x) graphically. Round to the nearest thousandth. Summarizing Translations of the Logarithmic Function Now that we have worked with each type of translation for the logarithmic function, we can summarize each in Table 4 to arrive at the general equation for translating exponential functions. Translations of the Parent Function y = logb(x) Translation Form Shift • Horizontally c units to the left • Vertically d units up Stretch and Compress • Stretch if ∣ a ∣ > 1 • Compression if ∣ a ∣ < 1 Reflect about the x-axis Reflect about the y-axis General equation for all translations y = logb (x + c) + d y = alogb(x) y = −logb(x) y = logb(−x) y = alogb(x + c) + d Table 4 translations of logarithmic functions All translations of the parent logarithmic function, y = logb(x), have the form f (x) = alogb(x + c) + d where the parent function, y = logb(x), b > 1, is • shifted vertically up d units. • shifted horizontally to the left c units. • stretched vertically by a factor of ∣ a ∣ if ∣ a ∣ > 0. • compressed vertically by a factor of ∣ a ∣ if 0 < ∣ a ∣ < 1. • reflected about the x-axis when a < 0. For f (x) = log(−x), the graph of the parent function is reflected about the y-axis. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 511 Example 10 Finding the Vertical Asymptote of a Logarithm Graph What is the vertical asymptote of f (x) = −2log3(x + 4) + 5? Solution The vertical asymptote is at x = −4. Analysis The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve 4 units to the left shifts the vertical asymptote to x = −4. Try It #10 What is the vertical asymptote of f (x) = 3 + ln(x − 1)? Example 11 Finding the Equation from a Graph Find a possible equation for the common logarithmic function graphed in Figure 15. f(x) 5 4 3 2 1 –1 –1 –2 –3 –5 –4 –3 –2 21 3 4 5 6 7 x Figure 15 Solution This graph has a vertical asymptote at x = −2 and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form: It appears the graph passes through the points (−1, 1) and (2, −1). Substituting (−1, 1), f (x) = −alog(x + 2) + k Ne xt, substituting in (2, –1), 1 = −alog(−1 + 2) + k Substitute (−1, 1). 1 = −alog(1) + k 1 = k Arithmetic. log(1) = 0. −1 = −alog(2 + 2) + 1 Plug in (2, −1). −2 = −alog(4) a = 2 _____ log(4) Arithmetic. Solve for a. This gives us the equation f (x) = − 2_ log(x + 2) + 1. log(4) Analysis Figure 15 . You can verify this answer by comparing the function values in Table 5 with the points on the graph in x f (x) x f (x) −1 1 4 −1.5850 0 0 5 −1.8074 1 −0.58496 6 −2 2 −1 7 −2.1699 3 −1.3219 8 −2.3219 Table 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 512 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #11 Give the equation of the natural logarithm graphed in Figure 16. f(x) –5 –4 –3 –2 4 3 2 1 –1 –1 –2 –3 –4 –5 321 4 5 x Figure 16 Q & A… Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph? Yes, if we know the function is a general logarithmic function. For example, look at the graph in Figure 16. The graph approaches x = −3 (or thereabouts) more and more closely, so x = −3 is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, {x \| x > −3}. The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as x → −3+, f (x) → −∞ and as x → ∞, f (x) → ∞. Access these online resources for additional instruction and practice with graphing logarithms. • Graph an Exponential Function and Logarithmic Function (http://openstaxcollege.org/l/graphexplog) • Match Graphs with Exponential and Logarithmic Functions (http://openstaxcollege.org/l/matchexplog) • Find the Domain of Logarithmic Functions (http://openstaxcollege.org/l/domainlog) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 SECTION EXERCISES 513 6.4 SECTION EXERCISES VERBAL 1. Th 2. ff 3. ff 4. f (x) = bx. x 5. ALGEBRAIC 6. f x=x+ 9. hx=x
+− −x ) 7. hx=( 10. fx=−x− 8. gx=x+− 11. fx=bx− 12. gx=−x 13. fx=x+ 14. fx=−x+ 15. gx=−x+− 17. fx=( x− 16. fx=−x ) 20. fx=−x+ 19. gx=x+− 18. hx= −x−+ xy 21. hx=x−+ 22. fx=x++ 23. gx=−x− 24. fx=x+− 25. hx=x− GRAPHICAL Figure 17 A B C x 26. fx=x 27. gx=x 28. hx=x y – – Figure 17 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 514 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Figure 18 – – – – Figure 18 29. fx= x 30. gx=x 31. hx= x 32. fx=xgx=x 34. fx=xgx=x 33. fx=xgx= x 35. fx=e xgx=x Figure 19 Figure 19 36. fx=−x+ 37. gx=−x+ 38. hx=x+ 39. fx=x+ 40. fx=x 41. fx=−x 42. gx=x++ 43. gx=−x+ 44. hx=− x+− 45. y=x 46. fx= Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 SECTION EXERCISES 515 47. fx=x 48. fx= – – – – – – – – – – TECHNOLOGY o fi 49. x−+=x−+ 50. x−+=−x−+ 51. x−=−x+ 52. x+= −x+ 53. −x=x++ EXTENSIONS 54. bb≠ b 55. fx= x gx) = −x 56. 57. fx=( x+ x− ) 58. x fx=x +x+ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 516 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Use the product power andor quotient rule to epand logarithmic epressions. • Use the product power andor quotient rule to condense logarithmic epressions. • aluate logarithms. • Use the change-of-base formula for logarithms. 6.5 LOGARITHMIC PROPERTIES Figure 1 The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan) In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances: • Battery acid: 0.8 • Stomach acid: 2.7 • Orange juice: 3.3 • Pure water: 7 (at 25° C) • Human blood: 7.35 • Fresh coconut: 7.8 • Sodium hydroxide (lye): 14 To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where a is the concentration of hydrogen ion in the solution pH = −log([H+]) = log ( 1 _____ ) ([H+] The equivalence of −log ([H+]) and log ( ) is one of the logarithm properties we will examine in this section. 1 _ [H+] Using the Product Rule for Logarithms Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove. logb(1) = 0 logb(b) = 1 For example, log5 1 = 0 since 50 = 1. And log5 5 = 1 since 51 = 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 517 Ne xt, we have the inverse property. logb(b x) = x b logb(x) = x, x > 0 For example, to evaluate log(100), we can rewrite the logarithm as log10(102), and then apply the inverse property logb (b x) = x to get log10(102) = 2. To evaluate e ln(7), we can rewrite the logarithm as e loge(7), and then apply the inverse property b log (x) = x to get eloge(7) = 7. b Finally, we have the one-to-one property. logbM = logbN if and only if M = N We can use the one-to-one property to solve the equation log3(3x) = log3(2x + 5) for x. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x : 3x = 2x + 5 x = 5 Set the arguments equal. Subtract 2x. But what about the equation log3(3x) + log3(2x + 5) = 2? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation. Recall that we use the product rule of exponents to combine the product of exponents by adding. We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below. Given any real number x and positive real numbers M, N, and b, where b ≠ 1, we will show logb(MN) = logb(M) + logb(N). Let m = logb(M) and n = logb(N). In exponential form, these equations are b m = M and b n = N. It follows that logb(MN) = logb(b mb n) = logb(b m + n) = m + n = logb(M) + logb(N) Substitute for M and N. Apply the product rule for exponents. Apply the inverse property of logs. Substitute for m and n. Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider logb(wxyz). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors: logb(wxyz) = logb(w) + logb(x) + logb(y) + logb(z) the product rule for logarithms The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms. logb(MN) = logb(M) + logb(N) for b > 0 How To… Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms. 1. Factor the argument completely, expressing each whole number factor as a product of primes. 2. Write the equivalent expression by summing the logarithms of each factor. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 518 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 1 Using the Product Rule for Logarithms Expand log3(30x(3x + 4)). Solution We begin by factoring the argument completely, expressing 30 as a product of primes. Ne xt we write the equivalent equation by summing the logarithms of each factor. log3(30x(3x + 4)) = log3(2 · 3 · 5 · x · (3x +4)) log3(30x(3x + 4)) = log3(2) + log3(3) + log3(5) + log3(x) + log3(3x + 4) Try It #1 Expand logb(8k). Using the Quotient Rule for Logarithms For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine a __ the quotient of exponents by subtracting: x = x a−b. The quotient rule for logarithms says that the logarithm of a b quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule. Given any real number x and positive real numbers M, N, and b, where b ≠ 1, we will show M ) = logb(M) − logb(N). logb ( __ N Let m = logb(M) and n = logb(N). In exponential form, these equations are bm = M and bn = N. It follows that bm bn ) ) = logb ( M logb ( __ __ N = logb(b m − n) = m − n = logb(M) − logb(N) Substitute for M and N. Substitute for m and n. Apply the inverse property of logs. Apply the quotient rule for exponents. For example, to expand log ( 2x2 + 6x _ 3x + 9 we get, ) , we must first express the quotient in lowest terms. Factoring and canceling log ( Factor the numerator and denominator. 2x(x + 3) ) ________ 3(x + 3) 2x2 + 6x _______ 3x + 9 ) = log ( 2x __ ) 3 = log ( Cancel the common factors. Ne xt we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule. log ( 2x __ ) = log(2x) − log(3) 3 = log(2) + log(x) − log(3) the quotient rule for logarithms The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms. M ) = logb(M) − logb(N) logb ( __ N How To… Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms. 1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms. 2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator. 3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 519 Example 2 Using the Quotient Rule for Logarithms 15x(x − 1) ) . __ (3x + 4)(2 − x) Expand log 2 ( Solution First we note that the quotient is factored and in lowest terms, so we apply the quotient rule. log2 ( 15x(x − 1)__ (3x + 4)(2 − x)) = log2(15x(x−1))− log2((3x + 4)(2 − x)) Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5. log2(15x(x − 1)) − log2((3x + 4)(2 − x)) = [log2(3) + log2(5) + log2(x) + log2(x − 1)] − [log2(3x + 4) + log2(2 − x)] = log2(3) + log2(5) + log2(x) + log2(x − 1) − log2(3x + 4) − log2(2 − x) Analysis There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for x = − 4 _ and x = 2. Also, since the argument of a logarithm must be positive, we note 3 as we observe the expanded logarithm, that x > 0, x > 1, x > − 4 _ , and x < 2. Combining these conditions is beyond the 3 scope of this section, and we will not consider them here or in subsequent exercises. Try It #2 Expand log3 ( 7x2 + 21x__ ) . 7x(x − 1)(x − 2) Using the Power Rule for Logarithms We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as x2? One method is as follows: logb(x2) = logb(x ⋅ x) = logb (x) + logb (x) = 2log b (x) Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may b
e able to change it to a power. For example, 100 = 102 — 1 __ −1 the power rule for logarithms The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base. logb(Mn) = nlogb(M) How To… Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm. 1. Express the argument as a power, if needed. 2. Write the equivalent expression by multiplying the exponent times the logarithm of the base. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 520 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 3 Expanding a Logarithm with Powers Expand log2(x5). Solution The argument is already written as a power, so we identify the exponent, 5, and the base, x, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. log2(x 5) = 5log2(x) Try It #3 Expand ln(x2). Example 4 Rewriting an Expression as a Power before Using the Power Rule Expand log3(25) using the power rule for logs. Solution Expressing the argument as a power, we get log3(25) = log3(52). Ne xt we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. log3(5 2) = 2log3(5) Try It #4 1 x2 ) . Expand ln ( _ Example 5 Using the Power Rule in Reverse Rewrite 4ln(x) using the power rule for logs to a single logarithm with a leading coefficient of 1. Solution Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression 4ln(x), we identify the factor, 4, as the exponent and the argument, x, as the base, and rewrite the product as a logarithm of a power: 4ln(x) = ln(x4). Try It #5 Rewrite 2log3(4) using the power rule for logs to a single logarithm with a leading coefficient of 1. Expanding Logarithmic Expressions Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example: logb ( 6x _ y ) = logb(6x) − logb(y) = logb(6) + logb(x) − logb(y) We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power: A ) = logb(AC −1) logb ( __ C = logb(A) + logb(C −1) = logb(A) + (−1)logb(C) = logb(A) − logb(C) We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product. With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 521 Example 6 Expanding Logarithms Using Product, Quotient, and Power Rules Rewrite ln ( x 4 y_ ) as a sum or difference of logs. 7 Solution First, because we have a quotient of two expressions, we can use the quotient rule: ln ( x 4y ) = ln(x 4y)− ln(7) ___ 7 Then seeing the product in the first term, we use the product rule: ln(x 4y) − ln(7) = ln(x 4) + ln(y) − ln(7) Finally, we use the power rule on the first term: ln(x4)+ ln(y) − ln(7) = 4ln(x) + ln(y) − ln(7) Try It #6 Expand log ( x2 y3 z4 ) . _ Example 7 Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression Expand log( √ — x ). log(√ — 1 __ x ) = log (x) 2 = 1 __ log(x) 2 Solution Try It #7 Expand ln( 3 √ — x2 ). Q & A… Can we expand ln(x2 + y2)? No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Example 8 Expanding Complex Logarithmic Expressions Expand log6 ( 64x3 (4x + 1) __ (2x − 1) ) . Solution We can expand by applying the Product and Quotient Rules. log6 ( 64x3(4x + 1) __ (2x − 1) ) = log6(64) + log6(x3) + log6(4x + 1) − log6(2x − 1) = log6(26) + log6(x3) + log6(4x + 1) − log6(2x − 1) = 6log6(2) + 3log6(x) + log6(4x + 1) − log6(2x − 1) Apply the Quotient Rule. Simplify by writing 64 as 26. Apply the Power Rule. Try It #8 Expand ln ( — (x − 1) (2x + 1) 2 √ ) . __ x 2 − 9 Condensing Logarithmic Expressions We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 522 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How To… Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm. 1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. 2. Ne xt apply the product property. Rewrite sums of logarithms as the logarithm of a product. 3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient. Example 9 Using the Product and Quotient Rules to Combine Logarithms Write log3(5) + log3(8) − log3(2) as a single logarithm. Solution Using the product and quotient rules This reduces our original expression to log3(5) + log3(8) = log3(5 · 8) = log3(40) Then, using the quotient rule log3(40) − log3(2) log3(40) − log3(2) = log3 ( 40 __ ) = log3(20) 2 Try It #9 Condense log(3) − log(4) + log(5) − log(6). Example 10 Condensing Complex Logarithmic Expressions Condense log2(x 2) + 1 __ log2(x − 1) − 3log2((x + 3)2). 2 Solution We apply the power rule first: log2(x 2) + 1 log2(x − 1) − 3log2((x + 3)2) = log2(x2) + log2( √ __ 2 Next we apply the product rule to the sum: — x − 1 ) − log2((x + 3)6) log2(x 2) + log2( √ — x − 1 ) − log2((x + 3)6) = log2(x2 √ — x − 1 ) − log2((x + 3)6) Finally, we apply the quotient rule to the difference: log2(x 2 √ — x − 1 )− log2((x + 3)6) = log2 ( — x − 1 x2 √ ________ (x + 3)6 ) Try It #10 Rewrite log(5) + 0.5log(x) − log(7x − 1) + 3log(x − 1) as a single logarithm. Example 11 Rewriting as a Single Logarithm Rewrite 2log(x) − 4log(x + 5) + 1 _ x log(3x + 5) as a single logarithm. Solution We apply the power rule first: 2log(x) − 4log(x + 5) + 1 _ x log(3x + 5) = log(x2) − log((x + 5)4) + log ( (3x + 5) x −1 ) Ne xt we apply the product rule to the sum: log(x2)− log((x + 5)4) + log ( (3x + 5) x −1 ) = log(x2)− log ( (x + 5)4(3x + 5) x −1 ) Finally, we apply the quotient rule to the difference: log(x2) − log ( (x + 5)4(3x + 5) x −1 ) = log ( x 2 ) __ (x + 5) 4 (3x + 5) x −1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 523 Try It #11 Condense 4(3log(x) + log(x + 5) − log(2x + 3)). Example 12 Applying of the Laws of Logs Recall that, in chemistry, pH = −log[H+]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH? Solution Suppose C is the original concentration of hydrogen ions, and P is the original pH of the liquid. Then P = −log(C). If the concentration is doubled, the new concentration is 2C. Then the pH of the new liquid is Using the product rule of logs pH = −log(2C) pH = −log(2C) = −(log(2) + log(C)) = −log(2) − log(C) Since P = −log(C), the new pH is pH = P − log(2) ≈ P − 0.301 When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301. Try It #12 How does the pH change when the concentration of positive hydrogen ions is decreased by half? Using the Change-of-Base Formula for Logarithms Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or e, we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs. To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms. Given any positive real numbers M, b, and n, where n ≠ 1 and b ≠ 1, we show logb(M) = logn(M) _ logn(b) Let y = logb(M). By taking the log base n of both sides of the equation, we arrive at an exponential form, namely b y = M. It follows that logn(b y) = logn(M) Apply the one-to-one property. ylogn(b) = logn(M) Apply the power rule for logarithms. y = logn(M) _ logn(b) Isolate y. logb(M) = logn(M) _ logn(b) Substitute for y. For example, to evaluate log5(36) using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log. log5(36) = log(36) _ log(5) Apply the change of base formula using base 10. ≈ 2.2266 Use a calculator to evaluate to 4 decimal places. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 524 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS the change-of-base formula The change-of-base formula can be used to evaluate a logarithm with any base. For any positive real numbers M, b, and n, where n ≠ 1 and b ≠ 1, logn(M) _ . logn(b) logb(M) = It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs. logb(M) = and logb(M) = ln(M)_ ln(b) logn(M) _ logn(b) How To… Given a logarithm with the form logb(M), use the change-of-base formula to rewrite it as a quotient of logs with any positive base n, where n ≠ 1. 1. Determine the new base n, remembering that the common log, log(x), has base 10, and the natural log, ln(x), has base e. 2. Rewrite the log as a quotient using the change-of-base formula a. The numerator of the quotient will be a logarithm with base n and argument M. b. The denominator of the quotient will be a logarithm
with base n and argument b. Example 13 Changing Logarithmic Expressions to Expressions Involving Only Natural Logs Change log5(3) to a quotient of natural logarithms. Solution Because we will be expressing log5(3) as a quotient of natural logarithms, the new base, n = e. We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5. logb(M) = ln(M)_ ln(b) log5(3) = ln(3)_ ln(5) Try It #13 Change log0.5(8) to a quotient of natural logarithms. Q & A… Can we change common logarithms to natural logarithms? Yes. Remember that log(9) means log10(9). So, log(9) = ln(9) _ . ln(10) Example 14 Using the Change-of-Base Formula with a Calculator Evaluate log2(10) using the change-of-base formula with a calculator. Solution According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e. log2(10) = ln(10)_ ln(2) ≈ 3.3219 Apply the change of base formula using base e. Use a calculator to evaluate to 4 decimal places. Try It #14 Evaluate log5(100) using the change-of-base formula. Access this online resource for additional instruction and practice with laws of logarithms. • The Properties of Logarithms (http://openstaxcollege.org/l/proplog) • Expand Logarithmic Expressions (http://openstaxcollege.org/l/expandlog) • Evaluate a Natural Logarithmic Expression (http://openstaxcollege.org/l/evaluatelog) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 SECTION EXERCISES 525 6.5 SECTION EXERCISES VERBAL 1. n √ — x 2. ALGEBRAIC ff 3. bxy 4. abc 5. b ( ) 6. ( x z w ) 7. ( ) 8. yx 9. +x+y 10. +a++b 12. a−d−c ) 13. −b ( 11. b−b 14. ff 15. ( xy z ) 19. ! r 16. ✓ ◆ ◆ 17. ✓ 18. — √xy− 20. p ✓ ◆ (√ y ) −y 21. 22. x+x 23. x−x 24. x+x+ 25. x− y+z 26. c+ a + b 27. e 28. =a=b ab ( 31. 29. 30. ) NUMERIC 32. )− ( 33. + 34. −+( ) o fi 35. 36. 37. 38. EXTENSIONS ( ) 39. 40. 42. x x++x+= nb bn= b>n> 41. x+−x−= 43. = Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 526 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Use like bases to solve exponential equations. • Use logarithms to solve exponential equations. • Use the definition of a logarithm to solve logarithmic equations. • Use the one-to-one property of logarithms to solve logarithmic equations. 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS Figure 1 Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr) In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions. Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions. Using Like Bases to Solve Exponential Equations The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where b > 0, b ≠ 1, bS = bT if and only if S = T. In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown. For example, consider the equation 34x − 7 = . To solve for x, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x : 32x _ 3 34x − 7 = 34x − 7 = 32x ___ 3 32x ___ 31 Rewrite 3 as 31. 34x − 7 = 32x − 1 4x − 7 = 2x − 1 2x = 6 x = 3 Use the division property of exponents. Apply the one-to-one property of exponents. Subtract 2x and add 7 to both sides. Divide by 3. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 527 using the one-to-one property of exponential functions to solve exponential equations For any algebraic expressions S and T, and any positive real number b ≠ 1, bS = bT if and only if S = T How To… Given an exponential equation with the form bS = bT, where S and T are algebraic expressions with an unknown, solve for the unknown. 1. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT. 2. Use the one-to-one property to set the exponents equal. 3. Solve the resulting equation, S = T, for the unknown. Example 1 Solving an Exponential Equation with a Common Base Solve 2x − 1 = 22x − 4. Solution Try It #1 Solve 52x = 53x + 2. 2x − 1 = 22x − 4 x − 1 = 2x − 4 x = 3 The common base is 2. By the one-to-one property the exponents must be equal. Solve for x. Rewriting Equations So All Powers Have the Same Base Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property. For example, consider the equation 256 = 4x − 5. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for x : 256 = 4x − 5 28 = (22)x − 5 28 = 22x − 10 8 = 2x − 10 18 = 2x x = 9 Rewrite each side as a power with base 2. Use the one-to-one property of exponents. Apply the one-to-one property of exponents. Add 10 to both sides. Divide by 2. How To… Given an exponential equation with unlike bases, use the one-to-one property to solve it. 1. Rewrite each side in the equation as a power with a common base. 2. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT. 3. Use the one-to-one property to set the exponents equal. 4. Solve the resulting equation, S = T, for the unknown. Example 2 Solving Equations by Rewriting Them to Have a Common Base Solve 8x + 2 = 16x + 1. Solution 8x + 2 = 16x + 1 (23)x + 2 = (24)x + 1 23x + 6 = 24x + 4 3x + 6 = 4x + 4 x = 2 Write 8 and 16 as powers of 2. To take a power of a power, multiply exponents . Use the one-to-one property to set the exponents equal. Solve for x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 528 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #2 Solve 52x = 253x + 2. Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base 1 __ 25x = 2 2 1 __ 5x = 2 1 __ 10 x = Write the square root of 2 as a power of 2. Use the one-to-one property. Solve for x. Example 3 Solve 25x = √ — 2 . Solution Try It #3 Solve 5x = √ — 5 . Q & A… Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problemsolving process? No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined. Example 4 Solving an Equation with Positive and Negative Powers Solve 3x + 1 = −2. Solution This equation has no solution. There is no real value of x that will make the equation a true statement because any power of a positive number is positive. Analysis Figure 2 shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution. y y = 3 x + 1 –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 x y = −2 1 2 3 4 5 They do not cross. Figure 2 Try It #4 Solve 2x = −100. Solving Exponential Equations Using Logarithms Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since log(a) = log(b) is equivalent to a = b, we may apply logarithms with the same base on both sides of an exponential equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 529 How To… Given an exponential equation in which a common base cannot be found, solve for the unknown. 1. Apply the logarithm of both sides of the equation. a. If one of the terms in the equation has base 10, use the common logarithm. b. If none of the terms in the equation has base 10, use the natural logarithm. 2. Use the rules of logarithms to solve for the unknown. Example 5 Solve 5x + 2 = 4x. Solution Solving an Equation Containing Powers of Different Bases 5x + 2 = 4x ln(5x + 2) = ln(4x) (x + 2)ln(5) = xln(4) xln(5) + 2ln(5) = xln(4) xln(5) − xln(4) = − 2ln(5) x(ln(5) − ln(4)) = − 2ln(5) 5 1 __ __ ) ) = ln ( xln ( 25 4 1 __ ln ( ) 25 _ 5 __ ) ln ( 4 x = There is no easy way to get the powers to have the same base . Take ln of both sides. Use laws of logs. Use the distributive law. Get terms containing x on one side, terms without x on the other. On the left hand side, factor out an x. Use the laws of logs. Divide by the coefficient of x. Try It #5 Solve 2x = 3x + 1. Q & A… Is there any way to solve 2x = 3x? Yes. The solution is 0. Equations Containing e One common type of exponential equations are those with base e. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base e on either side, we can use the n
atural logarithm to solve it. How To… Given an equation of the form y = Aekt, solve for t. 1. Divide both sides of the equation by A. 2. Apply the natural logarithm of both sides of the equation. 3. Divide both sides of the equation by k. Example 6 Solve an Equation of the Form y = Ae k t Solve 100 = 20e 2t. Solution 100 = 20e 2t 5 = e 2t ln(5) = 2t t = ln(5)___ 2 Divide by the coefficient of the power . Take ln of both sides. Use the fact that ln(x) and e x are inverse functions. Divide by the coefficient of t. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 530 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Analysis Using laws of logs, we can also write this answer in the form t = ln √ of the answer, we use a calculator. — 5 . If we want a decimal approximation Try It #6 Solve 3e 0.5t = 11. Q & A… Does every equation of the form y = Aekt have a solution? No. There is a solution when k ≠ 0, and when y and A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is 2 = −3et. Example 7 Solving an Equation That Can Be Simpliﬁed to the Form y = Ae k t Solve 4e2x + 5 = 12. Solution 4e2x + 5 = 12 4e2x = 7 7 __ e2x = 4 7 __ ) 2x = ln ( 4 Combine like terms. Divide by the coefficient of the power. Take ln of both sides. 7 1 __ __ ) ln ( x = 4 2 Solve for x. Try It #7 Solve 3 + e2t = 7e2t. Extraneous Solutions Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output. Example 8 Solving Exponential Functions in Quadratic Form Solve e2x − e x = 56. Solution e 2x − e x = 56 e 2x − e x − 56 = 0 (e x + 7)(e x − 8) = 0 Get one side of the equation equal to zero. Factor by the FOIL method. e x + 7 = 0 or e x − 8 = 0 If a product is zero, then one factor must be zero. e x = −7 or e x = 8 Isolate the exponentials. e x = 8 x = ln(8) Reject the equation in which the power equals a negative number. Solve the equation in which the power equals a positive number. Analysis When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation e x = −7 because a positive number never equals a negative number. The solution ln(−7) is not a real number, and in the real number system this solution is rejected as an extraneous solution. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 531 Try It #8 Solve e2x = e x + 2. Q & A… Does every logarithmic equation have a solution? No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions. Using the Deﬁnition of a Logarithm to Solve Logarithmic Equations We have already seen that every logarithmic equation logb(x) = y is equivalent to the exponential equation b y = x. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression. For example, consider the equation log2(2) + log2(3x − 5) = 3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x: log2(2) + log2(3x − 5) = 3 log2(2(3x − 5)) = 3 log2(6x − 10) = 3 23 = 6x − 10 8 = 6x − 10 18 = 6x x = 3 Apply the product rule of logarithms. Distribute. Apply the definition of a logarithm. Calculate 23. Add 10 to both sides. Divide by 6. using the definition of a logarithm to solve logarithmic equations For any algebraic expression S and real numbers b and c, where b > 0, b ≠ 1, logb(S) = c if and only if b c = S Example 9 Using Algebra to Solve a Logarithmic Equation Solve 2ln(x) + 3 = 7. Solution Try It #9 Solve 6 + ln(x) = 10. 2ln(x) + 3 = 7 2ln(x) = 4 ln(x) = 2 x = e2 Subtract 3. Divide by 2. Rewrite in exponential form. Example 10 Using Algebra Before and After Using the Deﬁnition of the Natural Logarithm Solve 2ln(6x) = 7. Solution 2ln(6x) = 7 7 __ ln(6x) = 2 7 __ 6x = e 2 7 1 __ __ e x = 2 6 Divide by 2. Use the definition of ln. Divide by 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 532 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #10 Solve 2ln(x + 1) = 10. Example 11 Solve ln(x) = 3. Solution Using a Graph to Understand the Solution to a Logarithmic Equation ln(x) = 3 x = e 3 Use the definition of the natural logarithm. Figure 3 represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words e3 ≈ 20. A calculator gives a better approximation: e3 ≈ 20.0855. y = 1n(x) y = 3 (e3, 3) ≈ (20.0855, 3) 4 8 12 16 20 24 28 x y 4 2 1 –1 –2 Figure 3 The graphs of y = ln(x ) and y = 3 cross at the point (e 3, 3), which is approximately (20.0855, 3). Try It #11 Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2x = 1000 to 2 decimal places. Using the One-to-One Property of Logarithms to Solve Logarithmic Equations As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where b ≠ 1, logb(S) = logb(T) if and only if S = T. For example, If log2(x − 1) = log2(8), then x − 1 = 8. So, if x − 1 = 8, then we can solve for x, and we get x = 9. To check, we can substitute x = 9 into the original equation: log2(9 − 1) = log2(8) = 3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown. For example, consider the equation log(3x − 2) − log(2) = log(x + 4). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x: log(3x − 2) − log(2) = log(x + 4) log ( ) = log(x + 4) 3x − 2 ______ 2 3x − 2 ______ 2 = x + 4 3x − 2 = 2x + 8 Apply the quotient rule of logarithms. Apply the one to one property of a logarithm. Multiply both sides of the equation by 2. x = 10 Subtract 2x and add 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 533 To check the result, substitute x = 10 into log(3x − 2) − log(2) = log(x + 4). log(3(10) − 2) − log(2) = log((10) + 4) log(28) − log(2) = log(14) 28 __ ) = log(14) 2 log ( The solution checks. using the one-to-one property of logarithms to solve logarithmic equations For any algebraic expressions S and T and any positive real number b, where b ≠ 1, logb(S) = logb(T) if and only if S = T Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution. How To… Given an equation containing logarithms, solve it using the one-to-one property. 1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form logbS = logbT. 2. Use the one-to-one property to set the arguments equal. 3. Solve the resulting equation, S = T, for the unknown. Example 12 Solving an Equation Using the One-to-One Property of Logarithms Solve ln(x2) = ln(2x + 3). Solution ln(x2) = ln(2x + 3) x2 = 2x + 3 x2 − 2x − 3 = 0 (x − 3)(x + 1) = 0 Use the one-to-one property of the logarithm. Get zero on one side before factoring. Factor using FOIL. x − 3 = 0 or x + 1 = 0 If a product is zero, one of the factors must be zero. x = 3 or x = −1 Solve for x. Analysis There are two solutions: 3 or −1. The solution −1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive. Try It #12 Solve ln(x2) = ln(1). Solving Applied Problems Using Exponential and Logarithmic Equations In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm. One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Table 1 lists the half-life for several of the more common radioactive substances. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 534 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Substance gallium-67 cobalt-60 Use nuclear medicine manufacturing technetium-99m nuclear medicine americium-241 construction Half-life 80 hours 5.3 years 6 hours 432 years carbon-14 uranium-235 archeological dating 5,715 years atomic power 703,800,000 years Table 1 We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay: A(t) = A0 e ln(0.5) _____ T t t__ T A(t) = A0e ln(0.5) t _ A(t
) = A0 (e ln(0.5)) T t _ 1 __ ) A(t) = A0 ( T 2 where • A0 is the amount initially present • T is the half-life of the substance • t is the time period over which the substance is studied • y is the amount of the substance present after time t Example 13 Using the Formula for Radioactive Decay to Find the Quantity of a Substance How long will it take for ten percent of a 1,000-gram sample of uranium-235 to decay? Solution t t y = 1000 e ln(0.5) __________ 703,800,000 900 = 1000 e ln(0.5) __________ 703,800,000 0.9 = e ln(0.5) __________ 703,800,000 t ln(0.9) = ln ( e ln(0.5) __________ 703,800,000 t ) ln(0.9) = ln(0.5) __ t 703,800,000 t = 703,800,000 × ln(0.9) _ ln(0.5) years After 10% decays, 900 grams are left. Divide by 1000. Take ln of both sides. ln(eM) = M Solve for t. t ≈ 106,979,777 years Analysis Ten percent of 1,000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams. Try It #13 How long will it take before twenty percent of our 1,000-gram sample of uranium-235 has decayed? Access these online resources for additional instruction and practice with exponential and logarithmic equations. • Solving Logarithmic Equations (http://openstaxcollege.org/l/solvelogeq) • Solving Exponential Equations with Logarithms (http://openstaxcollege.org/l/solveexplog) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 SECTION EXERCISES 535 6.6 SECTION EXERCISES VERBAL 1. 2. 3. ALGEBRAIC 4. = 8. = = = 5. 9. 12. −v−= −v 13. ⋅x= 6. 10. = + = 7. 11. = + = 14. = 17. 20. 23. x−= = x+= x− 15. 18. 21. 24. = ()= = x+= x− 16. 19. 22. 25. = = += e r+−= − fi 26. (+)= 27. (+)= 28. ( )= 29. 30. −x= −x )=( ( ) 32. −x= x−x x 31. (+)=(+) 35. (+)+( )= 33. 36. 39. 42. ()+( (/)+( )= )= (+) (+)+( ()= )=() 34. 37. 40. 43. ()+(+)= )= ()+( (+) ( () ( )=() 38. () ) =() +(+) =() (+) (+)= ( 44. 41. )=() GRAPHICAL xTh 45. 48. x−= − x−= 51. x−−= − 54. x+= −x 57. −x−= 46. 49. 52. 55. 58. x+= +−x= −x= −x −x= x− x−x+= 47. x= 50. 53. 56. −+−x= − −x−x= x− x+= x+ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 536 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 59. ft 60. Th I ) D D= ( I I I= − ⋅ 61. Th = e tt TECHNOLOGY Th 62. 64. 66. t= t= e−t= 63. 65. ex= x−= 67. ex−+= 68. +x+= 69. −x−= + 70. P P= e−xx Hint 71. ThM E M= )E ( E E= · EXTENSIONS 72. x= x b b 73. 74. r ) kt A= a ( + k t 75. y= Ae kt tt T T= Ts+ T−Tse−ktTs T k tt Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 537 LEARNING OBJECTIVES In this section, you will: • odel eponential growth and decay. • Use ewtons aw of Cooling. • pply compound interest formulas and continuous growth formulas. • Use logistic-growth models. 6. 7 EXPONENTIAL AND LOGARITHMIC MODELS Figure 1 A nuclear research reactor inside the Neely Nuclear Research Center on the Georgia Institute of Technology campus. (credit: Georgia Tech Research Institute) We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling. Modeling Exponential Growth and Decay In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function: y = A0e kt where A0 is equal to the value at time zero, e is Euler’s constant, and k is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model. On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form y = A0e kt where A0 is the starting value, and e is Euler’s constant. Now k is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 538 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in Figure 2 and Figure 3. It is important to remember that, although parts of each of the two graphs seem to lie on the x-axis, they are really a tiny distance above the x-axis. y 6 5 4 3 2 1 y = 2e3 x ,1 3 2e (0, 2) − 1 3 , 2 e –5 –4 –3 –2 –1–1 –2 21 3 4 y = 0 5 x y = 3e −2 x y ,− 1 2 3e y = 0 –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 (0, 3 Figure 2 A graph showing exponential growth. The equation is y = 2e 3x. Figure 3 A graph showing exponential decay. The equation is y = 3e −2x. Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is 4.01134972 × 1013. So, we could describe this number as having order of magnitude 1013. characteristics of the exponential function, y = A0 e kt An exponential function with the form y = A0e kt has the following characteristics: • one-to-one function • horizontal asymptote: y = 0 • domain: ( –∞, ∞) • range: (0, ∞) • x-intercept: none • y-intercept: (0, A0) • increasing if k > 0 (see Figure 4) • decreasing if k < 0 (see Figure 4) ( ) 1 k , A0e y = A0ekt k > 0 ( ) , – (0, A0) A0 e 1 k y ( ) 1 – k , A0e y = A0ekt k < 0 (0, A0) y = 0 t y = 0 y ( ) , A0 e 1 k t Figure 4 An exponential function models exponential growth when k > 0 and exponential decay when k < 0. Example 1 Graphing Exponential Growth A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time. Solution When an amount grows at a fixed percent per unit time, the growth is exponential. To find A0 we use the fact that A0 is the amount at time zero, so A0 = 10. To find k, use the fact that after one hour (t = 1) the population doubles from 10 to 20. The formula is derived as follows 20 = 10e k ⋅ 1 2 = e k ln2 = k Divide by 10 Take the natural logarithm so k = ln(2). Thus the equation we want to graph is y = 10e(ln2)t = 10(eln2)t = 10 · 2t. The graph is shown in Figure 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 539 y = 10e(ln 2)t y 320 280 240 200 160 120 80 40 1 2 3 4 5 6 t Figure 5 The graph of y = 10e (ln2)t. Analysis The population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude 104. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude 10 7, so we could say that the population has increased by three orders of magnitude in ten hours. Half-Life We now turn to exponential decay. One of the common terms associated with exponential decay, as stated above, is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay. To fi nd the half-life of a function describing exponential decay, solve the following equation: We find that the half-life depends only on the constant k and not on the starting quantity A0. The formula is derived as follows 1 __ A0 = A0e kt 2 1 __ A0 = A0e kt 2 1 __ = e kt 2 1 ) = kt ln ( __ 2 Divide by A0. Take the natural log. − ln(2) = kt Apply laws of logarithms. − ln(2) ____ k = t Divide by k. Since t, the time, is positive, k must, as expected, be negative. This gives us the half-life formula t = − ln(2) ____ k How To… Given the half-life, find the decay rate. 1. Write A = A0 ekt. 1 _ 2. Replace A by A0 and replace t by the given half-life. 2 3. Solve to find k. Express k as an exact value (do not round). ln(2) _ . t Note: It is also possible to find the decay rate using k = − Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 540 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 2 Finding the Function that Describes Radioactive Decay The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t. Solution This formula is derived as follows. A = A0e kt 0.5A0 = A0e k ⋅ 5730 0.5 = e5730k ln(0.5) = 5730k ln(0.5) ______ 5730 A = A0 e ( ln(0.5) ______ 5730 k = ) t The continuous growth formula. Substitute the half-life for t and 0.5A0 for f(t). Divide by A0. Take the natural log of both sides. Divide by the coefficient of k. Substitute for k in the continuous growth formula. ) t . We observe that the coefficient of t, ln(0.5) ______ 5730 The function that describes this continuous decay is f(t) = A0 e ( ln(0.5) _ 5730 ≈ −1.2097 is negative, as expected in the case o
f exponential decay. Try It #14 The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years. Radiocarbon Dating The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libb y, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years. Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries. As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated. Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t years is ) t ln(0.5) ______ 5730 A ≈ A0 e ( where • A is the amount of carbon-14 remaining • A0 is the amount of carbon-14 when the plant or animal began decaying. This formula is derived as follows: A = A0e kt 0.5A0 = A0e k ⋅ 5730 0.5 = e5730k ln(0.5) = 5730k ln(0.5) ______ 5730 A = A0 e ( ln(0.5) ______ 5730 k = ) t To find the age of an object, we solve this equation for t: A ) ln ( _ A0 _ − 0.000121 t = The continuous growth formula. Substitute the half-life for t and 0.5A0 for f (t). Divide by A0. Take the natural log of both sides. Divide by the coefficient of k. Substitute for r in the continuous growth formula. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 541 Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation A ≈ A0e−0.000121t we know the ratio of the percentage of ≈ e−0.000121t. We solve carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is r = this equation for t, to get A __ A0 t = ln(r) _ −0.000121 How To… Given the percentage of carbon-14 in an object, determine its age. 1. Express the given percentage of carbon-14 as an equivalent decimal, k. 2. Substitute for k in the equation t = and solve for the age, t. ln(r) _________ −0.000121 Example 3 Finding the Age of a Bone A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone? Solution We substitute 20% = 0.20 for k in the equation and solve for t : t = = ln(r) _ −0.000121 ln(0.20) _ −0.000121 Use the general form of the equation. Substitute for r. ≈ 13301 Round to the nearest year. The bone fragment is about 13,301 years old. Analysis The instruments that measure the percentage of carbon-14 are e xtremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as 13,301 years ± 1% or 13,301 years ± 133 years. Try It #15 Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains? Calculating Doubling Time For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time. Given the basic exponential growth equation A = A0e kt, doubling time can be found by solving for when the original quantity has doubled, that is, by solving 2A0 = A0e kt. The formula is derived as follows: Thus the doubling time is 2A0 = A0e kt 2 = e kt ln(2) = kt t = ln(2) _ k t = ln(2) _ k Divide by A0. Take the natural logarithm. Divide by the coefficient of t. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 542 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 4 Finding a Function That Describes Exponential Growth According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior. Solution The formula is derived as follows: t = 2 = ln(2) ___ k ln(2) ___ k ln(2) ___ 2 A = A0 e k = ln(2) _____ 2 t The doubling time formula. Use a doubling time of two years. Multiply by k and divide by 2. Substitute k into the continuous growth formula. The function is A0 e ln(2) t . _____ 2 Try It #16 Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account. Using Newton’s Law of Cooling Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature This formula is derived as follows: T(t) = Ae kt + Ts T(t) = Ab ct + Ts T(t) = Ae ln(bct) + Ts T(t) = Ae ctln(b) + Ts T(t) = Ae kt + Ts Laws of logarithms. Laws of logarithms. Rename the constant cln(b), calling it k. Newton’s law of cooling The temperature of an object, T, in surrounding air with temperature Ts will behave according to the formula T(t) = Ae kt + Ts where • t is time • A is the difference between the initial temperature of the object and the surroundings • k is a constant, the continuous rate of cooling of the object How To… Given a set of conditions, apply Newton’s Law of Cooling. 1. Set Ts equal to the y-coordinate of the horizontal asymptote (usually the ambient temperature). 2. Substitute the given values into the continuous growth formula T(t) = Ae kt + Ts to find the parameters A and k. 3. Substitute in the desired time to find the temperature or the desired temperature to find the time. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 543 Example 5 Using Newton’s Law of Cooling A cheesecake is taken out of the oven with an ideal internal temperature of 165°F, and is placed into a 35°F refrigerator. After 10 minutes, the cheesecake has cooled to 150°F. If we must wait until the cheesecake has cooled to 70°F before we eat it, how long will we have to wait? Solution Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation We know the initial temperature was 165, so T(0) = 165. T(t) = Ae kt + 35 165 = Ae k0 + 35 Substitute (0, 165). A = 130 Solve for A. We were given another data point, T(10) = 150, which we can use to solve for k. 150 = 130e k10 + 35 Substitute (10, 150). 115 = 130e k10 115 ___ 130 = e 10k Subtract 35. Divide by 130. 115 ___ 130 ln ( ) = 10k 115 ___ ln ( ) 130 _ 10 This gives us the equation for the cooling of the cheesecake: T(t) = 130e −0.0123t + 35. ≈ −0.0123 Divide by the coefficient of k. k = Take the natural log of both sides. Now we can solve for the time it will take for the temperature to cool to 70 degrees. 70 = 130e−0.0123t + 35 35 = 130e−0.0123t Substitute in 70 for T(t). Subtract 35. 35 ___ 130 = e−0.0123t Divide by 130. ln ( 35 ___ 130 Take the natural log of both sides ) = −0.0123t 35 ___ ) ln ( 130 _ ≈ 106.68 Divide by the coefficient of t. −0.0123 t = It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to 70°F. Try It #17 A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees? Using Logistic Growth Models Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the ex
ponential growth model is still useful over a short term, before approaching the limiting value. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 544 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants a, b, and c, the logistic growth of a population over time x is represented by the model f (x) = c _______ 1 + ae−b x The graph in Figure 6 shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases. f (x+a logistic growth The logistic growth model is where c _____ 1 + a • is the initial value Carrying capacity f (x) = c 1 + ae–bx ln(a) b ( )c , 2 Point of maximum growth Initial value of population x Figure 6 f (x) = c _ 1 + ae−b x • c is the carrying capacity, or limiting value • b is a constant determined by the rate of growth. Example 6 Using the Logistic-Growth Model An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population. For example, at time t = 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is b = 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed. Solution We substitute the given data into the logistic growth model c _______ 1 + ae−b x f (x) = This model predicts that, after ten days, the number of people who have had the flu is f (x) = Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is c = 1000. To find a, we use the formula that the number of cases at time t = 0 is = 1, from which it follows that a = 999. 1000 ____________ 1 + 999e−0.6030x ≈ 293.8. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, c = 1000. c _ 1 + a Analysis Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values. The graph in Figure 7 gives a good picture of how this model fits the data. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 545 y = 1000 1,000 cases on day 21 1,100 1,000 900 800 700 s e s a C 600 500 400 300 200 100 294 cases on day 10 1 case on day 0 20 cases on day 5 2 4 6 8 10 12 14 Days 16 18 20 22 24 26 Figure 7 The graph of f (x) = 1000 __ 1 + 999e −0.6030x Try It #18 Using the model in Example 6, estimate the number of cases of flu on day 15. Choosing an Appropriate Model for Data Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015. Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered. In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down. A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection. After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 546 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 7 Choosing a Mathematical Model Does a linear, exponential, logarithmic, or logistic model best fit the values listed in Table 1? Find the model, and use a graph to check your choice.386 2.197 2.773 3.219 3.584 3.892 4.159 4.394 Table 1 Solution First, plot the data on a graph as in Figure 8. For the purpose of graphing, round the data to two significant digits. y 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0. 10 Figure 8 x Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try y = aln(b x). Plugging in the first point, (1,0), gives 0 = alnb. We reject the case that a = 0 (if it were, all outputs would be 0), so we know ln(b) = 0. Thus b = 1 and y = aln(x). Ne xt we can use the point (9,4.394) to solve for a: y = aln(x) 4.394 = aln(9) a = 4.394 _____ ln(9) Because a = ≈ 2, an appropriate model for the data is y = 2ln(x). 4.394 _ ln(9) To check the accuracy of the model, we graph the function together with the given points as in Figure 9. y = 2 ln(x) x = 0 y 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0. 10 Figure 9 The graph of y = 2lnx. x We can conclude that the model is a good fit to the data. Compare Figure 9 to the graph of y = ln(x2) shown in Figure 10. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 547 y x = 0 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 y = ln(x2 10 Figure 10 The graph of y = ln(x 2) x The graphs appear to be identical when x > 0. A quick check confirms this conclusion: y = ln(x 2) = 2ln(x) for x > 0. However, if x < 0, the graph of y = ln(x 2) includes a “extra” branch, as shown in Figure 11. This occurs because, while y = 2ln(x) cannot have negative values in the domain (as such values would force the argument to be negative), the function y = ln(x 2) can have negative domain values. y y = ln(x2) 642 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 11 Try It #19 Does a linear, exponential, or logarithmic model best fit the data in Table 2? Find the model.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034 Table 2 Expressing an Exponential Model in Base e While powers and logarithms of any base can be used in modeling, the two most common bases are 10 and e. In science and mathematics, the base e is often preferred. We can use laws of exponents and laws of logarithms to change any base to base e. How To… Given a model with the form y = ab x, change it to the form y = A0e kx. 1. Rewrite y = ab x as y = aeln(b x). 2. Use the power rule of logarithms to rewrite y as y = ae xln(b) = aeln(b)x. 3. Note that a = A0 and k = ln(b) in the equation y = A0e kx. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 548 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 8 Changing to base e Change the function y = 2.5(3.1)x so that this same function is written in the form y = A0e kx. Solution The formula is derived as follows y = 2.5(3.1)x = 2.5e ln(3.1x ) = 2.5e xln3.1 Insert exponential and its inverse. Laws of logs. = 2.5e (ln3.1)x Commutative law of multiplication Try It #20 Change the function y = 3(0.5)x to one having e as the base. Access these online resources for additional instruction and practice with exponential and logarithmic models. • Logarithm Application – pH (http://openstaxcollege.org/l/logph) • Exponential Model – Age Using Half-Life (http://openstaxcollege.org/l/expmodelhalf) • Newton’s Law of Cooling (http://openstaxcollege.org/l/newtoncooling) • Exponential Growth Given Doubling Time (http://openstaxcollege.org/l/expgrowthdbl) • Exponential Growth – Find Initial Amount Given Doubling Ti
me (http://openstaxcollege.org/l/initialdouble) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 SECTION EXERCISES 549 6.7 SECTION EXERCISES VERBAL 1. halflife 2. 3. doubling time 4. . Th 5. NUMERIC 6. Thftt Tt=e −t+ft f x= +e−x 7. f 9. 11. hmic. Th 12. f(x=x eo fi 8. f 10. x f(x) TECHNOLOGY x f(x) x 13. 14. 15. 16. f(x) x f(x) x f(x) fit Pt= +e−t 17. 19. 18. f fi 20. e fi e fi ft 21. 22. e fi P Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 550 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS EXTENSIONS 23. 25. S M= ) ( S S 27. b x=e xbb≠ 24. Th Pt=Pe rtP r>t M 26. y c +ae−rx y= REAL-WORLD APPLICATIONS For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. 28. To the nearest hour, what is the half-life of the drug? 29. Write an exponential model representing the amount of the drug remaining in the patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram. 30. Using the model found in the previous exercise, find f (10) and interpret the result. Round to the nearest hundredth. For the following exercises, use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. 31. To the nearest day, how long will it take for half of 32. Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram. 34. The half-life of Radium-226 is 1590 years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits. 36. A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 5730 years.) the Iodine-125 to decay? 33. A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance? 35. The half-life of Erbium-165 is 10.4 hours. What is the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits. 37. A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 SECTION EXERCISES 551 ft ft 38. 39. 40. 41. 42. 43. 44. 45. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 552 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Build an exponential model from data. • Build a logarithmic model from data. • Build a logistic model from data. 6.8 FITTING EXPONENTIAL MODELS TO DATA In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from real-world observations. With regression analysis, we don’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events. Do not be confused by the word model. In mathematics, we often use the terms function, equation, and model interchangeably, even though they each have their own formal definition. The term model is typically used to indicate that the equation or function approximates a real-world situation. We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways regression is used to model real-world phenomena. Building an Exponential Model from Data As we’ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay. Recall that exponential functions have the form y = ab x or y = A0e kx. When performing regression analysis, we use the form most commonly used on graphing utilities, y = ab x. Take a moment to reflect on the characteristics we’ve already learned about the exponential function y = ab x (assume a > 0): • b must be greater than zero and not equal to one. • The initial value of the model is y = a. • If b > 1, the function models exponential growth. As x increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound. • If 0 < b < 1, the function models exponential decay. As x increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the x-axis. In other words, the outputs never become equal to or less than zero. As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable r, or r 2. (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the “goodness of fit” of the regression equation to the data. We more commonly use the value of r 2 instead of r, but the closer either value is to 1, the better the regression equation approximates the data. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 553 exponential regression Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form, y = ab x Note that: • b must be non-negative. • when b > 1, we have an exponential growth model. • when 0 < b < 1, we have an exponential decay model. How To… Given a set of data, perform exponential regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow an exponential pattern. 3. Find the equation that models the data. a. Select “ExpReg” from the STAT then CALC menu. b. Use the values returned for a and b to record the model, y = ab x. 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. Example 1 Using Exponential Regression to Fit a Model to Data In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. Table 1 shows results from the study[24]. The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol. BAC Relative Risk of Crashing BAC Relative Risk of Crashing 0 1 0.11 6.41 0.01 1.03 0.13 12.6 Table 1 0.03 1.06 0.15 22.1 0.05 1.38 0.17 0.07 2.09 0.19 0.09 3.54 0.21 39.05 65.32 99.78 a. Let x represent the BAC level, and let y represent the corresponding relative risk. Use exponential regression to fit a model to these data. b. After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth. 24 Source: Indiana University Center for Studies of Law in Action, 2007 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 554 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Solution a. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown in Figure 1: y 110 100 90 80 70 60 50 40 30 20 10 .02 .04 .06 .08 .10 .12 .14 .16 .18 .20 .22 x Figure 1 Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model, y = 0.58304829(2.20720213E10)x Co
nverting from scientific notation, we have: y = 0.58304829(22,072,021,300)x Notice that r 2 ≈ 0.97 which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 2: y 110 100 90 80 70 60 50 40 30 20 10 .02 .04 .06 .08 .10 .12 .14 .16 .18 .20 .22 x Figure 2 b. Use the model to estimate the risk associated with a BAC of 0.16. Substitute 0.16 for x in the model and solve for y. y = 0.58304829(22,072,021,300)x Use the regression model found in part (a). = 0.58304829(22,072,021,300)0.16 Substitute 0.16 for x. ≈ 26.35 Round to the nearest hundredth. If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 555 Try It #1 Table 2 shows a recent graduate’s credit card balance each month after graduation. Month 1 2 3 4 5 6 7 8 Debt ($) 620.00 761.88 899.80 1039.93 1270.63 1589.04 1851.31 2154.92 Table 2 a. Use exponential regression to fit a model to these data. b. If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating? Q & A… Is it reasonable to assume that an exponential regression model will represent a situation indefinitely? No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation). Building a Logarithmic Model from Data Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or decrease that helps us determine whether a logarithmic model is best. Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis, we use the form of the logarithmic function most commonly used on graphing utilities, y = a + bln(x). For this function • All input values, x, must be greater than zero. • The point (1, a) is on the graph of the model. • If b > 0, the model is increasing. Growth increases rapidly at first and then steadily slows over time. • If b < 0, the model is decreasing. Decay occurs rapidly at first and then steadily slows over time. logarithmic regression Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command “LnReg” on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form, y = a + bln(x) Note that: • all input values, x, must be non-negative. • when b > 0, the model is increasing. • when b < 0, the model is decreasing. How To… Given a set of data, perform logarithmic regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 556 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow a logarithmic pattern. 3. Find the equation that models the data. a. Select “LnReg” from the STAT then CALC menu. b. Use the values returned for a and b to record the model, y = a + bln(x). 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. Example 2 Using Logarithmic Regression to Fit a Model to Data Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century. Table 3 shows the average life expectancies, in years, of Americans from 1900–2010[25]. Year Life Expectancy (Years) 1900 47.3 1910 50.0 1920 54.1 1930 59.7 1940 62.9 1950 68.2 Year Life Expectancy (Years) 1960 69.7 1970 70.8 1980 73.7 1990 75.4 2000 76.8 2010 78.7 Table 3 a. Let x represent time in decades starting with x = 1 for the year 1900, x = 2 for the year 1910, and so on. Let y represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data. b. Use the model to predict the average American life expectancy for the year 2030. Solution a. Using the STAT then EDIT menu on a graphing utility, list the years using values 1–12 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern as shown in Figure 3: y 85 80 75 70 65 60 55 50 45 40 1 2 3 4 5 6 7 8 9 10 11 12 13 x Figure 3 Use the “LnReg” command from the STAT then CALC menu to obtain the logarithmic model, y = 42.52722583 + 13.85752327ln(x) Ne xt, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 4: 25 Source: Center for Disease Control and Prevention, 2013 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 557 y 85 80 75 70 65 60 55 50 45 40 1 2 3 4 5 6 7 8 9 10 11 12 13 x Figure 4 b. To predict the life expectancy of an American in the year 2030, substitute x = 14 for the in the model and solve for y: y = 42.52722583 + 13.85752327ln(x) Use the regression model found in part ( a). = 42.52722583 + 13.85752327ln(14) Substitute 14 for x. ≈ 79.1 Round to the nearest tenth If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030. Try It #2 Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table 4 shows the number of games sold, in thousands, from the years 2000–2010. Year 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 Number Sold (Thousands) 142 149 154 155 159 161 163 164 164 166 167 a. Let x represent time in years starting with x = 1 for the year 2000. Let y represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data. b. If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand. Table 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 558 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Building a Logistic Model from Data Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients. It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities: y = c _______ 1 + ae−b x Recall that: c _____ 1 + a is the initial value of the model. • • when b > 0, the model increases rapidly at first until it reaches its point of maximum growth rate, ( that point, growth steadily slows and the function becomes asymptotic to the upper bound y = c. ln(a) c ) . At _ _ , 2 b • c is the limiting value, sometimes called the carrying capacity, of the model. logistic regression Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command “Logistic” on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form Note that y = c _______ 1 + ae−b x • The initial value of the model is c _ . 1 + a • Output values for the model grow closer and closer to y = c as time increases. How To… Given a set of data, perform logistic regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow a logistic pattern. 3. Find the equation that models the data. a. Select “Logistic” from the STAT then CALC menu. b. Use the values returned for a, b, and c to record the model, y = c _ 1 + ae−b x . 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. Example 3 Using Logistic Regression to Fit a Model to Data Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table 5 shows the percentage of Americans with cellular service between the years 1995 and 2012[26]. 26 Source: The World Bankn, 2013 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 559 Year 1995 1996 1997 1998 1999 2000
2001 2002 2003 Americans with Cellular Service (%) 12.69 16.35 20.29 25.08 30.81 38.75 45.00 49.16 55.15 Year 2004 2005 2006 2007 2008 2009 2010 2011 2012 Americans with Cellular Service (%) 62.852 68.63 76.64 82.47 85.68 89.14 91.86 95.28 98.17 Table 5 a. Let x represent time in years starting with x = 0 for the year 1995. Let y represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data. b. Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent. c. Discuss the value returned for the upper limit c. What does this tell you about the model? What would the limiting value be if the model were exact? Solution a. Using the STAT then EDIT menu on a graphing utility, list the years using values 0–15 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 5: y 110 100 90 80 70 60 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x Figure 5 Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model, 105.7379526 ___________________ 1 + 6.88328979e−0.2595440013x Ne xt, graph the model in the same window as shown in Figure 6 the scatterplot to verify it is a good fit: y = y 110 100 90 80 70 60 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 560 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS b. To approximate the percentage of Americans with cellular service in the year 2013, substitute x = 18 for the in the model and solve for y: y = = 105.7379526 ____________________ 1 + 6.88328979e −0.2595440013x 105.7379526 _____________________ 1 + 6.88328979e −0.2595440013(18) Substitute 18 for x. Use the regression model found in part ( a). ≈ 99.3% Round to the nearest tenth According to the model, about 99.3% of Americans had cellular service in 2013. c. The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would be c = 100 and the model’s outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service! Try It #3 Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012. Year 1997 1998 1999 2000 2001 2002 2003 2004 Seal Population (Thousands) 3.493 5.282 6.357 9.201 11.224 12.964 16.226 18.137 Year 2005 2006 2007 2008 2009 2010 2011 2012 Seal Population (Thousands) 19.590 21.955 22.862 23.869 24.243 24.344 24.919 25.108 Table 6 a. Let x represent time in years starting with x = 0 for the year 1997. Let y represent the number of seals in thousands. Use logistic regression to fit a model to these data. b. Use the model to predict the seal population for the year 2020. c. To the nearest whole number, what is the limiting value of this model? Access this online resource for additional instruction and practice with exponential function models. • Exponential Regression on a Calculator (http://openstaxcollege.org/l/pregresscalc) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 SECTION EXERCISES 561 6.8 SECTION EXERCISES VERBAL 1. What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit. 2. What is a carrying capacity? What kind of model has a carrying capacity built into its formula? Why does this make sense? 3. What is regression analysis? Describe the process of performing regression analysis on a graphing utility. 4. What might a scatterplot of data points look like if it were best described by a logarithmic model? 5. What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation? GRAPHICAL For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph. y y 16 14 12 10 8 6 4 2 16 14 12 10 8 6 4 2 y 16 14 12 10 8 6 4 2 y 16 14 12 10 a) 6 7 8 9 10 1 2 3 4 x 6 7 8 9 10 1 2 3 4 x 5 (b) 6 7 8 9 10 x 5 (c) Figure 7 Figure 8 Figure 9 y 16 14 12 10 d) Figure 10 x 7 8 9 10 1 2 3 4 x 6 7 8 9 10 5 (e) Figure 11 6. y = 10.209e−0.294x 7. y = 5.598 − 1.912ln(x) 8. y = 2.104(1.479)x 9. y = 4.607 + 2.733ln(x) 10. y = 14.005 __________ 1 + 2.79e−0.812x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 562 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS NUMERIC 11. To the nearest whole number, what is the initial value of a population modeled by the logistic equation P(t) = 175 ____________ 1 + 6.995e−0.68t ? What is the carrying capacity? 12. Rewrite the exponential model A(t) = 1550(1.085)x as an equivalent model with base e. Express the exponent to four significant digits. 13. A logarithmic model is given by the equation h(p) = 67.682 − 5.792ln(p). To the nearest hundredth, for what value of p does h(p) = 62? 15. What is the y-intercept on the graph of the logistic model given in the previous exercise? TECHNOLOGY 14. A logistic model is given by the equation 90 P(t) = ________ 1 + 5e−0.42t . To the nearest hundredth, for what value of t does P(t) = 45? For the following exercises, use this scenario: The population P of a koi pond over x months is modeled by the function P(x) = 68 __ 1 + 16e−0.28x . 16. Graph the population model to show the population 17. What was the initial population of koi? over a span of 3 years. 18. How many koi will the pond have after one and a 19. How many months will it take before there are 20 half years? koi in the pond? 20. Use the intersect feature to approximate the number of months it will take before the population of the pond reaches half its carrying capacity. For the following exercises, use this scenario: The population P of an endangered species habitat for wolves 558 __ is modeled by the function P(x) = 1 + 54.8e−0.462x , where x is given in years. 21. Graph the population model to show the population 22. What was the initial population of wolves over a span of 10 years. transported to the habitat? 23. How many wolves will the habitat have after 3 years? 24. How many years will it take before there are 100 wolves in the habitat? 25. Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity. For the following exercises, refer to Table 7. x f (x) 1 1125 2 1495 3 2310 Table 7 4 3294 5 4650 6 6361 26. Use a graphing calculator to create a scatter diagram 27. Use the regression feature to find an exponential of the data. function that best fits the data in the table. 28. Write the exponential function as an exponential 29. Graph the exponential equation on the scatter equation with base e. diagram. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 SECTION EXERCISES 563 30. Use the intersect feature to find the value of x for which f (x) = 4000. For the following exercises, refer to Table 8. x f (x) 1 555 2 383 3 307 Table 8 4 210 5 158 6 122 31. Use a graphing calculator to create a scatter diagram 32. Use the regression feature to find an exponential of the data. function that best fits the data in the table. 33. Write the exponential function as an exponential 34. Graph the exponential equation on the scatter equation with base e. diagram. 35. Use the intersect feature to find the value of x for which f (x) = 250. For the following exercises, refer to Table 9. x f (x) 1 5.1 2 6.3 3 7.3 Table 9 4 7.7 5 8.1 6 8.6 36. Use a graphing calculator to create a scatter diagram 37. Use the LOGarithm option of the REGression of the data. feature to find a logarithmic function of the form y = a + bln(x) that best fits the data in the table. 38. Use the logarithmic function to find the value of the 39. Graph the logarithmic equation on the scatter function when x = 10. diagram. 40. Use the intersect feature to find the value of x for which f (x) = 7. For the following exercises, refer to Table 10. x f (x) 1 7.5 2 6 3 5.2 4 4.3 5 3.9 6 3.4 7 3.1 8 2.9 Table 10 41. Use a graphing calculator to create a scatter diagram 42. Use the LOGarithm option of the REGression of the data. feature to find a logarithmic function of the form y = a + bln(x) that best fits the data in the table. 43. Use the logarithmic function to find the value of the 44. Graph the logarithmic equation on the scatter function when x = 10. diagram. 45. Use the intersect feature to find the value of x for which f (x) = 8. For the following exercises, refer to Table 11. x f (x) 1 8.7 2 12.3 3 15.4 4 18.5 5 20.7 6 22.5 7 23.3 8 24 9 24.6 10 24.8 Table 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 564 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 46. Use a graphing calculator to create a scatter diagram of the data. 47. Use the LOGISTIC regression option to find a logistic growth model of the form y = best fits the data in the table. c _ 1 + ae−b x that 48. Graph the logistic equation on the scatter diagram. 49. To the nearest whole number, what is the predicted carrying capacity of the model? 50. Use the intersect feature to find the value of x for which the model reaches half its carrying capacity. For the following exercises, refer to Table 12. x f (x) 0 12 2 28.6 4 52.8 5 70.3 7 8 10 11 15 17 99.9 112.5 125.8 127.9 135.1 135.9 Table 12 51. Use a graphing calculator to create a scatter diagram of the data. 52. Use the LOGISTIC regression option to find a logistic growth model of the form y = best fits the data in the table. c ________ 1 + ae−b x that 53. Graph the logistic
equation on the scatter diagram. 54. To the nearest whole number, what is the predicted carrying capacity of the model? 55. Use the intersect feature to find the value of x for which the model reaches half its carrying capacity. EXTENSIONS 56. Recall that the general form of a logistic equation for a population is given by P(t) = c _ 1 + ae−bt , such that the initial population at time t = 0 is P(0) = P0. Show algebraically that c − P0 _ c − P(t) _ P0 P(t) e−bt. = 57. Use a graphing utility to find an exponential regression formula f (x) and a logarithmic regression formula g(x) for the points (1.5, 1.5) and (8.5, 8.5). Round all numbers to 6 decimal places. Graph the points and both formulas along with the line y = x on the same axis. Make a conjecture about the relationship of the regression formulas. 58. Verify the conjecture made in the previous exercise. Round all numbers to six decimal places when necessary. 59. Find the inverse function f −1 (x) for the logistic c _ 1 + ae−b x . Show all steps. function f (x) = the logistic model P(t) = 60. Use the result from the previous exercise to graph ________ 1 + 4e−0.5t along with its inverse on the same axis. What are the intercepts and asymptotes of each function? 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 565 CHAPTER 6 REVIEW Key Terms annual percentage rate (APR) the yearly interest rate earned by an investment account, also called nominal rate carrying capacity in a logistic model, the limiting value of the output change-of-base formula a formula for converting a logarithm with any base to a quotient of logarithms with any other base. common logarithm the exponent to which 10 must be raised to get x; log10(x) is written simply as log(x). compound interest interest earned on the total balance, not just the principal doubling time the time it takes for a quantity to double exponential growth a model that grows by a rate proportional to the amount present extraneous solution a solution introduced while solving an equation that does not satisfy the conditions of the original equation half-life the length of time it takes for a substance to exponentially decay to half of its original quantity logarithm the exponent to which b must be raised to get x; written y = logb(x) c _ logistic growth model a function of the form f (x) = 1 + a c ________ 1+ ae−b x where limiting value, and b is a constant determined by the rate of growth is the initial value, c is the carrying capacity, or natural logarithm the exponent to which the number e must be raised to get x; loge(x) is written as ln(x). Newton’s Law of Cooling the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature nominal rate the yearly interest rate earned by an investment account, also called annual percentage rate order of magnitude the power of ten, when a number is expressed in scientific notation, with one non-zero digit to the left of the decimal power rule for logarithms a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base product rule for logarithms a rule of logarithms that states that the log of a product is equal to a sum of logarithms quotient rule for logarithms a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms Key Equations definition of the exponential function f (x) = b x, where b > 0, b ≠ 1 definition of exponential growth compound interest formula continuous growth formula General Form for the Translation of the Parent Function f (x) = b x f (x) = ab x, where a > 0, b > 0, b ≠ 1 r ) kt, where _ A = a ( 1 + k A(t) is the account value at time t t is the number of years P is the initial investment, often called the principal r is the annual percentage rate (APR), or nominal rate n is the number of compounding periods in one year A(t) = ae rt, where t is the number of unit time periods of growth a is the starting amount (in the continuous compounding formula a is replaced with P, the principal) e is the mathematical constant, e ≈ 2.718282 f (x) = ab x + c + d Definition of the logarithmic function Definition of the common logarithm For x > 0, b > 0, b ≠ 1, y = logb(x) if and only if b y = x. For x > 0, y = log(x) if and only if 10y = x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 566 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Definition of the natural logarithm General Form for the Translation of the Parent Logarithmic Function f (x) = logb(x) The Product Rule for Logarithms The Quotient Rule for Logarithms The Power Rule for Logarithms The Change-of-Base Formula For x > 0, y = ln(x) if and only if ey = x. f (x) = alogb(x + c) + d logb(MN) = logb(M) + logb(N) M ) = logbM − logbN logb ( __ N logb(Mn) = nlogbM logbM = lognM _ lognb n > 0, n ≠ 1, b ≠ 1 One-to-one property for exponential functions Definition of a logarithm One-to-one property for logarithmic functions For any algebraic expressions S and T and any positive real number b, where bS = bT if and only if S = T. For any algebraic expression S and positive real numbers b and c, where b ≠ 1, logb(S) = c if and only if bc = S. For any algebraic expressions S and T and any positive real number b, where b ≠ 1, logbS = logbT if and only if S = T. If A = A0e kt, k < 0, the half-life is t = − . t = ln(2) _ k A ln ( ) _ A0 _ −0.000121 is the amount of carbon-14 when the plant or animal died, A0 A is the amount of carbon-14 remaining today, t is the age of the fossil in years If A = A0e kt, k > 0, the doubling time is t = T(t) = Ae kt + Ts, where Ts is the ambient temperature, A = T(0) − Ts, and k is the continuous rate of cooling. ln(2) ___ k Half-life formula Carbon-14 dating Doubling time formula Newton’s Law of Cooling Key Concepts 6.1 Exponential Functions • An exponential function is defined as a function with a positive constant other than 1 raised to a variable exponent. See Example 1. • A function is evaluated by solving at a specific value. See Example 2 and Example 3. • An exponential model can be found when the growth rate and initial value are known. See Example 4. • An exponential model can be found when the two data points from the model are known. See Example 5. • An exponential model can be found using two data points from the graph of the model. See Example 6. • An exponential model can be found using two data points from the graph and a calculator. See Example 7. • The value of an account at any time t can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See Example 8. • The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See Example 9. • The number e is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is e ≈ 2.718282. • Scientific and graphing calculators have the key [e x] or [exp(x)] for calculating powers of e. See Example 10. • Continuous growth or decay models are exponential models that use e as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See Example 11 and Example 12. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 567 6.2 Graphs of Exponential Functions • The graph of the function f (x) = b x has a y-intercept at (0, 1), domain (−∞, ∞), range (0, ∞), and horizontal asymptote y = 0. See Example 1. • If b > 1, the function is increasing. The left tail of the graph will approach the asymptote y = 0, and the right tail will increase without bound. • If 0 < b < 1, the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = 0. • The equation f (x) = b x + d represents a vertical shift of the parent function f (x) = b x. • The equation f (x) = b x + c represents a horizontal shift of the parent function f (x) = b x. See Example 2. • Approximate solutions of the equation f (x) = b x + c + d can be found using a graphing calculator. See Example 3. • The equation f (x) = ab x, where a > 0, represents a vertical stretch if ∣ a ∣ > 1 or compression if 0 < ∣ a ∣ < 1 of the parent function f (x) = b x. See Example 4. • When the parent function f (x) = b x is multiplied by −1, the result, f (x) = −b x, is a reflection about the x-axis. When the input is multiplied by −1, the result, f (x) = b−x, is a reflection about the y-axis. See Example 5. • All translations of the exponential function can be summarized by the general equation f (x) = ab x + c + d. See Table 3. • Using the general equation f (x) = ab x + c + d, we can write the equation of a function given its description. See Example 6. 6.3 Logarithmic Functions • The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function. • Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm. See Example 1. • Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm See Example 2. • Logarithmic functions with base b can be evaluated mentally using previous knowledge of powers of b. See Example 3 and Example 4. • Common logarithms can be evaluated mentally using previous knowledge of powers of 10. See Example 5. • When common logarithms cannot be evaluated mentally, a calculator can be used. See Example 6. • Real-world exponential problems with base 10 can be rewritten as a common logarithm and then evaluated using a calculator. See Example 7. • Natural logarithms can be evaluated using a calculator Example 8. 6.4 Graphs of Logarithmic Functions • To
find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for x. See Example 1 and Example 2. • The graph of the parent function f (x) = logb(x) has an x-intercept at (1, 0), domain (0, ∞), range (−∞, ∞), vertical asymptote x = 0, and • if b > 1, the function is increasing. • if 0 < b < 1, the function is decreasing. See Example 3. • The equation f (x) = logb(x + c) shifts the parent function y = logb(x) horizontally • left c units if c > 0. • right c units if c < 0. See Example 4. • The equation f (x) = logb(x) + d shifts the parent function y = logb(x) vertically • up d units if d > 0. • down d units if d < 0. See Example 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 568 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS • For any constant a > 0, the equation f (x) = alogb(x) • stretches the parent function y = logb(x) vertically by a factor of a if ∣ a ∣ > 1. • compresses the parent function y = logb(x) vertically by a factor of a if ∣ a ∣ < 1. See Example 6 and Example 7. • When the parent function y = logb(x) is multiplied by −1, the result is a reflection about the x-axis. When the input is multiplied by −1, the result is a reflection about the y-axis. • The equation f (x) = −logb(x) represents a reflection of the parent function about the x-axis. • The equation f (x) = logb(−x) represents a reflection of the parent function about the y-axis. See Example 8. • A graphing calculator may be used to approximate solutions to some logarithmic equations See Example 9. • All translations of the logarithmic function can be summarized by the general equation f (x) = alogb(x + c) + d. See Table 4. • Given an equation with the general form f (x) = alogb(x + c) + d, we can identify the vertical asymptote x = −c for the transformation. See Example 10. • Using the general equation f (x) = alogb(x + c) + d, we can write the equation of a logarithmic function given its graph. See Example 11. 6.5 Logarithmic Properties • We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See Example 1. • We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See Example 2. • We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See Example 3, Example 4, and Example 5. • We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See Example 6, Example 7, and Example 8. • The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See Example 9, Example 10, Example 11, and Example 12. • We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of- base formula. See Example 13. • The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and e as the quotient of natural or common logs. That way a calculator can be used to evaluate. See Example 14. 6.6 Exponential and Logarithmic Equations • We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown. • When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example 1. • When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example 2, Example 3, and Example 4. • When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example 5. • We can solve exponential equations with base e, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example 6 and Example 7. • After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 569 • When given an equation of the form logb(S) = c, where S is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation bc = S, and solve for the unknown. See Example 9 and Example 10. • We can also use graphing to solve equations with the form logb(S) = c. We graph both equations y = logb(S) and y = c on the same coordinate plane and identify the solution as the x-value of the intersecting point. See Example 11. • When given an equation of the form logbS = logbT, where S and T are algebraic expressions, we can use the one- to-one property of logarithms to solve the equation S = T for the unknown. See Example 12. • Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example 13. 6.7 Exponential and Logarithmic Models • The basic exponential function is f (x) = ab x. If b > 1, we have exponential growth; if 0 < b < 1, we have exponential decay. • We can also write this formula in terms of continuous growth as A = A0e kx, where A0 is the starting value. If A0 is positive, then we have exponential growth when k > 0 and exponential decay when k < 0. See Example 1. • In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See Example 2. • We can find the age, t, of an organic artifact by measuring the amount, k, of carbon-14 remaining in the artifact and using the formula t = to solve for t. See Example 3. ln(k) _ −0.000121 • Given a substance’s doubling time or half-life we can find a function that represents its exponential growth or decay. See Example 4. • We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See Example 5. • We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See Example 6. • We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See Example 7. • Any exponential function with the form y = ab x can be rewritten as an equivalent exponential function with the form y = A0e kx where k = lnb. See Example 8. 6.8 Fitting Exponential Models to Data • Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. • We use the command “ExpReg” on a graphing utility to fit function of the form y = ab x to a set of data points. See Example 1. • Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. • We use the command “LnReg” on a graphing utility to fit a function of the form y = a + bln(x) to a set of data points. See Example 2. • Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit. • We use the command “Logistic” on a graphing utility to fit a function of the form y = points. See Example 3. c _________ 1 + ae−b x to a set of data Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 570 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAPTER 6 REVIEW EXERCISES EXPONENTIAL FUNCTIONS 1. Determine whether the function y = 156(0.825)t represents exponential growth, exponential decay, or neither. Explain 2. The population of a herd of deer is represented by the function A(t) = 205(1.13)t, where t is given in years. To the nearest whole number, what will the herd population be after 6 years? 3. Find an exponential equation that passes through the points (2, 2.25) and (5, 60.75). 4. Determine whether Table 1 could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points. x f (x) 1 3 2 0.9 Table 1 3 0.27 4 0.081 5. A retirement account is opened with an initial deposit of $8,500 and earns 8.12% interest compounded monthly. What will the account be worth in 20 years? 7. Does the equation y = 2.294e−0.654t represent continuous growth, continuous decay, or neither? Explain. 6. Hsu-Mei wants to save $5,000 for a down payment on a car. To the nearest dollar, how much will she need to invest in an account now with 7.5% APR, compounded daily, in order to reach her goal in 3 years? 8. Suppose an investment account is opened with an initial deposit of $10,500 earning 6.25% interest, compounded continuously. How much will the account be worth after 25 years? GRAPHS OF EXPONENTIAL FUNCTIONS 9. Graph the function f (x) = 3.5(2)x. State the domain and range and give the y-intercept. 1 ) 10. Graph the function f (x) = 4 ( __ 8 x and its reflection about the y-axis on the same axes, and give the y-intercept. 11. The graph of f (x) = 6.5x is reflected about the y-axis and stretched vertically by a factor of 7. What is the equation of the new function, g (x) ? State its y-intercept, domain, and range. 12. The graph below shows transformations of the graph of f (x) =
2x. What is the equation for the transformation1–1 –2 –3 –6 –5 –4 –3 –2 21 3 4 5 6 x LOGARITHMIC FUNCTIONS 13. Rewrite log17(4913) = x as an equivalent exponential 14. Rewrite ln(s) = t as an equivalent exponential equation. equation. Figure 1 − 2 __ = b as an equivalent logarithmic 5 15. Rewrite a equation. 1 ) to exponential form. 17. Solve for xlog64(x) = ( _ 3 19. Evaluate log(0.000001) without using a calculator. 16. Rewrite e−3.5 = h as an equivalent logarithmic equation. 18. Evaluate log5 ( 20. Evaluate log(4.005) using a calculator. Round to the ) without using a calculator. 1 _ 125 nearest thousandth. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 571 21. Evaluate ln(e−0.8648) without using a calculator. 22. Evaluate ln ( 3 √ — 18 ) using a calculator. Round to the GRAPHS OF LOGARITHMIC FUNCTIONS 23. Graph the function g(x) = log(7x + 21) − 4. 25. State the domain, vertical asymptote, and end behavior of the function g (x) = ln(4x + 20) − 17. LOGARITHMIC PROPERTIES nearest thousandth. 24. Graph the function h(x) = 2ln(9 − 3x) + 1. 26. Rewrite ln(7r · 11st) in expanded form. 27. Rewrite log8(x) + log8(5) + log8(y) + log8(13) in ) in expanded form. 67 28. Rewrite logm ( ___ 83 1 30. Rewrite ln ( x5 ) as a product. __ 32. Use properties of logarithms to expand log ( r 2s11 t14 ) . _ compact form. 29. Rewrite ln(z) – ln(x) – ln(y) in compact form. 1 ) as a single logarithm. 31. Rewrite −logy ( __ 12 33. Use properties of logarithms to expand ln ( 2b √ ______ b + 1 ) . _____ b − 1 to a single logarithm. 37. Rewrite 512x − 17 = 125 as a logarithm. Then apply the change of base formula to solve for x using the common log. Round to the nearest thousandth. 125 __ −x − 3 = 53 by rewriting each side with a 39. Solve 1 ( ) _ 625 common base. 34. Condense the expression 5ln(b) + ln(c) + 35. Condense the expression 3log7v + 6log7w − ln(4 − a) _ 2 log 7 u _ 3 to a single logarithm. 36. Rewrite log3(12.75) to base e. EXPONENTIAL AND LOGARITHMIC EQUATIONS 38. Solve 2163x · 216x = 363x + 2 by rewriting each side with a common base. 40. Use logarithms to find the exact solution for 41. Use logarithms to find the exact solution for 7 · 17−9x − 7 = 49. If there is no solution, write no solution. 3e6n − 2 + 1 = −60. If there is no solution, write no solution. 42. Find the exact solution for 5e3x − 4 = 6 . If there is 43. Find the exact solution for 2e5x − 2 − 9 = −56. no solution, write no solution. If there is no solution, write no solution. 44. Find the exact solution for 52x − 3 = 7x + 1. If there is 45. Find the exact solution for e 2x − e x − 110 = 0. If no solution, write no solution. there is no solution, write no solution. 46. Use the definition of a logarithm to solve. 47. Use the definition of a logarithm to find the exact −5log7(10n) = 5. solution for 9 + 6ln(a + 3) = 33. 48. Use the one-to-one property of logarithms to find an exact solution for log8(7) + log8(−4x) = log8(5). If there is no solution, write no solution. 49. Use the one-to-one property of logarithms to find an exact solution for ln(5) + ln(5x2 − 5) = ln(56). If there is no solution, write no solution. 50. The formula for measuring sound intensity in decibels D is defined by the equation D = 10log I _ ) , where I is the intensity of the sound in watts ( I0 per square meter and I0 = 10−12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a large orchestra with a sound intensity of 6.3 · 10−3 watts per square meter? 52. Find the inverse function f −1 for the exponential function f (x. 51. The population of a city is modeled by the equation P(t) = 256, 114e0.25t where t is measured in years. If the city continues to grow at this rate, how many years will it take for the population to reach one million? 53. Find the inverse function f −1 for the logarithmic function f (x) = 0.25 · log2(x3 + 1). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 572 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS EXPONENTIAL AND LOGARITHMIC MODELS For the following exercises, use this scenario: A doctor prescribes 300 milligrams of a therapeutic drug that decays by about 17% each hour. 54. To the nearest minute, what is the half-life of the drug? 55. Write an exponential model representing the amount of the drug remaining in the patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 24 hours. Round to the nearest hundredth of a gram. For the following exercises, use this scenario: A soup with an internal temperature of 350° Fahrenheit was taken off the stove to cool in a 71°F room. After fifteen minutes, the internal temperature of the soup was 175°F. 56. Use Newton’s Law of Cooling to write a formula that 57. How many minutes will it take the soup to cool models this situation. to 85°F? For the following exercises, use this scenario: The equation N(t) = school who have heard a rumor after t days. 1200 __ 1 + 199e−0.625t models the number of people in a 58. How many people started the rumor? 59. To the nearest tenth, how many days will it be before the rumor spreads to half the carrying capacity? 60. What is the carrying capacity? For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic. 61. 62. x f (x) x f (x) 1 3.05 0.5 18.05 2 4.42 1 17 3 6.4 4 9.28 5 13.46 6 19.52 7 28.3 8 41.04 9 59.5 10 86.28 3 15.33 5 14.55 7 14.04 10 13.5 12 13.22 13 13.1 15 12.88 17 12.69 20 12.45 63. Find a formula for an exponential equation that goes through the points (−2, 100) and (0, 4). Then express the formula as an equivalent equation with base e. FITTING EXPONENTIAL MODELS TO DATA 64. What is the carrying capacity for a population modeled by the logistic equation P(t) = initial population for the model? 65. The population of a culture of bacteria is modeled by the logistic equation P(t) = 250, 000 ___________ 1 + 499e−0.45t ? What is the 14, 250 __ 1 + 29e−0.62t , where t is in days. To the nearest tenth, how many days will it take the culture to reach 75% of its carrying capacity? For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places. 66. 67. 68. x f (x) x f (x) x f (x) 1 409.4 2 260.7 3 170.4 4 110.6 5 74 6 44.7 7 32.4 8 19.5 9 12.7 0.15 36.21 0.25 28.88 0.5 24.39 0.75 18.28 1 16.5 1.5 12.99 2 9.91 2.25 8.57 2.75 7.23 3 5.99 0 9 2 22.6 4 44.2 5 62.1 7 96.9 8 113.4 10 133.4 11 137.6 15 148.4 10 8.1 3.5 4.81 17 149.3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 PRACTICE TEST 573 CHAPTER 6 PRACTICE TEST 1. The population of a pod of bottlenose dolphins is 2. Find an exponential equation that passes through modeled by the function A(t) = 8(1.17)t, where t is given in years. To the nearest whole number, what will the pod population be after 3 years? the points (0, 4) and (2, 9). 3. Drew wants to save $2,500 to go to the next 4. An investment account was opened with an World Cup. To the nearest dollar, how much will he need to invest in an account now with 6.25% APR, compounding daily, in order to reach his goal in 4 years? 5. Graph the function f (x) = 5(0.5)−x and its reflection across the y-axis on the same axes, and give the y-intercept. initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? 6. The graph below shows transformations of the 1 ) graph of f (x) = ( __ 2 transformation? x . What is the equation for the –4 –3 –2 y 4 3 2 1 0 –1–1 –2 –3 –4 –5 –6 –7 –8 21 3 4 5 6 7 8 x 7. Rewrite log8.5(614.125) = a as an equivalent exponential equation. 1 __ 8. Rewrite e = m as an equivalent logarithmic 2 equation. 9. Solve for x by converting the logarithmic equation 10. Evaluate log(10,000,000) without using a calculator. log 1 _ 7 (x) = 2 to exponential form. 11. Evaluate ln(0.716) using a calculator. Round to the 12. Graph the function g (x) = log(12 − 6x) + 3. nearest thousandth. 13. State the domain, vertical asymptote, and end 14. Rewrite log(17a · 2b) as a sum. behavior of the function f (x) = log5(39 − 13x) + 7. 15. Rewrite logt(96) − logt(8) in compact form. 17. Use properties of logarithm to expand ln (y 3z 2 · 3 √ — x − 4 ). 19. Rewrite 163x − 5 = 1000 as a logarithm. Then apply the change of base formula to solve for x using the natural log. Round to the nearest thousandth. 21. Use logarithms to find the exact solution for −9e10a − 8 −5 = −41. If there is no solution, write no solution. 16. Rewrite log8 ( a 1 __ b ) as a product. 18. Condense the expression 4ln(c) + ln(d) + logarithm. ln(a) _ 3 + ln(b + 3) _ 3 to a single 20. Solve ( 1 ) = ( _ 9 side with a common base. 1 _ 243 1 ) _ 81 x · −3x − 1 by rewriting each 22. Find the exact solution for 10e 4x + 2 + 5 = 56. If there is no solution, write no solution. 23. Find the exact solution for −5e−4x − 1 − 4 = 64. If 24. Find the exact solution for 2x − 3 = 62x − 1. If there is there is no solution, write no solution. no solution, write no solution. 25. Find the exact solution for e2x − e x − 72 = 0. If there is no solution, write no solution. 26. Use the definition of a logarithm to find the exact solution for 4log(2n) − 7 = −11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 574 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 27. Use the one-to-one property of logarithms to find an exact solution for log(4x2 − 10) + log(3) = log(51) If there is no solution, write no so
lution. 28. The formula for measuring sound intensity in decibels D is defined by the equation I ) D = 10log ( __ I0 where I is the intensity of the sound in watts per square meter and I0 = 10−12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a rock concert with a sound intensity of 4.7 · 10−1 watts per square meter? 30. Write the formula found in the previous exercise as an equivalent equation with base e. Express the exponent to five significant digits. 32. The population of a wildlife habitat is modeled __ 1 + 6.2e−0.35t , where t is by the equation P(t) = 360 given in years. How many animals were originally transported to the habitat? How many years will it take before the habitat reaches half its capacity? 29. A radiation safety officer is working with 112 grams of a radioactive substance. After 17 days, the sample has decayed to 80 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest day, what is the half-life of this substance? 31. A bottle of soda with a temperature of 71° Fahrenheit was taken off a shelf and placed in a refrigerator with an internal temperature of 35° F. After ten minutes, the internal temperature of the soda was 63° F. Use Newton’s Law of Cooling to write a formula that models this situation. To the nearest degree, what will the temperature of the soda be after one hour? 33. Enter the data from Table 2 into a graphing calculator and graph the resulting scatter plot. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic. x f (x) 1 3 2 8.55 3 11.79 4 14.09 5 15.88 Table 2 6 17.33 7 18.57 8 19.64 9 20.58 10 21.42 34. The population of a lake of fish is modeled by the logistic equation P(t) = __ 1 + 25e−0.75t , where t is time in years. To the nearest hundredth, how many years will it take the lake to reach 80% of its carrying capacity? 16, 120 For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places. 35. 36. 37. x f (x) x f (x) x f (x) 1 20 2 21.6 3 29.2 4 36.4 5 46.6 6 55.7 7 72.6 8 87.1 9 107.2 10 138.1 3 13.98 4 17.84 5 20.01 0 2.2 0.5 2.9 1 3.9 6 22.7 1.5 4.8 7 24.1 2 6.4 8 26.15 9 27.37 10 28.38 11 29.97 12 31.07 13 31.43 3 9.3 4 12.3 5 15 6 16.2 7 17.3 8 17.9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Try It Answers Chapter 1 Section 1.1 4 3 11 __ __ _ c. − 2. a. 4 (or 4.0), terminating b. 0. b. 1 1 1 1. a. _ 615384 , 3. a. rational and repeating repeating c. −0.85, terminating b. rational and terminating c. irrational d. rational and repeating e. irrational 4. a. positive, irrational; right b. negative, rational; left c. positive, rational; right d. negative, irrational; left e. positive, rational; right 5. N W I Q Q' 6. a. 10 b. 2 c. 4.5 d. 25 e. 26 a. − 35 __ 7 b. 0 c. √ — 169 — 24 d. √ e. 4.763763763... × × × × × × × × × × × 7. a. 11, commutative property of multiplication, associative property of multiplication, inverse property of multiplication, identity property of multiplication; b. 33, distributive property; 4 _ c. 26, distributive property; d. , commutative property of 9 addition, associative property of addition, inverse property of addition, identity property of addition; e. 0, distributive property, inverse property of addition, identity property of addition 8. 9. a. 5 b. 11 c. 9 Constants Variables a. 2πr(r + h) b. 2(L + W) c. 4y3 + y 2, π 2 4 r, h L, W y d. 26 121 _ π 3 10. a. 4 b. 11 c. d. 1,728 e. 3 11. 1,152 cm2 2 _ 12. a. −2y − 2z or −2(y + z) b. − 1 c. 3pq − 4p + q d. 7r − 2s + 6 t 13. A = P(1 + rt) Section 1.2 1. a. k15 b. ( 2 _ y ) 3. a. (3y)24 b. t35 c. (−g)16 2 _ 5k 3 5. a. 5 c. t 14 2. a. s7 b. (−3)5 c. (ef 2)2 1 _ 4. a. 1 b. c. 1 d. 1 2 _ b. c. − 8. a. 1 _ 1 1 _ _ 6. a. t −5 = t 5 b. 25 a18b21 e. r12 s8 1 1 _ _ f 3 c. (−3t)6 b. 7. a. g10h15 b. 125t3 c. −27y15 d. q24 _ p32 e. 1 f. b15 _ c3 v 6 _ 8u3 b. 625_ u32 e 4 1 _ _ f 4 d. x 3 c. 1 _ w105 d. 27r _ s b. 7.158 × 109 c. $8.55 × 1013 d. 3.34 × 10−9 e. 7.15 × 10−8 11. a. 703,000 b. −816,000,000,000 c. −0.00000000000039 12. a. −8.475 × 106 b. 8 × 10−8 c. 2.976 × 1013 d. 0.000008 d. −4.3 × 106 e. ≈ 1.24 × 1015 13. Number of cells: 3 × 1013; length of a cell: 8 × 10−6 m; total length: 2.4 × 108 m or 240,000,000 m 10. a. $1.52 × 105 16h10 _ 49 1 _ c20d12 9. a. e. — 4. Section 1.3 1. a. 15 b. 3 c. 4 d. 17 2. 5∣ x ∣ ∣ y ∣ √ 2yz Notice the absolute value signs around x and y? That's because their value must be positive. — x √ 2 3. 10 ∣ x ∣ _ 3y2 We do not need the absolute value signs for y 2 because that term will always be nonnegative. 5. b4 √ 10. a. −6 b. 6 c. 88 13. 28 x 23 __ 15 6 9. 14 − 7 √ 9 ) 5 = 35 = 243 12. x (5y) 9 __ 2 5 7. 0 8. 6 √ 9 11. ( √ 3ab 6. 13 √ 3 √ — 3 — — — — — Section 1.4 1. The degree is 6, the leading term is −x 6,and the leading coefficient is −1. 2. 2x 3 + 7x 2 − 4x − 3 3. −11x 3 − x 2 + 7x − 9 4. 3x 4 − 10x 3 − 8x 2 + 21x + 14 7. 4x 2 − 49 6. 16x 2 − 8x + 1 8. 6x 2 + 21xy − 29x − 7y + 9 5. 3x 2 + 16x − 35 Section 1.5 1. (b 2 − a)(x + 6) 2. (x − 6)(x − 1) 3. a. (2x + 3)(x + 3) b. (3x − 1)(2x + 1) 5. (9y + 10)(9y − 10) 6. (6a + b)(36a2 − 6ab + b2) 8. (5a − 1 ) − 1 __ (17a − 2) 4 7. (10x − 1)(100x 2 + 10x + 1) 4. (7x − 1)2 1. Section 1. xy 2 5. 2. (x + 5)(x + 6) __ (x + 2)(x + 4) Chapter 2 Section 2.1 1. x y = 1 __ x + 2 (x, y) 2 −2 y = 1 __ (−2) + 2 = 1 (−2, 1) 2 (−1) + 2 = 3 −1 y = 1 ( −1, 3 __ __ __ ) 2 2 2 0 y = 1 __ (0) + 2 = 2 2 (1) + 2 = 5 1 y = 1 __ __ 2 2 2 y = 1 __ (2) + 2 = 3 2 (0, 2) ( 1, 5 __ 2 ) (2, 3) 3. 1 4. 2(x − 7) __ (x + 5)(x − 3) y 5 4 3 2 1 –1–1 –2 (−2, 1) –5 –4 –3 –2 (2, 3) (0, 2) 21 3 4 5 x — 5 — 125 = 5 √ 3. √ 5 4. ( −5, ) _ 2 2. x-intercept is (4, 0); y-intercept is (0, 3). y 5 4 3 2 1 21 3 4 5 x –5 –4 –3 –2 –1–1 –2 –3 A-1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. A-2 Section 2.2 TRY IT ANSWERS Chapter 3 Section 3.1 3. x = 2. x = −3 10 _ 4. x = 1 3 1 1 _ _ and − , excluded values are − . 2 3 8. y = 4x − 3 1. x = −5 7 _ 17 5. x = − 1 2 _ _ 6. x = 7. m = − 3 3 10. Horizontal line: y = 2 11. y 9. x + 3y = 2 Parallel lines: equations are written in slope-intercept form. 12. y = 5x + 3 5 4 3 2 1 21 3 4 5 x –5 –4 –3 –2 –1–1 –2 Section 2.3 1. 11 and 25 4. L = 37 cm, W = 18 cm 5. 250 ft2 2. C = 2.5x + 3,650 3. 45 mi/h Section 2.4 — — −24 = 0 + 2i √ 1. √ 3. (3 − 4i) − (2 + 5i) = 1 − 9i 6 2. 5 _ − i 4. 2 5. 18 + i 6. −3 − 4i 7. −1 –5 –4 –3 –2 i 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 r Section 2.5 1. (x − 6)(x + 1) = 0; x = 6, x = −1 3. (x + 5)(x − 5) = 0; x = −5, x = 5 x = 7, x = −3 2 1 _ _ 4. (3x + 2)(4x + 1) = 0; x = − 5. x = 0, x = −10, , x = − 4 3 2 22 8. x = − _ 7. x = 3 ± √ 3 2. (x − 7)(x + 3) = 0; — 5 — 6. x = 4 ± √ 9. 5 units x = −1 1 _ x = 3 Section 2.6 2. 25 3. {−1} 1 1 1 __ _ _ , x = − 4. x = 0, x = 1. 4 2 2 2 _ 5. x = 1; extraneous solution: − 6. x = −2; extraneous 9 3 _ 8. x = −3, 3, −i, i solution: −1 7. x = −1, x = 2 10. x = −1, 0 is not a solution. 9. x = 2, x = 12 Section 2.7 1. [−3, 5] 2. (−∞, −2)∪[3, ∞) 3. x < 1 3 6. [ − _ 5. (2, ∞) 14 1 1 8. ( − 9 10. , ∞#) 7. 6 < x ≤ 9 or (6, 9] 4. x ≥ −5 k ≤ 1 or k ≥ 7; in interval notation, this would be (−∞, 1]∪[7, ∞). 9 8 7 6 5 4 3 2 1 21 3 4 5 876 9 10 x –2 –1–1 –2 –3 — 4. g(5) = 1 7. g(1) = 8 (Note: If two players had been tied for, say, 4th 5. m = 8 8. x = 0 or x = 2 1. a. Yes b. Yes place, then the name would not have been a function of rank.) 2. w = f (d) 3. Yes 3 x ___ 6. y = f (x) = √ 2 9. a. Yes, because each bank account has a single balance at any given time; b. No, because several bank account numbers may have the same balance; c. No, because the same output may 10. a. Yes, letter grade correspond to more than one input. is a function of percent grade; b. No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade. 12. No, because it does not pass the horizontal line test. 11. Yes 5. a. Values that are less than or equal to –2, or Section 3.2 1 1 __ __ 1. {−5, 0, 5, 10, 15} 2. (−∞, ∞) 3. ( −∞, , ∞#) ) ∪ ( 2 2 5 , ∞#) 4. [ − __ 2 values that are greater than or equal to –1 and less than 3; b. { x \|x ≤ –2 or –1 ≤ x < 3 } c. (–∞, –2] ∪ [–1, 3) 6. Domain = [1950, 2002]; Range = [47,000,000, 89,000,000] 7. Domain: (−∞, 2]; Range: (−∞, 0] 8. y 21 1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 Section 3.3 1. $2.84 − $2.31 ___________ = 5 years $0.53 _____ 5 years = $0.106 per year. 1 __ 2. 2 4. The local maximum appears to occur at (−1, 28), 3. a + 7 and the local minimum occurs at (5, −80). The function is increasing on (−∞, −1) ∪ (5, ∞) and decreasing on (−1, 5). f (x) 321 4 5 6 7 8 x (5, −80) (−1, 28) 40 20 –4 –3 –2 –1 –20 –40 –60 –80 –100 –120 Section 3.4 1. a. (fg)(x) = f (x)g(x) = (x − 1)(x 2 − 1f − g)(x) = f (x) − g(x) = (x − 1) − (x2 − 1) = x − x 2 b. No, the functions are not the same. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. TRY IT ANSWERS A-3 2. A gravitational force is still a force, so a(G(r)) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G(a(F)) does not make sense. 3. f ( g (1)) = f (3) = 3 and g ( f (4)) = g (1) = 3 4. g ( f (2)) = g (5) = 3 5. a. 8; b. 20 7. Possible answer: g(x) = √ 6. [−4, 0) ∪ (0, ∞) 4 + x2 ; h(x ____ 3 − x Section 3.5 1. b(t) = h (t) + 10 = − 4.9t 2 + 30t + 10 2. y 5 4 3 2 1 –1 –1 –2 g(x) f(x) 21 3 4 5 –5 –4 –3 –2 The graphs of f (x) and g (x) are shown here. The transformation is a horizontal shift. The function is shifted to the left by 2 units. x 3. h(x) 4. g(x) = 1 ____ x – 1 +
1 h(x) ! \|x – 2\|+ 4 10 8 6 4 2 –6 –5 –4 –3 –2 –1 –2 21 3 4 5 6 x 5. a. y 3 2 1 –1 –1 –2 –3 –4 –3 –2 b. 21 3 4 x y 4 3 2 1 21 3 4 x –4 –3 –2 –1 –1 –2 6. a. g(x) = −f (x) b. h(x) = f (−x) 7. x −2 0 g(x) −5 −10 −15 −20 4 2 x −2 h(x) 15 0 10 2 4 5 unknown 7. y f (x) = x2 h(x) = f (− x)= (− x)2 Notice: h(x) = f (−x) looks the same as f (x). –10 –8 –5 –4 –3 –2 5 4 3 2 1 0 –1–1 –2 –3 –4 –5 8. Even 21 3 4 5 9. x x g(x) 6 4 2 9 12 15 8 0 10. g(x) = 3x − 2 g(x) = −f (x) = −x2 1 __ 11. g(x) = f ( x ) so using the square root function we get ___ 3 g(x) = √ 1 _ x 3 Section 3.6 1. Using the variable p for passing, \| p − 80 \| ≤ 20 3. x = −1 or x = 2 2. f (x) = −\| x + 2 \| + 3 Section 3.7 3. Yes 2. Yes 1. h(2) = 6 f −1 is ( −∞, −2) and the range of function f −1 is (1, ∞). 5. a. f (60) = 50. In 60 minutes, 50 miles are traveled. b. f −1(60) = 70. To travel 60 miles, it will take 70 minutes. 4. The domain of function 7. x = 3y + 5 6. a. 3 b. 5.6 domain of f : [0, ∞); domain of f −1: (−∞, 2] 9. y 8. f −1(x) = (2 − x)2; y=x f –1(x) 21 3 4 5 76 (x) –3 –2 –1–1 –2 –3 Chapter 4 Section 4.1 4 − 3 1 _ −2 _ 0 − 2 1,868 − 1,442 __ 2,012 − 2,009 1. m = 2. m = = 1 __ ; decreasing because m < 0. = − 2 426 _ 3 = 142 people per year = 3. y = −7x + 3 5. 4. H(x) = 0.5x + 12.5 y 6. Possible answers include (−3, 7), (−6, 9), or (−9, 11) (0, 6) (4, 3) (8, 0) 8 10 x 42 6 y = 2x + 4 y = x 8. (16, 0) 9. a. f (x) = 2x; b. g(x) = − 1 __ x 2 x + 6 10. y = − 1 __ 3 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 y 10 8 6 4 2 –6 –4 –2 –2 –4 –6 –8 –10 y = 2x Section 4.2 1. C(x) = 0.25x + 25,000; The y-intercept is (0, 25,000). If the company does not produce a single doughnut, they still incur a cost of $25,000. 2. a. 41,100 b. 2020 3. 21.15 miles Section 4.3 1. 54° F 2. 150.871 billion gallons; extrapolation Chapter 5 Section 5.1 1. The path passes through the origin and has vertex at (−4, 7), so (h)x = − 7 __ 16 need to be about 4 but h(−7.5) ≈ 1.64; he doesn’t make it. 2. g(x) = x 2 − 6x + 13 in general form; g(x) = (x − 3)2 + 4 in standard form (x + 4)2 + 7. To make the shot, h(−7.5) would Download the OpenStax text for free at http://cnx.org/content/col11759/latest. A-4 TRY IT ANSWERS 8 _ 11 4. y-intercept at (0, 13), No x-intercepts 3. The domain is all real numbers. The range is f (x) ≥ 8 [ _ 11 5. 3 seconds; 256 feet; 7 seconds , ∞#) . , or 3. The degree is 6. The leading term is −x 6. Section 5.2 1. f (x) is a power function because it can be written as f (x) = 8x 5. The other functions are not power functions. 2. As x approaches positive or negative infinity, f (x) decreases without bound: as x → ±∞, f (x) → −∞ because of the negative coefficient. The leading coefficient is −1. 4. As x → ∞, f (x) → −∞; as x → −∞, f (x) → −∞. It has the shape of an even degree power 5. The leading term is function with a negative coefficient. 0.2x 3, so it is a degree 3 polynomial. As x approaches positive infinity, f (x) increases without bound; as x approaches negative infinity, f (x) decreases without bound. 6. y-intercept (0, 0); x-intercepts (0, 0), (−2, 0), and (5, 0) 7. There are at most 12 x-intercepts and at most 11 turning points. 8. The end behavior indicates an odd-degree polynomial function; there are 3 x-intercepts and 2 turning points, so the degree is odd and at least 3. Because of the end behavior, we know 9. The x-intercepts that the lead coefficient must be negative. are (2, 0), (−1, 0), and (5, 0), the y-intercept is (0, 2), and the graph has at most 2 turning points. Section 5.3 1. y-intercept (0, 0); x-intercepts (0, 0), (−5, 0), (2, 0), and (3, 0) 2. The graph has a zero of −5 with multiplicity 1, a zero of−1 with multiplicity 2, and a zero of 3 with even multiplicity. Section 5.6 1. End behavior: as x → ±∞, f (x) → 0; Local behavior: as x → 0, f (x) → ∞ (there are no x- or y-intercepts). 2. y The function and the asymptotes are shifted 3 units right and 4 units down. As x → 3, f (x) → ∞, and as x → ±∞, f (x) → −4. The function is f (x) = ______ (x − 3)2 − 4. 1 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x y = –4 x = 3 3. 4. The domain is all real numbers except x = 1 and x = 5. 12 __ 11 5. Removable discontinuity at x = 5. Vertical asymptotes: x = 0, x = 1. 6. Vertical asymptotes at x = 2 and x = –3; horizontal asymptote at y = 4. squared function, we find the rational form. 7. For the transformed reciprocal f (x) = 1 ______ (x − 3)2 − 4 = 1 − 4(x − 3)2 ___________ (x − 3)2 = 1 − 4(x2 − 6x + 9) _______________ (x − 3)(x − 3) = −4x2 + 24x − 35 ______________ x2 − 6x + 9 Because the numerator is the same degree as the denominator we know that as x → ±∞ , f (x) → −4; so y = –4 is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is x = 3, because as x → 3, f (x) → ∞. We then set the numerator equal to 0 and find the x-intercepts are at (2.5, 0) and (3.5, 0). Finally, we evaluate the function at 0 and find 3. y 10 –4 –3 –2 –1 1 2 x –10 –20 –30 –40 –50 4. Because f is a polynomial function and since f (1) is negative and f (2) is positive, there is at least one real zero between x = 1 and x = 2. 8. 5. f (x) = − 1 _ (x − 2)3(x + 1)2(x − 4) 6. The minimum occurs 8 at approximately the point (0, −6.5), and the maximum occurs at approximately the point (3.5, 7). –10 –8 –6 –4 Section 5.4 y 6 5 4 3 2 1 –2 –1 –2 –3 –4 –5 642 8 10 the y-intercept to be at ( 0, − 35 ) . _ 9 1 _ Horizontal asymptote at y = . 2 Vertical asymptotes at x = 1 4 and x = 3. y-intercept at ( 0, _ ) . 3 x-intercepts at (2, 0) and (–2, 0). (–2, 0) is a zero with multiplicity 2, and the graph bounces off the x-axis at this point. (2, 0) is a single zero and the graph crosses the axis at this point. x 1. 4x 2 − 8x + 15 − 2. 3x 3 − 3x 2 + 21x − 150 + 78 _ 4x + 5 1,090 _ x + 7 3. 3x 2 − 4x + 1 Section 5.5 1. f (−3) = −412 2. The zeros are 2, −2, and −4. 1 _ 3. There are no rational zeros. 4. The zeros are −4, , and 1. 2 5. f (x positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros. 7. 3 meters by 4 meters by 7 meters 6. There must be 4, 2, or x 2 − 2x + 10 Section 5.7 1. f −1( f(x)) = f −1 ( ) − 5 = (x − 5) + 5 = x x + 5 _____ 3 ) = 3 ( x + 5 _____ 3 (3x − 5) + 5 __________ = 3 3. f −1(x) = √ x − 1 — 3x __ 3 = x , x ≥ 0 5. f −1(x) = 2x + 3 ______ x − 1 and f ( f −1 (x)) = f (3x − 5) = 2. f −1(x ______ 2 4. f −1(x) = Section 5.8 1. 128 ____ 3 9 __ 2. 2 3. x = 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. TRY IT ANSWERS A-5 Chapter 6 Section 6.1 — 3. About 1.548 billion people; by the year 2031, 1. g(x) = 0.875x and j(x) = 1095.6−2x represent exponential functions. 2. 5.5556 India's population will exceed China's by about 0.001 billion, or 1 4. (0, 129) and (2, 236); N(t) = 129(1.3526)t million people. — 2 )x ; Answers may vary due to 5. f (x) = 2(1.5)x 6. f (x) = √ 2 ( √ round-off error. the answer should be very close to 1.4142(1.4142)x. 7. y ≈ 12 · 1.85x 8. About $3,644,675.88 10. e−0.5 ≈ 0.60653 11. $3,659,823.44 12. 3.77E-26(This is calculator notation for the number written as 3.77 × 10−26 in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.) 9. $13,693 Section 6.2 1. f(x) f(x (–1, 0.25) –5 –4 –3 –2 –1–1 –2 (1, 4) (0, 1) 21 3 4 5 x The domain is ( −∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. The domain is ( −∞, ∞); the range is (3, ∞); the horizontal asymptote is y = 3. 2. f(x) (–1, 3.25) 10 8 6 4 2 –5 –4 –3 –2 –1–2 –4 f(x) = 2 x – 1 + 3 (1, 4) y = 3 (0, 3.5) 21 3 4 5 x 3. x ≈ −1.608 4. f(x) The domain is ( −∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. 1 f(x) = (4)x 2 (1, 2) (0, 0.5) y = 0 21 1–1 –2 (–1, 0.125) –5 –3 –2 –4 5. g(x) = 1.25 –x (–1, 1.25) g(x) 5 4 3 2 1 (0, 1) (1, 0.8) –10 –8 –6 –4 –2–1 42 6 8 10 y = 0 x The domain is ( −∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. 1 __ 6. f(x) = − ex − 2; the domain is ( −∞, ∞); the range is 3 (− ∞, 2); the horizontal asymptote is y = 2. Section 6.3 1. a. log10(1,000,000) = 6 is equivalent to 106 = 1,000,000 b. log5(25) = 2 is equivalent to 52 = 25 2. a. 32 = 9 is equivalent to log3(9) = 2 b. 53 = 125 is equivalent to log5(125) = 3 1 1 is equivalent to log2 ( _ _ ) = −1 c. 2−1 = 2 2 1 121 = 121 1 _ 3. log121(11) = (recalling that √ 2 _ = 11) 2 — 1 4. log2 ( _ ) = −5 5. log(1,000,000) = 6 6. log(123) ≈ 2.0899 32 8. It is not 7. The difference in magnitudes was about 3.929. possible to take the logarithm of a negative number in the set of real numbers. , 11 5 (1, 0) 42 6 8 f (x) = log (x) x = 0 10 1 5 Section 6.4 1. (2, ∞) 3. 2. (5, ∞) f(x) 4 3 2 1 –2–1 –2 –3 –10 –8 –6 –4 4. x = −4 (−1, 1) y 5 4 3 2 1 –6 –5 –4 –3 –2 –1 (−3, 0) –1 –2 –3 –4 –5 The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. x The domain is (−4, ∞), the range (−∞, ∞), and the asymptote x = –4. x = 0 f (x) = log3(x + 4 ) y = log3(x) 5 4 6 (3, 1) x 21 3 (1, 0) The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. 5. y f(x) = log2(x) + 2 x = 0 (0.5, 1) (0.25, 0) (2, 1) (1, 0) y = log2(x) x 6. y x = 0 y = log4(x) (4, 1) (1, 0) 1 f(x) = log4(x) 2 (16, 1) x The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. 7. 8 10 –10 –8 –6 –4 –2–1 –2 –3 –4 –5 y 4 3 2 1 –10 –8 – – 46 –2–1 –2 42 x = 0 6 8 10 The domain is (2, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 2. x The domain is (−∞, 0), the range is (−∞, ∞), and the vertical asymptote is x = 0. 9. x ≈ 3.049 11. f (x) = 2ln(x + 3) − 1 10. x = 1 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. A-6 Section 6.5 TRY IT ANSWERS 1. logb(2) + logb(2) + logb(2) + logb(k) = 3logb(2) + logb(k) 2. log3(x + 3) − log3(x − 1) − log3(x − 2) 4. −2ln(x) 5. log3(16) 7. 2 1 __ __ ln(x − 1) + ln(2x + 1) − ln(x + 3) − ln(x − 3) ln(x) 8. 2 3 6. 2log(x) + 3log(y) − 4l
og(z) 3. 2ln(x) 9. log ( 3 ⋅ 5 ____ 4 ⋅ 6 5 __ ) ; can also be written log ( ) by reducing the fraction 8 — to lowest terms. x 10. log ( ) (2x + 3)4 ; this answer could also be written log ( 5(x − 1)3 √ ___________ (7x − 1) x12(x + 5)4 ________ 11. log 12. The pH increases by about 0.301. 13. ln(8) ____ ln(0.5) 14. ln(100) _____ ≈ ln(5) 4.6051 _____ 1.6094 = 2.861 4 x3(x + 5) ) _______ . (2x + 3) Section 6.6 1. x = −2 2. x = −1 2 4. The equation has no solution. 11 11 __ __ ) or ln ( 6. t = 2ln ( ) 3 3 7. t = ln ( 1 1 __ _ ) = − ln(2) — 2 2 √ 10. x = e5 − 1 11. x ≈ 9.97 ln(0.8) ______ ln(0.5) 1 __ 3. x = 2 5. x = ln(3) _ 2 _ ) ln ( 3 8. x = ln(2) 9. x = e4 12. x = 1 or x = −1 13. t =#703,800,000 × years ≈ 226,572,993 years. Section 6.7 1. f (t) = A0 e −0.0000000087t 2. Less than 230 years; 229.3157 to be exact 3. f (t) = A0 e ( ln(2) 5. 895 cases on day 15 6. Exponential. y = 2e 0.5x 4. 6.026 hours 7. y = 3e (ln 0.5)x ) t ____ 3 Section 6.8 1. a. The exponential regression model that fits these data is y = 522.88585984(1.19645256)x. b. If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year. 2. a. The logarithmic regression model that fits these data is y = 141.91242949 + 10.45366573ln(x) b. If sales continue at this rate, about 171,000 games will be sold in the year 2015. 3. a. The logistic regression model that fits these data is y = 25.65665979 _____________________ 1 + 6.113686306e−0.3852149008x . b. If the population continues to grow at this rate, there will be c. To the nearest whole number, about 25,634 seals in 2020. the carrying capacity is 25,657. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Answers CHAPTER 1 Section 1.1 3. Th 1. . Th d diffff 9. − 11. 23. − 35. 43. z− 51. x 49. x_ 55. 45. y+ −+ 31. − 41. −b+ 13. − 25. 17. 29. − 39. −y− 15. 27. 47. −b+ 33. ± 21. 7. − 5. − 37. 19. 53. 57. g+−= 59. 65. 61. 67. 63. Section 1.2 1. ××× 3. 5. 7. 17. 25. a 35. y x 19. 27. b c 29. ab d 37. 39. a 9. 11. 21. ×− 13. 15. 23. 31. m c b 41. 33. q p y z 43. 45. 47 × 49. 51. m 53. a 55. n a c 57. a b c 59. Section 1.3 1. t. Th — 13. √ _____ 3. Th — 11. √ — 19. √ 21. √ 15. 17. √ 9. 5. 7. — — 23. √ — 31. √ — 25. 33. √ — 27. — √ 35. x 29. − — +√ __________ 37. √ — p 45. y √ — 39. m √ — m 41. b√ — a 43. x — 47. √ d ______ d — x−√ √ x 53. — 49. √ — — +√ x ____________ −#x 55. n√ — 51. −w√ — w 57. — √ m m 59. d 61. — x √ ______ — √ mnc ______ acmn 69. 63. z √ — 65. 67 — #−# −√ _________ 71. √ — — x+√ ___________ 73. — √ ____ Section 1.4 9. 5. 11. x +x+ 15. b− b+b−b+ 1. Th Th 3. 7. 13. w +w+ 17. x −x− 21. v −v+ 25. y −y+ 29. y −y+ 35. −m+ 41. y −y −y+ 45. a−b 47. t −tu+u 49. t +x +t−tx−x 51. r +rd−d 55. t −t +t+ 37. q − 39. t +t −t −t+ 27. p+p+ 31. c − 57. a +ac−ac −c 23. n−n+n− 43. p−p −p+ 19. b−b+ 53. x −x−m 33. n− Section 1.5 15. p+p− 5. m 7. m 9. y 11. a−a+ 1. Th x −y x −y =x+yx−y 3. x 13. n − n+ 17. h+h− 21. t+t − 27. d+d− 25. p+p− 31. n+ 29. b+cb−c 35. p− 37. x+x −x+ 39. a+a−a+ 41. x−x +x+ 43. r+sr −rs+s 45. c+− −c− 47. x+ x+ 51. x−x+ 55. x+x− 59. 53. x+x− 57. z +az+az−a 19. d−d− 23. x+x− ___ − z− 49. z− 33. y+ x+x−x+ Section 1.6 1. 3. B-1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-2 5. 7. b+ 9. x+ x+ 11. a+ a− d − d − 17. 19. 15. c− c+ x− x+ 31. y− y+ 23. y+ y+ n− n− t+ t+ b+ b− y −y+ y −y− b+a ab x+ x− 13. 21. 29. 37. 43. 51. 25. p+ p+ x+y xy 27. d+ d+ 35. a− a−a− 33. 39. z +z+ z −z− 41. x+xy+y x+xy+y+ 45. 53. +ab b x+ x− 47. a−b 49. c +c− c +c+ 55. y+ 57. ANSWERS 31. 33. − x y − – – – – – – – – – – 35. x − y 37. x − y Chapter 1 Review Exercises 1. − 3. 5. y= 7. m 11. 13. 15. a 17. 19. a 9. x y 27. 25. √ — 33. − — √ 35. x +x + − – – – – – 39. 23. — — 31. √ 21. × 29. √ 37. x −x+ 43. a+ab−b 49. a−a+ 55. p+p−p+ −p− 59. p+ 39. k−k− 45. p 51. x+ 47. a 53. h−k 41. x +x +x+ 57. q−pq+pq+p 61. x+ 65. m+ 63. x− m 41 – – – – – – – 67. x+y xy 69. Chapter 1 Practice Test 1. 3. x= 9. 11. x 13. 5. 15. √ x — 7. 17. √ — 19. q −q −q 23. x+x− 21. n−n +n− 25. c−c +c+ 27. z− z− 29. a+b b CHAPTER 2 Section 2.1 1. xy 3. y 5. x y 7. x y y− y( ) 15. y=x− 17. d=√ 19. d=√ ) 25. − 27. 29. y= 21. d≈ 23. ( − 9. x #−#x 11. y=−x 13. y= = — — 43. d= 45. d= 47. − 49. x=y=− 51. x=y= 55. −= 61. 53. x=−y= 57. 59. 63. 54 ft 3. Th Section 2.2 1. 5. x 7. x= 9. x= 11. x= 13. x= 17. x≠−x=− 19. x≠x= 21. x≠x=− ) :(, = − ) 31. :(,) :(, m= 15. x=− ; ; 23. 29. 33. ; 27. :(, 35. = = 39. =; :(, x+ ___ : 43. y=− y=− :(,) :(,) ) 51. y=− 49. ) x+ y= 25. m=− =; = ; = − 41. :; ___ y=− x+ 45. ; 37. 47. 53. x+y= Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 55. y 57. y ANSWERS B-3 – – – – –– – – – – x – – – – –– – – – – x 61. y = −3x − 5 5. xx = ax+b=d 9. x=− x=− 15. x=− 11. x=x=− 7. x=x= 13. x=− x= ___ 17. x=x=− 19. x=−x= 21. x=x=− 27. x=±√ — 23. x=x=− 29. z= z=− m= C x+ B Th A 67.y=− B 65. = 69. 71. 30 ft m=− 58. 63. = − + Section 2.3 1. 3. −x 9. +m 7. 17. −x 13. +P 15. 21. 23. 27. 25. B=+x 31. r= 11. 29. R= 5. v+ 19. 33. W= 35. f= 39. h= P−L = pq p+q A b+b = = − + = 37. m=− 41. = =160 ft 47. 49. h= V πr 45. A= 53. C=π 43. 51. r =√ V πh Section 2.4 33. 35. 39. x= — −±√ — −±√ ± x≈−x≈ 45. = 53. ± 25. x=−x= — ±√ 31. x= 37. — ±√ ± 41. x= 43. x= 47. = 49. = 51. x≈x≈ 55. ax +bx+c= b x=− c x + a a c b b =− x+ + x+ a a a ) ( x+ b a x+b a =± √ = b a b− ac a b− ac a — −b±√b ac a − x= xx+= Th 57. 61. x−x−x+=300. Thx x= 63. 59. 1. 3. ii Section 2.6 5. −+i 7. +i 11. i 13. 9. − + i −−−− −− − − − − r −−−− 15. −i 17. −+i 19. −i 23. −+i 25. −−i 27. 31. −i 33. + i 39. 41. − 43. i 47. i 49. 51. −i 35. i 45. ( — i ) √ + 53. −i i −− − − − − r 21. +i 29. − i — 37. +i√ =− − i 55. 1. 3. − . Th − 5. 7. x= 9. x= 11. x=x= 15. y= − 23. t= 21. x= 17. m=− 13. x=−− 19. x= 27. x=− 33. x=− 39. x=− 45. x=−− 29. x=− 35. x=− 41. x= 47. 25. x= 31. x=− 37. x=−− 43. 49. Section 2.5 1. 3. a ∙ b= ff a=b= Section 2.7 1. 3. −∞∞ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-4 ANSWERS 5. x= y = x−e left y ____ __ ] 9. [ − 7. ( −∞ ∞) 11. −∞ 13. ( −∞− 16. x > −6 and x > −2 x > −2, (−2, +∞) x < − 3 or x ≥ 1 (−∞, −3) ∪ [1, ∞) All real numbers (−∞, ∞) Take the intersection of two sets. Take the union of the two sets. 5,9) 18. 23. 21. ] ( [10, ,9/2] ( [ 1 All real numbers (−∞, ∞) (−∞, −4] ∪ [8, +∞) 1 ) 27. [2,5] 29. No solution [−11, −3] y 45. 65. −∞+∞ 67 69. x=− 71. − 73. ≤T≤≤T≤ Chapter 2 Review Exercises 1. xy− 3. y= x+ 9. ( ) 5. √ — =√ — 7. 11 – – – – – – – 13. x= 15. x= 17. 19. y= x+ 21. y= x+ 23. 25. 27. x=− ± — i√ 12 10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10 –12 42 6 8 10 x 47. It is never less than zero. No solution. y 12 10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10 –12 31. +i 33. i 35. −−i 42 6 8 10 x 39. x=−i 45. x=− 51. x=±√ — 41. x=−− — −±√ 47. x= 53. x= 29. − − — 37. −−i√ 43. x= − 49. x= — 55. x=±√ 57. x=− ____ ] [ − 63. − 59. x= 65. ____ 61. −∞ ) 67. ( − 69. x=∞ 2 4 6 8 10 x 51. Where the blue line is above the red line; always. All real numbers. ( −∞, −∞) y 12 10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10 –12 42 6 8 10 25. 30. 34. 40. 43. [−10, 12] 42. [6, 12] (−∞, −1) ∪ (3, ∞) y 12 10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10 –12 2 4 6 8 10 x 32. ( −∞, − _ ) ∪ (4, ∞) 36. No solution 38. (−5, 11) 49. Where the blue line is above the red line; point of intersection is x = −3. (−∞, −3) y 12 10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10 –12 53. 55. 57. 59. 61. 63. (−1, 3) (−∞, 4) {x\|x < 6} {x\|−3 ≤ x < 5} (−2, 1] (−∞, 4] Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-5 Chapter 2 Practice Test 1. y= x. − y __ __ 73. f−= f=f=f= 75. f−= 77. Th 79. Th − 13. − 5. −∞ 7. x=− — ±√ 11. x= 17. y= x− ± √ 25. x= 19. i — 27. 9. x≠−x=− 15. y=− x− i − 29. x= 21. − 23. x=− – – – – – – – 81. Th − y 83. Th 0, 10 y x – x CHAPTER 3 Section 3.1 1. 3. 5. 9. 11. 7. 13. 17. 15. 23. 21. 27. f−=−f=−f−a=−a−−fa=−a+ fa+h=a+h− f−a=√ +f= −a −fa+h= 19. 25. +a +−fa=−√ 29. f−=√ — — — — −a−h + 31. f−=f=− √ f−a=∣ −a−∣−∣ −a+∣−fa=−∣ a−∣+∣ a+∣ fa+h=∣ a+h−∣−∣ a+h+∣ 33. gx−ga _ x−a 37. a. f = b. x=− =x + a+ x ≠ a 35. a. f −= b. x= __ 39. a. r=− t 43. 49. 41. b. f−= c. t= 45. 47. 53. a. f = b. f x=−x=− 51. 55. 57. 59. 65. 69. f−=f−=f=f=f= 71. f−=f−=f=f=f= 67. f x=x= 61. 63. 85. Th − 87. Th − – 89. a.g=b.Th f 100 s 91. a. fts 200 ft. b. Thfts 350 ft. Section 3.2 — √ 1. Th — 3. Thxfx= x −∞∞ x fx= √ ∞ xy −∞∞ x 5. 9. −∞ )∪ ( − 15. ( −∞− ∞#) 17. −∞−∪−∪∞ 19. −∞−∪−∪∞ 11. −∞∞ 13. −∞∞ 7. −∞∞ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-6 ANSWERS 23. ∞ 21. −∞ 25. −∞−∪−∪∞ 27. 29. − 31. − 33. −∞∞ ] [ −− 35. [ −− ] ] ∪[ ] ∪[ 37. −∞∞ 39. −∞∞ y 41. −∞∞ y – – – – – – – – – – 43. −∞∞ y x x x – – – – – – – 45. −∞∞ y x – – – – – – – – – – – – – – – – – 47. f−=f−=f−=f= 49. f−=−f=f=f= 51. f −=−f=f=f= 53. −∞∪∞ 55 Th Th 57. 59. fx= — x− √ 61. a. Th b. Th c. Th Section 3.3 8. − 6. x+h 1. 3. Th 10. _________ 14. +h x+h− 31. x+a 33. 25. 29. + −∞−∪∞ h+h+ ≈− b+ − + 23. 20. 18. 12. 26. 16. 35. 27. − 37. −∞∪ ∪∞ − 40. 42. −− 44. a. − b. − 45. −−∞∞ 47. −−−− −∞ −−−−∞ −−−−∞ − 49. 51. 57. 53. 55. b= ≈− ≈ Section 3.4 1. g f g 3. f x= x+gx= x− Thf (gx=f x−=x−+=x g ( f x=g x+=x+−=x f∘g=g∘f 5. f+gx=x+−∞∞ f−gx=x +x−−∞∞ fgx=−x −x +x +x−∞∞ f ( g ) x= _ 7. f+gx= x +x −x −∞− √ x+x+ −∞∪∞ x ∪−√ ∪√ ∞ √ — — — — f−gx= x+x− −∞∪∞ x — — fgx=x+−∞∪∞ f ( g ) x=x +x −∞∪∞ _ x−∞ 9. f+gx=x +√ f−gx=x −√ fgx=x √ f ( x g ) x= _ x− √ b. fgx=#x −x+ d. g ∘ gx=x− e.f ∘ f−= x−∞ x−∞ ∞ — — 11. a.fg= c.gfx=x − 13. fgx=√ — x++gfx=x+√ — x + — — x + √ ; gfx= x x+ √ 15. fgx= x x __ x≠gfx=x−x≠ 17. fgx= x++ — −x √ __ ) b. ( −∞ 21. a.g∘f x=− 19. fghx= 23. a. ∪∞x=−b.∞c.∞ 25. ∞ 31. fx= 27. fx=xgx=x− x ; gx=x+ 29. fx= x ; gx= √ x− x− x+ x ; gx=x+ 35. fx=√ 33. fx= x ; gx= √ — — — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-7 37. fx= √ — x gx=x− 29. 39. fx=x gx= x− 41. fx=√ x ; gx= — x− x+ 43. 55. 67. 45. 57. 69. 75. fg= gf= 47. 59. 71. 51. 63. 49. 61. 73. fg=gf=− 53. 65. 77. fgx=x+x+ – – – –
– 87. f∘g =g∘f = 85. 81. f∘gx=g∘f x= 79. g∘gx=x+ 83. −∞∞ 89. f∘g =g∘f = 93. At=π ( √ 97. a.NTt=t +t− b. ≈ t+)A=π ( √ 95. A=π 91. — )=π — Section 3.5 1. 3. 5. f −xxfx f−x= fx f−x= −fx x++ 9. gx= 7. gx= ∣ x−∣− 11. Thfx+ f 13. Thfx− 15. Thfx+ f f 19. Th f fx+−hift 21. −∞− f −∞ 17. Thfx− 23. ∞ 25. y 27 – – – – – – – – – – – – – – – — — fx= ∣ x−∣− fx= √ x+− fx= x− fx= ∣ x+∣− — fx= − √ x fx= −x++ fx= √ 31. 33. 35. 37. 39. 41. 43. 45. 49. g x Thg f f −x + 47. 51. Th 53. 55. Thg Thg f f y Thg 59. 57. f 63. gx= x+− hift 67. hift y gx= ∣ −x ∣ 61. gx= x−+ 65. 69. hift hift – – – – – – – – – – – 71. y Th 73. hift – – – – – – – – – Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-8 ANSWERS Th 25. Th Th – – – – – 29. – – – – 27 31 – – – – – – – 75. Th fx= √xhift y 77. 79. 81. y a x Section 3.6 1. ∣ A ∣ =B A B A −B 3. x 5. Thx x ∣ x−∣= 7. ∣ x−∣≥ 9. Thx 11. − 13. −− 15. − 17. y 19. y – – x – – – – – 23. x – – – – – 21 33. − y 35 37. Tha yTh yx= 39. ∣ p−∣≤ 41. ∣ x−∣≤ SECTION 3.7 1. y y 5. y=f −x 7. 11. 9. x 3. f x= f −x=x− 13. f −x=−x f −x= 17. f −x= 15. 19. f −x= 23. f −x= 29. f −xx= p+ + f −x= f −x= 21. 25. 31. r + 27. f −x= + 33. 35. 37. f x−∞f −x=√x − — f x∞f −x=√x+ fgx=xgfx=x — 39. 41. 43. 45. 47. 49. 51. 53. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-9 55. y f – f –– – – –– – 69. f −x=+ – – – – – 57. 59. 61. − 63. x 65. 67. x f −1(x) 71. 73. f −x = x − d _ t= td= Th x 49. – – – – 53. y – – – – – – y Chapter 3 Review Exercises 1. 3. 5. f−=−f=− f−a=−a−a; −fa=a−a; fa +h=−a−ah −h+a +h 7. 11. 9. 13. – – – – 23 15. 19. 17. − −+a−a −+a =−a +a≠ 21. −∞−∪(−∪(∞ 25. 27. ∞ −∞ 29. −−∞−∞ 31. −− 33. 35. f∘g x=−x, g∘f x=−−x 37. f∘g x= √ x + g∘f x= _ — x+ √ 39. f∘g x= 41. f∘g x= +x _ +x x + = x + ∞) — x √ )∪ ( − ( −∞− )∪(∞) 43. gx= 47. fx=√ — x 45 51. y – – – – – – – – – – – – – – – – 55. fx=∣ x −∣ 57. 59. 61. x ∣ x +∣+ 63. fx= 65. fx=−∣ x−∣+ x − 69. f −x= 71. f −x=√ — x − 73. Th – – – – – – – 67. y – – – – – – – – – 75. 5. Th y 3. − 15. 9. −a+b+b≠a Chapter 3 Practice Test 1. — 7. a−a 11. √ 13. 17. x+ 21. −∞−∞ 23. − 27. fx={ \|x\| x ≤ x > 29. x = 31. x − 33. f −x=− 19. f−x 25. f= −x_ CHAPTER 4 Section 4.1 3. dt = −t 1. 5. Thaa ya xa. Th Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-10 ANSWERS 9. 11. 27. − 31. y = x− 7. 17. 23. 13. 15. 21. 29. _ x − 19. 25. 33. y = − x+ 37. 39. 41. yx 43. y−x ( ) 45. yx− 47. m=−m=− 49. m=−m= 35. y = −x− 51. m=− m=− 55. y=− __ t+ 57. 53. y =x− 61. y =x− 63. y =− 65. 59. y=− __ x+ 67. 69. x 99. y − −−− −− − − − − − 103. y – – x – – – x 101. aa = b = b. qp = p− 107. x = a 105. x = − __ d ____ c−a 109. y= x− ad ____ c−a 111. y=x− 113. x< x> 115. 71. – – – – – 75. – – – – – 79. – – – – – 83 – – – – – – – 73. x – – – – – 77. x – – – – – 81 – – – – – – – 85. y = 87. x = − 89. gx = −x+ 91. fx = x− 93. gx = − __ 95. fx = x− 97. fx = −x+ x+ x 117. Th ts. Th 119. Th− 121. Section 4.2 9. 11. 5. 7. 19. Wt=t+ 25. Ct=−t 23. 27. 1. 3. 15. − 13. Pt=t + 17. ft 21. −15, 0) Th x 29. 33. fi 35. y =t − 37. fi 39. 43. 41. 45. a. b. c. d. e. Pt=+t f. 47. a. Cx=x + b. The fl c. 49. a. Pt=t + 51. a. Rt= −t + b. b. c. 53. 55. 57. 31. y =−t + Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-11 Section 4.3 1. ft 3. 5. Th 9. 7. 11. 13. 15. 17. 19. 21. 29. y=−x+r=− 23. 25. y=x+r= 27. y=−x+ r=− 31. y=x−r= −(−− fi 37. y=x+ 41. y=−x+ 33. −−− 35. 39. y=x− Chapter 4 Review Exercises 3. 1. 9. y=x− 11. 13. 15. −− 17. m−m=− 19. y=−x+ 5. y=−x+ 7. 23. 25. 27. y=−x+ 29. a. b. c. Pt= t+ 31. 33. y= x 21. – – – – – y – – – – – – – 37. y 35 Year 39. 41. y=−x+ r=− 43. 45. Chapter 4 Practice Test 1. 3. 5. y=−x− 7. y=−x− 9. 15. y=−x+ 13. −− 11. 21. 19. 23. y=x+ 25. a. b. c. y=−t+ 17. =− y= y – – –– – – – x 29. 31. y=x+ 33. r= 27. y x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-12 ANSWERS CHAPTER 5 Section 5.1 ) − 17. − 1. 3. a= 5. 7. gx=x+ − −− − ) ( − 9. fx=( x+ ) − 11. kx= x−−− 13. fx= ( x− ( ) − 15. − x= x=− − 19. − −x=− 23. −∞∞ 21. −∞∞∞ −∞ 27. fx=x+x+ 29. fx=x−x+ 31. fx=− __ __ 35. −) x= +√ 33. fx=x−x+ 37. ( ) − x= — +√ )( ( 25. −∞∞−∞ − √ __ x+ x − − − − − −) 39. ( x= +√ )( ( −# √ ) — — y −−−−− − − − − − x 41. fx=x+x+ 43. fx=−x−x− 45. fx=− __ x− x+ 47. fx=x+x+ 49. fx=−x+x 51. Th h. Th 53. Thhift 55. Th 57. −∞∞ −∞ 59. −∞∞ ∞ 61. fx=x+ 63. fx=−x− 65. fx=x+x− 67. 71. Th 69. 73. 75. Section 5.2 13. − 15. − 17. x→∞fx→∞x→−∞fx→∞ 1. Th . Th 3. x fxx fx 5. Th 7. 9. 11. 19. x→−∞fx→−∞x→∞fx→−∞ 21. x→−∞fx→−∞x→∞fx→−∞ 23. x→∞fx→∞x→−∞fx→−∞ 25. yt− 27. y−x− 29. yx− 31. 33. 37. 39. 41. 43. 45. 47. x→−∞fx→∞ x→∞fx→∞ 49. x→−∞fx→∞ x→∞fx→−∞ 35. x fx x fx − − − − 51. y x x→−∞fx→∞ x→∞fx→∞ − − 53. y x x→−∞fx→−∞ x→∞fx→∞ y −− − − − − − x x −− − − − − − − −− − 55. y x− x→−∞fx→−∞ x→∞fx→∞ 57. y− x− x→−∞fx→∞ x→∞fx→∞ y y −−−− − − − − − − x −−−−−− − − − − − x x y Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-13 59. yx − x→−∞fx→−∞ x→∞fx→∞ 61. fx=x− 63. f x=x−x+ x 65. f x=x+ 67. Vm=m+ m+ m+ 69. Vx=x−x+ x y −−−−− − − − − − Section 5.3 x 9. − 7. −− 5. Th 3. a b 1. x x fx=$ a b 11. − 15. −− 19. − 23. −− 25. f =−f = 27. f =f =− 29. f =f =− 31. − 33. − 35. − − 13. − 17. −( _ ) )( −√ 21. ( √ 37. ) — — 39. 41. 43. x1, 0 −4, 0 y (0, 4); ax→−∞g x→ −∞x→∞g x→∞ 45. x3, 0 2, 0 y (0, −x→−∞k x→ −∞x→∞k x→∞ kx gx −−−−− − − x −−−−−− − − − − − − − − − − x 47. x0, 0−2, 0 y(0, 0); ax→−∞ n x→∞x→∞ n x→−∞ nx −−−−− − − − − − x x+ x− x−x+ x+ 51. f x= 49. f x=− 53. −− 55. − 33. 35. 57. f x=− x+ x−x− 59. f x= x−x−x+ 61. f x=−x−x− 63. f x=−x+ x+ x− 65. f x=− x−x−x+ 67. −−− 69. : −− 71. − 73. f x=x−x+ 77. f x=x−x+ x+ π x+ x+ x+ 79. f x= 75. f x=x−x+ x Section 5.4 1. Th 3. x+ + x+ x− 5. x+ x+ 9. x−+ x− 7. x− x− x+ 11. x−+ x− x+ 13. x −x+ x−x+ 15. x+ x+ + x− 17. x−x+ − x+ 19. x−x+ − 23. x−x+ − 27. x+ x+ x− x+ x+ 21. x+ x+ 25. x−x+ 29. x−x+ 31. x−x+ 41. 35. x−x+ x− 39. x−x− 43. 33. x−x+ x− 37. x−x+ x−x+ x+ x+ 45. x−x+ x+ 49. x+ x+ − 51. x+ x+ 53. x−x+ x−− 57. x−x+ x−+ 55. x−x+ x−x+ x−x+ + i x−i 47. x−x+ x+ x+ 61. + 59. + −i x+ i −i x−i 71. x− 63. x+ ix−+ 69. x+ Section 5.5 65. x + 67. x+ 73. x − 1. Th 3. 7. 9. 5. p + p 19. .+, .. p, , 15. + p + 23. 1 1 27. ± ± p, 11. 13. ,+ , + −− − , , ± ± ± 3 5 5 2 1 8 ± (+)( ± 5 4 ± ( ± , , , , , 17. 21. 25. ± 29. 31. ( ) + (+)( 3p − , , , 1 , 2 ± ± 15 15 2 1 ± (+)) p ! ) , ± 5 8 ± ))( p ! 3p )(+ , 1 4 15 4 , 1 8 ± 15 8 , ± ± − (+) ✓ ( ( ))( (+)) ◆ 3 1 , 3 2 , ± 3 4 , ± Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-14 ANSWERS 44. f x y= − −−− 38. f x 40. f x −− − − − − − − x −−−−−− − − − − − x −− − 42. f x x − − − − − − −− − − − − 46. f x −−−−−− − − − − − x −−−−−− − − − − − x 47. 49. 53. 55. 51. − _ , − _3 1 2, 4 2 fx= x +x −x− fx=− __ x−x 57. 61. 59. Section 5.6 3. Th 5. es. Th 1. Th 7. x=− 9. x=−− 11. x=− y= x=− x=−y= x=− 15. x=− y=x=− 17. x=−y= x=− 13. 19. x= y=− x= 21. ) 23. xy( − + 25. x→− f x→−∞x→− f x→∞ x→±∞f x→ 27. x→+ f x→−∞ x→− f x→∞; 29. x→±∞f x→− − f x→−∞ x→− f x→∞ − + x→− f x→∞ x→− + x→− f x→−∞x→±∞f x→ 31. y=x+ 35. x= y= 37. x= y= y 33. y=x y −− − − − − −− − − − − −−− y= x − x= x x= 39. x=− ) )( − y=( px −− − − − − − −−− x= − y= x 43. x=− y= )( ( − ) f x −− − − − − − − −−−− x= − x= y= x 47. x= y=x+−( ) ) ( hx 41. x= y= sx x= −− − − −−− y= x 45. x=− y= − ax − −−− y= x x=− −− − − − − 49. x=− y= −( − ) wx x= x= − y=x+ x= x − −−−− y= x − − − − −− − − − − −− − − − − Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-15 51. fx= 55. fx= ⋅ x− x− ________ x− x− x+ x+ 53. fx= x+x− x+x+ 57. fx= x− x− x− 59. fx=− x− x− 61. fx= ∙ x+x− x− 63. fx=− x− ____________ x+x− 65. x=y= x y x y − − 31. f−x=√ y — x− 33. f−x=√ y — x − − −−−− − − − − − x − −−−− − − − − − x — x− 37. f−x=√ y — x+− x − −− −− − − − − − x 35. f−x 67. x=−y= − − − x y x y − − − − − − − 39. 69. x=−y= −0 − −0 −0 − x y x y ∞#) 71. ( fx 73. −∞∪∞ fx − −−−− − − − − − 43. −∪ ∞ f x − − − −− − − − − − 47. − y −−− − – − − − − − – – – – –– – – – – x – – – – –– – – – – x 75. () 77. () 79. (−) 81. Ct= +t +t 83. ft 87. 85. Section 5.7 1. diffict oxy 3. 5. f−x=√ 7. f−x=√ x+− x + — — 9. f−x=√ — −x — 13. f−x= 17. f−x= √ x− −x ______ ∞ 11. f−x= ± √ √ 15. f−x= x− — −x 19. f−x= 21. f−x=−x 27. f−x= x− ______ x+ 23. f−x= 29. f−x=√ x+ ______ x — x+− x−+ __________ 25. f−x= x+ ______ −x x x x 41. −∪∞ f x −−−−− − − − − − 45. −∞−ċ− f x − − − −− − − − − − 49. −−− y −−− − – − − − − − x−b a −h x x x 53. f−x= √ 57. th = √ x−b 51. f−x= a cx−b 55. f−x= a−x A , ≈ π A+π −2, 3.99 ft π 59. r A=√ 63. rA=√ T 61. lT=( ) __ π 65. rV=√ ≈ 3.26 ft V ≈ 5.64 ft π Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-16 ANSWERS Section 5.8 1. Th 5. y=x 3. x 19. y=x√ 17. y=xzw 9. y=x 11. y= 13. y= — z x 23. y= xz — wt √ 31. y= 25. y= 33. y= 27. y= 29. y= 35. y= 37. y= 31. −( − ) x=y= y y = − − − −− − − − − − x x = 33. − ( )x=− y= y − − −−− − − − − − x = − y = x x = 35. y=x− 37. f−x=√ — x + 39. f −x=√ — x+− 41. f−x= x+− 47. ≈ x≥ − x Chapter 5 Practice Test 43. y= 45. y= 3. x→−∞ 7. y=x 15. y= — x √ xz w 21. y= 41. y= x y −−−− − − − − − − __ 45. y= x y − −−−− − − − − − 39. y= __ 43. y= √ — x y − − − −− − − − − − x 47. ≈ 49. ≈ 51. 53. 55. x 1. f x→∞ x→∞f x→∞ − 5. f x=x− 7. 9. − 13. { −− } 11. x+x+x++ x− 15. −− 17. f x=− x−x−x+ 19. 21. −( ) x=−y= y 23. f−x=x−+x≥ x+ x− 25. f−x= −−−− − − − − − x = − − 27. y= y = x x = CHAPTER 6 Section 6.1 1. 3. t. Th nominal 11. ft 13. 5. 7. 9. Chapter 5 Review Exercises 1. f x = x − − − −− fx −−−− −− − − − − x 3. f x= x++ 5. 7. 9. 11. x→−∞f x→−∞ x→∞f x→∞ 13. −− − 15. 17. 19. x + x+ 21. x −x+− 23. x −x−x+x −x− 25. { −− 27. { } } 29. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-17 t infl 15. 17. 21. f x=( __ ) 27. r 33. P=At⋅( + n ) ( __ ) 29. 19. f x=x x ≈x 35. 25. 23. −nt − 39. 41. 45. f −=− 47. f −≈− 51. y=⋅x 53. y≈⋅x 31. 37. 43. 49. f ≈ 55. y≈⋅x n −a a( + a[ ( + 57. = At−a a r ___ ) a −] = r ___ ) =(
+ = a r In=( + n ) 59. ff x=a⋅( ) x b b> 1. Thn> fx=a⋅( ) x b 61. 67. x =ae−nx=ae−nx 63. =ab−x=aen− r ) − − 65. Section 6.2 1. x. Th 3. gx=−xy 5. gx) =−x+y 7. gx=( ) y x , ✓ ◆ : ;:= :(,);:= 13. :(,);:= 15. 17. : 19. 21. 0, 1 1024 ◆ ✓ ;:= 9. 11. 23. y 25. y −− − − − − − − − − fx= x x − − −fx= − x −− − − − − − − −f x=x − x f x=−x + 27. hx= hx −−−−− x 29. x→∞fx→−∞ x→−∞f x→−1 31. x → ∞fx→ x→ −∞fx→∞ 33. fx=x− 35. fx=x − 37. fx=−x 39. y=−x+ 41. y=−x+ 43. g≈ 47. x≈− 45. h−=− x 51. Thgx) =( ) y b fx=bxb> 49. x≈− fx=b ) x( b x yf −x 53. Thgxhx fxn x( bn ) bx b> fx=b fx−n 55. Section 6.3 1. b b. Th ybybx d txby 3. b y=x x 5. Th b ee (x) 7. ac=b 15. e n=w 21. n= 27. x= 35. x=e 43. 45. 47. 49. 51. − 53. 55. − 57. 17. ck=d 23. y ( 25. h=k 31. x= 19. y=x 13. a= 11. b=a 29. x= 41. 9. x y= )=x 33. x= 39. (x) 37. 65. − 63. 71. 61. ≈ − ≈ 67. 59. 73. 69. x=x= fx=x). Thn n=n = n Thx=f x=x 75. x x=x=e x=e . Th x=e= 79. 77. = e Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-18 Section 6.4 5. 1. y=xab ba 3. hift y ff 7. ( −∞, )−∞∞ 9. ( − ∞)−∞∞ 11. ∞x= ∞#) x=− 13. ( − 15. −∞x=− 17. ( ∞#) x= x→( ) 19. −∞x=− x→−+fx→−∞x→∞fx→∞ 21. ∞−∞∞x= x( )y 23. −∞, 0); ra−∞∞ x=x−ey 25. ∞−∞∞x= xey 27. 33. + fx→−∞s x → ∞fx → ∞ 31. 29. y y 35. 37. −− − − − − f x= x x − − gx= x f x= e x − − −− − − − − x gx= x 39. fx 41. y −− − − − − − − x −−−−−− −− − − − − − x − − 45. 47. 49. 51. 53. f x=−x− f x=x+ x= x≈ x≈− 43. gx −−−− − −− − − − − x Thfx= 55. b≠bx=−_ 57. xgx=−x b x x+ x− > x+ x− fx= > x−∞− ∞Th−∞−∪∞ Section 6.5 _ n bx n = 1. . Thbx 3. b +b +bx+by 5. b−b 7. −k 11. b 13. b 15. x+y−z () 19. 17. () 21. ) () x ( xz ) — y √ = b )= () ( ()+ 9. xy () 25. 31. 27. 23. 29. 33. = ( ≈ a−b b 39. a b− ≈− 37. ≈ x= x+−x−=( x = x+ _____ )= x− () 35. 41. 42. = = = x+ _____ x− x+ _____ − x− x− x+ _______ _____ − x− x− x+−x+ _____________ x− x− _____ x− = x= +− −= −x= bnn 1. Th bn= nn nb = nb Section 6.6 1. 3. Th . Th x= 7. 9. 5. x= x=− x= 19. 13. x= () () 15. 11. 17. x=− x= + 21. () () () Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-19 31. x=− 32. 61. y 39. x= 41. x= 63. 67. 69. 71. 73. 65. ≈ ≈ ≈ ≈− _ k ) t=(( y A ) t=(( _ − T−Ts k ) T−Ts ) x 75. Section 6.7 5. 1. s. Th 3. Th t diff orders of magnitude 7. f ≈ 9. 13. 11. f x=x 15. 25. 33. x= 27. 35. x= x= 37. x= x= 43. x= 45. x= ≈ e y x 47. x −− − − − − − e+ ≈ x= y x 49. x=− y −−−−−−− − − − − x – 51. −− − − − − − 53. 55. y x= ≈ y x − − − − − −− − − − − − − − −− − − − − − − 57. x= y ≈ fx x x y x − − − 59. y −− x 17. Pt − − − − − −−−− −− t fx=ex x 19. 23. 21. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. B-20 ANSWERS 27. y=bx bb≠ 1. Th __ __ y=bx y=xb ey=exb y=exb S ) ( 25. M= S S M=( ) S S M =( ) S SM =S 29. A=e−tA≈ 33. f t=e−t; 35. r≈− 37. f t=etftP ≈ 39. f t=et 41. ≈ 31. 43. ≈ 45. Section 6.8 1. 3. t fi . Th 5. y 7. 13. p≈ 19. 21. 11. P = 15. y 17. 9. y x y 31. x 23. 25. 27. f x=x 29. y x 33. f x=e−x 35. f x=x≈ 37. y=+x 39. y x 43. f ≈ 45. f x=x≈ 47. f x= +e−x 49. 41. y x 51. y 53. y x x 55. f x=x≈ g x= y =x 57. f x=x 59. f −x= c a−( _ −) x b Chapter 6 Review Exercises 5. 1. 3. y =x 7. y 9. y −−−−−−−− − x 11. gx=−xy 15. ab= − 19. =− 13. x = 17. x = Download the OpenStax text for free at http://cnx.org/content/col11759/latest. ANSWERS B-21 33. y 35. y =x y x 37. y = +e−x y x 21. e−=− 23. gx x −−−−− − − − − − − − − 25. x> − x =− x→−+ f x→−∞ x→∞f x→∞ 27. xy z 29. ( xy ) 31. y 33. +b+ b+−b− 35. ( v w — u √ ) 39. x =− 41. 43. 37. x =# 45. x = 51. 47. a=e− 53. f−x= __ 49. x =± x− — √ 55. f t=tf ≈g 57. x 59. 61. y x 63. y =x; y =e–x 65. 67. y =−x y Chapter 6 Practice Test x 1. 5. y 3. y f−x= −x fx= −x 7. a= 9. x = 11. ≈− 13. x< x = x→−f x→−∞ x→−∞f x→∞ −−−−− − x 15. t 17. y+z+ x− 19. x = _______ + __ ≈ 23. 25. x = 21. a= + ________ — 27. x =± √ ____ 29. f t=e−t; 31. Tt=e−t+T≈°F Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Index A absolute maximum 205, 267 absolute minimum 205, 267 absolute value 189, 247 absolute value equation 137, 151, 250 absolute value function 187, 247, 250 absolute value inequality 146, 147 addition method 582, 587, 674 Addition Principle 802, 828 addition property 143 algebraic expression 11, 66 angle of rotation 733, 751 annual interest 797 annual percentage rate (APR) 474, 567 annuity 797, 828 apoapsis 742 area 104, 151 area of a circle 361 arithmetic sequence 771, 772, 774, 775, 790, 791, 828 arithmetic series 791, 828 arrow notation 415, 454 associative property of addition 8, 66 associative property of multiplication 8, 66 asymptote 700 augmented matrix 636, 640, 641, 653, 674 average rate of change 196, 267 axes of symmetry 700 axis of symmetry 344, 347, 454, 722, 723 B base 6, 66 binomial 42, 66, 396, 828 binomial coefficient 812 binomial expansion 813, 815, 828 Binomial Theorem 813, 814, 828 break-even point 589, 674 C carrying capacity 546, 567 Cartesian coordinate system 74, 151 Celsius 254 center of a hyperbola 700, 751 center of an ellipse 685, 751 central rectangle 700 change-of-base formula 525, 526, 567 circle 607, 609 co-vertex 685, 687, 700 coefficient 41, 42, 66, 361, 405, 454 coefficient matrix 636, 638, 655, 674 column 625, 674 column matrix 626 combination 807, 812, 828 combining functions 210 common base 529 common difference 771, 790, 828 common logarithm 496, 567 common ratio 781, 793, 828 commutative 211 commutative property of addition 8, 66 commutative property of multiplication 8, 66 complement of an event 823, 828 completing the square 124, 125, 151 complex conjugate 116, 151 Complex Conjugate Theorem 409 complex number 112, 113, 151 complex plane 112, 151 composite function 209, 210, 211, 267 compound inequality 145, 151 compound interest 474, 567 compression 292, 486, 508 conditional equation 87, 151 conic 684, 699, 748 conic section 751 conjugate axis 700, 751 consistent system 578, 674 constant 11, 41, 42, 66 constant function 187 constant of variation 448, 454 constant rate of change 309 continuous 376 continuous function 371, 454 coordinate plane 717 correlation coefficient 327, 334 cost function 209, 588, 674 Cramer’s Rule 663, 666, 670, 674 cube root 362 cubic function 187, 439 D decompose a composite function 217 decomposition 615 decreasing function 201, 267, 282 decreasing linear function 283, 334 degenerate conic sections 729 degree 42, 66, 366, 454 dependent system 579, 587, 599, 674 dependent variable 160, 267 Descartes’ Rule of Signs 410, 454 determinant 663, 665, 666, 674 difference of squares 46, 66 directrix 717, 720, 722, 743, 747, 748, 751 discriminant 127, 151 distance formula 80, 151, 701, 717 distributive property 8, 66 diverge 794, 828 dividend 395 Division Algorithm 395, 396, 403, 454 divisor 395 domain 160, 168, 180, 181, 267 domain and range 180 domain and range of inverse functions 258 domain of a composite function 216 domain of a rational function 420 doubling time 539, 543, 567 E eccentricity 743, 751 electrostatic force 199 elimination 608 ellipse 608, 685, 686, 687, 689, 692, 716, 743, 747, 751 ellipsis 758 end behavior 363, 424, 454 endpoint 198 entry 625, 674 equation 13, 66, 166 equation in quadratic form 138 equation in two variables 76, 151 equations in quadratic form 151 even function 233, 267 event 819, 828 experiment 819, 828 explicit formula 759, 775, 784, 828 exponent 6, 66 exponential 482 exponential decay 466, 472, 481, 539, 541, 544, 554 exponential equation 528 exponential function 466 exponential growth 466, 469, 539, 543, 545, 554, 567 exponential notation 6, 66 extraneous solution 134, 151, 532, 567 extrapolation 324, 325, 334 F factor by grouping 51, 66 Factor Theorem 404, 454 factorial 766 factoring 119 Fahrenheit 254 feasible region 611, 674 finite arithmetic sequence 776 finite sequence 759, 828 foci 685, 687, 700, 751 focus 685, 717, 720, 722, 742, 747, 748 focus (of an ellipse) 751 focus (of a parabola) 751 FOIL 44, 114 formula 13, 66, 166 function 160, 189, 267 function notation 162 Fundamental Counting Principle 804, 828 Fundamental Theorem of Algebra 408, 409, 454 G Gauss 594, 636 Gaussian elimination 594, 639, 674 general form 345 general form of a quadratic function 345, 347, 454 geometric sequence 781, 783, 793, 828 geometric series 793, 828 global maximum 388, 389, 454 global minimum 388, 389, 454 graph in two variables 76, 151 greatest common factor 49, 66, 119 C-1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. C-2 H INDEX inverse property of multiplicative inverse of a power rule for logarithms 521, half-life 535, 539, 541, 567 Heaviside method 617 horizontal asymptote 417, 418, 423, 454 horizontal compression 237, 267 horizontal line 95, 297, 298, 334 horizontal line test 173, 267 horizontal reflection 229, 230, 267 horizontal shift 225, 267, 484, 505 horizontal stretch 237, 267 hyperbola 699, 702, 703, 704, 707, 708, 711, 717, 744, 746, 751 I identity equation 87, 151 identity matrix 649, 653, 674 identity property of addition 9, 66 identity property of multiplication 9, 66 imaginary number 111, 112, 151, 454 inconsistent equation 87, 151 inconsistent system 579, 586, 598, 674 increasing function 201, 267, 282 increasing linear function 283, 334 independent system 578, 579, 674 independent variable 160, 267 index 36, 66 index of summation 789, 790, 828 inequality 610 infinite geometric series 794 infinite sequence 759, 828 infinite series 794, 828 input 160, 267 integers 2, 5, 66 intercepts 79, 151 Intermediate Value Theorem 386, 454 interpolation 324, 334 intersection 821 interval 142, 151 interval notation 142, 151, 180, 184, 201, 267 inverse function 255, 267, 436, 439 inverse matrix 653, 655 inverse of a radical function 42 inverse of a rational function 444 inverse property of addition 9, 66 multiplication 9, 66 inverse variation 449, 454 inversely proportional 449, 454 invertible function 438, 454 invertible matrix 649, 663 irrational numbers 3, 5, 66 J joint variation 451, 454 L latus rectum
717, 722, 751 leading coefficient 42, 66, 366, 454 leading term 42, 66, 366, 454 least common denominator 60, 66, 89 least squares regression 325, 334 linear equation 87, 151 Linear Factorization Theorem 409, 454 linear function 280, 294, 309, matrix 649, 674 multiplicity 380, 454 mutually exclusive events 822, 828 N n factorial 766, 828 natural logarithm 498, 531, 567 natural numbers 2, 5, 67, 160 Newton’s Law of Cooling 544, 567 nominal rate 474 nondegenerate conic section 729, 731, 751 nonlinear inequality 610, 674 nth partial sum 789, 828 nth term of a sequence 828 nth term of the sequence 759 O odd function 233, 267 one-to-one 482, 494, 519, 525 one-to-one function 170, 257, 334 267 linear growth 466 linear inequality 151 linear model 310, 322 linear relationship 322 local extrema 200, 267 local maximum 200, 267, 389 local minimum 200, 267, 389 logarithm 494, 567 logarithmic equation 533 logarithmic model 557 logistic growth model 546, 567 logistic regression 560 long division 394 lower limit of summation 789, 790 M magnitude 189, 224 main diagonal 638, 674 major and minor axes 687 major axis 685, 689, 751 matrix 625, 626, 630, 636, 674 matrix multiplication 630, 650, 655 matrix operations 626 maximum value 344 midpoint formula 82, 151 minimum value 344 minor axis 685, 689, 751 model breakdown 324, 334 modulus 189 monomial 42, 67 Multiplication Principle 803, 804, 828 multiplication property 143 multiplicative inverse 651 order of magnitude 540, 567 order of operations 6, 67 ordered pair 75, 151, 160, 181 ordered triple 594 origin 75, 151, 248 outcomes 819, 828 output 160, 267 P parabola 344, 350, 611, 716, 721, 723, 742, 745, 751 parallel 96 parallel lines 96, 298, 299, 334 parent function 505 partial fraction 615, 674 partial fraction decomposition 615, 674 Pascal’s Triangle 814 perfect square trinomial 45, 67 periapsis 742 perimeter 104, 151 permutation 804, 828 perpendicular 96 perpendicular lines 97, 299, 334 pH 518 piecewise function 189, 190, 267, 762 point-slope form 285, 334 point-slope formula 98, 705 polar equation 743, 751 polar form of a conic 748 polynomial 41, 42, 67, 405 polynomial equation 133, 151 polynomial function 365, 376, 383, 387, 455 power function 361, 455 525, 567 principal nth root 36, 67 principal square root 31, 67 probability 819, 828 probability model 819, 828 product of two matrices 630 product rule for logarithms 519, 521, 567 profit function 589, 674 properties of determinants 669 Proxima Centauri 540 Pythagorean Theorem 80, 127, 152 Q quadrant 74, 152 quadratic 138, 619, 621 quadratic equation 119, 120, 123, 125, 152 quadratic formula 125, 126, 152, 355 quadratic function 187, 347, 349 quotient 395 quotient rule for logarithms 520, 567 R radical 31, 67 radical equation 134, 135, 152 radical expression 31, 67 radical functions 438 radicand 31, 67, 134 radiocarbon dating 542 range 160, 267 rate of change 196, 267, 280 rational equation 89, 90, 152 rational exponents 37 rational expression 58, 67, 89, 615, 621 rational function 419, 426, 429, 455 rational number 2, 5, 67, 89 Rational Zero Theorem 405, 455 real number line 4, 67 real numbers 4, 67 reciprocal 96, 255, 362 reciprocal function 188, 415 recursive formula 764, 774, 783, 829 reflection 487, 510 regression analysis 554, 557, 560 regression line 326 relation 160, 267 relative 200 remainder 395 Remainder Theorem 403, 455 removable discontinuity 422, 455 restricting the domain 262 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. INDEX C-3 revenue function 588, 674 Richter Scale 493 roots 345, 455 row 625, 674 row matrix 626 row operations 638, 642, 652, 653, 654, 674 row-echelon form 638, 640, 674 trinomial 42, 67 turning point 369, 384, 455 U union of two events 821, 829 upper limit of summation 789, 790, 829 upper triangular form 594 row-equivalent 638, 674 V variable 11, 67 varies directly 448, 455 varies inversely 449, 455 vertex 344, 455, 685, 717, 723 vertex form of a quadratic function 346, 455 vertical asymptote 417, 420, 424, 455 vertical compression 234, 268 vertical line 95, 297, 298, 334 vertical line test 171, 268 vertical reflection 229, 230, 268 vertical shift 222, 223, 268, 293, 483, 506, 544 vertical stretch 234, 268, 292, 508 vertices 685, 687 volume 104, 152 volume of a sphere 361 W whole numbers 2, 5, 67 X x-axis 74, 152 x-coordinate 75, 152 x-intercept 79, 152, 296 Y y-axis 74, 152 y-coordinate 75, 152 y-intercept 79, 152, 281, 282, 291, 296 Z zero-product property 120, 152 zeros 345, 377, 380, 405, 455 S sample space 819, 829 scalar 628 scalar multiple 628, 674 scalar multiplication 628 scatter plot 322 scientific notation 25, 26, 27, 67 sequence 758, 771, 829 series 789, 829 set-builder notation 142, 183, 184, 268 sigma 789 slope 92, 93, 152, 281, 282, 284, 334 slope-intercept form 281, 334 smooth curve 371, 455 solution set 87, 152, 595, 674 solving systems of linear equations 580, 582 square matrix 626, 663 square root function 188 square root property 123, 152 standard form 95 standard form of a quadratic function 346, 347, 455 stretch 486 substitution method 581, 674 summation notation 789, 790, 829 surface area 436 synthetic division 398, 407, 455 system of equations 637, 638, 640, 641, 655 system of linear equations 315, 578, 580, 581, 675 system of nonlinear equations 605, 675 system of nonlinear inequalities 611, 675 system of three equations in three variables 666 T term 758, 772, 829 term of a polynomial 41, 42, 67 term of a polynomial function 365, 455 transformation 222, 292 translation 721 transverse axis 700, 751 Download the OpenStax text for free at http://cnx.org/content/col11759/latest.
introduced. We move on, in Section 2.4, to simplifying trigonometric expressions and proving that a trigonometric equation is an identity. Then, finally, Section 2.5 introduces definitions for circular functions of varying radii, along with applications. 32 The Trigonometric Functions 2.1 Right Triangle Trigonometry Learning Objectives In this section you will:  Identify the trigonometric functions.  Learn the trigonometric function values for 30 degrees, 45 degrees and 60 degrees.  Solve right triangles and related application problems. As we shall see in the sections to come, many applications in trigonometry involve finding the measures of the interior angles, and the lengths of the sides, of right triangles. Recall that a right triangle is a triangle containing one right angle and two acute angles. In this section, we will define a new group of functions known as trigonometric functions that will assist us in determining unknown angle measures and/or side lengths for right triangles. Noting that two right triangles are similar if they have one congruent acute angle, we will use properties of similar triangles to establish trigonometric function values for three special angles: 30°, 45° and 60°. Similar Triangles We begin with a definition from geometry. Recall that two triangles are similar if they have the same shape or, more specifically, if their corresponding angles are congruent. Additionally, two triangles are similar if and only if their corresponding sides are proportional. In the following triangles, , , and . Thus, triangle ABC is similar to triangle RST and ARBSCT.ABBCCARSSTTRBACSRT 2.1 Right Triangle Trigonometry 33 Trigonometric Functions We consider the generic right triangle below with acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side c (the side opposite the right angle) is called the hypotenuse. The six commonly used trigonometric functions are defined below. The Trigonometric Functions: Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then  The sine of θ, denoted , is defined by .  The cosine of θ, denoted , is defined by  The tangent of θ, denoted , is defined by  The cotangent of θ, denoted , is defined by  The secant of θ, denoted , is defined by . . . .  The cosecant of θ, denoted , is defined by . sinoppositehypotenusesinbccosadjacenthypotenusecosactanoppositeadjacenttanbacotadjacentoppositecotabsechypotenuseadjacentseccacschypotenuseoppositecsccb 34 The Trigonometric Functions The following are important properties of the trigonometric functions. 1. For all right triangles with the same acute angle θ, because they are similar, the values of the resulting trigonometric functions of θ will be identical. This is a result of the property of equivalent proportions of corresponding sides within similar triangles. 2. Cosecant, secant and cotangent are reciprocal functions of sine, cosine and tangent, respectively. Thus, if we know the sine, cosine and tangent values for an angle, we can easily determine the remaining three trigonometric functions. In particular, Example 2.1.1. Use the following triangle to evaluate , , , , and . Solution. From the definitions of trigonometric functions, 1cscsin1seccos1cottansincostancscseccotoppositehypotenuseadjacenthypotenuseoppositeadjacent4sin53cos54tan3435 2.1 Right Triangle Trigonometry 35 The reciprocals of these three function values result in the remaining three trigonometric function values: Example 2.1.2. Verify that the following triangle is similar to the triangle in Example 2.1.1. Then evaluate the trigonometric function values for the angle corresponding to α. Solution. The side lengths of the second triangle are proportional to the corresponding side lengths of the first triangle by a scale factor of 3: Thus the triangles are similar, with the angle β being equal in measure to α. To evaluate the trigonometric function values for β, we save a bit of writing by using the abbreviations opp, adj and hyp in place of opposite, adjacent and hypotenuse, respectively. The trigonometric function values for this similar triangle will be 15cscsin415seccos313cottan4912153345.opphypadjhypoppadj12344sin153559333cos1535512344tan933312915   36 The Trigonometric Functions Using reciprocal properties, the remaining three values are We note that the trigonometric function values are identical for the two similar triangles in Example 2.1.1 and Example 2.1.2, and observe that trigonometric ratios are not affected by the value of the scale factor. Pythagorean Theorem The Pythagorean Theorem will be useful in our next task: determining trigonometric function values for 30°, 45° and 60° angles. The Pythagorean Theorem: The square of the hypotenuse in a right triangle is equal to the sum of the squares of the two shorter sides. In particular, in a right triangle with hypotenuse c and the shorter sides of lengths a and b, Trigonometric Functions of 30°, 60°, 90° Triangles We begin by finding the values of trigonometric functions for 30°. We sketch a 30°, 60°, 90° right triangle with hypotenuse of length x and envision this triangle as being half of a 60°, 60°, 60° equilateral triangle with sides of length x. 355csc344355sec333333cot344222cabxxx 30 60xl 30 60 2x 2.1 Right Triangle Trigonometry 37 Noting that the altitude of the equilateral triangle bisects its base, it follows that the shortest side of the 30°, 60°, 90° triangle has length . We apply the Pythagorean Theorem to determine the length, l, of the third side of the 30°, 60°, 90° triangle in terms of x. Using the resulting side lengths, along with the definitions of the trigonometric functions, we have Taking the reciprocals of these three function values results in the remaining three trigonometric function values: We note that these trigonometric function values apply to any 30° angle. The reader is encouraged to determine the trigonometric function values for 60° angles. 2x22222222243432xxxxlxlxll30303021sin2323cos221tan332xxxxxx3030303030301csc2sin12seccos31cot3tanx 30 60 2x 32x 38 The Trigonometric Functions Trigonometric Functions of 45°, 45°, 90° Triangles To find the values of the trigonometric functions for 45°, we sketch a 45°, 45°, 90° right isosceles triangle with hypotenuse h and remaining two sides each of length x. Using the Pythagorean Theorem, the hypotenuse can be written in terms of x as follows. The resulting trigonometric function values for 45° are After rationalizing denominators and adding trigonometric functions for 60°, we summarize the trigonometric function values for these special cases in the following table. 2222222xxhxhhx4545454545454545451sin221cos22tan11csc2sin1sec2cos1cot1tanxxxxxxhxx 45 45xx 45 45 2x 2.1 Right Triangle Trigonometry 39 Trigonometric Function Values for 30°, 45° and 60° θ 30° 45° 60° 2 1 1 2 Solving Right Triangles We will use these values in the next four examples to determine measures of missing angles and sides. This is sometimes referred to as solving right triangles. Example 2.1.3. Find the measure of the missing angle and the lengths of the missing sides of: Solution. The first and easiest task is to find the measure of the missing angle. Since the sum of the angles of a triangle is 180°, we know that the missing angle has measure . We now proceed to find the lengths of the remaining two sides of the triangle. Let c denote the length of the hypotenuse of the triangle. From , we get sincostancscseccot12323323332222223212323333180309060307cosc3030307cos17cos7sec.ccc 40 The Trigonometric Functions Since , we arrive at the length of the hypotenuse: . At this point, we have two ways to proceed to find the length of the side opposite the 30° angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is , so we could find the missing side by applying the Pythagorean Theorem and solving the following for b: Alternatively, we could use the definition, , to get . Choosing the latter, we find The triangle with all of its recorded data follows. Example 2.1.4. A right triangle has one angle of 60° and a hypotenuse of 20. Find the unknown side lengths and missing angle measure. Solution. Again, we begin with finding the measure of the missing angle. The sum of the angles of a triangle is 180°, from which it follows that the missing angle measure is . We assign the missing side lengths as a, for the side adjacent to 60°, and b, for the side opposite 60°. 3023sec31433c14332221437.3boppositetanadjacent30tan7b307tan37373.3b180609030 2.1 Right Triangle Trigonometry 41 Since , we find We find length a using the Pythagorean Theorem, although the same result could be achieved through solving for a. The triangle with all of its data is recorded below. 60sin20b6020sin3202103.b60cos20a22221032040030010.aaaba20 601020 60 30 103 42 The Trigonometric Functions Solving Applied Problems Right triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. The following example uses trigonometric functions as well as the concept of an ‘angle of inclination’. The angle of inclination, commonly known as the angle of elevation, of an object refers to the angle whose initial side is some kind of horizontal base-line (say,
the ground), and whose terminal side is the line-of-sight to an object above the base-line. This is represented schematically below. The angle of inclination (elevation) from the base line to the object is θ. Example 2.1.5. The angle of inclination, from a point on the ground 30 feet away from the base of a water tower, to the top of the water tower, is 60°. Find the height of the water tower to the nearest foot. Solution. We can represent the problem situation using a right triangle as shown. If we let h denote the height of the tower, then we have . From this we get Hence, the water tower is approximately 52 feet tall. Example 2.1.6. In order to determine the height of a California redwood tree, two sightings from the ground, one 200 feet directly behind the other, are made. If the angles of inclination were 45° and 30°, respectively, how tall is the tree to the nearest foot. 60tan30h6030tan30351.96h 2.1 Right Triangle Trigonometry 43 Solution. Sketching the problem situation below, we find ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the first observation point. Using trigonometric functions, we get a pair of equations: and . Since , the first equation gives , or . Substituting this into the second equation gives Clearing fractions, we get . The result is a linear equation for h, so we proceed to expand the right hand side and gather all the terms involving h to one side. Hence, the tree is approximately 273 feet tall. Example 2.1.7. How long must a ladder be to reach a windowsill 50 feet above the ground if the ladder is resting against the building at an angle of 75° with the ground? 45tanhx30tan200hx45tan11hxxh30tan2003or .2003hhhh32003hh320033320033320033320032003273.2133hhhhhhhh 44 The Trigonometric Functions Solution. We know that the angle of inclination, or elevation, is 75° and that the opposite side is 50 feet in length. The length of the hypotenuse, h, will give us the necessary length for the ladder to reach a height of 50 feet. Using the trigonometric function for sine of 75°, we have Noting that trigonometric function values for 75° are not included in the table for special cases, use of a calculator is necessary to find an approximate value for . It is good practice to verify that the calculator is set to the correct mode, in this case degrees, before proceeding with calculations. We have found that the height of the ladder is approximately 51.8 feet. This section leads us to the Unit Circle and an alternate definition for trigonometric functions. Through this new definition we will expand the domain for trigonometric functions to include angle measures outside the interval . 757550sin50sin500.965925851.76hhhh75sin0,90h50 75 2.1 Right Triangle Trigonometry 45 2.1 Exercises 1. For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle. 2. The tangent of an angle compares which sides of the right triangle? 3. What is the relationship between the two acute angles in a right triangle? In Exercises 4 and 5, use the given triangle to evaluate each trigonometric function of angle A. 4. Find , , , , and . sinAcosAtanAcscAsecAcotA 46 The Trigonometric Functions 5. Find , , , , and . In Exercises 6 – 13, find the measurement of the missing angle and the lengths of the missing sides. 6. Find θ, b, and c. 7. Find θ, a, and c. sinAcosAtanAcscAsecAcotA 2.1 Right Triangle Trigonometry 47 8. Find α, a, and b. 9. Find β, a, and c. 10. Find θ, a, and c. 48 11. Find α, b, and c. The Trigonometric Functions 12. Find θ, a, and c. 13. Find β, b, and c. In Exercises 14 – 25, assume that θ is an acute angle in a right triangle. 14. If and the side adjacent to θ has length 4, how long is the hypotenuse? 15. If and the hypotenuse has length 5280, how long is the side adjacent to θ? 16. If and the side opposite θ has length 117.42, how long is the hypotenuse? 17. If and the hypotenuse has length 10, how long is the side opposite θ? 1278.123595 2.1 Right Triangle Trigonometry 49 18. If and the hypotenuse has length 10, how long is the side adjacent to θ? 19. If and the side opposite θ has length 306, how long is the side adjacent to θ? 20. If and the side opposite θ has length 4, how long is the side adjacent to θ? 21. If and the hypotenuse has length 10, how long is the side opposite θ? 22. If and the side adjacent to θ has length 2, how long is the side opposite θ? 23. If 24. If and the side opposite θ has length 14, how long is the hypotenuse? and the hypotenuse has length 3.98, how long is the side adjacent to θ? 25. If and the side adjacent to θ has length 31, how long is the side opposite θ? 26. Find x. 27. Find x. 537.530158738.22.0542 The Trigonometric Functions 50 28. Find x. 29. Find x. 30. A tree standing vertically on level ground casts a 120 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is 21.4°. Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem. 31. The broadcast tower for radio station WSAZ (Home of “Algebra in the Morning with Carl and Jeff”) has two enormous flashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is 7.970° and to the second light is 7.125°. Find the distance between the lights to the nearest foot. 32. In this section, we defined the angle of inclination (also known as the angle of elevation) and in this exercise we introduce a related angle – the angle of depression (also known as the angle of declination). The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal. This is represented schematically below. 2.1 Right Triangle Trigonometry 51 The angle of depression from the horizontal to the object is θ. (a) Show that if the horizontal is above and parallel to level ground then the angle of depression (from observer to object) and the angle of inclination (from object to observer) will be congruent because they are alternate interior angles. (b) From a fire tower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a fire off in the distance. The angle of depression to the fire is 2.5°. How far away from the base of the tower is the fire? (c) The ranger in part (b) sees a Sasquatch running directly from the fire towards the fire tower. The ranger takes two sightings. At the first sighting, the angle of depression from the tower to the Sasquatch is 6°. The second sighting, taken just 10 seconds later, gives the angle of depression as 6.5°. How far did the Sasquatch travel in those 10 seconds? Round your answer to the nearest foot. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the fire tower from his location at the second sighting? Round your answer to the nearest minute. 33. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is 50° and the angle of depression to the base of the tree is 10°. What is the height of the tree? Round your answer to the nearest foot. 34. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The first sighting had an angle of depression of 8.2° and the second sighting had an angle of depression of 25.9°. How far had the boat traveled between the sightings? 35. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a 43° angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground? 52 The Trigonometric Functions 36. A 33 foot ladder leans against a building, so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 37. A 23 foot ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 38. The angle of elevation to the top of a building in New York City is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. 39. The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building. 40. Assuming that a 370 foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be 60°, how far from the base of the tree am I? 41. Let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that and . The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Chapter 4. 42. Let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that and . sincossincosseccsctancot 2.2 Determining Cosine and Sine Values from the Unit Circle 53 2.2 Determining Cosine and Sine Values from the Unit Circle Learning Objectives In this section you will:  Sketch oriented arcs on the Unit Circle.  Determine the cosine and sine values of an angle from a point on the Unit Circle.  Learn and apply the Pythagorean identity.  Apply the Reference Angle Theorem.  Learn the cosine and sine values fo
r the common angles: 0°, 30°, 45°, 60° and 90°, or for their equivalent radian measures.  Learn the signs of the cosine and sine functions in each quadrant. We have already defined the Trigonometric Functions as functions of acute angles within right triangles. In this section, we will expand upon that definition by redefining the cosine and sine functions using the Unit Circle. Thus, the new designation ‘Circular Functions’ will often be used in place of ‘Trigonometric Functions’ as we adopt this new definition. The Unit Circle Consider the Unit Circle, , as shown, with the angle θ in standard position and the corresponding arc measuring s units in length. By the definition established in Section 1.2, and the fact that the Unit Circle has radius 1, the radian measure of θ is so that, once again blurring the distinction between an angle and its measure, we have . In order to identify real numbers with oriented angles, we make good use of this fact by essentially ‘wrapping’ the real number line around the Unit Circle and associating to each real number t an oriented arc on the Unit Circle with initial point . 221xy1ssrss1,0xy11  s 54 The Trigonometric Functions Given a real number and the vertical line containing the (vertical) interval , we ‘wrap’ the interval around the Unit Circle in a counter-clockwise fashion. The resulting arc has a length of t units. Therefore, the corresponding angle has radian measure equal to t. If , we wrap the interval clockwise around the Unit Circle. Since we have defined clockwise rotation as having negative radian measure, the angle determined by this arc has the negative radian measure equal to t. Note that if , we are at the point on the x-axis which corresponds to an angle with radian measure 0. Thus, we identify each real number t with the corresponding angle having radian measure of t. Example 2.2.1. Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers. 1. 2. 3. 4. Solution. 1. The arc associated with is the arc on the Unit Circle which subtends the angle in radian measure. Since is of a revolution, we have an arc which begins at the point and proceeds counter-clockwise up to midway through Quadrant II. 2. Since one revolution is radians, and is negative, we graph the arc which begins at and proceeds clockwise for one full revolution. 0t1x0,t0,t0t,0t0t1,034t2t2t117t34t3434381,022t1,0xy11 t t 0txy11 t t 0txy11 34t 2.2 Determining Cosine and Sine Values from the Unit Circle 55 3. Like , is negative, so we begin our arc at and proceed clockwise around the Unit Circle. With and , we find rotating 2 radians clockwise from the point lands us in Quadrant III between and . To more accurately place the endpoint, we proceed as we did in Example 1.1.3, successively halving the angle measure until we find , which tells us our arc extends, clockwise, almost a quarter of the way into Quadrant III. 4. Since 117 is positive, the arc corresponding to begins at and proceeds counter-clockwise. As 117 is much greater than , we wrap around the Unit Circle several times before finally reaching our endpoint. We approximate as 18.62 which tells us we complete 18 revolutions counter-clockwise with 0.62, or just shy of of a revolution, remaining. In other words, the terminal side of the angle which measures 117 radians in standard position is just short of being midway through Quadrant III. The Cosine and Sine as Circular Functions This leads to our new definition for the cosine and sine functions. Consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. 2t2t1,01.5723.141,0251.968117t1,02117258xy11 2txy11 2txy11 117t 56 The Trigonometric Functions By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written , while the y-coordinate of P is called the sine of θ, written .15 The reader is encouraged to verify that these rules that match an angle with its cosine and sine satisfy the definition of a function: for each angle θ, there is only one associated value of and only one associated value of . It is important to note that any angle that is not labeled as being in degrees is, by default, assumed to be in radians. In the following example, the angles in number 2 and number 4 are radian measures: radians and radians, respectively. Example 2.2.2. Find the cosine and sine of the following angles. 1. 2. 3. 4. 5. Solution. 1. To find and , we plot the angle in standard position and find the point on the terminal side of θ which lies on the Unit Circle. Since 270° represents of a counter-clockwise revolution, the terminal side of θ lies along the negative y-axis. Hence, the point we seek is so that and . 15 The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 4.2. cossincossin627045660270cos270sin270340,1270cos0270sin1 2.2 Determining Cosine and Sine Values from the Unit Circle 57 2. The angle represents one half of a clockwise revolution so its terminal side lies on the negative x-axis. The point on the Unit Circle that lies on the negative x-axis is , from which and . 3. When we sketch in standard position, we see that its terminal side does not lie along any of the coordinate axes. We let denote the point on the terminal side of θ which lies on the Unit Circle. By definition, and . If we drop a perpendicular line segment from P to the x-axis, we obtain a 45°, 45°, 90° right isosceles triangle whose legs have lengths x and y units. From the properties of isosceles triangles, it follows that . 1,0cos1sin045,Pxy45cosx45sinyyx 58 The Trigonometric Functions lies on the Unit Circle, so . Substituting into this equation yields , or Now, lies in the first quadrant where , so . Since , we can also conclude that . Finally, we have and . 4. As before, the terminal side does not lie on any of the coordinate axes so we proceed using a triangle approach. Letting denote the point on the terminal side of θ which lies on the Unit Circle,we drop a perpendicular line segment from P to the x-axis to form a 30°, 60°, 90° right triangle. ,Pxy221xyyx221x12.22x,Pxy0x22xyx22y2cos452452sin2,Pxy 2.2 Determining Cosine and Sine Values from the Unit Circle 59 Noting that we have half of an equilateral triangle with sides of length 1, we find , so . Since lies on the Unit Circle, we substitute into to get , or . In the first quadrant , so . 5. Plotting in standard position, we find θ is not a quadrantal angle and set about using a triangle approach. Once again, we get a 30°, 60°, 90° right triangle and, after computations similar to part 4 of this example, we find and . It is not by accident that the last three angles in Example 2.2.2 are 30°, 45° and 60° (or , and , respectively). In Section 2.1 we used right triangles to obtain these same cosine and sine values for 30°, 45° and 60°. In this section, the Unit Circle approach to calculating trigonometric function values allows us to expand the domain imposed by acute angles within a right triangle to include negative angles, and other angles outside the interval . Knowing which quadrant an angle θ terminates in will help us determine whether and are positive or negative, as indicated below. 12y1sin62,Pxy12y221xy234x32x0x3cos62x60601cos2x603sin2y6430,90cossin 60 The Trigonometric Functions Sign of Cosine and Sine in Each Quadrant The Pythagorean Identity In Example 2.2.2, it was quite easy to find the cosine and sine of the quadrantal angles, but for non- quadrantal angles, the task is more involved. In these latter cases, we made good use of the fact that the point lies on the Unit Circle, . If we substitute and into , we get the identity . An unfortunate convention, from a function notation perspective, is to write as and as . We will follow this convention. Thus, our identity results in the following theorem, one of the most important results in trigonometry. Theorem 2.1. The Pythagorean Identity: For any angle θ, . The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the distance formula and the equation for a circle are ultimately derived. The word ‘identity’ reminds us that, regardless of the angle θ, the equation in Theorem 2.1 is always true. If one of or is known, Theorem 2.1 can be used to determine the other, up to a sign. If, in addition, we know where the terminal side of θ lies when in standard position, we can remove the ambiguity of the sign and completely determine the missing value as the next example illustrates. ,cos,sinPxy221xycosxsiny221xy22cossin12cos2cos2sin2sin22cossin1cossinxy I II III IV cos0sin0 cos0sin0 cos0sin0 cos0sin0 2.2 Determining Cosine and Sine Values from the Unit Circle 61 Example 2.2.3. Using the given information about θ, find the indicated value. 1. If θ is a Quadrant II angle with , find . with , find . , find . 2. If 3. If Solution. 1. When we substitute into the Pythagorean identity, , we obtain . Solving, we find . Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. In Quadrant II, the x-coordinates are negative. Hence, . 2. Substituting into gives . Since we are given that , we know θ is a Quadrant III angle. Since x and y are negative in Quadrant III, both sine and cosine are negative in Quadrant III. Hence, we conclude . 3. When we substitute into , we find . 3sin5cos325cos5sinsin1cos3sin522cossin129cos1254cos54cos55cos522cossin1225sin553225sin5sin1
22cossin1cos0xy1 35 3cos,5 The Trigonometric Functions 62 Symmetry Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of . We plot θ in standard position and, as usual, let denote the point on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies radians short of one half revolution. In Example 2.2.2, we determined that and . This means that the point on the terminal side of the angle , when plotted in standard position, is . From the figure, it is clear that the point can be obtained by reflecting the point about the y-axis. Hence, and . Reference Angles In the above scenario, angle is called the reference angle for the angle . 56,Pxy63cos621sin62631,22,Pxy31,2253cos6251sin62656 2.2 Determining Cosine and Sine Values from the Unit Circle 63 In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis.   If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x- axis. If θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point , then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α, as seen in the following illustration. Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle cos,sin 64 The Trigonometric Functions We have just outlined the proof of the following theorem. Theorem 2.2. Reference Angle Theorem: Suppose α is the reference angle for θ. Then and , where the sign, + or –, is determined by the quadrant in which the terminal side of θ lies. In light of Theorem 2.2, it pays to know the cosine and sine values for certain common angles. In the following table, we summarize the values which we consider essential. Cosine and Sine Values of Common Angles θ degrees θ radians 0° 30° 45° 60° 90° 0 1 0 0 1 Example 2.2.4. Find the cosine and sine of the following angles. 1. 2. 3. 4. coscossinsincossin63212422223123222251165473xy 1,0 13,22 22,22 31,22 0,1 2.2 Determining Cosine and Sine Values from the Unit Circle 65 Solution. 1. We begin by plotting in standard position and find its terminal side overshoots the negative x-axis to land in Quadrant III. Hence, we obtain a reference angle α by subtracting: Since θ is a Quadrant III angle, and . The Reference Angle Theorem yields: and . 2. The terminal side of , when plotted in standard position, lies in Quadrant IV, just shy of the positive x- axis. To find the reference angle α, we subtract: Since θ is a Quadrant IV angle, and , so the Reference Angle Theorem gives: and . 3. To plot , we rotate clockwise an angle of from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angle of radians with respect to the negative x- 22518022518045.cos0sin0225452coscos2225452sinsin2116116622.cos0sin0113coscos662111sinsin6625454544 66 The Trigonometric Functions axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: and . 4. Since the angle measures more than , we find the terminal side of θ by rotating one full revolution followed by an additional radians. Since θ and α are coterminal, and . The reader may have noticed that, when expressed in radian measure, the reference angle for a non- quadrantal angle is easy to spot. Reduced fraction multiples of π with a denominator of 6 have as a reference angle. Those with a denominator of 4 have as their reference angle, and those with a denominator of 3 have as their reference angle. 52coscos44252sinsin44273623723371coscos33273sinsin332643 2.2 Determining Cosine and Sine Values from the Unit Circle 67 The next example summarizes all of the important ideas discussed in this section. Example 2.2.5. Suppose α is an acute angle with . 1. Find and use this to plot α in standard position. 2. Find the sine and cosine of the following angles. (a) (b) (c) (d) 5cos13sin232 68 Solution. The Trigonometric Functions 1. Proceeding as in Example 2.2.3, we substitute into and find . Since α is an acute (and therefore Quadrant I) angle, is positive. Hence, . To plot α in standard position, we begin our rotation from the positive x-axis to the ray which contains the point . 2. (a) To find the cosine and sine of , we first plot θ in standard position. We can imagine the sum of the angles as a sequence of two rotation: a rotation of π radians followed by a rotation of α radians.16 We see that α is the reference angle for θ, so by the Reference Angle Theorem, and . Since the terminal side of θ falls in Quadrant III, both and are negative. Hence, and . (b) Rewriting as , we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: and . 16 Since π + α = α + π, θ may be plotted by reversing the order of rotations given here. Try it! 5cos1322cossin112sin13sin12sin13512cos,sin,13135coscos1312sinsin13cossin5cos1312sin13225cos1312sin13 2.2 Determining Cosine and Sine Values from the Unit Circle 69 (c) Taking a cue from the previous problem, we rewrite as . The angle represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get and . (d) To plot , we first rotate radians and follow up with α radians. The reference angle here is not α, so the Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let be the point on the terminal side of θ which lies on the Unit Circle so that and . Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the following figure are equal. Hence, . Similarly, we find . 3335cos1312sin1322,Qxycosxsiny12cos13x5sin13y 70 The Trigonometric Functions We close this section by noting that we can easily extend the functions cosine and sine to real numbers by identifying a real number t with the angle radians. Using this identification, we define and . In practice this means expressions like and can be found by regarding the inputs as angles in radian measure or real numbers; the choice is the reader’s. If we trace the identification of real numbers t with angles θ in radian measure to its roots, as explained at the beginning of this section, we can spell out this correspondence more precisely. For each real number t, we associate an oriented arc t units in length with initial point and endpoint . tcoscostsinsintcos2sin1,0cos,sinPtt 2.2 Determining Cosine and Sine Values from the Unit Circle 71 2.2 Exercises In Exercises 1 – 5, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 1. 2. 3. 4. 5. In Exercises 6 – 9, use the given sign of the cosine and sine functions to find the quadrant in which the terminal point determined by t lies. 6. 8. and and 7. 9. and and 10. Use the numbers 0, 1, 2, 3 and 4 to complete the following table of cosine and sine values for common angles. (This exercise serves as a memory tool for remembering these values.) Θ 0 In Exercises 11 – 30, find the exact value of the cosine and sine of the given angle. 11. 15. 12. 16. 13. 17. 14. 18. 56tt6t2t12tcos0tsin0tcos0tsin0tcos0tsin0tcos0tsin0tcossin226224223222220432233476 72 19. 23. 27. The Trigonometric Functions 20. 24. 28. 21. 25. 29. 22. 26. 30. In Exercises 31 – 40, use the results developed throughout the section to find the requested value. 31. If with θ in Quadrant IV, what is ? 32. If with θ in Quadrant I, what is ? 33. If with θ in Quadrant II, what is ? 34. If with θ in Quadrant III, what is ? 35. If with θ in Quadrant III, what is 36. If with θ in Quadrant IV, what is 37. If and , what is ? 38. If and , what is 39. If and , what is ? ? ? ? 40. If and , what is ? 54433253742361324363461031177sin25cos4cos9sin5sin13cos2cos11sin2sin3cos28cos53sin25sin52cos10cos10522sinsin0.4232coscos0.982sin 2.3 The Six Circular Functions 73 2.3 The Six Circular Functions Learning Objectives In this section you will:  Determine the values of the six circular functions from the coordinates of a point on the Unit Circle.  Learn and apply the reciprocal and quotient identities.  Learn and apply the Generalized Reference Angle Theorem.  Find angles that satisfy circular function equations. In this section, we extend the definition of cosine and sine as points on the Unit Circle to include the remaining four circular functions: tangent, cotangent, secant and cosecant. The Circular Functions The Circular Functions: Suppose θ is an angle plotted in standard position and is the point on the terminal side of θ which lies on the Unit Circle. The circular functions are defined as follows.  The sine of θ, den
oted , is defined by  The cosine of θ, denoted , is defined by . .  The tangent of θ, denoted , is defined by , provided  The cosecant of θ, denoted , is defined by , provided . .  The secant of θ, denoted , is defined by , provided .  The cotangent of θ, denoted , is defined by , provided . In Section 2.2, we defined and for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions.17 17 In Section 2.1 we also showed cosine and sine to be functions of an angle residing in a right triangle so we could just as easily call them trigonometric functions. You will find that we do indeed use the phrase ‘trigonometric function’ interchangeably with ‘circular function’. ,Pxysinsinycoscosxtantanyx0xcsc1cscy0ysec1secx0xcotcotxy0ycossin 74 The Trigonometric Functions Historical Note: While we left the history of the name ‘sine’ as an interesting research project in Section 2.2, the names ‘tangent’ and ‘secant’ can be explained using the diagram below. Consider the acute angle θ in standard position. Let denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let denote the point on the terminal side of θ which lies on the vertical line . The word ‘tangent’ comes from the Latin meaning ‘to touch’. The line is a tangent line to the Unit Circle since it intersects, or touches, the circle at only one point, namely . Dropping perpendiculars from P and Q creates the pair of similar triangles ∆OPA and ∆OQB. Thus which gives , where this last equality comes from the definition of the tangent of θ. We have just shown that for acute angles θ, is the y-coordinate of the point on the terminal side of θ which lies on the tangent line . The word ‘secant’ means ‘to cut’. A secant line is any line that cuts through a circle at two points. The line containing the terminal side of θ is a secant line that intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OPA is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have, from similar triangles, that or . Hence, for an acute angle θ, is the length of the line segment which lies on the secant line determined by the terminal side of θ and ‘cuts off’ the tangent line . Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for calculus which we’ll explore in the Exercises. Of the six circular functions, only cosine and sine are defined for all angles. Since and in their definitions as circular functions, it is customary to rephrase the remaining four ,Pxy1,'Qy1x1x1,0'1yyx'tanyyxtan1x11hx1sechxsec1xcosxsiny 2.3 The Six Circular Functions 75 circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with and y with in the definitions presented at the beginning of this section. Reciprocal and Quotient Identities Theorem 2.3. Reciprocal and Quotient Identities:     , provided ; if then is undefined. , provided , provided , provided ; if ; if ; if then is undefined. then is undefined. then is undefined. Before using Theorem 2.3 in an example, the following mnemonic may help with remembering the signs of the trigonometric functions in each quadrant. We assign the first letter of each word in the phrase “All Students Take Calculus” to Quadrants I, II, III and IV, respectively. Note that cosine, sine and tangent are All positive in Quadrant I, the Sine alone is positive in Quadrant II, then Tangent alone is positive in Quadrant III and the Cosine alone is positive in Quadrant IV. It is high time for an example. Example 2.3.1. Find the indicated value, if it exists. 1. 2. 3. cossinsintancosyxcos0cos0tancoscotsinxysin0sin0cot11seccosxcos0cos0sec11cscsinysin0sin0csc60sec7csc43cotxyASTC The Trigonometric Functions 76 4. 5. 6. Solution. , where θ is any angle coterminal with . , where and θ is a Quadrant IV angle. , where and . 1. From Theorem 2.3, the reciprocal identity for secant will help us out here. 2. We apply the reciprocal identity for cosecant and note that . 3. Since radians is not one of the common angles from Section 2.2, we resort to the calculator for a decimal approximation. We use the quotient identity for cotangent and check that our calculator is in radian mode. Noting that , this problem could also be solved as follows. tan32coscsc5sintan3321260601seccos1272sin4271csc74sin412223333coscotsin7.0151cottan 2.3 The Six Circular Functions 77 4. If θ is coterminal with , then and . Attempting to compute results in , so is undefined. 5. We are given that . From , it follows that . Then we have the following. 6. It is given that . From the quotient identity for tangent, we know . Be careful! We can NOT assume any values for and . We CAN assume that . 331cottan7.015323coscos023sinsin12sintancos10tancsc51cscsin1sin52221cos154cos52cos52cos525cos5 Pythagorean identity cosθ>0 in Quadrant IVtan3sin3cossincossin3cos 78 The Trigonometric Functions While the reciprocal and quotient identities presented in Theorem 2.3 allow us to always convert problems involving tangent, cotangent, secant and cosecant to problems involving cosine and sine, it is not always convenient to do so.18 The tangent and cotangent values of the common angles are summarized in the following chart. 18 As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt. 222222sin3cos1sincos31sinsin131sinsin1910sin199sin103sin103sin10310sin10Pythagorean identitysinθ<0 in Quadrant III 2.3 The Six Circular Functions 79 Tangent and Cotangent Values of Common Angles θ degrees θ radians 0° 30° 45° 60° 90° 0 1 0 0 undefined 1 1 0 1 undefined 0 Finding Angles that Satisfy Cosine and Sine Equations Our next example asks us to solve some very basic trigonometric equations.19 Example 2.3.2. Find all of the angles which satisfy the given equation. 1. 2. 3. Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in each of these problems. This choice will be justified later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian measure will be the only appropriate angle measure so it is worth the time to become fluent in radians now. 1. If , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at . This means θ is a Quadrant I or IV angle with reference angle . 19 We will study trigonometric equations more formally in Chapter 6. Enjoy these relatively straightforward exercises while they last! cossinsincostancossincot632121333342222312323133321cos21sin2cos01cos212x3 80 The Trigonometric Functions One solution in Quadrant I is , and since all other Quadrant I solutions must be coterminal with , we find for integers k.20 Proceeding similarly for the Quadrant IV case, we find the solution to is , so our answer in this quadrant is for integers k. 2. If , then when θ is plotted in standard position, its terminal side intersects the Unit Circle at . From this, we determine θ is a Quadrant III or Quadrant IV angle with reference angle . 20 Recall in Section 1.2, two angles in radian measure are coterminal if and only if they differ by an integer multiple of 2π. Hence to describe all angles coterminal with a given angle, we add 2πk for integers k = 0, ±1, ±2, ···. 3323k1cos253523k1sin212y6 2.3 The Six Circular Functions 81 In Quadrant III, one solution is , so we capture all Quadrant III solutions by adding integer multiples of 2π: . In Quadrant IV, one solution is so all the solutions here are of the form for integers k. 3. The angles with are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y-axis. While, technically speaking, isn’t a reference angle, we can nonetheless use it to find our answers. If we follow the procedure set forth in the previous examples, we find and for integers k. While this solution is correct, it can be shortened to for integers k. (Can you see why this works from the diagram?) One of the key items to take from Example 2.3.2 is that, in general, solutions to trigonometric equations consist of infinitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra ‘When in doubt, write it out!’ This is especially important when checking answers to the Exercises. For example, in number 2, another Quadrant IV solution to is . Hence, the family of Quadrant IV answers to number 2 above could have been written for integers k. While on the surface this family may look different than the stated solution of for integers k, we leave it to the reader to show they represent the same list of angles. 76726k1161126kcos0222k322k2k1sin2626k1126k 82 The Trigonometric Functions Finding Angles that Satisfy Other Circular Function Equations Before determining angles in equations of the other four circular functions, we introduce the Generalized Reference Angle Theorem. This theorem results from coupling the reciprocal and quotient identities, Theorem 2.3, with the Reference Angle Theorem, Theorem 2.2. Theorem 2.4. Generalized Reference Angle Theorem: The values of the circular functions of an angle, if they exi
st, are the same, up to a sign, as the corresponding circular functions of the reference angle. More specifically, if α is the reference angle for θ, then , , , , and . The sign, + or –, is determined by the quadrant in which the terminal side of θ lies. We put Theorem 2.4 to good use in the following example. Example 2.3.3. Find all angles which satisfy the given equation. 1. Solution. 2. 3. 1. To solve , we convert to a cosine and get , or . This is the same equation we solved in Example 2.3.2, number 1, so we know the answer is or for integers k. 2. Noting that , we see . According to Theorem 2.4, we know the solutions to must, therefore, have a reference angle of . Our next task is to determine in which quadrants the solutions to this equation lie. sinsincoscostantancsccscsecseccotcotsec2tan3cot1sec212cos1cos223k523ksin3tan3cos3tan33tan33 2.3 The Six Circular Functions 83 Since tangent is defined as the ratio of points on the Unit Circle with , tangent is positive when x and y have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III. In Quadrant I we get the solutions for integers k, and for Quadrant III we get for integers k. While these descriptions of the solutions are correct, they can be combined as for integers k. The latter form of the solution is best understood looking at the geometry of the situation in the following diagram.21 3. We see that has a cotangent of 1, which means the solutions to have a reference angle of . 21 See Example 2.3.2, number 3, for another example of this kind of simplification of the solution. yx,xy0x23k423k3k4cot14 84 The Trigonometric Functions To find the quadrants in which our solutions lie, we note that for a point on the Unit Circle where . If is negative, then x and y must have different signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is , and for Quadrant IV we get , for integers k. Can these be combined? Indeed they can! One such way to capture all the solutions is for integers k. Suppose we are asked to solve an equation such as . As we have already mentioned, the distinction between t as a real number and as an angle radians is often blurred. Indeed, we solve in the exact same manner22 as we did in Example 2.3.2 number 2. Our solution is only cosmetically different in that the variable used is t rather than θ: or for integers k. As we progress in our study of the trigonometric functions, keep in mind that any properties developed which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers. 22 Well, to be pedantic, we would be technically using reference numbers or reference arcs instead of reference angles, but the idea is the same. cotxy,xy0ycot324k724k34k1sin2tt1sin2t726tk1126tk 85 2.3 Exercises In Exercises 1 – 20, find the exact value or state that it is undefined. The Trigonometric Functions 1. 5. 9. 13. 17. 2. 6. 10. 14. 18. 3. 7. 11. 15. 19. 4. 8. 12. 16. 20. In Exercises 21 – 44, find all angles which satisfy the given equation. 21. 24. 27. 30. 33. 36. 39. 42. 22. 25. 28. 3 1. 3 4. 37. 40. 43. 23. 26. 29. 3 2 .csc( θ) 35. 38. 41. 44. tan4sec65csc64cot311tan63sec2csc313cot2tan1175sec3csc3cot531tan2sec47csc47cot62tan3sec7csc23cot41sin23cos2sin02cos23sin2cos1sin13cos2cos1.001tan3sec()213cot3tan0sec1csc2cot0tan1sec012cscsec1tan3csc2cot1 86 The Trigonometric Functions In Exercises 45 – 61, solve the equation for t. Give exact values. (See the comments following Example 2.3.3.) 45. 48. 51. 54. 57. 60. 46. 49. 52. 55. 58. 61. 47. 50. 53. 56. 59. 62. Explain why the fact that does not necessarily mean and . (See the solution to number 6 in Example 2.3.1.) cos0t2sin2tcos3t1sin2t1cos2tsin2tcos1tsin1t2cos2tcot1t3tan3t23sec3tcsc0tcot3t3tan3t23sec3t23csc3t31tan3sin3cos1 2.4 Verifying Trigonometric Identities 87 2.4 Verifying Trigonometric Identities Learning Objectives In this section you will:  Learn and apply the Pythagorean identities and conjugates.  Simplify trigonometric expressions.  Prove that a trigonometric equation is an identity. We have already seen the importance of identities in trigonometry. Our next task is to use the reciprocal and quotient identities found in Theorem 2.3, coupled with the Pythagorean identity found in Theorem 2.1, to derive the new Pythagorean-like identities for the remaining four circular functions. The Pythagorean Identities Theorem 2.1 states that, for any angle θ, . Through manipulating this identity, we will obtain two alternate versions relating secant and tangent, followed by cosecant and cotangent. To obtain an identity relating secant and tangent, we start with and, assuming , divide both sides of the equation by . The result is the Pythagorean identity . We next look for an identity relating cosecant and cotangent. We assume that and divide both sides of by . 22cossin122cossin1cos02cos22222222222cossin1cossin1coscoscossin11coscos1tansec from quotient & reciprocal identities 221tansecsin022cossin12sin 88 The Trigonometric Functions Thus, we have , our third Pythagorean Identity,. The three Pythagorean Identities, along with some of their other common forms, are summarized in the following theorem. Theorem 2.5. The Pythagorean Identities: 1. 2. 3. Common Alternate Forms: , provided Common Alternate Forms: , provided Common Alternate Forms: and and and Trigonometric identities play an important role, both in trigonometry and calculus. We’ll use them in this book to find the values of the circular functions of an angle and to solve equations. In calculus, they are needed to rewrite expressions in a format that enables or simplifies integration. In this next example, we make good use of Theorem 2.3 and Theorem 2.5. Example 2.4.1. Verify the following identities. Assume that all quantities are defined. 1. 3. 5. 2. 4. 6. 22222222222cossin1cossin1sinsinsincos11sinsincot1cscfrom quotient & reciprocal ident ities22cot1csc22cossin1221sincos221cossin221tanseccos022sectan122sec1tan221cotcscsin022csccot122csc1cot1sincsctansinsecsectansectan1sec11tancossin336sectan1sin1sinsin1cos1cossin 2.4 Verifying Trigonometric Identities 89 Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1. To verify we start with the left side, using the reciprocal identity for cosecant. 2. We start with the right hand side of . 3. We begin with the left hand side of the equation. 4. While both sides of the equation contain fractions, the left side affords us more opportunities to use our identities. 1sincsc111cscsinsintansinsectan1sinsecsincossincos from reciprocal identity for secantfrom quotient identity for tangent2222sectansectansecsectantansectansectan1 binomial multiplicationPythagorean iden tity 90 The Trigonometric Functions 5. The right hand side of the equation seems to hold promise. We find a common denominator. 6. It is debatable which side of the equation is more complicated. One thing which stands out is that the denominator on the left hand side is , while the numerator on the right hand side is . This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity . 1seccossin1tan1cos1coscossincos1coscoscossincoscoscos1cossin22231sin31sin331sin1sin1sin1sin33sin33sin1sin6sin1sin6sincos1sin6coscos6sectan Pythagorean identityreciprocal and quotient identities1cos1cos1cos 2.4 Verifying Trigonometric Identities 91 In the preceding example, number 6, we see that multiplying by produces a difference of squares that can be simplified to one term using Theorem 2.5. This is exactly the same kind of phenomenon that occurs when we multiply expressions such as by . For this reason, the quantities and are called ‘Pythagorean conjugates’. The following list includes other Pythagorean conjugates. Pythagorean Conjugates       : : : and and and and and : and : : 221cossinsin1cos1cos1cossin1cos1cossin1cossinsin1cossinsin1cossinbinomial multiplicationPyth agorean identity1cos1cos12121cos1cos1cos1cos221cos1cos1cossin1sin1sin221sin1sin1sincossec1sec122sec1sec1sec1tansectansectan22sectansectansectan1csc1csc122csc1csc1csc1cotcsccotcsccot22csccotcsccotcsccot1 92 The Trigonometric Functions Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling just to become proficient in the basics. Like many things in life, there is no short-cut here. There is no complete algorithm for verifying identiti
es. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) follows and ample practice is provided for you in the Exercises. Strategies for Verifying Identities  Try working on the more complicated side of the identity.  Use the Reciprocal and Quotient Identities in Theorem 2.3 to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify any resulting complex fractions.  Add rational expressions with unlike denominators by obtaining common denominators.  Use the Pythagorean identities in Theorem 2.5 to exchange sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or differences of squares to one term.  Multiply numerator and denominator by Pythagorean conjugates in order to take advantage of the Pythagorean identities in Theorem 2.5.  If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work. Most importantly, keep in mind that we are not solving equations. To verify identities, we choose one side of the identity and work with that side until it matches the other side. Verifying identities is an important skill and we will work with identities again in Chapter 4, as more tools become available. Time spent now in developing some proficiency will be useful throughout the course. 2.4 Verifying Trigonometric Identities 93 2.4 Exercises In Exercises 1 – 47, verify the identity. Assume that all quantities are defined. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. cossec1tancossinsincsc1tancot1csccoscot2sinsectancos2coscsccotsin1sinsectancos1coscsccotsin2cossec1sin2sincsc1cos2seccos1tan2cscsin1cot2tancotsec12cottancsc1224cos4sin4229cossin832tantansectan252sin1cossin41022sec1tansec23costantansincos4224secsectantancos11seccos11secsin11cscsin11csc1cottan11cottan11tancossin1tancossintancotseccsccscsincotcoscossectansincostancotcsc 94 The Trigonometric Functions 31. 33. 35. 37. 39. 41. 43. 45. 47. 32. 34. 36. 38. 40. 42. 44. 46. In Exercises 48 – 51, verify the identity. You may need to review the properties of absolute value and logarithms before proceeding. 48. 50. 49. 51. sintancotsec2112csc1cos1cos112csccotsec1sec1112sectancsc1csc1112cotcsccotcsccotcossinsincos1tan1cot1sectansectan1sectansectan1csccotcsccot1csccotcsccot21secsectan1sin21secsectan1sin21csccsccot1cos21csccsccot1coscos1sin1sincossincsccot1cos21sinsectan1sinlnseclncoslncsclnsinlnsectanlnsectanlncsccotlncsccot 2.5 Beyond the Unit Circle 95 2.5 Beyond the Unit Circle Learning Objectives In this section you will:  Determine the values of the six circular functions from the coordinates of a point on a circle, centered at the origin, with any radius r.  Solve related application problems.  Describe the position of a particle experiencing circular motion. Recall that in defining the cosine and sine functions in Section 2.2, we assigned to each angle a position on the Unit Circle. Here we broaden our scope to include circles of radius r centered at the origin. Determining Cosine and Sine Consider for the moment the acute angle θ drawn below in standard position. Let be the point on the terminal side of θ which lies on the circle , and let be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two triangles, ∆OPA and ∆OQB. These triangles are similar23. Thus, it follows that , from which . We similarly find . Since, by definition, and , we get the coordinates of Q to be and . By reflecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles. 23 Do you remember why? ,Qxy222xyr','Pxy'1xrrx'xrx'yry'cosx'sinycosxrsinyr 96 The Trigonometric Functions Not only can we describe the coordinates of Q in terms of and , but since the radius of the circle is , we can also express and in terms of the coordinates of Q. Throughout this textbook, by convention, the radius r of a circle is treated as positive as it relates to solving for trigonometric values. These results are summarized in the following theorem. Theorem 2.6. If is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle then and . Moreover, Note that in the case of the Unit Circle we have , so Theorem 2.6 reduces to our Unit Circle definitions of and . Example 2.5.1 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point . Find and . 2. In Example 1.3.4 in Section 1.3, we approximated the radius of the circle of revolution at 40.7608° North Latitude on Earth to be 2999 miles. Justify this approximation if the radius of the circle of revolution at the Equator is approximately 3960 miles. Solution. 1. Using Theorem 2.6 with and , we find so that cossin22rxycossin,Qxy222xyrcosxrsinyr2222cos= and sin.xxyyrrxyxy221rxycossin4,2Qcossin4x2y22422025r 2.5 Beyond the Unit Circle 97 2. Assuming the Earth is a sphere, a cross-section through the poles produces a circle of radius 3960 miles. Viewing the Equator as the x-axis, the value we seek is the x-coordinate of the point indicated in the figure. Using Theorem 2.6, we get . Using a calculator in degree mode, we find . Hence, the radius of the circle of revolution at North Latitude 40.7608° is approximately 2999 miles. Position of a Particle in Circular Motion Theorem 2.6 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point when , the angle θ is in standard position. By definition, , which we rewrite as . According to Theorem 2.6, the location of the object on the circle is found using the equations and . With , we have and . Hence, at time t, the object is at the point . We have just argued the following. cos and sin42cos sin2525255cos sin.55xyrr,Qxy3960cos40.7608x3960cos40.76082999,0r0ttt,Qxycosxrsinyrt costxr sintyr cos,sinttrr 98 The Trigonometric Functions Equations for Circular Motion: Suppose an object is traveling on a circular path of radius r centered at the origin with constant angular velocity ω. If corresponds to the point , then the x and y coordinates of the object are functions of t and are given by and . Here, indicates a clockwise direction. Example 2.5.2. Suppose we are in the situation of Example 1.3.4. Find the equations of motion of Salt Lake Community College as the Earth rotates. Solution. From Example 1.3.4, we take miles and . Hence, using and , the equations of motion are Note that x and y are measured in miles and t is measured in hours. Determining the Other Four Circular Functions We have generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. Using Theorem 2.6 in conjunction with Theorem 2.3, we generalize the remaining circular functions in kind. 0t,0r costxr sintyr02999r12 hours costxr sintyr2999cos and 2999sin.1212xtyt 2.5 Beyond the Unit Circle 99 Theorem 2.7. Suppose is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle . Then the circle has radius r and     , provided , provided . . , provided , provided . . Example 2.5.3. 1. Suppose the terminal side of θ, when plotted in standard position, contains the point . Find the values of the six circular functions of θ. 2. Suppose θ is a Quadrant IV angle with . Find the values of the five remaining circular functions of θ. Solution. 1. The radius of the circle containing the point is With , and , we apply Theorems 2.6 and 2.7 to find the values of the six circular functions of θ.24 24 For convenience, the sketch shows 0 < θ < 2π. In reality, θ may be any angle, plotted in standard position, which contains the point Q (3,–4) on its terminal side. ,Qxy222xyr22cscxyryy0y22secxyrxx0xtanyx0xcotxy0y3,4Qcot43,4Q2222345.rxy3x4y5r 100 The Trigonometric Functions 2. We look for a point which lies on the terminal side of θ 25, when θ is plotted in standard position. We are given that θ is a Quadrant IV angle, so we know and . Also, . Since , we may choose26 and , from which The five remaining circular function values follow. 25 Again, θ may be any angle, plotted in standard position, with Q on its terminal side. 26 We may choose any values x and y so long as x > 0, y < 0 and x ∕ y = –4. For example, we could choose x = 8 and y = –2. The fact that all such points lie on the terminal side of θ is a consequence of the fact that the terminal side of θ is the portion of the line with slope –1 ∕ 4 which extends from the origin into Quadrant IV. 45sin csc5435cos sec5343tan cot.34yrryxrrxyxxy,Qxy0x0ycot
4xy4414x1y22224117.rxy117sin17174417cos17171tan417csc17117sec4yrxryxryrx 2.5 Beyond the Unit Circle 101 We close this section by noting that we have not yet discussed the domains and ranges of the circular functions. In Chapter 3, we will graph the circular functions. This will provide a visual platform for the introduction of the domain and range of each circular function. 102 2.5 Exercises The Trigonometric Functions In Exercises 1 – 8, let θ be the angle in standard position whose terminal side contains the given point. Find the values of the six circular functions of θ. 1. 5. 2. 6. 3. 7. 4. 8. In Exercises 9 – 10, determine the radius of the circle of revolution at the given latitude on Earth. Assume that the radius of the circle of revolution at the Equator is approximately 3960 miles. 9. 55.39° North Latitude 10. 44.29° South Latitude In Exercises 11 – 12, determine the radius of the circle of revolution at the latitude corresponding to the given position on Earth. You may use the Internet for determining latitudes. Assume that the radius of the circle of revolution at the Equator is approximately 3960 miles. 11. Sydney, Australia 12. Nome, Alaska In Exercises 13 – 16, find the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that corresponds to a position along the positive x-axis. 13. A point on the edge of a yo-yo which is 2.25 inches in diameter and spins at 4500 revolutions per minute. 14. A point on the edge of a yo-yo used in the trick ‘Around the World’ in which the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. Assume the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle. 15. A point on the edge of the circular disk in a computer hard drive. The circular disk has diameter 2.5 inches and spins at a rate of 7200 RPM (revolutions per minute). 16. A passenger on the Giant Wheel at Cedar Point Amusement Part. The Giant Wheel is a circle with diameter 128 feet. It completes two revolutions in 2 minutes and 7 seconds. 1,5A3,1B6,2C10,12D7,24P3,4Q5,9R2,11T0t 2.5 Beyond the Unit Circle 103 In Exercises 17 – 30, use the given information to find the exact values of the remaining circular functions of θ. 17. 19. 21. 23. 25. 27. 29. with θ in Quadrant II. with θ in Quadrant I. 18. 20. with θ in Quadrant III. with θ in Quadrant IV. with θ in Quadrant III. 22. with θ in Quadrant II. with θ in Quadrant IV. 24. with θ in Quadrant II. with θ in Quadrant III. 26. with θ in Quadrant I. with . with . 28. 30. with . with . 31. In deriving the equations for circular motion, we assumed that at the object was at the point . If this is not the case, we can adjust the equations of motion by introducing a time delay. If is the first time the object passes through the point , show, with the help of your classmates, the equations of motion are and . 3sin512tan525csc24sec71091csc91cot23tan2sec4cot51cos3cot202csc52tan1032sec253220t,0r00t,0r 0()costtxr 0()sinttyr 104 CHAPTER 3 GRAPHS OF THE TRIGONOMETRIC FUNCTIONS Chapter Outline 3.1 Graphs of the Cosine and Sine Functions 3.2 Properties of the Graphs of Sinusoids 3.3 Graphs of the Tangent and Cotangent Functions 3.4 Graphs of the Secant and Cosecant Functions Introduction In Chapter 3, we graph the six trigonometric functions and learn about important properties of each function such as domain, range, period, and whether the function is even or odd. We begin with graphs of the cosine and sine functions in Section 3.1, which lead into further discussion of their designation as sinusoids in Section 3.2. We notice the similarities between sine and cosine graphs, along with horizontal shifts that will turn a sine graph into a cosine graph, or vice versa. Graphs of sinusoids and their applications, including harmonic motion, in Section 3.2 are followed by graphs and properties of tangent and cotangent functions in Section 3.3. We end with Section 3.4, graphing secant and cosecant functions and observing properties of each of these functions, as well as the relationship between the two. Throughout Chapter 3, applications are introduced. Particular attention is paid to the graphing techniques that allow us to graph transformations of each function. This is an essential chapter in our textbook as it provides many of the tools we will need in applying identities and formulas, as well as solving trigonometric equations, in future chapters. 3.1 Graphs of the Cosine and Sine Functions 105 3.1 Graphs of the Cosine and Sine Functions Learning Objectives In this section you will:  Graph the cosine and sine functions and their transformations. Identify the period.  Learn the properties of the cosine and sine functions, including domain and range.  Determine whether a function is even or odd. We return to our discussion of the circular (trigonometric) functions as functions of real numbers and turn our attention to graphing the cosine and sine functions in the Cartesian Plane. Before proceeding with graphs of these trigonometric functions, we note that the graphs of both functions are continuous and smooth. Geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes, vertical asymptotes, corners or cusps. As we shall see, the graphs of both functions meander nicely and don’t cause any trouble. Graph of the Cosine Function To graph the cosine function, we use x as the independent variable and y as the dependent variable.27 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. We graph by making a table using some of the common values of x in the interval . This generates a portion of the cosine graph, which we call the fundamental cycle of . 27 The use of x and y in this context is not to be confused with the x- and y-coordinates of points on the Unit Circle which define cosine and sine. Using the term ‘trigonometric’ function as opposed to ‘circular’ function may help to avoid confusion. cosyx0,2cosyx 106 Graphs of the Trigonometric Functions x 0 1 0 π –1 x 2π 0 1 Noting that is defined for all real numbers x, we plot the points from the table to guide us in sketching the graph of on the interval . The fundamental cycle of . A few things about the graph above are worth mentioning: 1. This graph represents only part of the graph of . To get the entire graph, imagine copying and pasting this graph end to end infinitely in both directions (left and right) along the x-axis. 2. The vertical scale here has been greatly exaggerated for clarity and aesthetics. Below is an accurate to-scale graph of showing several cycles with the fundamental cycle plotted thicker than the others. cosyx,cosxxcosyx,cosxx0,1542252,424222,42323,022,02742272,42342232,422,1,1cosyx,cosxxcosyx0,2cosyxcosyxcosyx 3.1 Graphs of the Cosine and Sine Functions 107 The graph of is usually described as ‘wavelike’ and, indeed, many of the applications involving the cosine and sine functions feature the modeling of wavelike phenomena. An accurately scaled graph of . Graph of the Sine Function We can plot the fundamental cycle of the graph of similarly. x 0 0 1 π 0 x –1 2π 0 The fundamental cycle of . cosyxcosyxsinyxsinyx,sinxxsinyx,sinxx0,0542252,424222,42323,122,12742272,42342232,422,0,0sinyx 108 Graphs of the Trigonometric Functions As with the graph of , we can provide an accurately scaled graph of with the fundamental cycle highlighted. An accurately scaled graph of . It is no accident that the graphs of and are so similar. The graph of is a result of the graph of being shifted units to the left. Try it! Period of the Cosine and Sine Functions Not only can we obtain a graph of the sine function by shifting the graph of the cosine function units to the left, we can shift the graph of by 2π units to the left and obtain a graph that is equivalent to the original graph of . The same can be said for shifts of 4π, 6π, 8π, ··· units to the left. We say that the cosine function is periodic, as defined below. Periodic Functions: A function f is said to be periodic if there is a real number p so that for all real numbers x in the domain of f. The smallest positive number p, if it exists, is called the period of f. We see that by the definition of periodic functions, is periodic since for any integer k. To determine the period of f, we need to find the smallest positive real number p so that for all real numbers x or, said differently, the smallest positive real number p such that for all real numbers x. We know that for all real numbers x but the question remains if any smaller real number will do the trick. Suppose and for all real numbers x. Then, in cosyxsinyxsinyxcosyxsinyxcosyxsinyx22cosyxcosyxfxpfxcosfxxcos2cosxkxfxpfxcoscosxpxcos2cosxx0pcoscosxpx 3.1 Graphs of the Cosine and Sine Functions 109 particular, so that . From this we know that p is a multiple of 2π and, since the smallest positive multiple of 2π is 2π itself, we have the result. Similarly, we can show is periodic with 2π as its period. Having period 2π essentially means that we can completely understand everything about the functions and by studying one interval of length 2π, say . Even/Odd Properties of the Cosine and Sine Functions While we will explore the even and odd properties of the cosine and sine functions further in Section 4.1, for now it is worth revisiting the graphical test for symmetry as introduced in college algebra. Recall that we refer to a function as even
if its graph is symmetric about the y-axis and a function as odd if its graph is symmetric about the origin. Observe the symmetry of the ‘accurately scaled’ graphs of the cosine and sine functions from earlier in this section. The graph of is symmetric about the y-axis. As will be proved algebraically in Section 4.1, the cosine function is an even function. The graph of is symmetric about the origin. The sine function is an odd function. From a previous algebra class, you may recall that, for an even function f, . If f is odd then . This is true for all values of x, and applies to the trigonometric functions as well. For the cosine function, an even function, . The sine function is odd, and thus . Domain and Range of the Cosine and Sine Functions The two functions and , both trigonometric functions of x, are defined for all real values of x. This is evident in the preceding graphs of cosine and sine. Or, going back to our Unit Circle definitions for cosine and sine, whether we think of identifying the real number t with the angle radians, or think of wrapping an oriented arc around the Unit Circle to find coordinates on the Unit Circle, it should be clear that both cosine and sine are defined for any real input number t. Thus, the domain of both and is . cos0cos0pcos1psingxxcosfxxsingxx0,2cosyxsinyxfxfxfxfxcoscosxxsinsinxxcosfxxsingxxtcosfttsingtt, 110 Graphs of the Trigonometric Functions Looking at the graphs of the cosine and sine functions, we see that the range includes all real numbers between –1 and 1, inclusive. Revisiting the Unit Circle, and represent x- and y-coordinates, respectively, of points on the Unit Circle. As points on the Unit Circle, they take on all of the values between –1 and 1, inclusive. In other words, the range of both and is the interval . Following is a summary of properties of both functions. Theorem 3.1. Properties of the Cosine and Sine Functions: The function  has domain  has range The function  has domain  has range   is continuous and smooth is even   is continuous and smooth is odd  has period 2π  has period 2π Graphs of Transformations of the Cosine and Sine Functions Now that we know the basic shapes of the graphs of and , we can use transformations to graph more complicated curves. The following theorem, borrowed from college algebra, will guide us in transformations of graphs of cosine and sine functions. cosfttsingttcosfttsingtt1,1cosftt,1,1singtt,1,1cosyxsinyx 3.1 Graphs of the Cosine and Sine Functions 111 Theorem 3.2. Transformations of Periodic Functions: Suppose f is a periodic function. If and , then to graph 1. Divide the period of the function by ω to determine the period of the transformed function g. For sine and cosine, the period is 2π, so is the period of g. Note that ω is the number of times the cycle of g repeats on . 2. Find the vertical shift by determining the value of B. The graph will be shifted up by B if and down by if . The line will be the ‘baseline’. 3. Determine the vertical scaling, , also known as the amplitude. If , note that the graph will be reflected about the baseline . 4. The horizontal shift , also known as the phase shift, can be determined by rewriting as . This results in a horizontal shift to the left if or right if . In transforming one cycle of or , we need to keep track of the movement of some key points. We choose to track the points with x-values , , , and . These quarter marks correspond to quadrantal angles, and as such, mark the location of the zeros and the local extrema of these functions over exactly one period. Example 3.1.1. Graph one cycle of the function . Solution. We start with the following graph of the fundamental cycle of . The locations of quarter marks are denoted with dashed blue lines, and intersect the graph at the key points , , , and . The baseline of is marked by a dashed pink line. 0A0gxAfxBfx20,20BB0ByBA0AyBxx00cosyxsinyx023222sin31fxxsinyx0,0,12,03,122,00y 112 Graphs of the Trigonometric Functions The steps in Theorem 3.2 help us determine transformations for graphing . Step 1: We first determine the period. The period for is 2π. For coefficient of x is 3, so the graph of the function will cycle three times as fast as other words, it will repeat itself three times on the interval , with one cycle being , the . In completed in a period of . Step 2: We next determine the vertical shift. The graph of has a vertical shift of 1. By shifting the baseline for the graph of up by 1, we have a baseline of for the graph of . Step 3: The vertical scaling, or amplitude, of is 1. For the function , the sine function is multiplied by 2 so the function values are twice as much. Thus the amplitude is 2. Step 4: The horizontal shift is 0. Finally, we are ready to graph one cycle of . Since the horizontal shift is 0, we can leave the first quarter mark at . We then divide the period of by 4 to determine a distance of between quarter marks. The positions for quarter marks and the baseline of are used as a guide in graphing . 2sin31fxxsinyx2sin31fxxsinyx0,2232sin31fxxsinyx1y2sin31fxxsinyx2sin31fxx2sin31fxx0x2361y2sin31fxxxy0 sinyx 3.1 Graphs of the Cosine and Sine Functions 113 The graph of will maintain the shape of the sine function. With an amplitude of 2, the local maximums and minimums will occur 2 units above and 2 units below the baseline, respectively. We can plot 5 key points of this function using the quarter mark locations and baseline as a guide, and finish the graph by connecting the points with a continuous smooth curve. Example 3.1.2. Graph one cycle of the function . Solution. We begin with a graph of the cosine function, noting the key points of , , , and . 2sin31fxx3cos2fxx0,1,02,13,022,1xy0xyxy 0,1 ,13 2,13 ,36 ,12 2sin31fxx 114 Graphs of the Trigonometric Functions Step 1: The period of is 2π. For , the coefficient of x is 2 so the period is Step 2: The vertical shift of is 0. Thus, the baseline will remain at . Step 3: The amplitude is . Since , the graph will be reflected about the baseline . Step 4: To determine the horizontal shift, the function can be rewritten as The horizontal shift is , indicating a shift to the right of units. Now we are ready to graph this transformation of . From the horizontal shift of , the first quarter mark will be at . The period is π, so the distance between quarter marks will be . We denote the positions of the quarter marks and the baseline . The amplitude is 3. While the basic shape is that of the cosine, the graph will be reflected about the baseline , resulting in reversed positions for the local maximum and minimum values. After determining the 5 key points for one cycle of of . , we connect the points with a continuous smooth curve, replicating the shape cosyx3cos2fxx223cos2fxx0y33300y3cos23cos22fxxx22cosyx22x40y0y3cos2fxxcosyxxy01-1 cosyx 3.1 Graphs of the Cosine and Sine Functions 115 Above, we have graphed one complete period of the function . This graph could easily be extended in either direction. In the next example we sketch the graph of a sine function over the larger interval of two periods. Example 3.1.3. Graph two full periods of the function and state the period. Identify the maximum and minimum y-values and their corresponding x-values. Solution. To sketch one period of the function through transformations of the the fundamental cycle of , we note the following. 3cos2fxx4sin2gxx4sin2gxxsinyxxyxyxy 3,04 ,3 5,04 3,323cos2fxx ,32 116 Graphs of the Trigonometric Functions 1. The period is , so the distance between quarter marks will be . 2. There is a vertical shift of –2, resulting in a baseline of . 3. The amplitude is 4. 4. There is no horizontal shift, so the first quarter mark is at . To sketch two periods, we extend the graph for one period, as follows. The period of 2 appears twice in the preceding graph. A maximum y-value of 2 corresponds with x- values of and . The minimum y-value is and corresponds with x-values of and . The functions in this section are examples of sinusoids. Roughly speaking, a sinusoid is the result of performing transformations to the basic graph of or . Sinusoids will be looked at extensively in Section 3.2. 2221422y0x4sin2ygxx321261232cosfxxsingxxxy2-112-2-6-2xy 3.1 Graphs of the Cosine and Sine Functions 117 3.1 Exercises 1. Why are the cosine and sine functions called periodic functions? 2. How does the graph of compare with the graph of ? Explain how you could horizontally translate the graph of to obtain the graph of . 3. For the function , what constants affect the range and how do they affect the range? In Exercises 4 – 15, graph one cycle of the given function. State the period of the function. 4. 7. 10. 13. 5. 8. 11. 14. 6. 9. 12. 15. In Exercises 16 – 27, graph two full periods of each function and state the period. Identify the maximum and minimum y-values and their corresponding x-values. 16. 19. 22. 25. 17. 20. 23. 26. 18. 21. 24. 27. sinyxcosyxsinyxcosyxcosfxABxCD3sinyxsin3yx2cosyxcos2yxsin3yxsin2yx11cos323yxcos324yxsin24yx2cos4132yx31cos2232yx4sin2yx2sinfxx2cos3fxx3sinfxx4sinfxx2cosfxxcos2fxx12sin2fxx4cosfxx63cos5fxx3sin845yx2sin3214yx5sin5202yx 118 Graphs of the Trigonometric Functions 28
. Show that a constant function f is periodic by showing that for all real numbers x. Then show that f has no period by showing that you cannot find a smallest number p such that for all real numbers x. Said another way, show that for all real numbers x for ALL values of , so no smallest value exists to satisfy the definition of period. 117fxfxfxpfxfxpfx0p 3.2 Properties of the Graphs of Sinusoids 119 3.2 Properties of the Graphs of Sinusoids Learning Objectives In this section you will:  Learn the properties of graphs of sinusoidal functions, including period, phase shift, amplitude and vertical shift.  Use properties to graph sinusoidal functions.  Write an equation of the form or from the graph of a sinusoidal function.  Solve applications of sinusoids, including harmonic motion. Sinusoids can be characterized by four properties: period, amplitude, phase shift, and vertical shift. 1. We have already discussed period, that is, how long it takes for the sinusoid to complete one cycle. The standard period of both and is 2π, but horizontal scalings will change the period of the resulting sinusoid. 2. The amplitude of the sinusoid is a measure of how ‘tall’ the wave is, as indicated in the figure below. The amplitude of the standard cosine and sine functions is 1, but vertical scalings can alter this. 3. The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We have seen that a phase (horizontal) shift of to the right takes to since . As the reader can verify, a phase shift of to the left takes to . 4. The vertical shift of a sinusoid is assumed to be 0, but we will state the more general case. C=Acos++BωxxS=Asin++Bωxxcosfxxsingxx2cosfxxsingxxcossin2xx2singxxcosfxx 120 Graphs of the Trigonometric Functions The following theorem shows how to find these four fundamental quantities from the formula of a given sinusoid. Theorem 3.3. For , the functions  have period  have amplitude and  have phase shift  have vertical shift B We note that in some scientific and engineering circles, the quantity ϕ mentioned in Theorem 3.3 is called the phase angle of the sinusoid. Since our interest in this book is primarily with graphing sinusoids, we focus our attention on the horizontal shift induced by ϕ. Graphs of Sinusoids The proof of Theorem 3.3 is left to the reader. The parameter ω, which is stipulated to be positive, is called the (angular) frequency of the sinusoid and is the number of cycles the sinusoid completes over a 2π interval. We can always ensure using the even property of the cosine function, or the odd property of the sine function, from which Theorem 3.3 using the functions following two examples. Example 3.2.1. Graph . and and . We now test out in the Solution. Using Theorem 3.3, we first write the function f in the form prescribed in the theorem. From Theorem 3.3, , , and , resulting in the following. 0cosCxAxBsinSxAxB2A0coscosxxsinsinxx3cos12xfx13sin222gxx3cos12xfx3cos123cos122xfxx3A221B 3.2 Properties of the Graphs of Sinusoids 121  The period of f is .  The amplitude is .  The phase shift is , indicating a shift to the right 1 unit.  The vertical shift is , indicating a shift up 1 unit, and a baseline of . The graph shows one cycle of . Using the period, amplitude, phase shift and vertical shift, in conjunction with techniques from Section 3.1, the graph of is sketched as a transformation of . Key points are emphasized on each graph. The baseline for is shown as a dashed line. Example 3.2.2. Graph . Solution. Before applying Theorem 3.3, we use the odd property of the sine function to write in the required form. 224233A2121B1y3cos12xfxyfxcosyxyfx13sin222gxxgxxy cosyx yfx 122 Graphs of the Trigonometric Functions We next determine that , , and . The properties follow from Theorem 3.3.  The period of g is  The amplitude is . .  The phase shift is , indicating a shift right units.  The vertical shift is up units. We graph via transformations of the sine function, using the above properties as a guide. Before graphing, we note that . Since , the graph of must be reflected about the baseline. 13sin22213sin22213sin2 2213sin222gxxxxxfrom odd property of sine12A232B2211222223213sin222gxx12A0Aygxxy ygx sinyx 3.2 Properties of the Graphs of Sinusoids 123 Remember that the cycle graphed through transformations of the sine function in Example 3.1.2 is only one portion of the graph of . Indeed, another complete cycle begins at , and a third at . Note that whatever cycle we choose is sufficient to completely determine the sinusoid. Determining an Equation from the Graph of a Sinusoid Example 3.2.3. Below is the graph of one complete cycle of a sinusoid . One cycle of . 1. Find a cosine function whose graph matches the graph of . 2. Find a sine function whose graph matches the graph of . Solution. 1. We fit the data to a function of the form by determining A, ω, ϕ and B.  Since one cycle is graphed over the interval , its period is . According to Theorem 3.3, , so that . ygx2x32xyfxyfxyfxyfxcosCxAxB1,5516263xy 51,2 55,2 11,22 71,22 32,2 124 Graphs of the Trigonometric Functions  To find the amplitude, we note that the range of the sinusoid is . The midpoint of the range is , indicating a baseline of . After marking the graph with the baseline, we see that the amplitude is .  Next, we see that the phase shift is , so we have or .  Finally, we refer to the baseline to verify a vertical shift of . Our final answer is . 2. Most of the work to fit the data to a function of the form is done.  The period, amplitude and vertical shift are the same as part 1. Thus, , and .  The trickier part is finding the phase shift. To that end, we imagine extending the graph of the given sinusoid as in the figure below so that we can identify a cycle beginning at . 35,221212y51222A11312B12cos332CxxsinSxAxB32A12B71,22xybaseline 71,22 55,2 131,22 38,2 191,22xybaseline 51,2 55,2 11,22 71,22 32,2 3.2 Properties of the Graphs of Sinusoids 125 Taking the phase shift to be , we get , or Hence, our answer is . Note that each of the answers given in Example 3.2.3 is one choice out of many possible answers. For example, when fitting a sine function to the data, we could have chosen to start at taking . In this case, the phase shift is so for an answer of . Alternately, we could have extended the graph of to the left and considered a sine function starting at , and so on. Each of these formulas determine the same sinusoidal curve and the formulas are equivalent using identities. Applications of Sinusoids Sinusoids are used to model a fair number of behaviors that possess a wavelike motion, such as sound, voltage, and spring action. The following examples look at circular motion that can be expressed as a sinusoidal function. Example 3.2.4. A circle with radius 3 feet is mounted with its center 4 feet off the ground. The point closest to the ground is labeled P, as shown below. Sketch a graph of the height above the ground of the point P as the circle is rotated. Find a function h that gives the height in terms of the angle x of rotation. 7272727237.6712sin362Sxx11,222A12612sin362Sxxyfx51,22 126 Graphs of the Trigonometric Functions Solution. Sketching the height, we note that it will start 1 foot above the ground, then increase up to 7 feet above the ground, and continue to oscillate 3 feet above and below the center value of 4 feet. Although we could use a transformation of either the cosine or sine function, we start by looking for characteristics that would make one function easier to use than the other. Since this graph starts at its lowest value, when , using the cosine function would not require a horizontal shift. Thus, we choose to model the graph with a cosine function. We note that a standard cosine graph starts at the highest value so we do need to incorporate a vertical reflection. We see that the graph oscillates 3 feet above and below the horizontal center of the graph. The basic cosine graph has an amplitude of 1, so this graph has been vertically stretched by 3. Finally, to move the center of the circle up to a height of 4 feet, the graph has been vertically shifted up by 4. Putting these transformations together, we find that . 0x3cos4hxxxh(x)baseline 3.2 Properties of the Graphs of Sinusoids 127 Example 3.2.5. The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. Solution. The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes. Because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve: , for time t and height H. A phase shift is not required. With a diameter of 135 meters, the wheel has a radius of 67.5 meters. The height will oscillate with amplitude 67.5 meters above and below the horizontal center of the wheel. Passengers board 2 meters above the ground level, so the center of the wheel must be located meters above ground level. The horizontal midline of the oscillation will be at 69.5 meters. Putting this all together, we have  Period:  Ampli
tude: ; (due to the vertical reflection of the cosine curve)  Phase Shift:  Vertical Shift: An equation for the rider’s height, with t in minutes and H in meters, is cosHtAtB67.5269.52301567.5A67.5A0069.5B67.5cos69.515Htt 128 Graphs of the Trigonometric Functions Harmonic Motion One of the major applictions of sinusoids in Science and Engineering is the study of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. Here, we restrict our attention to modeling a simple spring system. Before we jump into the mathematics, there are some Physics terms and concepts we need to discuss.  In Physics, ‘mass’ is defined as a measure of an object’s resistance to straight-line motion whereas ‘weight’ is the amount of force (pull) gravity exerts on an object. An object’s mass cannot change,28 while its weight could change. An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon, but its mass is the same in both places.  In the English system of units, ‘pounds’ (lbs.) is a measure of force (weight), and the corresponding unit of mass is the ‘slug’. In the SI system, the unit of force is ‘Newtons’ (N) and the associated unit of mass is the ‘kilogram’ (kg).  We convert between mass and weight using the formula29 . Here, w is the weight of the object, m is the mass and g is the acceleration due to gravity. In the English system, and in the SI system, . Hence, on Earth a mass of 1 slug weighs 32 lbs. and a mass of 1 kg weighs 9.8 N.30 Suppose we attach an object with mass m to a spring as detected below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrim’ when the weight of the object is perfectly balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. Usually denoted by the letter k, the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke’s Law . If the object is released above or below the equilibrium position, or if the object is released with an upward or downward velocity, the object will bounce up and down on the end of the spring until some 28 Well, assuming the object isn’t subjected to relativistic speeds . . . 29 This is a consequence of Newton’s Second Law of Motion F=ma where F is force, m is mass and a is acceleration. In our present setting, the force involved is weight which is caused by the acceleration due to gravity. 30 Note that 1 pound = 1 slug foot / second2 and 1 Newton = 1 kg meter / second2. wmg232feetsecondg29.8 meterssecondgFkd 3.2 Properties of the Graphs of Sinusoids 129 external force stops it. If we let denote the object’s displacement from the equilibrium position at time t, then    means the object is at the equilibrium position, means the object is above the equilibrium position, means the object is below the equilibrium position. The function is called the ‘equation of motion’ of the object.31 at the above the below the equilibrium position equilibrium position equilibrium position If we ignore all other influences on the system except gravity and the spring force, then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indefinitely. In this case, we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we ignore friction caused by surrounding medium, which in our case is air). In the following theorem, which comes from Differential Equations,   is a function of the mass m of the object, the spring constant k, the initial displacement of the object and initial velocity of the object. means the object is released from the equilibrium position, means the object is released above the equilibrium position and means the object is released below the equilibrium position. 31 To keep units compatible, if we are using the English system, we use feet (ft.) to measure displacement. If we are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). xt0xt0xt0xtxt0xt0xt0xtxt0x0v00x00x00x 130 Graphs of the Trigonometric Functions  means the object is released from rest, means the object is heading upwards and means the object is heading downwards.32 Theorem 3.4. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. If the initial displacement from the equilibrium position is and the initial velocity of the object is , then the displacement x from the equilibrium position at time t is given by where and   and . It is a great exercise in ‘dimensional analysis’ to verify that the formulas in Theorem 3.4 work out so that ω has units and A has units ft. or m, depending on which system we choose. Example 3.2.6. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If an object is attached to the spring and released 3 feet below the equilibrium position from rest, find the equation of motion of the object, . When does the object first pass through the equilibrium position? Is the object heading upwards or downwards at this instant? 2. If the object is attached to the spring and released 3 feet below the equilibrium position with an upward velocity of 8 feet per second, find the equation of motion of the object, . What is the longest distance the object travels above the equilibrium position? When does this first happen? Solution. In order to use the formulas in Theorem 3.4, we first need to determine the spring constant k and the mass m of the object. We know the object weighs 64 lbs. and stretches the spring 8 ft. Using Hooke’s Law with and , we get 32 The sign conventions here are carried over from Physics. If not for the spring, the object would fall towards the ground, which is the natural or positive direction. Since the spring force acts in direct oppostition to gravity, any movement upwards is considered negative. 00v00v00v0x0vsinxtAtkm2200vAx0sinAx0cosAv1sxtxt64F8d 3.2 Properties of the Graphs of Sinusoids 131 To find m, we use with lbs. and . We get slugs. We can now proceed to apply Theorem 3..4. 1. To find the equation of motion, , we must determine values for A, ω and ϕ. We begin with ω. Now, since the object is released 3 feet below the equilibrium position, from rest, and . We use these values, along with , to find A. To determine the phase ϕ, we use and , along with a formula from Theorem 3.4. Hence, the equation of motion is . 648lbs.8ft.Fkdkk Hooke's Lawwmg64w2.32ftgs2msinxtAt822km from Theorem 3.4 since =8 lbs./ft. and =2 slugs km03x00v22200220323vAx from Theorem 3.403x3A0sin3sin3sin12Ax from Theorem 3.43sin22xtt 132 Graphs of the Trigonometric Functions Next, to find when the object passes through the equilibrium position, we solve . The object first passes through the equilibrium position at the smallest positive t value, which in this case is seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it first passes through it. To check this answer, we graph one cycle of . Since our applied domain in this situation is , and the period of , we graph over the interval is . Remembering that means the object is below the equilibrium position and means the object is above the equilibrium position, the fact our graph is crossing through the t-axis at confirms our answer. 2. The only difference between this problem and the previous problem is that we now release the object with an upward velocity of 8 ft./s. We still have and , but now we have , the negative indicating the velocity is directed upwards. Here, we get 0xt3sin202sin202 for integers 42tttkk after the usual analysis0.784txt0txt222xt0,0xt0xt4t203x08vtx 3sin22xtt 3.2 Properties of the Graphs of Sinusoids 133 We use and to determine ϕ. We will need to identify ϕ using the arcsine since is not a common angle. With the sine being positive, ϕ is in Quadrant I or Quadrant II. Knowing whether the cosine is positive or negative will determine which of those quadrants ϕ resides in. Since we know , and , the formula for from Theorem 3.4. will help us find the cosine. This tells us that ϕ is in Quadrant II, so we have . Hence, . Now, since the amplitude is 5, the object will travel at most 5 feet above the equilibrium position. This happens when , the negative sign once again signifying that the object is above the equilibrium position. 2200228325.vAx5A03x0sin5sin33sin5Ax from Theorem 3.4355A208v0v0cos(5)(2)cos84cos5Av3arcsin535sin2arcsin5xtt5xt35sin2arcsin553sin2arcsin15tt 134 Graphs of the Trigonometric Functions Going through the usual machinations, we get for integers k. The smallest of these values is when , that is, seconds after the start of the motion. 13arcsin254tk0k13arcsin1.107254t 3.2 Properties of the Graphs of Sinusoids 135 3.2 Exercises In Exercises 1 – 12, state the period, phase shift, amplitude and vertical shift of the given function. Graph one cycle of the function. 1. 4. 7. 10. 2. 5. 8. 11. 3. 6. 9. 12. 13. Write an equation of the form for the sine function whose graph is shown below. 14. Write an equation of the form for the cosine function whose graph is shown below. 3sinyxsin3yx2cosyxcos2yxsin3yxsin2yx11cos323yxcos324yxsin24yx2cos4132yx31cos2232yx4si
n2yxsinSxAxBcosCxAxB 136 Graphs of the Trigonometric Functions 15. Write an equation of the form for the cosine function whose graph is shown below. 16. Write an equation of the form for the sine function whose graph is shown below. 17. Write an equation of the form for the cosine function whose graph is shown below. cosCxAxBsinSxAxBcosCxAxB 3.2 Properties of the Graphs of Sinusoids 137 18. Write an equation of the form for the sine function whose graph is shown below. 19. Write an equation of the form for the cosine function whose graph is shown below. 20. Write an equation of the form for the sine function whose graph is shown below. sinSxAxBcosCxAxBsinSxAxB 138 Graphs of the Trigonometric Functions In Exercises 21 – 22, verify the identity by using technology to graph the right and left hand sides. 21. 22. In Exercises 23 – 26, graph the function with the help of technology and discuss the given questions with your classmates. 23. 24. 25. 26. . Is this function periodic? If so, what is the period? . What appears to be the horizontal asymptote of the graph? . Graph on the same set of axes and describe the behavior of f. . What’s happening as ? 27. A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function gives a person’s height in meters above the ground t minutes after the wheel begins to turn. a. Find the period, amplitude and vertical shift of . b. Find a formula for the height function . c. How high off the ground is a person after 5 minutes? 28. Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached. a. Find the spring constant k in and the mass of the object in slugs. b. Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. When is the first time the object passes through the equilibrium position? In which direction is it heading? 22cossin1xxcossin2xxcos3sinfxxxsinxfxxsinfxxxyx1sinfxx0xhththtlbs.ft. 3.2 Properties of the Graphs of Sinusoids 139 c. Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Find when the object passes through the equilibrium position heading downwards for the third time. 140 3.3 Graphs of the Tangent and Cotangent Functions Learning Objectives In this section you will:  Graph the tangent and cotangent functions and their transformations. Identify the period and vertical asymptotes.  Learn the properties of the tangent and cotangent functions, including domain and range; determine whether a function is even or odd. We now turn our attention to the graphs of the tangent and cotangent functions. Graph of the Tangent Function When constructing a table of values for the tangent function, we recall that . We can use our common values for the fundamental cycles of and in determining values of the tangent. It follows that the tangent function, , is undefined at and , both points at which . sintancosxxxcosyxsinyxtanyx2x3/2xcos0x 141 x undefined undefined cosxsinxsintancosxxx,tanxx01000,0422221,1420134222213,14100,054222215,14320174222217,1421002,0 142 To determine the behavior of the graph of when x is close to or , we will look at some values for on both sides of and on both sides of .  The following chart shows some values for when x is less than, but close to . We note that and include approximate values of the tangent for the indicated radian measures of x. We note that values of are positive, getting larger and larger, as x approaches from the left. The result is a vertical asymptote at . x 1.5 1.55 1.56 1.57 14 48 93 1256 undefined Mathematical notation:  When x is greater than, but close to , as x approaches from the right, the values of get smaller and smaller, approaching negative infinity: . x 1.58 1.59 1.6 1.7 undefined –109 –52 –34 –8 Noting that , we look at values for when x is close to . x 4.6 4.65 4.7 4.71 9 16 81 419 undefined tanyx232tanx2x32xtanx221.571tanx22x1.5712tanxAs , tan.2xx22tanxAs , tan2xx1.5712tanx324.712tanx3234.7122tanx3As , tan.2xx 143 x 4.72 4.73 4.75 4.8 undefined –131 –57 –27 –11 Thus we have vertical asymptotes at and at . We also know something about the behavior of the graph as it approaches these vertical asymptotes from each side. Plotting this information followed by the usual ‘copy and paste’ produces the following two graphs. The graph of over . The graph of , with fundamental cycle highlighted. From the graph, it appears the tangent function is periodic with period π. This is, in fact, the case as we will prove in Section 4.2, following the introduction of the sum identity for tangent. We take as our fundamental cycle for the interval . 34.7122tanx3As , tan2xx2x32xtanyx0,2tanyxtanyx2,2 144 From the graph, it appears that the domain of the tangent function, , includes all real numbers x except for . These are the x-values where and are the only numbers for which is undefined. Thus, the domain of is all real numbers x, excluding for any integer k. The range of , as observed from the graph, includes all real numbers. Graph of the Cotangent Function It should be no surprise that the graph of the cotangent function behaves similarly to the graph of the tangent function. Plotting over the interval results in the graph below. x x undefined undefined undefined The graph of over . tanyx2,32,52,xcos0xtanyxtanyx2xktanyxcotyx0,2cotx,cotxxcotx,cotxx05415,1441,143203,0220,027417,143413,142cotyx0,2 145 It clearly appears that the period of is π, which is indeed the case and will be revisited in Section 4.2. The vertical asymptotes in the interval are , and . We take as one fundamental cycle the interval . A more complete graph of is below, with the fundamental cycle highlighted. The graph of . We see, from the graph, the apparent domain of the cotangent function is all real numbers x except for . These are the x-values where and are the only numbers for which is undefined. Thus, the domain of is all real numbers x, excluding , for any integer k. The range of includes all real numbers. Note that on the intervals between their vertical asymptotes, both and are continuous and smooth. In other words, they are continuous and smooth on their domains. Both functions are odd, as you can see by the symmetry of the graphs about the origin, and as we will verify algebraically in Section 4.1. The following theorem summarizes the properties of the tangent and cotangent functions. cotx0,20xx2x0,cotyxcotyx0,,2,xsin0xcotyxcotyxxkcotyxtanyxcotyx 146 Theorem 3.4. Properties of the Tangent and Cotangent Functions:  The function – has domain – has range is continuous and smooth on its domain is odd – – – has period π  The function – has domain – has range – is continuous and smooth on its domain – is odd – has period π Graphs of Transformations of the Tangent and Cotangent Functions Graphing transformations of the tangent and cotangent functions is similar to graphing transformations of the sine and cosine. Theorem 3.2, in which , can be used for both, but there are a few differences to be aware of.  The period of both tangent and cotangent is π, and so the period of transformations will be .  The vertical scaling is , but tangent and cotangent do not have amplitude since they do not possess the wavelike characteristics of the sine and cosine functions.  The vertical shift is B, but a baseline is not defined for transformations of the tangent and cotangent functions.  Since we do not have a baseline, vertical scaling should be completed before the vertical shift.  The horizontal shift, , is unchanged and will be our last step in graphing transformations of the tangent and cotangent. tanJxx:, is any integer2xxkk,cotKxx:, is any integerxxkk,gxAfxBA Example 3.3.1. Graph one cycle of the following functions. Find the period. 147 1. Solution. 2. 1. We will graph the function through a series of transformations to the fundamental cycle of the graph of . Before proceeding, we rewrite the function in a format that will simplify this process: The fundamental cycle of is restricted to the interval . In tracking transformations of the graph of , we will start with the quarter marks , , , and . The first and last of these quarter marks are place markers for vertical asymptotes, but it will be important to track vertical asymptotes as well as points in the transformation process. Hence, we start with the two vertical asymptotes and , and the three points , and .  We begin by determining the period of . Since the coefficient of x is , the period is . Stretching the period π of the tangent function to the period 2π of the transformed function will result in quarter marks of , , 0, and . 3tan2xfx3cot142gxx3tan2xfxtanyxfx1tan032fxxtanyx2,2tanyx2x40422x2x4,10,04,11tan032fxx1221222xy tanyx 148  The vertical scaling is , so the graph will not be stretched in the vertical direction. However, tells us the graph will be reflected about the x-axis, as follows.  The vertical shift is 3, indicating a shift up by 3 units.  Finally, the horizontal shift is 0, so the graph will not be shifted left or right. One cycle of . 11103tan2xf
xxyxy 149 2. We graph the function using transformations of . The fundamental cycle of the cotangent function is on the domain . We use the quarter marks , , , and . Before proceeding, the function can be written as follows.  The period of is , resulting in quarter marks 0, 1, 2, 3 and 4, shown below.  The vertical scaling is . This will stretch the graph vertically by a factor of 3. We multiply the y-coordinates of the points at the second and fourth quarter marks, i.e. and , by 3 to obtain the transformed points and . The middle point, , does not move. Note these transformations in the graph to the right.  Next, the graph is shifted up by 1 unit.  Lastly, a horizontal shift of –2 moves the graph two units to the left. 3cot142gxxcotyx0,04234gx3cot214gxxgx44331,13,11,33,32,0xy 1,1 3,1xy cotyxxy 1,3 3,3 2,0 150 One cycle of . Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit the more extensive coverage here that was given to sinusoidal functions. The ambitious reader is invited to explore further results from this section. We next move on to graphs of secant and cosecant functions. 3cot142gxxxy 0,1 1,4 1,2 151 3.3 Exercises In Exercises 1 – 6, graph one cycle of the given function. State the period of the function. 1. 4. 2. 5. 3. 6. In Exercises 7 – 15, graph two full periods of each function. State the period and asymptotes. 7. 10. 13. 8. 11. 14. 9. 12. 15. In Exercises 16 – 17, find an equation for the graph of each function. 16. tan3yx12tan34yx1tan213yxcot6yx111cot5yx13cot2132yxtanfxxcotfxx2tan432fxxtan2fxxtan2fxx4tanfxxtan4fxxtanfxx3cot2fxx 17. 152 18. Verify the identity by using technology to graph the right and left hand sides. 19. Graph the function with the help of technology. Graph on the same set of axes and describe the behavior of f. 20. The function marks the distance in the movement of a light beam from a police car across a wall for time x, in seconds, and distance , in feet. a. Graph the function on the interval . b. Find and interpret the vertical stretching factor, the period and any asymptotes. c. Evaluate and and discuss the function’s values at those inputs. tantanxxtanfxxxyx20tan10fxxfxfx0,51f2.5f 153 3.4 Graphs of the Secant and Cosecant Functions In this section you will: Learning Objectives  Graph the secant and cosecant functions and their transformations. Identify the period and vertical asymptotes.  Learn the properties of the secant and cosecant functions, including domain and range; determine whether a function is even or odd. Finally, we turn our attention to graphing the secant and cosecant functions. Graph of the Secant Function To get started, we graph the secant function using our table of values for the fundamental cycle of . We can take reciprocals of these cosine values since . x undefined undefined cosyx1seccosxxcosx1seccosxx,secxx0110,14222,2420342223,2411,1542225,24320742227,242112,1 154 The domain of the secant function excludes all odd multiples of since these are the values of x for which at . In our table based on the fundamental cycle of , the secant is undefined and . These are both x-values at which vertical asymptotes occur. To determine the behavior of the graph of when x is close to or , we will look at some values for on both sides of and on both sides of .  The following chart shows some values for when x is less than, but close to . We note that and include approximate values of the secant for the indicated radian measures of x. x 1.5 1.55 1.56 1.57 14 48 93 1256 undefined We note that values of are positive, getting larger and larger, as x approaches from the left:  We next look at values of when x is greater than, but close to . x 1.58 1.59 1.6 1.7 undefined –109 –52 –34 –8 As x approaches from the right, the values of get smaller and smaller, approaching negative infinity:  Using a similar analysis, which we leave to the reader, . 2cos0xcosyx2x32xsecyx232secx2x32xsecx221.5711.57121seccosxxsecx2As , sec.2xxsecx21.57121seccosxx2secxAs , sec2xx3as , sec, and23as , sec.2xxxx 155 Plotting points and asymptotes, with graph behavior echoing the above results, we have the following two graphs. The second graph is an extension of the first. The graph of over . The graph of , with fundamental cycle highlighted. In the above illustration, the dotted graph of is included for reference. It is helpful to graph the secant function by starting with a graph of the cosine function and sketching vertical asymptotes at each x-value for which . The points where are also points where . After drawing the asymptotes and marking the points where , a rough graph of can quickly be completed by sketching the ‘U’ shapes of the secant function. Since is periodic with period 2π, it follows that is also periodic with period 2π.1 Due to the close relationship between the cosine and secant, the fundamental cycle of the secant function is the same as that of the cosine function. We previously noted that the domain of the secant function excludes all odd multiples of . The range of , as observed graphically, includes all real numbers y 1 Provided sec(α) and sec(β) are defined, sec(α) = sec(β) if and only if cos(α) = cos(β). Hence, sec(x) inherits its period from cos(x). secyx0,2secyxcosyxcos0xcos1xsec1xsec1xsecyxcosxsecx2secyx 156 such that or , or equivalently . By thinking of the secant function as being the reciprocal of the cosine function, a similar result can be obtained algebraically. Graph of the Cosecant Function As one would expect, to graph , we begin with and take reciprocals of the corresponding y-values. Here, we encounter issues at , and . These are locations of vertical asymptotes. Proceeding with an analysis of graph behavior near these asymptotes, we graph the fundamental cycle of followed by an extended graph of . A dotted graph of is included for reference. x undefined undefined undefined 1y1y1ycscyxsinyx0xx2xcscyxcscyxsinyxsinx1cscsinxx,cscxx004222,24211,12342223,240542225,2432113,12742227,2420 157 The fundamental cycle of . The graph of . Since and are merely phase shifts of each other, so too are and . As with the tangent and cotangent functions, both and are continuous and smooth on their domains. The following theorem summarizes the properties of the secant and cosecant functions. Note that all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively. cscyxcscyxsinyxcosyxcscyxsecyxsecyxcscyx 158 Theorem 3.5. Properties of the Secant and Cosecant Functions:  The function – has domain – has range is continuous and smooth on its domain is even – – – has period 2π  The function – has domain – has range is continuous and smooth on its domain is odd – – – has period 2π Graphs of Transformations of the Secant and Cosecant Functions In the next example, we discuss graphing more general secant and cosecant functions Example 3.4.1. Graph one cycle of the following functions. State the period of each. 1. Solution. 2. 1. Before graphing we will graph to use as a guide. Using the technique from Section 3.1, we will start with a graph of the cosine function and apply appropriate transformations. First, however, it helps to rewrite in a format suggested by Theorem 3.2.  The period is .  The vertical shift is 1, for a baseline of . secFxx:, is any integer2xxkk:1,11,yycscGxx:, is any integerxxkk:1,11,yy12sec2fxxcsc53xgx12sec2fxx12cos2yx12cos2yx2cos201yx221y 159  The amplitude is and, since , the graph will be reflected about the baseline.  There is no horizontal shift. One cycle of . Next, to graph , we observe the following.  Points where the graph of the cosine function crosses the baseline are vertical asymptotes of the secant function.  Maximum values of the cosine function occur at the lowest points on the secant curve.  Minimum values of the cosine function occur at points that are maximum values for the secant.  Transformations of the secant function retain the familiar “U” shape between vertical asymptotes. Below, we use as a guide in graphing one cycle of . 222012cos2yx12sec2fxx12cos2yxfxxybaseline 12cos2yx 160 One cycle of . Since one cycle is graphed on the interval , the period is . 2. We graph the function by first graphing the corresponding sine function, , and then using the sine function as a guide. The first step is to rewrite the sine function in the form suggested by Theorem 3.2. We proceed to graph as a transformation of , referring to Theorem 3.2 as necessary.  The period is .  The vertical shift is for a baseline of . 12sec2fxx0,0csc53xgxsin53xysin5315sin3315sin3315sin33xyxxx from odd property of sine15sin33yx sinyx225353yxybaseline 161  The amplitude is and, since , the graph will be reflected about the baseline.  Since , the graph will be shifted to the right by 1 unit. Locating quarter marks and corresponding points results in the following graph of . One cycle of . We use the transformed sine graph as a guide in sketching one cycle of . One cycle of . 11331031xxsin53xysin53xycsc53xgxcsc53xgxxybaseline 12cos2yx
 51,3 3,22 52,3 54,23 53,3 53yxybaseline 12cos2yx 162 We find the period to be . While real world applications of secants and cosecants are limited, at least in comparison to the large number of available sinusoidal applications, a couple of examples are included in the Exercises. We conclude Chapter 3 with the expectation of putting to good use the properties and graphs of the trigonometric functions that have been introduced in this chapter. 312 163 3.4 Exercises In Exercises 1 – 6, graph one cycle of the given function. State the period of the function. 1. 4. 2. 5. 3. 6. In Exercises 7 – 18, graph two full periods of each function. State the period and asymptotes. 7. 10. 13. 16. 8. 11. 14. 17. 9. 12. 15. 18. In Exercises 19 – 22, find an equation for the graph of each function. 19. sec2yxcsc3yx11sec323yxcsc2yxsec324yxcsc24yxsecfxxcscfxx2sec14fxx6csc3fxx2cscfxx1csc4fxx4sec3fxx7sec5fxx9csc10fxx2csc14fxxsec23fxx7csc54fxx 164 20. 21. 22. 165 23. Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x, measured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 and x is measured negative to the left and positive to the right. The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance , in kilometers, from the fisherman to the boat is given by the function . a. What is a reasonable domain for ? b. Graph on the domain. c. Find and discuss the meaning of any vertical asymptotes on the graph of . d. Calculate and interpret . Round to the nearest hundredth. e. Calculate and interpret . Round to the nearest hundredth. f. What is the minimum distance between the fisherman and the boat? When does this occur? 24. A laser rangefinder is locked on a comet approaching Earth. The distance , in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by . a. Graph on the interval . b. Evaluate and interpret the information. c. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? d. Find and discuss the meaning of any vertical asymptotes. dx1.5secdxxdxdxdx3d6dgx250,000csc30gxxgx0,305g 166 Trigonometric Identities and Formulas CHAPTER 4 TRIGONOMETRIC IDENTITIES AND FORMULAS Chapter Outline 4.1 The Even/Odd Identities 4.2 Sum and Difference Identities 4.3 Double Angle Identities 4.4 Power Reduction and Half Angle Formulas 4.5 Product to Sum and Sum to Product Formulas 4.6 Using Sum Identities in Determining Sinusoidal Formulas Introduction In Chapter 4, we begin exploring new trigonometric identities and formulas which provide us with varying ways to represent the same trigonometric expression. Section 4.1 advances our study of even/odd identities and includes algebraic proofs of these identities. In Section 4.2, we discover the essential sum and difference identities upon which many of the remaining identities and formulas depend. Section 4.3 introduces the double angles identities, from which the power reduction and half angle formulas are derived in Section 4.4. Along with the sum and difference identities, the half angle formulas will help us obtain exact values for trigonometric functions of some ‘non-common’ angles. The product to sum and sum to product formulas introduced in Section 4.5 will allow further manipulation of trigonometric expressions. Finally, Section 4.6 revisits the formulas of sinusoids first introduced in Section 3.2. With the assistance of the sum and difference identities, we may now write sinusoidal formulas in a standard format that will simplify graphing. Throughout Chapter 4, attention will be paid to finding exact values of trigonometric functions, writing trigonometric expressions in varying formats, and verifying trigonometric identities. These skills will be important in the future study of calculus. 167 4.1 The Even/Odd Identities Learning Objectives In this section you will:  Learn the even/odd identities.  Use the even/odd identities in simplifying trigonometric expressions.  Use the even/odd identities in verifying trigonometric identities. In Section 2.4, we saw the utility of the Pythagorean identities, Theorem 2.5, along with the reciprocal and quotient identities from Theorem 2.3. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we formally introduce the even/odd identities1, while recalling their graphical significance from Chapter 3. The Even/Odd Identities Theorem 4.1. The Even/Odd Identities: For all applicable angles θ,       In light of the reciprocal and quotient identities, Theorem 2.3, it suffices to show and . The remaining four circular functions can be expressed in terms of and so the proofs of their even/odd identities are left as exercises. Consider an angle θ plotted in standard position. Let be the angle coterminal with θ with . (We can construct the angle by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured to the right.) Since θ and are coterminal, and . 1 As mentioned at the end of Section 2.3, properties of the circular functions, when thought of as functions of angles in radian measure, hold equally well if we view these functions as functions of real numbers. Not surprisingly, the even/odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. sinsincsccsccoscossecsectantancotcotcoscossinsincossin0002000coscos0sinsin 168 We now consider the angles and . Since θ is coterminal with , there is some integer k so that . Therefore, Since k is an integer, so is –k, which means –θ is coterminal with . Hence, and . Let P and Q denote the points on the terminal sides of and , respectively, which lie on the Unit Circle. By definition, the coordinates of P are and the coordinates of Q are . Since and sweep out congruent central sectors of the Unit Circle, it follows that the points P and Q are symmetric about the x-axis. Thus, and . Since the cosines and sines of and are the same as those for θ and –θ, respectively, we get and , as required. Simplifying Expressions The even/odd identities are readily demonstrated using any of the common angles noted in Section 2.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. Example 4.1.1. Use identities to fully simplify the expression: . 0002k0022.kk00coscos0sinsin0000cos,sin00cos,sin0000coscos00sinsin00coscossinsin1sin1sinxx Solution. We begin with the odd identity of the sine function. 169 Verifying Identities This section ends with the proof of a trigonometric identity. Looking back at Section 2.4, where we began verifying identities, we can now add the even/odd identities to the Pythagorean, reciprocal and quotient identities as tools in proving that identities are true. Example 4.1.2. Verify the identity: . Solution. We begin with the left, more complicated, side. 2222 sinsin 1s cosin1sin1sin1sin1s1sinincosxxxxxxxxxxsince difference of squares22sincoscossinsincos22222222sincossincossincossincossincossincossincossincossincossincos sincossincos1coseven/odd identitiesdifference of squares sin 170 4.1 Exercises 1. We know is an even function, and and are odd functions. What about , and ? Are they even, odd or neither? Why? 2. Examine the graph of on the interval . How can we tell whether the function is even or odd by only observing the graph of ? In Exercises 3 – 8, use identities to fully simplify the expression. 3. 5. 7. 4. 6. 8. In Exercises 9 – 14, use the even/odd identities to verify the identity. Assume all quantities are defined. 9. 11. 13. In Exercises 15 – 18, prove or disprove the identity. 15. 17. 10. 12. 14. 16. 18. 19. Verify the even/odd identities for tangent, cosecant, secant and cotangent. cosgxxsinfxxtanhxx2cosGxx2sinFxx2tanHxxsecfxx,secfxxsincoscscxxxcsccoscotxxxcottansecttt323sincsccos2coscosttttttancotxxsincosseccsctancotxxxxxxsin32sin23cos5cos544tt22tan1tan1tt55csccscsec6sec6tt9779cotcot112cotcsc1cos1cosxxxx2tansincossecxxxxsecsintancotxxxx1sincoscos1sinxxxx 171 4.2 The Sum and Difference Identities In this section you will: Learning Objectives  Learn the sum and difference identities for cosine, sine and tangent.  Use the sum and difference identities to find values of trigonometric functions.  Use the sum and difference identities in verifying trigonometric identities.  Learn and apply the cofunction identities. We begin with a theorem introducing the sum and difference identities for the cosine function, followed by a proof of the theorem. The Sum and Difference Identities for the Cosine Function Theorem 4.2: Sum and Difference Identities for Cosine.   We first prove the result for differences. As in the proof of the even/odd identities, we can reduce the proof for general angles and to angles and , coterminal with and , respectively, each of which measure between
0 and 2π radians. Since and are coterminal, as are and , it follows that is coterminal with . Consider the following case where . coscoscossinsincoscoscossinsin00000000 172 Since the angles POQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B.1 Using the distance formula to determine distances QP and BA, we have or, after squaring both sides: Expanding the left hand side, then using the Pythagorean identities , we get the following: . and Turning our attention to the right hand side, we will use : Putting in all together, we get which simplifies as follows: . Since and , and , and , are all coterminal pairs of angles, we have 1 In the picture, the triangles POQ and AOB are congruent. However, α0 – β0 could be 0 or it could be π, neither of which makes a triangle. Or, α0 – β0 could be larger than π, which makes a triangle, just not the one we’ve drawn. You should think about these three cases. 222200000000coscossinsincos1sin0220002000002cos1sincoscossinsin02200cossin12200cossin1220000coscossinsin2222000000002222000000000000cos2coscoscossin2sinsinsincossincossin2coscos2sinsin22coscos2sinin.s220000cossin1222200000000002200000000cos1sin0cos2cos1sin1cossin2co.s22cos00000022coscos2sinsi2cn2os00000000000000000022cos22coscos2sinsin2cos2coscos2sinsincoscoscossinsin after swapping sides after subtracting 2 from each side after dividing through by -20000 173 For the case where , we can apply the above argument to the angle to obtain the identity Applying the even identity of cosine, we get from which it follows that . To verify the sum identity for cosine, we use the difference identity along with the even/odd identities: We put these newfound identities to good use in the following example. Example 4.2.1. 1. Find the exact value of . 2. Verify the identity . Solution. 1. In order to use Theorem 4.2 to find , we need to write 15° as a sum or difference of angles whose cosines and sines we know. One way to do so is to write . coscoscossinsin.0000000000coscoscossinsin.000000coscoscos,coscoscossinsincoscoscoscossinsincoscossinsincoscossinsincoscossinsin. difference identity for cosine even identity of cosine odd identity of sinecos15cossin2cos15154530 174 2. This is a straightforward application of Theorem 4.2. The identity verified in Example 4.2.1, namely , is the first of the celebrated cofunction identities. These identities were first hinted at in the 2.1 Exercises, problem 41. The Cofunction Identities From , we get which says, in words, that the ‘co’sine of an angle is the sine of its ‘co’mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises. cos15cos4530cos45cos30sin45sin3023212222624 difference identity for cosinecoscoscossinsin2220cos1sinsincossin2sincos2sincos222cos 175 Theorem 4.3. Cofunction Identities: For all applicable angles θ,       With the cofunction identities in place, we are now in the position to derive the sum and difference identities for sine. The Sum and Difference Identities for the Sine Function We begin with the sum identity. We can derive the difference identity for sine by rewriting as and using the sum identity and the even/odd identities. Again, we leave the details to the reader. Theorem 4.4. Sum and Difference Identities for Sine: For all angles α and β,   Example 4.2.2. 1. Find the exact value of . 2. If α is a Quadrant II angle with and β is a Quadrant III angle with , find . 3. Derive a formula for in terms of and . sincos2cossin2seccsc2cscsec2tancot2cottan2 ssincos2cos2coscossinsin22incos2from difference identity for cosine sinsinsinsincoscossinsinsincoscossin19sin125sin13tan2sintantantan Solution. 176 1. As in Example 4.2.1, we need to write the angle as a sum or difference of common angles. The denominator of 12 suggests a combination of angles with denominators 3 and 4. One such combination is from which we have 2. Using the difference identity for sine, . We know that , but need to find , and . To find , we use along with a Pythagorean identity. We next use a different Pythagorean identity, along with , to find . 191219163121212434194sinsin123444sincoscossin343432122222624 sum identity for sinesinsincoscossin5sin13coscossincos5sin13222225cos113144cos16912cos1312cos1cossin13  from since is a Quadrant II angle tan2cos 177 We need to determine , knowing that and . We now have all the pieces needed to find . 3. We can start expanding . 22222 112secsec5sec515cos1cotans5csefrom since is a Quadrant III angle reciprocal identity for secantsin1cos5tan2sintancos1sin252sinsintan5cos from the quotient identity sinsinsincoscossin511221313552913529565tan 178 The last step is to replace with and with . Naturally, this result is limited to those cases where all of the tangents are defined. The Sum and Difference Identities for the Tangent Function The formula developed in Example 4.2.2 for can be used to find a formula for by rewriting the difference as a sum, . The reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for sine, cosine and tangent. sintancossincoscossincoscossinsin quotient identity for tangent sum identitiesinsincosc1sincoscossincoscos1coscossinsincoscossincoscossincoscoscooscoscoscossinsincoscoscoscossinsincoscossins1coss for cosine and sine goal: & sincossincostansincostantantantan1tantantantantan 179 Theorem 4.5. Sum and Difference Identities: For all applicable angles α and β,    In the statement of Theorem 4.5, we have combined the cases for the sum ‘+’ and difference ‘–‘ of angles into one formula. The convention is that if you want the formula for the sum ‘+’ of two angles, use the top sign in the formula; for the difference ‘–‘ use the bottom sign. For example, . We finish this section by, as promised in Section 3.3, proving algebraically that the period of the tangent function is π. Recall that a function f is periodic if there is a real number p so that for all real numbers t in the domain of f. The smallest positive number p, if it exists, is called the period of f. To prove that the period of is π, we appeal to the sum identity for tangents. This tells us that the tangent is a periodic function and that the period of is at most π. To show that it is exactly π, suppose p is a positive real number so that for all real numbers x. For , we have which means p is a multiple of π. The smallest positive multiple of π is π itself, so we have established that the period of the tangent function is π. sinsincoscossincoscoscossinsintantantan1tantantantantan1tantanftpfttanJxxtantantan1tantantan01tan0tanJxxxxxxxJxtanxtantanxpx0xtantan0tan0tantan0xpxpp from We leave it to the reader to prove algebraically that the period of the cotangent function is also π.2 180 2 Certainly, mimicking the proof for the period of tan(x) is an option. For another approach, consider transforming tan(x) to cot(x) using identities. 181 4.2 Exercises In Exercises 1 – 15, use the sum and difference identities to find the exact value. You may have need of the quotient, reciprocal or even/odd identities as well. 1. 4. 7. 10. 13. 2. 5. 8. 11. 14. 3. 6. 9. 12. 15. 16. If α is a Quadrant IV angle with , and , where , find (a) (d) (b) (e) (c) (f) 17. If , where , and β is a Quadrant II angle with , find (a) (d) (b) (e) (c) (f) 18. If , where , and , where , find (a) (b) (c) 19. If , where , and , where , find (a) (b) (c) cos75sec165sin105csc195cot255tan37513cos1211sin1213tan127cos1217tan12sin1211cot125csc12sec125cos510sin102cossintancossintancsc302tan7cossintancossintan3sin50212cos13322sincostan5sec3224tan732cscseccot 182 21. 23. 25. 27. In Exercises 20 – 32, verify the identity. 20. 22. 24. 26. 28. 29. 30. 31. 32. 33. Verify the cofunction identities for tangent, secant, cosecant and cotangent. 34. Verify the difference identities for sine and tangent. coscossinsintancot2sinsin2sincossinsin2cossincoscos2coscossin1cottansin1cottancos1tantancos1tantancoscos2sinsin
tansincossincostansincossincossinsinsincos1cossinththhtthhhcoscoscos1sincossinththhtthhh2tantantansec1tantanthththhth 183 4.3 Double Angle Identities Learning Objectives In this section you will:  Learn the double angle identities for sine, cosine and tangent.  Find trigonometric values of double angles.  Verify identities involving double angles. In Section 4.2, the sum identities for the trigonometric functions were introduced: Double Angle Identities Using the sum identities, in the case where , we let to attain the double angle identities in the following theorem. Theorem 4.6. Double Angle Identities: For all applicable angles θ,    The three different forms for can be explained by our ability to exchange squares of cosine and sine via a Pythagorean identity. We verify that : sinsincoscossincoscoscossinsintantantan.1tantansin22sincos2222cossincos22cos112sin22tantan21tancos22cos212sin 184 Trigonometric Values of Double Angles Now that we have established the double angle identities, we put them to good use in determining trigonometric values of double angles. Example 4.3.1. 1. Suppose lies on the terminal side of θ when θ is plotted in standard position. Find and . Determine the quadrant in which the terminal side of the angle 2θ lies when it is plotted in standard position. 2. If for , find an expression for in terms of x. Solution. 1. Using , from Theorem 2.6 in Section 2.5, with and , we find . Hence, and . It follows that and 222222222cos2coscoscossinsincosscos2cossincoin1sinsin12sin.ssin1 sum identity for cosine verifies first form: from 3,4Pcos2sin2sinx22sin2222xyr3x4y225rxy3cos5xr4sin5yr2222cos2cossin3455725 from double angle identitysin22sincos4325524.25 from double angle identity 185 Since both the cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III. 2. If your first reaction to is that x should be the cosine of θ, then you have indeed learned something. However, context is everything. Here, x is just a variable. It does not necessarily represent the x-coordinate of a point on the Unit Circle. Here, x represents the quantity , and what we wish to know is how to express in terms of x. We will see more of this kind of thing in Chapter 5 and, as usual, this is something we need for calculus. We start with the double angle identity for sine: We need to write in terms of x to finish the problem. There are two different methods that come readily to mind, both of which are good to know. The first is purely algebraic, using the Pythagorean identity: The second method, preferred by many, provides a visual approach for determining triangle with acute angle θ and, noting that . We sketch a right we label the hypotenuse with length 1 and the side opposite θ with length x. We then use the Pythagorean Theorem to determine the length of the side adjacent to θ. sinxsinsin2sin22sincos2cssinoxxfrom the problem statement that cos222222cossin1cos1cos1cos1cos022xxx Pythagorean identity from the problem statement since cosoppositehypotenusesin,1xx222222adjacent lengthadjacent lengthadjacent length111xxx1x  186 This results in the following triangle. From the triangle we see that . Then, back to solving for , we have a final answer is . Verifying Identities that Include Double Angles Establishing trigonometric identities using the double angle identities is our next task. As before, starting with the more complicated side of an equation is usually a good strategy. Example 4.3.2. 1. Verify the identity . 2. Verify the identity . 221cos11xxsin22sin221xx22tansin21tan2tan2cottan1x 21x 187 Solution. 1. We start with the right hand side of the identity. 2. In this case, we begin with the left side of the equation. Part 2 of the previous example is a case where the more complicated side of the initial equation appeared on the right, but we chose to start with the left side. Beginning with the right side would have required some thinking ahead, possibly working backwards. Try it! When using identities to simplify a trigonometric expression, solve a trigonometric equation or verify a trigonometric identity, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated, although generally one of the paths will result in a simpler solution. In verifying identities, the strategies established in Section 2.4 will help, but there is no substitute for practice. 22222tan2tan1tansecsin2cos1cossin2coscos2sincossin2from Pythagorean identityfrom reciprocal & quotient identitiesfrom double angle identity for sine222tantan21tan12tantan11tantan21tantan2cottandouble angle identitygoal: numerator of 2reciprocal identity for cotangent 188 One last note before we move on to Section 4.4. While double angle identities could be established for secant, cosecant and cotangent, the identities already established in this section may be used in their place. Recall that secant, cosecant and cotangent are reciprocal identities of cosine, sine and tangent, respectively. Thus, for example, and so any of the three double angle identities for cosine may be used in determining . 1sec2cos2sec2 189 4.3 Exercises 1. Use the double angle identity to verify the double angle identity . 2. Use the double angle identities for and to verify the double angle identity for . In Exercises 3 – 12, use the given information about θ to find the exact values of (a) 3. 5. 7. 9. where where where where 11. where (b) (c) 4. 6. 8. 10. 12. where where where where where In Exercises 13 – 22, verify the identity. Assume all quantities are defined. 13. 15. 17. 19. 14. 16. 18. 20. 22cos2cossin2cos22cos1cos2sin2tan2sin2cos2tan27sin2532228cos530212tan532csc423cos5024sin53212cos133225sin132sec5322tan2222tansin21tan221tancos21tan22sincostan22cos12cossin1sin22cossin1sin211tan21tan1tancottancsc22cossinsec2cossincossin 190 21. 22. 23. Suppose θ is a Quadrant I angle with . Verify the following formulas. (a) (b) (c) 24. Discuss with your classmates how each of the formulas, if any, in Exercise 23 change if we assume θ is a Quadrant II, III or IV angle. 25. Suppose θ is a Quadrant I angle with . Verify the following formulas. (a) (c) (b) (d) 26. Discuss with your classmates how each of the formulas, if any, in Exercise 25 change if we assume θ is a Quadrant II, III or IV angle. 27. If for , find an expression for in terms of x. 28. If for , find an expression for in terms of x. 112coscossincossincos2112sincossincossincos2sinx2cos1x2sin221xx2cos212xtanx21cos1x2sin1xx22sin21xx221cos21xxsin2x22cos2tan7x22sin2 191 4.4 Power Reduction and Half Angle Formulas Learning Objectives In this section you will:  Learn and apply the power reduction formulas for sine and cosine.  Learn and apply the half angle formulas for sine, cosine and tangent. In Section 4.3, the double angle identities allowed us to write as powers of sine and/or cosine. In calculus, we have occasion to do the reverse; that is, reduce the power of sine and cosine. Power Reduction Formulas Solving the identity for and the identity for result in the aptly-named ‘power reduction’ formulas below. Theorem 4.7. Power Reduction Formulas: For all angles θ,   Example 4.4.1. Rewrite as a sum and/or difference of cosines to the first power. Solution. We begin with a straightforward application of Theorem 4.7. cos22cos212sin2sin2cos22cos12cos21cos2sin221cos2cos222sincos 192 Another application of the power reduction formulas is the half angle formulas. Half Angle Formulas To start, we apply the power reduction formula to : We can obtain a formula for by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine. By using a quotient identity, we obtain a half angle formula for tangent. These formulas are summarized below. 2222 21cos21cos2sincos2211cos2411cos2441cos2211442111cos448811cos488from power reduction formulasreplacing with in power reduction formula2cos221cos22cos221cos.2cos2 193 Theorem 4.8. Half Angle Formulas. For all applicable angles θ,    where the choice of ± depends on the quadrant in which the terminal side of lies. Example 4.4.2. 1. Use a half angle formula to find the exact value of . 2. Suppose with . Find . 3. Use the identity , verified in Example 4.3.2, to derive the identity . Solution. 1. To use the half angle formula, we note that . 1cossin221coscos221costan21cos2cos1503cos5sin222tansin21tansintan21cos30152 194 Back in Example 4.2.1, we found by using the difference identity for cosine. In that case, we determined . The reader is encouraged to prove these two expressions are equal. 2. If , then , which means . 3. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the
identity , 30cos15cos21cos3023122312222232 from half angle formula for cosine positive since 15 is in Quadrant Icos1562cos1540022sin021cossin223152315525810255 half angle formula for sine22tansin21tan 195 and we will manipulate it into the identity . If we are to use to derive an identity for , it seems reasonable to proceed by replacing each occurrence of θ with . sintan21cos22tansin21tantan2222222tansin21tan2tan2sin221tan22tan2sinsec2sin2tanc22os2 replacing with from a Pythagorean identity 1cos22sin2tan22sintan1cos2sintan21cos from reciprocal identity for secant from power reduction formula for cosine 196 4.4 Exercises In Exercises 1 – 15, use half angle formulas to find the exact value. You may have need of the quotient, reciprocal or even/odd identities as well. 1. 4. 7. 10. 13. 2. 5. 8. 11. 14. 3. 6. 9. 12. 15. In Exercises 16 – 29, use the given information about θ to find the exact values of (a) 16. 18. 20. 22. 24. 26. 28. (b) (c) 17. 19. 21. 23. 25. where where where where where where where where where where , where θ is in Quadrant IV 27. , where θ is in Quadrant III , where θ is in Quadrant II 29. , where θ is in Quadrant II cos75sin105cos67.5sin157.5tan112.57cos12sin12cos85sin87tan811cos1211sin125tan123tan123tan8sin2cos2tan27sin2532228cos530212tan532csc423cos5024sin53212cos133225sin132sec5322tan224tan312sin13csc7sec4 197 30. Without using your calculator, show that . 31. Let θ be a Quadrant III angle with . Show that this is not enough information to determine the sign of by first assuming and then assuming . Compute in both cases. 2362241cos5sin273232sin2 198 4.5 Product to Sum and Sum to Product Formulas Learning Objectives In this section you will:  Learn and apply the Product to Sum Formulas.  Learn and apply the Sum to Product Formulas. This section begins with an example that uses identities from Sections 4.2 and 4.3 to write a trigonometric expression as the sum of trigonometric expressions. Example 4.5.1. Express as a polynomial in terms of . Solution. The double angle identity expresses as a polynomial in terms of . We are now asked to find such an identity for . Finally, we exchange for , courtesy of a Pythagorean identity, and get Thus, can be expressed as the polynomial . Having just shown how we could rewrite as the sum of powers of , it might occur to you that similar operations could be applied to or to rewrite the expressions as sums of powers of . This will be of use in calculus, as will the formulas yet to be presented in this section. cos3cos2cos22cos1cos2coscos3232cos3cos2cos2cossin2sin2cos1cos2sincossin2coscos2sincosfrom sum identity for cosine from double angle identities2sin21cos3232333cos32coscos2sincos2coscos21coscos2coscos2cos2cos4cos3cos.cos334cos3coscos3coscos4cos5cos 199 Product to Sum Formulas Our next batch of identities, the Product to Sum Formulas1, are easily verified by expanding each of the right hand sides in accordance with the Sum and Difference Identities. The details are left as exercises. Theorem 4.9. Product to Sum Formulas. For all angles α and β,    Example 4.5.1. Write as a sum. Solution. Identifying and , we use the Product to Sum Formula for . Sum to Product Formulas Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Chapter 6. These are easily verified using the Product to Sum Formulas and, as such, their proofs are left as exercises. 1 These are also known as the Prosthaphaeresis Formulas and have a rich history. Conduct some research on them as your schedule allows. 1sinsincoscos21coscoscoscos21sincossinsin2cos2cos626coscos1cos2cos6cos26cos26211cos4cos82211cos4cos822 even property of cosine 200 Theorem 4.10. Sum to Product Formulas. For all angles α and β,    Example 4.5.2. Write as a product. Solution. Using the Sum to Product Formula for , with and , yields the following. Where the last equality is courtesy of the odd identity for sine, . The reader is reminded that all of the identities presented in this chapter which regard the circular functions as functions of angles in radian measure apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. sinsin2sincos22coscos2coscos22coscos2sinsin22sinsin3sinsin333sinsin32sincos222sincos22sincos2sinsin 201 4.5 Exercises In Exercises 1 – 6, write the given product as a sum. You may need to use an even/odd identity. 1. 4. 2. 5. 3. 6. In Exercises 7 – 12, write the given sum as a product. You may need to use an even/odd or cofunction identity. 7. 10. 8. 11. 9. 12. In Exercises 13 – 20, verify the identity. Assume all quantities are defined. 13. 15. 17. 18. 19. 20. 14. 16. (HINT: Use result for 16.) 21. In Example 4.5.1, we wrote as a polynomial in terms of . In Exercise 16, we had you verify an identity which expresses as a polynomial in terms of . Can you find a polynomial in terms of for ? ? Can you find a pattern so that could be written as a polynomial in cosine for any natural number n? 22. In Exercise 15, we had you verify an identity which expresses as a polynomial in terms of . Can you do the same for ? What about for ? If not, what goes wrong? 23. Verify the Product to Sum Identities. 24. Verify the Sum to Product Identities. cos3cos5sin2sin7sin9coscos2cos6sin3sin2cossin3cos3cos5sin2sin7sin5cos6sin9sinsincoscossin48sincos44cos2348coscos44cos233sin33sin4sin42cos48cos8cos133sin44sincos4sincos2432sincos2cos22cos4cos64232sincos2cos22cos4cos68642cos8128cos256cos160cos32cos1cos3coscos4coscoscos5cos6cosnsin3sinsin5sin4 202 4.6 Using Sum Identities in Determining Sinusoidal Formulas Learning Objectives In this section you will:  Write a trigonometric function of sines and cosines in general sinusoidal format: or . The motivation for this section lies in the occasions when a trigonometric function is defined in terms of both sines and cosines. By rewriting the function as a sine function or as a cosine function, properties of the function, such as amplitude and period, will become more apparent. This will be a tool used in future courses such as differential equations. To get started, if we use the sum identity for cosine, we can expand to yield Similarly, using the sum identity for sine, is equivalent to . . Making these observations allows us to recognize (and graph) functions as sinusoids which, at first glance, don’t appear to fit the forms of either or . Example 4.6.1. Consider the function . 1. Find a formula for in the form 2. Find a formula for in the form Check your answers analytically using identities. Solution. for for . . 1. The key to this problem is to use the expanded forms of the sinusoid formulas and match up corresponding coefficients. Equating with the expanded form of , we get or cosCxAxBsinSxAxBcosCxAxBcoscossinsinCxAxAxBsinSxAxBsincoscossinSxAxAxBCxSxcos23sin2fxxxfxcosCxAxB0fxsinSxAxB0cos23sin2fxxxcosCxAxBcos23sin2coscossinsinxxAxAxB 203 . By matching up corresponding coefficients and constants, we get and . To determine A and ϕ, a bit more work is involved. Rewriting the equation will help. On the left hand side, the coefficient of is 1, while on the right hand side it is . Since this equation is to hold for all real numbers, we must have that . Similarly, we find by equating the coefficients of that . We now have a system of nonlinear equations that will allow us to determine values for A and ϕ. Choosing , we have and or, after some rearrangement, and . One such angle which satisfies this criteria is . Hence, one way to write as a sinusoid is . We can easily check our answer using the sum formula for cosine. This verifies that is equivalent to . cos3sincoscossinsinAA+20+2ωBωxxxx20Bcos2sin2cos2sin2xxxx13cossinAAcos2xcosAcos1Asin2xsin3A22222222222cos1cossinsi113113132n3AAAAAAAAAAAA Pythagorean Identity from and multiplying through by =2A2cos12sin31cos23sin23fx2cos23fxx2cos232cos2cossin2sin33132cos2sin222cos23sin2fxxxxxxxx2cos23fxxcos23sin2fxxx 204 2. Proceeding as before, we equate with the expanded form of to get or Once again, we may take and . To determine A and ϕ, we begin by rewriting the equation. We equate the coefficients of , then , on either side and get and . Using the Pythagorean identity as before, we get . Then , or , and , which means . One such angle which meets these criteria is . Hence, we have . We check our work analytically, using the sum formula for sine. Thus, is equivalent to . cos23sin2fxxxsinSxAxBcos23sin2sincoscossinxxAxAxBcos3sincossinsincosAA+20+2ωBωxxxx20Bcos2sin2cos2sin2xxxx1si-3cosnAAcos2xsin2xsin1Acos3A2
2cossin12A2sin11sin22cos33cos25652sin26fxx52sin26552sin2coscos2sin66312sin2cos2223sin2cos2cos23sin2fxxxxxxxxxx52sin26fxxcos23sin2fxxx 205 Graphing the three formulas for on a graphing calculator of computer graphing program will result in three identical graphs, verifying our analytical work. It is worth mentioning that, had we chosen instead of as we worked through the preceding example, our final answers would have looked different. The reader is encouraged to rework the example using and to then use identities to show that the formulas are all equivalent. It is important to note that in order for the technique presented in the example to fit a function into the form of one of the general equations, or arguments of the cosine and sine function must match. That is, while , the is a sinusoid, is not.1 The general equations of sinusoids will be explored further in the exercises. 1 This graph does, however, exhibit sinusoid-like characteristics. Check it out! fx2A2A2AcosCxAxBsinSxAxBcos23sin2fxxxcos23sin3gxxx 206 4.6 Exercises In Exercises 1 – 10, use Example 4.6.1 as a guide to show that the function is a sinusoid by rewriting it in the forms and for and . 1. 3. 5. 7. 9. 2. 4. 6. 8. 10. 11. In Exercises 1 – 10, you should have noticed a relationship between the phases ϕ for and . Show that if , then where . 12. Let ϕ be an angle measured in radians and let be a point on the terminal side of ϕ when it is drawn in standard position. Use Theorem 2.6 and the sum identity for sine to show that (with ) can be written as . cosCxAxBsinSxAxB0022sin2cos1fxxx33sin33cos3fxxxsincos2fxxx13sin2cos222fxxx23cos2sinfxxx333cos2sin2622fxxx13cos5sin522fxxx63cos36sin33fxxx5252sincos22fxxx3sin33cos66xxfxCxSxsinfxAxBcosfxAxB2,PabsincosfxaxbxB022sinfxabxB 207 CHAPTER 5 THE INVERSE TRIGONOMETRIC FUNCTIONS Chapter Outline 5.1 Properties of the Inverse Cosine and Sine Functions 5.2 Properties of the Inverse Tangent and Cotangent Functions 5.3 Properties of the Inverse Secant and Cosecant Functions 5.4 Calculators and the Inverse Circular Functions Introduction Chapter 5 introduces the valuable inverse (circular) trigonometric functions. The first three sections are devoted to defining these inverse functions and identifying properties of each. Emphasis is on determining function values and rewriting expressions containing inverse trigonometric functions. Section 5.1 focuses on the inverse cosine and sine functions. In Section 5.2, the inverse tangent and cotangent functions are added, followed by the inverse secant and cosecant in Section 5.3. Throughout the first three sections, relationships between the inverse functions are developed. Because of the necessity for using inverse trigonometric functions in solving real-world applications, the calculation of degree or radian measure is often desired. Section 5.4 includes techniques for determining approximate values of inverse trigonometric functions through technology. In addition to real-world applications of inverse trigonometric functions, Section 5.4 includes finding domains and ranges, and verifying domains and ranges through graphing technology. Chapter 5 is essential to the understanding of trigonometric functions and their uses. The inverse trigonometric functions will be needed in solving trigonometric equations and solving triangles, the focus of the next two chapters. 208 5.1 Properties of the Inverse Cosine and Sine Functions In this section you will: Learning Objectives  Learn and be able to apply properties of the inverse cosine and sine functions, including domain and range.  Find exact values of inverse cosine and sine functions, and of their composition with other trigonometric functions.  Convert compositions of trigonometric and inverse cosine or sine functions to algebraic expressions. In this chapter, we concern ourselves with finding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domain of each circular function to obtain a one-to-one function. The Inverse Cosine Function We first consider . Choosing the interval allows us to keep the range as along with the properties of being smooth and continuous. Restricting the domain of to . Recall that the inverse of a function f is typically denoted f –1. The notation for the inverse of is denoted as either or , read ‘arc-cosine of x’.1 To understand the ‘arc’ in arccosine, recall that an inverse function, by definition, reverses the process of the original function. The function takes a real number input t, associates it with the angle radians, and returns the value . Digging deeper, we have that is 1 The obvious pitfall here is our convention of writing (cos(x))2 as cos2(x), (cos(x))3 as cos3(x) and so on. It is easy to confuse cos–1(x) as (cos(x))–1, which is equivalent to sec(x), not the inverse of cos(x). Always be careful to check the context of cos–1(x)! cosfxx0,1,1cosfxx0,cosfxx11cosfxx1arccosfxxcosftttcoscoscostxy 209 the x-coordinate of the terminal point on the Unit Circle of an oriented arc of length whose initial point is . Hence, we may view the inputs to as oriented arcs and the outputs as x- coordinates on the Unit Circle. The function f –1, then, would take x-coordinates on the Unit Circle and return oriented arcs, hence the ‘arc’ in arccosine. Below are the graphs of and , where we obtain the latter from the former by reflecting it across the line . This is achieved by switching the x and y coordinates. , The Inverse Sine Function We restrict in a similar manner, although the interval of choice is . The inverse of , denoted , is read ‘arcsine of x’. t1,0cosfttcosfxx1arccosfxxyxcosfxx0x1arccosfxxsingxx,22singxx1arcsingxxxyxyxy 210 , We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 5.1. Properties of the Arccosine and Arcsine Functions  Properties of – Domain: – Range: – – if and only if and provided – – arccosine is neither even nor odd provided  Properties of – Domain: – Range: – – – – arcsine is odd if and only if and provided provided singxx22x1arcsingxxarccos()Fxx1,10,arccosxt0tcostxcosarccosxx11xarccoscosxx0xarcsinGxx1,1,22arcsinxt22tsintxsinarcsinxx11xarcsinsinxx22xxyxy 211 Everything in Theorem 5.1 is a direct consequence of the fact that for and are inverses of each other, as are for and . It’s about time for an example. Example 5.1.1. Find the exact values of the following. 2. 3. 4. 6. 7. 8. 1. 5. Solution. 1. To find , we need to find the real number t (or, equivalently, an angle measuring t radians) with and . We know that meets these criteria, so . 2. The value of is a real number t between and with . The number we seek is . Hence, . 3. We begin by observing that is equivalent to . The number lies in the interval with . Our answer is . cosfxx0xarccosFxxsingxx22xarcsinGxx1arccos22arcsin212cos211sin2arccoscos6111coscos613coscos53sinarccos51arccos21cos2t0t3t1arccos232arcsin2222sin2t4t2arcsin2412cos22arccos212cos2t0,2cos2t123cos24 212 4. To find , we seek the number t in the interval with . The answer is so that . 5. Since , we could simply invoke Theorem 5.1 to get . However, in order to make sure we understand why this is the case, we choose to work the example through using the definition of arccosine. Working from the inside out, . Now, is the real number t with and . We find , so that 6. Since does not fall between 0 and π, Theorem 5.1 does not apply. We are forced to work through from the inside out starting with . From the previous problem, we know . Hence, 7. One way to simplify is to use Theorem 5.1 directly. Since is between –1 and 1, we have and we are done. 11sin2,221sin2t6t11sin2606arccoscos663arccoscosarccos623arccos20t3cos2t6t3arccoscosarccos62.611611113coscoscos6213cos2611113coscoscos62.613coscos535133coscos55 213 However, as before, to really understand why this cancellation occurs, we let . Then, by definition, . Hence, 8. To evaluate , as in the previous example, we let so that for some t where . Since , we can narrow this down a bit and conclude that , so that t corresponds to an angle in Quadrant II. We move on to finding . A geometric approach is to sketch the angle t, along with its corresponding reference angle . We then introduce a ‘reference triangle’ in Quadrant II. Since , the reference triangle will have . We label the adjacent side with length 3 and the hypotenuse with length 5. The Pythagorean Theorem can be used to find the length y of the opposite side. 13cos5t3cos5t13coscoscos53.5t3sinarccos53arccos5t3cos5t0tcos0t2t3sinarccossin5tt3cos5t3cos5txy5y3t t 3,y 214 Then . And, since sine is positive in Quadrant II, . Finally, Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that as opposed to , as observed i
n part 6 of the previous example. Example 5.1.2. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. 1. Solution. 2. 1. We begin this solution by letting , so that . We sketch a right triangle representing . To find the length of the opposite side, y, in terms of x, the Pythagorean Theorem yields 222235164yyy As a length, y is positive. 4sin55yt4sin5t3sinarccossin54.5t11arccoscos66116tanarccosxcos2arcsinxarccostxcostxadjacenthypotenusecos1xt222222111xyyxyx1xty 215 This results in To determine the values of x for which this equivalence is valid, we consider our substitution . The domain of is , or . Additionally, since the tangent is not defined when the cosine is 0, we need to discard . Hence, is valid for x in . Note that for x in , we have , resulting in values of the tangent being less than or equal to zero. This corresponds correctly in sign with the values obtained for since . 2. We proceed as in the previous problem by writing so that so that t lies in the interval with . We aim to express in terms of x. We have three choices for rewriting : , or . Since we know , it is easiest to use the last form.2 2 To use the first or second form, we could use a right triangle, like we did in part 1, to evaluate sin(t) in terms of x. 2 tanarccostan1xtxxoppositesince tangent is adjacentarccostxarccosx1,111x0x21tanarccosxxx1,00,11,0arccos2x21xx10xarcsintx,22sintxcos2arcsincos2xtcos2t22cossintt22cos1t212sintsinxt22cos2arcsincos212sin12xttx1xt 21x 216 To find the restrictions on x, we revisit our substitution . Since is defined only for , the equivalence is valid only on . Even though the expression we arrived at in part 2 of the last example, namely , is defined for all real numbers, the equivalence is valid for only . This is similar to the fact that while the expression x is defined for all real numbers, the equivalence is valid only for . For this reason, it pays to be careful when we determine the intervals where such equivalences are valid. arcsintxarcsinx11x2cos2arcsin12xx1,1212x2cos2arcsin12xx11x2xx0x 217 5.1 Exercises In Exercises 1 – 18, find the exact value. 1. 4. 7. 10. 13. 16. 2. 5. 8. 11. 14. 17. In Exercises 19 – 44, find the exact value or state that it is undefined. 19. 22. 25. 28. 31. 34. 37. 20. 23. 26. 29. 32. 35. 38. 3. 6. 9. 12. 15. 18. 21. 24. 27. 30. 33. 36. 39. arcsin113sin212sin21arcsin2arcsin01arcsin212sin23arcsin2arcsin11cos113cos22arccos211cos2arccos01arccos212cos213cos2arccos11sinarcsin212sinsin213sinsin5sinarcsin0.425sinarcsin412coscos21cosarccos215coscos13cosarccos0.998cosarccosarcsinsin6arcsinsin313sinsin4111sinsin64arcsinsin31coscos42arccoscos313coscos21coscos65arccoscos41sinarccos2 40. 43. 218 41. 44. 42. In Exercises 45 – 54, rewrite the quantities as algebraic expressions of x and state the domain on which the equivalence is valid. 45. 48. 51. 54. 46. 49. 52. 47. 50. 53. 55. If for , find an expression for in terms of x. 56. Show that for . 57. Discuss with your classmates why . 58. Why do the functions and have different ranges? 59. Since the functions and are inverse functions, why is not equal to ? 13sincos55cosarcsin1315sinsin1344sin2arcsin53cos2arcsin5sinarccosx1tansinx1sin2cosxsinarccos2xsinarccos5x1cossin2xsin2arcsin7x13sin2sin3xcos2arcsin4x11sinsincosxxsin2x22sin2arcsinarccos2xx11x11sin3021sinfxx1cosgxxcosyx1cosyx1coscos66 219 5.2 Properties of the Inverse Tangent and Cotangent Functions In this section you will: Learning Objectives  Learn and apply properties of the inverse tangent and cotangent functions, including domain and range.  Find exact values of inverse tangent and cotangent functions, and of their composition with other trigonometric functions.  Convert compositions of trigonometric and inverse tangent or cotangent functions to algebraic expressions. The Inverse Tangent Function We restrict to its fundamental cycle on to obtain the inverse of tangent. The inverse, reflect , named arctangent, is also denoted . In the following graphs, we about the line to obtain . Note that the vertical asymptotes and from the graph of become the horizontal asymptotes and on the graph of . , tanfxx,221arctanfxx1tanxtanfxxyx1arctanfxx2x2xtanfxx2y2y1arctanfxxtanfxx22x1arctanfxxxyxy 220 The Inverse Cotangent Function Next, we restrict to its fundamental cycle on to obtain the arccotangent: or . Once again, the vertical asymptotes and of the graph of become the horizontal asymptotes and of the graph of when the graph of the cotangent is reflected about the line to obtain the graph of the arccotangent. , cotgxx0,1arccotgxx11cotgxx0xxcotgxx0yy1arccotgxxyxcotgxx0x1arccotgxxxyxy 221 Theorem 5.2. Properties of the Arctangent and Arccotangent Functions  Properties of – Domain: – Range: – as , ; as , – – – – if and only if and for for all real numbers x provided – arctangent is odd  Properties of – Domain: – Range – as , ; as , – – – if and only if and for for all real numbers x – – arccotangent is neither even nor odd provided Example 5.2.1. Find the exact values of the following. 1. 2. 3. 4. arctanFxx,,22xarctan2xxarctan2xarctanxt22ttantx1arctanarccotxx0xtanarctanxxarctantanxx22xarccotGxx,0,xarccotxxarccot0xarccotxt0tcottx1arccotarctanxx0xcotarccotxxarccotcotxx0xarctan3arccot3cotarccot513sintan4 Solution. 222 1. We know is the real number t between and with . We find , so . 2. The real number lies in the interval with . We get . 3. We can apply Theorem 5.2 directly and obtain . However, working it through provides us with yet another opportunity to understand why this is the case. If we let , then for some t, . Hence, 4. We start simplifying by letting . Then for some t, . Since , we know, in fact, . One way to proceed is to use the Pythagorean identity , since this relates the reciprocals of and and is valid for all t under consideration.1 Along with this identity, we use , from , to solve for . 1 It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 5.1.1. Were the identities we used there valid for all t under consideration? arctan322tan3t3tarctan33arccot3t0,cot3t5arccot36cotarccot55arccot5tcot5t0tcotarccot5cot5.t13sintan413tan4t3tan4t22ttan0t02t221cotcsctttantsint4cot3t3tan4tcsct 223 With , we choose so that . Hence, Example 5.2.2. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. 1. Solution. 2. 1. If we let , then and . We look for a way to express in terms of x. Before we get started using identities, we note that is undefined, for any integer k, when The only members of this family which lie in are , which means the values of t under consideration for are . 222221cotcsc41csc325csc95csc3ttttt02t5csc3t3sin5t13sintansin43.5ttan2arctanx1coscot2xarctantx22ttantxtan2arctantan2xttan2t2, or2.42tktk,224ttan2t,,,244442 224 Returning to , we note the double angle identity is valid for all of the values of t under consideration. Hence, we get To find where this equivalence is valid, we check back with our substitution . Since the domain of is all real numbers, the only exclusions come from , the values of t we discarded earlier. We exclude corresponding values of x: Hence, the equivalence holds for all x in . 2. To get started, we let so that where . Then and, since is always defined, there are no additional restrictions on t. Our goal is to then express in terms of x. Using the identity , which is valid for t in , we can write . Thus With and , the Pythagorean identity provides a path for expressing in terms of x. arctan2t22tantan21tanttt22tan2arctantan22tan1tan2.1xtttxxarctantxarctanx4ttantan41.xt22tan2arctan1xxx,11,11,1cot2txcot2tx0t1coscot2cosxtcostcostcoscotsinttt0,coscotsinttt1coscot2coscotsin2sin.xtttxtcot2tx1cscsintt221cotcscttsint 225 Since t is between 0 and π, . Thus, and . Finally, This is true for all real numbers x since is defined for all real numbers x and we encountered no additional restrictions on t. 222221cotcsc12csccsc41ttxttxcsc0t2csc41tx21sin41tx12coscot22sin2.41xxtxx1cot2x 226 5.2 Exercises In Exercises 1 – 14, find the exact value. 1. 4. 7. 10. 13. 2. 5. 8. 11. 14. In Exercises 15 – 48, find the exact value or state that it is undefined. 15. 18. 21. 24. 27. 30. 33. 36. 16. 19. 22. 25. 28. 31. 34. 37. 3. 6. 9. 12. 17. 20. 23. 26. 29. 32. 35. 38. arctan31tan113tan3arctan03arctan31tan11tan31cot3arccot13arccot31cot013cot3arccot1arccot31tantan11tantan35tanar
ctan12tanarctan0.9651tantan3cotarccot11cotcot37cotarccot241cotcot0.00117cotarccot4arctantan31tantan41tantanarctantan212tantan3arccotcot31cotcot4arccotcot1cotcot22arccotcot3sinarctan21sincot5cosarctan71coscot3 227 40. 43. 46. 41. 44. 47. 39. 42. 45. 48. In Exercises 49 – 55, rewrite the quantities as algebraic expressions of x and state the domain on which the equivalence is valid. 49. 52. 55. 50. 53. 51. 54. 56. If for , find an expression for in terms of x. 125tansin51tanarccos2tanarccot1212cotarcsin1313cotcos21cottan0.253tanarctan3arccos51sin2tan2cos2arccot5arctan2sin21costanx1sin2tanxcos2arctanx1costan3xcosarcsinarctanxx1tan2sinx1sinarctan2xtan7x2211sin222 228 5.3 Properties of the Inverse Secant and Cosecant Functions In this section you will: Learning Objectives  Learn and apply properties of the inverse secant and cosecant functions, including domain and range.  Find exact values of inverse secant and cosecant functions, and of their composition with other trigonometric functions.  Convert compositions of trigonometric and inverse secant or cosecant functions to algebraic expressions. The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were first discussed in Section 3.4, are given below with the fundamental cycles highlighted. The graph of Fundamental cycle of The graph of Fundamental cycle of It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of and restricts the domain of the function so that it is one-to- one. The same is true for cosecant. Thus, in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely , and another piece to cover the bottom, namely . secyxsecyxcscyxcscyx,11,1,,1xyxyxyxy 229 The Inverse Secant Function For , we restrict the domain to , corresponding to that of cosine, and reflect about the line to obtain the graph of . on The Inverse Cosecant Function We restrict to , in correspondence with the sine, and reflect about the line to obtain . on secfxx0,,22secfxxyx1arcsecfxxsecfxx0,,221arcsecfxxcscgxx,00,22yx1arccscgxxcscgxx,00,221arccscgxxxyxyxyxy 230 Note that and may also be written as and , respectively. For both arcsecant and arccosecant, the domain is , which can be written as . Theorem 5.3. Properties of the Arcsecant and Arccosecant Functions  Properties of – Domain: – Range: – as , ; as , – – – – if and only if or and provided provided provided or – arcsecant is neither even nor odd  Properties of – Domain: – Range: – as , ; as – – – – if and only if provided provided , or and provided or – arccosecant is odd Example 5.3.1. Find the exact values of the following. 1. 2. 3. 4. arcsecxarccscx1secx1cscx,11,:1xxarcsecFxx:1,11,xx0,,22xarcsec2xxarcsec2xarcsecxt02t2tsectx1arcsecarccosxx1xsecarcsecxx1xarcsecsecxx02x2xarccscGxx:1,11,xx,00,22xarccsc0xxarccsc0xarccscxt02t02tcsctx1arccscarcsinxx1xcscarccscxx1xarccsccscxx02x02xarcsec21csc215secsec4cotarccsc3 231 Solution. 1. Since , we can use Theorem 5.3 and have 2. Once again, with , Theorem 5.3 comes to our aid giving 3. Since doesn’t fall between 0 and or between and π, we cannot use the inverse property stated in Theorem 5.3. We can, nevertheless, begin by working inside out which yields 4. One way to begin to simplify is to let . Then and, since this is negative, we have that t lies in the interval . We are after , knowing , so we use the Pythagorean identity . 211arcsec2arccos2.321111csc2sin2.65422111215secsecsec241cos23.4 Note: from Theorem 5.3cotarccsc3arccsc3tcsc3t,02cottarccsc3t221cotcsctt22221cotcsc1cot3cot8cot22ttttt 232 Since , , so we get . Example 5.3.2. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. 1. Solution. 2. 1. We begin simplifying by letting . Then, for t in , and we seek a formula for . Since is defined for all t values under consideration, there are no additional restrictions on t. To relate to , we use the identity , which is valid for all t under consideration. If t belongs to then ; if, on the other hand, t belongs to then . As a result, we get a piecewise defined function for . Now we need to determine what these conditions on t mean for . When , it follows that , and when , we have . With no further restrictions on t, we can express as an algebraic expression of x. 02tcot0tcotarccsc322tanarcsecx1coscsc4xtanarcsecxarcsectxsectx0,,22tanttantsecttant221tansectt222221tansec1tantan1tttxtx0,2tan0t,2tan0ttant22 1, if 0t<2tan1, if 2xtxtsecxt02t1x2t1xtanarcsecx 233 2. To simplify , we start by letting . Then for t in , and we now set about finding an expression for Since is defined for all t, there are no additional restrictions on t. From get . To find , we can make use of the identity . . , we Since t belongs to , we know , so we choose . (The absolute value here is necessary since x could be negative.) To find the values of x for which this equivalence is valid, we look back at our original substitution . The domain of requires its argument x to satisfy , and so the domain of will require . With no additional restrictions on t, the equivalence holds for all x in . 22 1, if 1tanarcsec1, if 1xxxxx1coscsc4x1csc4txcsc4tx,00,221coscsc4cosxtcostcsc4tx1sin4txcost22cossin1tt2222222cossin11cos14161cos16161cos4tttxxtxxtx,00,22cos0t2161cos4xtx1csc4tx1cscx1x1csc4x41x4141 or 4111 or 44xxxxx21161coscsc44xxx11,,44 234 5.3 Exercises In Exercises 1 – 16, find the exact value. 1. 4. 7. 10. 13. 16. 2. 5. 8. 11. 14. In Exercises 17 – 52, find the exact value or state that it is undefined. 17. 20. 23. 26. 29. 32. 35. 18. 21. 24. 27. 30. 33. 36. 3. 6. 9. 12. 15. 19. 22. 25. 28. 31. 34. 37. arcsec21csc21sec21csc223arcsec323arccsc31sec1arccsc1arcsec21sec2123sec31sec11csc2arccsc223arccsc3arccsc11secsec2secarcsec111secsec21secsec0.75secarcsec1171csccsc223cscarccsc312csccsc2cscarccsc1.0001cscarccsc41secsec44arcsecsec315secsec61secsec25arcsecsec3arccsccsc615csccsc412csccsc3arccsccsc211arccsccsc6111secsec12 38. 41. 44. 47. 50. 235 39. 42. 45. 48. 51. 40. 43. 46. 49. 52. In Exercises 53 – 55, rewrite the quantities as algebraic expressions of x and state the domain on which the equivalence is valid. 53. 54. 55. 56. If for , find an expression for in terms of x. 57. Show that for as long as we use as the range of 58. Show that . . for as long as we use as the range of 9arccsccsc81sincsc31cossec55tanarcsec31cotcsc53secarccos2112secsin13secarctan10110seccot10cscarccot93cscarcsin512csctan3cosarcsec3arctan2113sin2csc5125cos2sec7secarctanxcscarccosxsecarctan2tanarctan2xxsec4x024tan41arcsecarccosxx1x0,,22arcsecfxx111cscsinxx1x,00,221cscfxx 236 5.4 Calculators and the Inverse Circular Functions In this section you will: Learning Objectives  Use technology to evaluate approximate values of the inverse trigonometric functions.  Find domains and ranges of inverse trigonometric functions.  Use inverse trigonometric functions to solve real-world applications. In the sections to come, we will have need to approximate values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin-1, cos-1 and tan-1, respectively. If we are asked for an arccotangent, arcsecant or arccosecant, we often need to employ some ingenuity, as the next example illustrates. Using a Calculator to Find Values Example 5.4.1. Use a calculator to approximate the following values to four decimal places. 1. 2. 3. 4. Solution. 1. Since , we can use a property from Theorem 5.2 to rewrite as . After verifying that our calculator or other graphing tool is in radian mode, we find 2. Noting that , we can use a property from Theorem 5.3 to write arccot21sec51cot23arccsc220arccot21arctan21arccot2arctan20.4636 radians.51111sec5cos51.3694 radians. 237 3. Since the argument –2 is negative, we cannot directly apply Theorem 5.2 to help us find . We will, however, be able to use by first establishing a relationship between and .  By definition, the real number satisfies with . Since , we know more specifically that , so t corresponds to an angle β in Quadrant IV with .  We next visualize the angle radians and note that θ is a Quadrant II angle with . This means θ is exactly π units away from β, and we get Hence, radians. 4. To approximate , noting that , we can use Theorem 5.3: 1cot211tan21cot211tan
211tan2t1tan2t22ttan0t02t1tan21cot21tan211tan22.6779.1cot22.67793arccsc231232arccscarcsin230.7297 radians.xy 1cot(2) radians   238 Domain and Range of Inverse Trigonometric Functions Example 5.4.2. Find the domain and range of the following functions. Check your answers using graphing technology. 1. 2. 3. Solution. 1. Since the domain of is , we can find the domain of by setting the argument of the arccosine, in this case , between –1 and 1. So the domain of is . To determine the range of f, we select three key points on the graph of , and . We use transformations to track these points to : , and on the graph of . Plotting these values tells us that the range of f is . The following graphing calculator screen confirms a domain of and range of . arccos25xfx13tan4fxxarccot2xgxarccosFxx11xarccos25xfx5x11515515555xxxarccos25xfx5,5arccosFxx1,0,21,05,20,05,2arccos25xfx,225,5,22 239 2. To find the domain of , we note the domain of is all real numbers. The only restrictions, if any, on the domain of come from the argument, 4x, and since 4x is defined for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can choose the key point along with horizontal asymptotes and from the graph of . We find that the graph of differs by a horizontal compression with a factor of 4 and a vertical stretch with a factor of 3. It is the latter which affects the range, producing a range of . We confirm the domain of and range of using graphing technology. arccos25xyfx13tan4fxx1tanFxx13tan4fxx0,02y2y1tanyFxx13tan4yfxx33,22,33,22 240 3. To find the domain of , we proceed as above. Since the domain of is , and is defined for all x, we get the domain of g is as well. As for the range, we note that the range of , like that of , is limited by a pair of horizontal asymptotes, in this case and . We graph starting with and first performing a horizontal expansion by a factor of 2, followed by a vertical shift upwards by π. This latter transformation is the one which affects the range, making it . To check this graphically using technology, it may be necessary to create a piecewise defined function for that makes use of the arctangent function.  Using Theorem 5.2, we have that when , or, in this case, when . Hence, for , we have . 13tan4yfxxarccot2xgxarccotGxx,2x,arccotGxx1tanFxx0yyarccot2xygxarccotyGxx,2arccot2xgx2arccotarctan2xx02x0x0x2arctangxx 241  When , we can use the same argument in Example 5.4.1, part 3, that gave us to give us . Thus, for ,  What about ? We know , and neither of the formulas for g involving arctangent will produce this result. Finally, we have a piecewise function to use when graphing with technology such as graphing calculators. The graph confirms a domain of and range of . 02x111cot2tan22arccotarctan2xx0xarccot22arctan2arctan2.xgxxx0x0arccot0gygx2arctan2 when 0arccot when 022arctan when 0xxxygxxxxarccot2xygx,,2 242 Applications of Inverse Trigonometric Functions The inverse trigonometric functions are typically found in applications where the measure of an angle is required. One such scenario is presented in the following example. Example 5.4.3.1 The roof on the house below has a 6/12 pitch. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using trigonometric functions of right triangles, we find the angle of inclination, labeled θ below, satisfies . Since θ is an acute angle, we can use the arctangent function and we find (using a calculator in degree mode) 1 Thanks to Dan Stitz for this problem. 61tan122 6 feet 12 feet  243 1arctan226.57. 244 5.4 Exercises In Exercises 1 – 6, use a calculator to evaluate each expression. Express answers to the nearest hundredth. 1. 4. 2. 5. 3. 6. In Exercises 7 – 18, find the domain of the given function. Write your answers in interval notation. 7. 10. 13. 16. 8. 11. 14. 17. 9. 12. 15. 18. 19. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. 20. At Cliffs of Insanity Point, the Great Sasquatch Canyon is 7117 feet deep. From that point, a fire is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the fire? Express your answer using degree measure rounded to one decimal place. 21. Shelving that is being built at the college library is to be 14 inches deep. An 18-inch rod will be attached to the wall and to the underside of the shelf, at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 1cos0.4arcsin0.233arccos51cos0.81tan6arctan61sin5fxx131cos2xfx2arcsinsin2fxx21arccos4fxxarctan4fxx122cot9xfxx1tanln21fxxarccot21fxx1sec12fxxarccsc5fxx3arcsec8xfx12cscxfxe 245 22. A parasailor is being pulled by a boat on Lake Powell. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 23. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the rifle with a tranquilizer dart is 300 feet, find the smallest value of θ for which the corresponding point on the ground is in range of the rifle. Round your answer to the nearest hundredth of a degree. 24. Suppose a 13-foot ladder leans against the side of a house, reaching to the bottom of a second-floor window 12 feet above the ground. What angle does the ladder make with the house? Round your answer to the nearest tenth of a degree. 246 CHAPTER 6 TRIGONOMETRIC EQUATIONS Chapter Outline 6.1 Solving Equations Using the Inverse Trigonometric Functions 6.2 Solving Equations Involving a Single Trigonometric Function 6.3 Solving Equations of Multiple Trigonometric Functions/Arguments Introduction Chapter 6 focuses on solving trigonometric equations through the implementation of tools and formulas accumulated throughout the preceding chapters. In Section 6.1, inverse trigonometric functions are used to find acute angles within right triangles, as well as to find exact solutions of trigonometric equations. Section 6.2 applies identities and inverse functions in solving equations of a single trigonometric function. The process continues in Section 6.3 for equations that include multiple trigonometric functions and/or differing arguments. This essential chapter is primarily used to develop algebraic skills necessary for solving equations. In Chapter 7, these skills will be put to use in the many applications of triangles made possible through the Law of Sines and the Law of Cosines. 247 6.1 Solving Equations Using the Inverse Trigonometric Functions In this section you will: Learning Objectives  Use inverse trigonometric functions to solve right triangles  Use inverse trigonometric functions to solve for angles in trigonometric equations. Solving for Angles in Right Triangles In Section 5.4, we used inverse trigonometric functions to solve real-world applications. Here, we apply the same technique in solving for acute angles within right triangles. Using inverse trigonometric functions, we can determine an angle, being given only the value of a trigonometric function, such as cosine in the following example. Thus, inverse trigonometric functions truly serve as inverses with the angle representing the inverse of the trigonometric function. Example 6.1.1. Solve the following triangle for the angle θ. Solution. Because we know the lengths of the hypotenuse and the side adjacent to the angle θ, it makes sense for us to use the cosine function. 19cos129cos120.7227 radians or 41.4096 from properties of the arccosine 248 Knowing the measure of one acute angle in a right triangle, we can easily determine the measure of the second acute angle. In the example above, the measure of the angle opposite the side of length 9 would be, approximately, . Note that the exact measure is . Solving for Angles in Trigonometric Equations In Section 2.3, we learned how to solve equations like and . In each case, we appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of common angles. If, on the other hand, we had been asked to find all angles with or to solve for real numbers t, we would have been hard pressed to do so. With the introduction of the inverse trigonometric functions, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root func
tion can be used to solve certain quadratic equations. The equation is a lot like in that it has friendly ‘common value’ answers . The equation , on the other hand, is a lot like . We know there are answers but we can’t express them using ‘friendly’ numbers.1 To solve , we make use of the square root function and write . We can certainly approximate these answers using a calculator, but as far as exact answers go, we leave them as . In the same way, we will use the arcsine function to solve , as seen in the following example. Example 6.1.2. Solve the following equations. 1. Find all angles θ for which . 2. Find all real numbers t for which . 1 This is all, of course, a matter of opinion. Many find just as nice as . 1809041.409648.590419sin121sin2tan1t1sin3tan2t24x1sin22x27x1sin327x7x7x1sin31sin3tan2t72 249 3. Solve for x. Solution. 1. If , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at . Geometrically, we see that this happens at two places: in Quadrant I and in Quadrant II. If we let α denote the acute solution to the equation, then all of the solutions to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all solutions in Quadrant II. Noting that is not the sine of any of the common angles, we use the arcsine function to express our answers. The real number is defined so it satisfies with . Hence, radians. Since the solutions in Quadrant I are all coterminal with α, we get part of our solution to be Turning our attention to Quadrant II, we get one solution to be . Hence, the Quadrant II solutions are 5sec3x1sin313y131arcsin3t02t1sin3t1arcsin321arcsin2, for integers .3kkkxy 13arcsin radians 13xy 13  250 Our final answer is that the solution to is or for all integers k. 2. We may visualize the solutions to as angles θ with . Since tangent is negative only in Quadrants II and IV, we focus our efforts there. We note that none of the common angles have tangent –2, so we need to use the arctangent, or inverse tangent, function to express our answers. The real number satisfies and . If we let radians, we see that all of the Quadrant IV solutions to are coterminal with β. Moreover, the solutions from Quadrant II differ by exactly π units from the solutions in Quadrant IV, so all the solutions to are of the form Switching back to the variable t, we record our final answer to as for integers k. 21arcsin2, for integers .3kkk1sin31arcsin23k1arcsin23ktan2ttan21tan2ttan2t02t1tan2tan2ttan2t1tan2 for some integer .kkktan2t1tan2tkxy 1tan2 radiansxy   251 3. The last equation we are asked to solve, , poses an immediate problem. We are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Adopting an angle approach, we consider and note that our solutions lie in Quadrants II and III. Since isn’t the secant of any of the common angles, we’ll need to express our solutions in terms of the arcsecant function. The real number is defined so that with . If we let , we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use . Since all angle solutions are coterminal with β or –β, we get our solutions for to be, for integers k, or or Switching back to the variable x, we record our final answer to as or for integers k. 5sec3x5sec3535arcsec3x2x5sec3x5arcsec35arcsec35sec32k2k5arcsec23k5arcsec2.3k5sec3x5arcsec23xk5arcsec23xk 252 Note that, with , it follows from Theorem 5.3 that . Another way to write our solution to is or for integers k. The reader is encouraged to check the answers found in Example 6.1.2, both analytically and with technology. 51353arcsecarccos355sec3x3arccos25xk3arccos25xkxy 35arccos radiansxy 35arccos radians 35arccos radians 253 6.1 Exercises In Exercises 1 – 2, find the angle θ in the given right triangle. Express your answers using degree measure rounded to two decimal places. 1. 2. In Exercises 3 – 5, find the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 3. lengths 3, 4 and 5 4. lengths 5, 12 and 13 5. lengths 336, 527 and 625 In Exercises 6 – 25, solve the equation using the techniques discussed in Example 6.1.2. Then approximate the solutions which lie in the interval to four decimal places. 6. 9. 12. 15. 18. 21. 24. 7. 10. 13. 16. 19. 22. 25. 8. 11. 14. 17. 20. 23. 0,27sin11x2cos9xsin0.569xcos0.117xsin0.008x359cos360xtan117xcot12x3sec2x90csc17xtan10x3sin8x7cos16xtan0.03xsin0.3502xsin0.721xcos0.9824xcos0.5637x1cot117xtan0.6109x 254 6.2 Solving Equations Involving a Single Trigonometric Function In this section you will: Learning Objectives  Write complete solutions to equations containing a single trigonometric function.  Evaluate exact solutions in the interval [0,2π). In this section, we continue solving some basic equations involving trigonometric functions. Below we summarize the techniques that were first introduced in Section 2.3. Note that we use the letter u as the argument of each circular function for generality. Strategies for Solving Basic Equations Involving Trigonometric Functions  To solve or for , first solve for u in the interval and add integer multiples of the period 2π. If or , there are no real solutions.  To solve or for respectively, and solve as above. If , convert to cosine or sine, or , there are no real solutions.  To solve for any real number c, first solve for u in the interval and add integer multiples of the period π.  To solve for , convert to tangent and solve as above. If , the solution to is for integers k. Using the above guidelines, we can comfortably solve and find the solution or for integers k. How do we solve something like ? This equation has the form , so we know the solutions take the form or for integers k. Then, since the argument of sine is 3x, we have or for integers k. To cosucsinuc11c0,21c1csecuccscuc1c1c11ctanuc,22cotuc0c0ccot0u2uk1sin2x26xk526xk1sin32x1sin2u26uk526uk326xk5326xk 255 solve for x, we divide both sides1 of the equations by 3, and obtain our solution of or for integers k. In the remainder of this section, we look at examples of equations which contain a single trigonometric function. The solutions provide practice with, and extensions of, the technique applied in solving . We will add to the general solution of each equation specific solutions that fall in the interval . Equations Involving Cosines or Sines Example 6.2.1. Solve the equation , giving the exact solutions which lie in . Solution. The solutions to are or for integers k. Since the argument of cosine here is 2x, this means or for integers k. Solving for x gives or for integers k. To determine which of our solutions lie in , we substitute the integer values into and : k … –2 – Don’t forget to divide the 2πk by 3 as well! 2183xk52183xk1sin32x0,23cos22x0,23cos2u526uk726uk5226xk7226xk512xk712xk0,20,1,2,k512xk712xk512xk191271251217122912712xk171251271219123112 256 The solutions in the interval correspond to and . They are , , and . In the preceding example, the solutions and can be checked analytically by substituting them into the left hand side of the original equation, .  Starting with , we have  Similarly, we find, for , This confirms the solution of or for integers k. Equations Involving Tangents or Cotangents Next, we look at an example of a cotangent function. Example 6.2.2. Solve , giving the exact solutions which lie in . 0,20k1k512x71217121912512xk712xk3cos22x512xk55cos2cos21265cos63.2kk since the period of cosine is 2712xk77cos2cos21267cos63.2kk512xk712xkcot30x0,2 257 Solution. Since has the form , we know . So, in this case, for integers k. Solving for x yields . We next determine which of our solutions lie in . k … –1 0 1 2 3 4 5 6 … … … The solutions in are , , , , and , corresponding to through . To check the solution of , we start with the left side of . This confirms our solution of for integers k. Equations Involving Secants or Cosecants We look next at an equation involving a cosecant function which we will solve through conversion to its reciprocal, a sine function. Example 6.2.3. Solve , giving the exact solutions which lie in . cot30xcot0u2uk32xk63xk0,263xk6625676321161360,26x25676321160k5k63xkcot30xcot3cot632cot20.kk since the period of cotangent is 63xk1csc23x0,2 258 Solution. Noting that this equation has the form , we rewrite it as and find or for integers k. Since the argument of cosecant here is , or . We solve the first of these equations for x. Solving the other equation, , produces . Putting these two solutions together, we have or for integers k. Despite the infinitely many solutions of , we find that none of them lie in . This can be verified graphically by plotting and , and observing that the two functions do not intersect at all over the interval . csc2u2sin2u24uk324uk13x1234xk13234xk123412341523453241564xkxkxkxkxk add to both sides combine the like terms and multiply both sides by 3413234xk2164xk1
564xk2164xk1csc23x0,21csc3yx2y0,2 259 The reader is encouraged to check the solutions of Example 6.2.3 as we did following the first two examples in this section. We next solve an equation that at first glance does not fit the profile of equations thus far in this section. Example 6.2.4. Solve . List the solutions which lie in the interval . Solution. The complication in solving an equation like comes not from the argument of secant, which is just x, but rather from the fact that secant is being squared. Thus, we begin by extracting square roots to get . Converting to cosines, we have .  For  For , we get or for integers k. , we get or for integers k. If we take a step back and think of these families of solutions geometrically, we see we are finding the measures of all angles with a reference angle of . As a result, these solutions can be combined and we may write our solutions as and for integers k. The solutions in the interval come from the values and as indicated in the following table. 2sec4x0,22sec4xsec2x1cos2x1cos2x23xk523xk1cos2x223xk423xk33xk23xk0,20k1kxy 1csc3yx 2y 260 k … –1 0 1 2 … … … … … The solutions in the interval are , , and . Solutions that Include Inverse Trigonometric Functions Our remaining two examples require inverse trigonometric functions in their solutions. Otherwise, the solution process is similar to that of the first four examples. Example 6.2.5. Solve and determine solutions that lie in the interval . Solution. The equation has the form , whose solution is . Hence, , so for integers k. To determine which of our answers lie in the interval , we first need to get an idea of the value of . While we could easily find an approximation using a calculator,2 we proceed analytically. Since , it follows that . Multiplying through by 2 gives . We are now in a position to argue which of the solutions lie in .  For , we get , so we discard this answer and all answers where . 2 Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? 3xk233437323xk32353830,23x234353tan32x0,2tan32xtan3uarctan3ukarctan32xk2arctan32xk0,22arctan330arctan3022arctan302arctan32xk0,20k2arctan30x2arctan32xk0k 261  Next, we turn our attention to and get . Starting with the inequality , we add 2π and get . This means lies in .  Advancing k to 2 produces . Once again, we get from that . Since this is outside the interval , we discard and all solutions of the form for . Thus, the only solution of in the interval is . A similar process determines the solutions of the following equation involving a sine function. Example 6.2.6. Solve and find solutions in the interval . Solution. To solve , we first note that the equation has the form , which has the family of solutions or for integers k. Since the argument of sine here is 2x, we get or which gives or for integers k. To determine which of these solutions lie in , we first need to get an idea of the value of . Once again, we could use a calculator but we adopt an analytic route here. 1k2arctan32x2arctan302arctan3222arctan32x0,22arctan34x2arctan3032arctan3440,22arctan34x2arctan32xk2ktan32x0,22arctan323.7851xsin20.87x0,2sin20.87xsin0.87uarcsin0.872ukarcsin0.872uk2arcsin0.872xk2arcsin0.872xk1arcsin0.872xk1arcsin0.8722xk0,21arcsin0.872x0arcsin0.87210arcsin012.8724 by definition after multiplying through by 262  Starting with the family of solutions , we use the same kind of arguments as in our solution to Example 6.2.5 and find only the solutions corresponding to and lie in : and .  Next, we move to the family . Here we need to get a better estimate of . Proceeding with the usual arguments, we find the only solutions which lie in correspond to and , namely and All told, we have found four solutions to in : , and . . , We will continue solving equations containing trigonometric functions in Section 6.3, with the added complexity of multiple trigonometric functions and/or arguments. 1arcsin0.872xk0k1k0,21arcsin0.872x1arcsin0.872x1arcsin0.8722xk1arcsin0.872210arcsin0.872410arcsin0.87241arcsin0.8722241arcsin0.8742222 from above multiply through by -1 add 0,20k1k1arcsin0.8722x31arcsin0.8722xsin20.87x0,21arcsin0.872x1arcsin0.872x1arcsin0.8722x31arcsin0.8722x 263 6.2 Exercises In Exercises 1 –18, find the exact solutions of the equation and then list those solutions which are in the interval . 1. 4. 7. 10. 13. 16. 2. 5. 8. 11. 14. 17. In Exercises 19 – 26, solve the equation. 19. 22. 25. 20. 23. 26. 3. 6. 9. 12. 15. 18. 21. 24. 27. With the help of your classmates, determine the number of solutions to in . Then find the number of solutions to , and in . A pattern should emerge. Explain how this pattern would help you solve equations like . Now consider , and . What do you find? Replace with –1 and repeat the whole exploration. 0,2sin50x1cos32x3sin22xtan61xcsc41xsec32x3cot23xcos99x2sin32x5cos06x1sin232x72cos34xcsc0xtan21x2tan3x24sec3x21cos2x23sin4xarccos2x2arcsin2x4arctan310x6arccot250x4arcsec2x12arccsc23x229arcsin0x229arccos0x1sin2x0,21sin22x1sin32x1sin42x0,21sin112x1sin22x31sin22x51sin22x12 264 6.3 Solving Equations of Multiple Trigonometric Functions/Arguments In this section you will: Learning Objectives  Solve equations containing multiple trigonometric functions and/or arguments.  Evaluate exact solutions in the interval [0,2π). Each of the examples in Section 6.2 featured one trigonometric function. If an equation involves two different trigonometric functions, or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and algebra to reduce the equation to a form similar to the equations in Section 6.2. Equations with Different Powers of the Same Function Example 6.3.1. Solve the equation and list the solutions which lie in the interval . Solution. We resist the temptation to divide both sides of by (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. We get or . Solving for , we find or .  The solution to is , with and being two solutions which lie in .  To solve , we use the arcsine function to get or for integers k. We find the two solutions here which lie in to be and . 323sinsinxx0,2323sinsinxx2sinx323223sinsin3sinsin0sin3sin10xxxxxxFactor out from both terms.2sinx2sin0x3sin10xsinxsin0x1sin3xsin0xxk0xx0,21sin3x1arcsin23xk1arcsin23xk0,21arcsin3x1arcsin3x 265 We can visualize the solutions by graphing and . It may be necessary to ‘zoom in’ close to and to verify that the graphs do in fact intersect four times. interval close to 0 close to π To summarize, the solutions to in the interval are , , and .1 Equations Containing Multiple Trigonometric Functions In the next example, we make use of a Pythagorean identity. Example 6.3.2. Solve the equation and list the solutions in . Solution. Analysis of reveals two different trigonometric functions, so an identity is in order. Since , we have 1 Note that we are not counting as a solution since it is not in the interval . In the forthcoming solutions, remember that while may be a solution to the equation, it isn’t counted among the solutions in . 33sinyx2sinyx0xx0,2323sinsinxx0,20x1arcsin3x1arcsin3xx2sectan3xx0,22sectan3xx22sec1tanxx2x0,22x0,2 33sinyx 2sinyxxy 33sinyx 2sinyx 33sinyx 2sinyx 266 This gives or . Since , we have or .  From , we get for integers k. The solutions which lie in are and .  To solve , we employ the arctangent function and get for integers k. Using the same sort of argument we saw in Example 6.2.5, we get and as solutions in . The points of intersection on the following graph of and correspond to the four solutions of on the interval . As discussed above, these solutions include , , and . 222222sec1tansectan31tantan3tantan20201t0an2xxxxxxxxuuuuux since for 1u2utanuxtan1xtan2xtan1x4xk0,234x74xtan2xarctan2xkarctan2xarctan2x0,22secyxtan3yx2sectan3xx0,2arctan2x34xarctan2x74xxy 2secyx tan3yx 267 Equations Containing Multiple Arguments of the Same Function Some trigonometric equations can be solved by treating them as quadratic equations. Before proceeding in this manner with the following example, we will need to apply a double angle identity. Example 6.3.3. Solve the equation . Solution. In the equation we have the same trigonometric function, namely cosine, on both sides, but the arguments differ. Using the identity , we obtain an equation quadratic in form and can proceed as we have done in the past. This gives or . Since , we have or .  Solving , we get or for integers k.  From , we get for integers k. The answers which lie in are , and . Try graphing and to verify that the curves intersect in three places on , and that the x-coordinates of these points confirm our results. Next, we look at an example that uses a technique similar to Example 6.3.3, but relies on an identity established in Example 4.5.1. Example 6.3.4. Solve the equation . Solution. From Example 4.5.1, we know that . This transforms our equation into a polynomial in terms of .
cos23cos2xxcos23cos2xx2cos22cos1xx222cos22cos1coscos23cos22cos13cos22cos3cos1023102110xxuxxxxxxxuuuu since for 12u1ucosux1cos2xcos1x1cos2x23xk523xkcos1x2xk0,20x353cos2yx3cos2yx0,2cos313cosxx3cos34cos3cosxxxcosx 268 We get , or and . Since , our solutions would result from , . Both 2 and –2 are outside the range of cosine. Thus, the only real solution to in is , so that for integers k. The only solutions which lie are and . Equations Containing Different Functions and Different Arguments The following example includes different trigonometric functions and different arguments. Example 6.3.5. Solve for x. Solution. In examining the equation , not only do we have different functions involved, namely sine and cosine, we also have different arguments to contend with, namely 2x and x. Using the identity makes all of the arguments the same and we proceed as we would solving any nonlinear equation, We gather all of the nonzero terms on one side of the equation and factor. 3332coscos313cos4cos3cos13cos4cos16cos041604404220uxxxxxxxxuuuuuuu for 0u2u2ucosuxcos0xcos2xcos2xcos313cosxxcos0x2xk0,22x32xsin23cosxxsin23cosxxsin22sincosxxxsin23cos2sincos3cos2sincos3cos0cos2sin30xxxxxxxxxx 269 Then or . From , we obtain for integers k. From , we get or for integers k. The answers which lie in are , , and . The last example in this section, which also includes different functions and arguments, tests our memory a bit and introduces another solution technique. Example 6.3.6. Solve the following equation for x: . Solution. Unlike the previous problem, there seems to be no quick way to get the circular functions or their argument to match in the equation . If we stare at it long enough, however, we realize that the left hand side is the expanded form of the sum formula for . Hence, our original equation is equivalent to . We proceed in solving for x. cos0x3sin2xcos0x2xk3sin2x23xk223xk0,22x32323sincoscossin122xxxxsincoscossin122xxxxsin2xxsincoscossin122sin123sin12xxxxxxx3sin12x3sin123222433xxkxk since sine is equal to 1 after multiplying both sides by 23 270 The solution of for integers k has the following two solutions in the interval : and . When solving trigonometric equations, try something! Practice will help, but trying different solution techniques will improve your problem solving skills. After working through the examples in this section, spend some time with the problems in the Exercises. Try checking your solutions through viewing the intersection of graphs, as in Example 6.3.1 and Example 6.3.2. 433xk0,23x53x 271 6.3 Exercises In Exercises 1 –40, solve the equation, giving the exact solutions which lie in . 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 0,2sincosxxsin2sinxxsin2cosxxcos2sinxxcos2cosxxcos225cosxx3cos2cos20xxcos25sin2xx3cos2sin2xx22sec3tanxx2tan1secxx2cot3csc3xxsec2cscxx2coscsccot6cotxxxxsin2tanxx42cot4csc7xx2cos2csc0xx3tan3tanxx23tansec2xx3coscosxxtan22cos0xx32csccsc4csc4xxx22tan1tanxxtansecxxsin6coscos6sinxxxxsin3coscos3sinxxxxcos2cossin2sin1xxxx3cos5cos3sin5sin32xxxxsincos1xxsin3cos1xx2cos2sin1xx3sin2cos21xxcos23sin22xx33sin33cos333xxcos3cos5xxcos4cos2xx 37. 39. 272 38. 40. sin5sin3xxcos5cos2xxsin6sin0xxtancosxx 273 CHAPTER 7 BEYOND RIGHT TRIANGLES Chapter Outline 7.1 Solving Triangles with the Law of Sines 7.2 Applications of the Law of Sines 7.3 The Law of Cosines Introduction Chapter 7 introduces some new tools that will help in solving obtuse triangles, and in solving real-life applications. In Section 7.1, triangles are defined as Angle-Side-Angle, Angle-Angle-Side and Side-Side- Angle. The Law of Sines is introduced to assist with solving triangles of these types. Section 7.2 further promotes the Law of Sines as a tool in finding the area of a triangle and in applying the Law of Sines to various applications. In Section 7.3, the Law of Cosines is used to solve triangles of the types Side- Angle-Side and Side-Side-Side. Solutions to many applications are made possible through the Law of Cosines. Heron’s Formula, which is proved through the Law of Cosines, provides a simple method for finding the area of a triangle in which the lengths of all three sides are known. This is a critical chapter in developing aptitude for solving real-life trigonometric applications. It is followed by Chapter 8, in which we delve into the more theoretical, but still very application-oriented, polar coordinates and complex numbers. 274 7.1 Solving Triangles with the Law of Sines Learning Objectives In this section you will:  Use the Law of Sines to solve oblique triangles.  Distinguish between ASA, AAS and ASS triangles.  Determine the existence of, and values for, multiple solutions of oblique triangles.  Determine when given criteria will not result in a triangle. Trigonometry literally means ‘measuring triangles’ and with Chapters 1 – 6 under our belts, we are more than prepared to do just that. The main goal of this chapter is to develop theorems which allow us to solve triangles – that is, find the length of each side of a triangle and the measure of each of its angles. Solving Right Triangles We have had some experience solving right triangles. The following example reviews what we know. Example 7.1.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundredth of a degree. Solution. For definitiveness, we label the triangle below.  To find the length of the missing side a, we use the Pythagorean Theorem to get , which then yields units. Now that all three sides of the triangle are known, there are several ways we can find α and β using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to the data given to us in the problem. In this case, the lengths 4 and 7 were given. 22247a33ab = 4c = 7a   275  We want to relate the lengths 4 and 7 to α. Since and α is an acute angle, radians. Converting to degrees, we find .  We see that , so radians and we have . Note that we could have used the measure of angle α to find the measure of angle β, using the fact that α and β are complements, from which . A few remarks about Example 7.1.1 are in order. 1. First, we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs , and are called angle-side opposite pairs. 2. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it minimizes the chances of propagated error.3 3. Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being.4 The Law of Sines The Pythagorean Theorem along with the definitions of the trigonometric functions in Section 2.1 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? Any triangle that is not a right triangle is an oblique triangle. In certain cases, we can use the Law of Sines to solve oblique triangles. 1 as well as the measure of said angle 2 as well as the length of said side 3 Your Science teachers should thank us for this. 4 Don’t worry! Radians will be back before you know it! 4cos74arccos755.154sin74arcsin734.8590,a,b,c 276 Theorem 7.1. The Law of Sines: Given a triangle with angle-side opposite pairs , and , the following ratios hold Or, equivalently, The proof of the Law of Sines can be broken into three cases. 1. For our first case, consider the triangle below, all of whose angles are acute, with angle- side pairs , and . If we drop an altitude from vertex B, we divide the triangle into two right triangles: and . If we call the length of the altitude h (for height), we get that and so that . After some rearrangement of the last equation, we get . If we drop an altitude from vertex A, we can proceed as above using the triangles and to get , completing the proof for this case. ,a,b,csinsinsinabcsinsinsinabcABC,a,b,cABQBCQsinhcsinhasinsinhcasinsinacABQACQsinsinbcbcaABC   caABChQ  bcABCQh'   277 2. For our next case, consider the triangle below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: and . Proceeding as before, we get and so that . Dropping an altitude from vertex B also generates two right triangles, and . We know that so that . Since , so in fact we have . Proceeding to , we get so , . Putting this together with the previous equation, we get , and we are finished with this case. 3. The remaining case is when is a right triangle. In this case, the definitions of trigonometric functions from Section 2.1 can be used to verify the Law of Sines and this verification is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least three measurements of angles and/or sides, including at least one of t
he sides. Also, note that we need to be given, or be able to find, at least one angle-side opposite pair. We will investigate three possible oblique triangle problem situations. AAS (Angle-Angle-Side) Here, we know the measurements of two angles and a side that is not between the known angles. Example 7.1.2. Solve the triangle: , units, . Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. ABCABQACQsinhbsinhcsinsinbcABQBCQ'sin'hc'sin'hc'180sin'sin'sinhcBCQ'sinha'sinhasinsincaABC1207a45 278 Solution. Knowing an angle-side opposite pair, namely and , we may proceed in using the Law of Sines. Now that we have two angle-side pairs, it is time to find the third. To find γ, we use the fact that the sum of the measures of the angles in a triangle is 180°. Hence, . To find c, we have no choice but to use the derived value , yet we can minimize the propagation of error here by using the given angle-side opposite pair . The Law of Sines gives us The exact value of could be found using the difference identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus ‘exact’ here means . 1207a7sin45sin1207sin45sin12072232725.72 u45nits3bbbb since 180120451515,a7sin15sin1207sin15sin1202.09 unitscccsin157sin15sin120c 279 ASA (Angle-Side-Angle) In this case, we know the measurements of two angles and the included side. Example 7.1.3. Solve the triangle: , , units. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. Solution. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since . As in the previous example, we are forced to use a derived value in our computations since the only angle-side pair available is . The Law of Sines gives To find b we use the angle-side pair which yields ASS (Angle-Side-Side) Knowing the measurement of two sides and an angle that is not between the known sides proves to be more complex than our first two scenarios. While we can use the Law of Sines to solve any oblique triangle, some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as ASS, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solutions. 85305.25c180853065,c5.25sin85sin655.25sin85sin655.77 unitsaaa,c5.25sin30sin655.25sin30sin652.90 unitsbbb 280 Example 7.1.4. Solve the triangle: , unit, units. Solution. Since we are given and c, we use the Law of Sines to find the measure of γ. Since the range of the sine function is , there is no real number with . Geometrically, we see that side a is just too short to make a triangle. The following examples keep the same value for the measure of α and the length of c while varying the length of a. We will discuss the preceding case in more detail after we see what happens in the next three examples. Example 7.1.5. Solve the triangle: , units, units. Solution. Using the Law of Sines, we get Now γ is an angle in a triangle which also contains the angle . This means that γ must measure between 0° and 150° in order to fit inside the triangle with α. The only angle that satisfies this requirement and has is . In other words, we have a right triangle. We find the measure of β to be and then determine b using the Law of Sines. 301a4c,asin30sin41sin4sin30sin21,1sin2302a4csin30sin42sin2sin30sin130sin190180309060 281 In this case, the side a is precisely long enough to form a unique right triangle. Example 7.1.6. Solve the triangle: , units, units. Solution. Proceeding as we have in the previous two examples, we use the Law of Sines to find γ. In this case, we have Since γ lies in a triangle with , we must have . There are two angles γ that fall in this range and have : radians, approximately 41.81°, and radians, approximately 138.19°. sin30sin6022sin60sin3023212233.46 unitsbbbb303a4csin30sin434sin30sin32sin33001502sin32arcsin32arcsin3 282  In the case radians ≈ 41.81°, we find5 . Using the Law of Sines with the angle-side opposite pair and β, we find units.  In the case radians ≈ 138.19°, we repeat the exact same steps and find β ≈ 11.81° and b ≈ 1.23 units.6 Both triangles are drawn below. Example 7.1.7. Solve the triangle: , units, units. Solution. For this last problem, we repeat the usual Law of Sines routine to find that 5 To find an exact expression for β, we convert everything back to radians: radians, radians and radians. Hence, radians ≈ 108.19°. 6 An exact answer for β in this case is radians ≈ 11.81°. 2arcsin31803041.81108.19,a3sin108.195.70sin30b2arcsin3304a4csin30sin44sinsin301sin23062arcsin3180252arcsinarcsin63632arcsin36 283 Then γ must inhabit a triangle with , so we must have . Since the measure of γ must be strictly less than 150°, there is just one angle which satisfies both required conditions, namely . So and, using the Law of Sines one last time, Some remarks are in order. 1. If we are given the measures of two of the angles in a triangle, say α and β, the measure of the third angle γ is uniquely determined using the equation . Knowing the measures of all three angles of a triangle completely determines its shape. 2. If, in addition to being given the measures of two angles, we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides in order to determine the size of the triangle. This is true for the previously described AAS and ASA cases. 3. If we are given the measure of just one of the angles along with the lengths of two sides, only one of which is adjacent to the given angle, we have the ASS case.7 As we saw in Examples 7.14 – 7.17, this information may describe one right triangle, one oblique triangle, two oblique triangles, or no triangle. 7 In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case. 300150301803030120sin30sin12044sin120sin3043212436.93 unitsbbbb180 The four possibilities in the ASS case are summarized in the following theorem. 284 Theorem 7.2. Suppose and are intended to be angle-side pairs in a triangle where α, a and c are given. Let .     If If If If , then no triangle exists which satisfies the given criteria. , then so exactly one (right) triangle exists which satisfies the criteria. , then two distinct triangles exist which satisfy the given criteria. , then γ is acute and exactly one triangle exists which satisfies the given criteria. Theorem 7.2 is proved on a case-by-case basis.  If then . If a triangle were to exist, the Law of Sines would have so that , which is impossible. In the figure below, we see geometrically why this is the case. , no triangle Simply put, if the side a is too short to connect to form a triangle. This means if , we are always guaranteed to have at least one triangle, and the remaining parts of the theorem tell us what kind and how many triangles to expect in each case.  If , then and the Law of Sines gives so that . Here, as required. ,a,csinhcahah90hacacahsinacsinsincasinsin1caaaahahahahsinacsinsinacsinsin1caaa90 285 ,  Moving along, now suppose . As before, the Law of Sines gives . Since , or which means there are two solutions to : an acute angle which we’ll call , and its supplement . We need to argue that each of these angles ‘fit’ into a triangle with α. o Since and are angle-side opposite pairs, the assumption in this case gives . Since is acute, we must have that α is acute as well. This means that one triangle can contain both α and , giving us one of the triangles promised in the theorem. o If we manipulate the inequality a bit, we have , which gives and . This proves a triangle can contain both of the angles α , giving us the second triangle predicted in the theorem.  To prove the last case in the theorem, we assume . Then , which forces γ to be an acute angle. Hence, we get only one triangle in this case, completing the proof. , two triangles ah90hacsinsincahasincasin1casinsinca00180,a0,cca0000018018001801800180hacac 286 , one triangle One last comment before we end this discussion. In the Angle-Side-Side case, if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. We will next move on to Section 7.2 where we use the Law of Sines to solve application problems. ac 287 7.1 Exercises In Exercises 1 – 12, find the length of side x. Round to the nearest tenth. 1. 3. 5. 2. 4. 6. 7. 9. 288 8. 10. 11. Notice that x is an obtuse angle. 12. In Exercises 13 – 32, solve for the remaining side(s) and angle(s) if possible. As in the text, , and are angle-side opposite pairs. 13. 15. 17. , , , , , , 14. 16. 18. , , , , , , ,a,b,c13175a73.254.1117a958533.33a956233.33a11735a42b11745a42b 19. 21. 23. 25. 27. 29. 31. , , , , , , , , , , , , , , 289 20. 22. 24. 26. 28. 30. 32. , , , , , , , , , , , , , , 33. Find the radius of the circle. Round to the nearest tenth. 34. Find the diameter of the circle. Round to the nearest tenth. 35. Prove that the Law of Sines holds when is a right triangle. 36. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. 68.788a92b4217a23.5b68.770a90b307a14b4239a23.5b535328.01c657a100b74.63c3.05a
10216.75b13c10216.75b18c1023516.75b29.1383.95314.15b120614c5025a12.5bABC 290 37. Discuss with your classmates why the Law of Sines cannot be used to find the angles in a triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 36. Given and , choose four different values for a so that a) the information yields no triangle b) the information yields exactly one right triangle c) the information yields two distinct triangles d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have , , and your choice of a yield only one triangle where that unique triangle has three acute angles. 3010b3010b 291 7.2 Applications of the Law of Sines Learning Objectives In this section you will:  Find the area of an oblique triangle using the sine function.  Solve applied problems using the Law of Sines. Following our practice with solving triangles for missing values in Section 7.1, we begin Section 7.2 by using some of those values to find the area of an oblique triangle. Finding the Area of an Oblique Triangle The following theorem introduces a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. Theorem 7.3. Suppose , and are the angle-side opposite pairs of a triangle. Then the area enclosed by the triangle is given by Example 7.2.1. Find the area of the triangle in which , units, and . Solution. This is the triangle from Example 7.1.2 in which we found all three angles and all three sides. To minimize propagated error, we choose , from Theorem 7.3, because it uses the most pieces of given information. We are given and , and we calculated in Example 7.1.2. Using these values, we find ,a,b,c111sinsinsin222Abcacab1207a451sin2Aac7a457sin15sin120c 292 The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 7.3. Solving Applied Problems Using the Law of Sines The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering and physics involve three dimensions and motion. Example 7.2.2. Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 15 degrees, whereas the angle of elevation measured by the second station is 35 degrees. Find the altitude of the aircraft and round your answer to the nearest tenth of a mile. Solution. To find the altitude, or height, of the aircraft, we first sketch a triangle which reflects the information given to us in the problem. We then use the triangle to determine the distance from one station to the aircraft. Letting a represent the distance from the first station to the aircraft, we look for an angle-side opposite pair from which we can determine the distance a. We know the measure of two angles in the triangle, but the measure of the angle opposite the side of length 20 miles is missing. Noting that the angles in a triangle add up to 180 degrees, we find the unknown angle measure to be . This gives us an angle-side opposite pair with known values and allows us to set up a Law of Sines relationship. 7sin1517sin452sin1205.18 square unitsA1801535130 293 The distance a, from the first station to the aircraft, is about 14.98 miles. Now that we know a, we can use right triangle relationships to solve for the height, h, of the aircraft. The aircraft is at an altitude of approximately 3.9 miles. Example 7.2.3. Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is 30° and at the second point is 45°. Assuming a straight coastline, find the distance from the second observation to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the first observation point labeled as P and the second as Q. sin130sin3520sin13020sin3520sin35sin13014.98aaaasin15sin1514.98sin153.88hahahhh 15 14.98 milesa 294 In order to use the Law of Sines to find the distance d from Q to the island, we first need to find the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 45° are supplemental, so that . We can now find β. By the Law of Sines, we have Next, to find the point on the coast closest to the island, which we’ve labeled as C, we need to find the perpendicular distance from the island to the coast.1 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island. Using the right triangle definition of sine, we get 1 Do you see why C must lie to the right of Q? 180451351803018030135155sin30sin155sin30sin159.66 milesddd 295 Hence, the island is approximately 6.83 miles from the coast. To find the distance from Q to C, we note that so by symmetry , we get miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point. We close this section with the encouragement that, by working through the many problems in the Exercises, you will become proficient in applying the Law of Sines to real-world applications, and will be ready to move on to the Law of Cosines in Section 7.3. sin45sin4529.6626.83 milesydydyy1809045456.83xy 296 7.2 Exercises In Exercises 1 – 6, find the area of each triangle. Round each answer to the nearest tenth. 1. 3. 5. 2. 4. 6. 297 7. Find the area of the triangles. As in the text, , and are angle-side opposite pairs. (a) (b) (c) , , , , , units units units, units 8. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in the figure. Determine the distance of the boat from station A and the distance of the boat from shore. Round your answers to the nearest whole foot. 9. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot. 10. A man and a woman standing 3.5 miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot. 11. Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. 12. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1040 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. ,a,b,c13175a535328.01c5025a12.5b 298 13. Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be 75° and radios Sally immediately to find the angle of inclination from her position to the craft is 50°. How high off the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.) 14. A yield sign measures 30 inches on all three sides. What is the area of the sign? Grade: The grade of a road is much like the pitch of a roof in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 15 – 17, we first have you change road grades into angles and then use the Law of Sines in an application. 15. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7% grade means that the road (hypotenuse) makes about a 4° angle with the horizontal. (It will not be exactly 4°, but it is pretty close. 16. What grade is given by a 9.65° angle made by the road and the horizontal? 17. Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road.1 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6°. Use the Law of Sines to find the height of the tree. (Hint: First show that the tree makes a 94° angle with the road.) 1 The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 299 Bearings (Another Classic Application): In the next several exercises we introduce and w
ork with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N40°E (read “40° east of north”) is a bearing which is rotated clockwise 40° from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of . Similarly, S50°W would point into Quadrant III along the terminal side of because we started out pointing due south (along ) and rotated clockwise 50° back to 220°. Counter- clockwise rotations would be found in the bearings N60°W (which is on the terminal side of ) and S27°E (which lies along the terminal side of ). These four bearings are drawn in the plane below. The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.) 18. Find the angle θ in standard position with which corresponds to each of the bearings given below. (a) due west (b) S83°E (c) N5.5°E (d) due south (e) N31.25°W (f) S72°41’12”W (g) N45°E (h) S45°W 502202701502970360 300 19. The Colonel spots a campfire at a bearing N42°E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the fire to be N20°W from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot. 20. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53°W which brings her to the Muffin Ridge Observatory. From there, she knows a bearing of S65°Ewill take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquatch Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 21. The captain of the SS Bigfoot sees a signal flare at a bearing of N15°E from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of N75°W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50°E, find the distances from the flare to each vessel, rounded to the nearest tenth of a mile. 22. Carl spies a potential Sasquatch nest at a bearing of N10°E and radios Jeff, who is at a bearing of N50°E from Carl’s position. From Jeff’s position, the nest is at a bearing of S70°W. If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot. 23. A hiker determines the bearing to a lodge from her current position is S40°W. She proceeds to hike 2 miles at a bearing of S20°E at which point she determines the bearing to the lodge is S75°W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 24. A watchtower spots a ship off shore at a bearing of N70°E. A second tower, which is 50 miles from the first at a bearing of S80°E from the first tower, determines the bearing to the ship to be N25°W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. 25. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55°. From a point five stories below the original observer, the angle of inclination to the gargoyle is 20°. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.) 301 26. Use the cases and diagrams in the proof of the Law of Sines (Theorem 7.1) to prove the area formulas given in Theorem 7.3. Why do those formulas yield square units when four quantities are being multiplied together? 302 7.3 The Law of Cosines Learning Objectives In this section you will:  Use The Law of Cosines to solve oblique triangles.  Solve SAS and SSS Triangles.  Use Heron’s Formula to find the area of a triangle.  Solve applied problems using the Law of Cosines. In Section 7.1, we developed the Law of Sines (Theorem 7.1) to enable us to solve triangles in the ‘Angle-Side-Angle’ (ASA), the ‘Angle-Angle-Side’ (AAS) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the ‘Side-AngleSide’ (SAS) and ‘Side-Side-Side’ (SSS) cases.1 The Law of Cosines We state and prove the Law of Cosines theorem below. Theorem 7.4. The Law of Cosines. Given a triangle with angle-side opposite pairs , and , the following equations hold or, solving for the cosine in each equation, we have To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin, and with side b positioned along the positive x-axis. 1 Here, ‘Side-Angle-Side’ means that we are given two sides and the included angle – that is, the given angle adjacent to both of the given sides. ,a,b,c2222cosabcbc2222cosbacac2222coscabab222cos2bcabc222cos2acbac222cos2abcab 303 From this set-up, we immediately find that the coordinates of A and C are and . From Theorem 2.6, we know that since the point lies on a circle of radius c, the coordinates of B are . (This would be true even if α were an obtuse or right angle so although we have drawn the case where α is acute, the following computations hold for any angle α drawn in standard position where .) We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, we get The remaining formulas given in Theorem 7.4 can be shown by simply reorienting the triangle to place a different vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that is an angle-side opposite pair and b and c are the sides adjacent to α. The same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula, which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the 0,0A,0Cb,Bxy,cos,sinBxyBcc0180222222222222222222222222222222cossin0cossincossincos2cossincossin2cos12cosc2ossin1acbcacbcacbcacbcbcabcbcabcbcabcbc since cos,a 304 Pythagorean Theorem. If we have a triangle in which , then so we get the familiar relationship . What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.2 Example 7.3.1. Solve the triangle in which , units, and units. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. Solution. We are given the lengths of two sides, and , and the measure of the included angle, . With no angle-side opposite pair to use, we apply the Law of Cosines to find b. In order to determine the measures of the remaining angles α and γ, we are forced to use the derived value for b. There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that, unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute angle is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles is positive, the sine of an angle alone is not enough to determine if the angle in question is acute or obtuse. We proceed with the Law of Cosines. When using the Law of Cosines, It’s always best to find the measure of the largest unknown angle first, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to find α first. To that end, we use the formula for from Theorem 7.4 and substitute , and . 2 This shouldn’t come as too much of a shock. All of the theorems in Trigonometry can ultimately be traced back to the definition of the circular functions along with the distance formula and, hence, the Pythagorean Theorem. 90coscos900222cab507a2c7a2c502222222cos72272cos505328cos505.92 unitsbacacbbb,bcos7a5328cos50b2c 305 Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and π radians. This matches the range of the arccosine function, so we have At this point, we could find γ using the Law of Cosines, to minimize the propagation of error. Or we can take the shortcut of using angle measures that have already been determined. We sketch the triangle below. As mentioned earlier in the preceding example, once we’d determined b it was possible to use the Law of Sines to determine the remaining angles. However, noting that this was the ambiguous ASS case, proceeding with the Law of Sines would require caution. It is advisable to first find the smallest of the unknown angles, since we are guaranteed it will be acute.3 In this case, we would find γ since the side 3 There can only be one obtuse angle in the triangle, and if there is one, it must be the largest. 222222
cos25328cos5027cos25328cos50227cos50cos5328cos50bcabc after simplifying27cos50arccos radians5328cos50114.99180180114.995015.01 306 opposite γ is smaller than the side opposite the other unknown angle, α. Using the Law of Sines with the angle-side opposite pair , we get The usual calculations produce and we find Example 7.3.2. Solve for the angles in the triangle with sides of length units, units and units. Solution. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we find β first, since it is opposite the longest side, b. We then determine β through finding the arccosine of . As in the previous problem, now that we have obtained an angle-side opposite pair we could proceed using the Law of Sines. The Law of Cosines, however, offers us a rare opportunity to find the remaining angles using only the data given to us in the statement of the problem. Using the problem data, we get and . ,bsin50sin25328cos5015.011801805015.01114.994a7b5c222222cos245724515acbac151arccos radians5101.54,b5arccos radians 44.42729arccos radians 34.0535 307 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may differ slightly from those posted in the Examples and the Exercises. Example 7.3.2 is a great example of this, in that the approximate values we record for the measures of the angles sum to 180.01°, which is geometrically impossible. Solving Applied Problems Using the Law of Cosines Next, we have an application of the Law of Cosines. Example 7.3.3. A researcher wishes to determine the width of a vernal pond as drawn below. From a point P, he finds the distance to the western-most point of the pond to be 950 feet, while the distance to the northern-most point of the pond from P is 1000 feet. If the angle between the two lines of sight is 60°, find the width of the pond. 308 Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to find the length of the missing side opposite the given angle. Calling this length w (for width), we get We next take the square root to get Heron’s Formula In Section 7.2, we used the proof of the Law of Sines to develop Theorem 7.3 as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula – Heron’s Formula. Theorem 7.5. Heron’s Formula. Suppose a, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let . Then the area A, enclosed by the triangle, is given by We prove Theorem 7.5 using Theorem 7.3. Using the convention that the angle γ is opposite the side c, we have from Theorem 7.3. In order to simplify computations, we start by manipulating the expression for A2. The Law of Cosines tells us , so substituting this into our equation for A2 gives 222950100029501000cos60952,500w952500976 feetw12sabcAssasbsc1sin2Aab22222222221sin21sin41cos4cossin1Aababab since 222cos2abcab 309 At this stage, we recognize the last factor as the semiperimeter, . To complete the proof, we note that 222222222222222222222222222222222221cos41421444444162abAababcababcababababcababababcaba Law of Cosines2222222222222222222222162216221616bcababccaabbaabbccaabbaabbccababccabca difference of squares perfect square trinomials16162222babcabcbcaacbabcabcbcaacbabcabc difference of squares122abcsabc2222abcsaaabcabca 310 Similarly, we find and . Hence, we get so that as required. We close with an example of Heron’s Formula. Example 7.3.4. Find the area enclosed by the triangle in Example 7.3.2. Solution. In Example 7.3.2, we are given a triangle with sides of length units, units and units. Using these values, we find . Using Heron’s Formula, we get 2acbsb2abcsc22222bcaacbabcabcAsasbscsAssasbsc4a7b5c145782s8848785841396469.80 square unitsAssasbsc 311 7.3 Exercises In Exercises 1 – 4, solve for the length of the unknown side. Round to the nearest tenth. 1. 3. 2. 4. In Exercises 5 – 8, find the measure of angle A. Round to the nearest tenth. 5. 6. 7. 312 8. 9. Find the measure of each angle in the triangle. Round to the nearest tenth. In Exercises 10 – 19, use the Law of Cosines to find the remaining side(s) and angles(s) if possible. 10. 12. 14. 16. 18. 11. 13. 15. 17. 19. In Exercises 20 – 25, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique. 20. 22. 24. 21. 23. 25. 7, 12, 59.3ab104, 25, 37bc153, 8.2, 153ac3, 4, 90ab120, 3, 4bc7, 10, 13abc1, 2, 5abc300, 302, 48abc5, 5, 5abc5, 12, 13abc18, 63, 20ab37, 45, 26abc16, 63, 20ab22, 63, 20ab42, 117, 88bc7, 170, 98.6c 313 In Exercises 26 – 29, find the area of the triangle. Round to the nearest hundredth. 26. 27. 28. 29. 30. Find the area of the triangles. (a) (b) (c) 7, 10, 13abc300, 302, 48abc5, 12, 13abc 314 31. A rectangular octagon is inscribed in a circle with a radius of 8 inches. Find the perimeter of the octagon. 32. A rectangular pentagon is inscribed in a circle of radius 12 cm. Find the perimeter of the pentagon. 33. The hour hand on my antique Seth Thomas schoolhouse clock is 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o’clock. Round your answer to the nearest hundredth of an inch. 34. A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern- most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is 117°, what is the diameter of the crater? Round your answer to the nearest hundredth of a mile. 35. From the Pedimaxus International Airport a tour helicopter can fly to Cliffs of Insanity Point by following a bearing of N8.2°E for 192 miles and it can fly to Bigfoot Falls by following a bearing of S68.5°E for 207 miles.1 Find the distance between Cliffs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. 1 Please refer to the 7.2 Exercises for an introduction to bearings. 315 36. Cliffs of Insanity Point and Bigfoot Falls from Exercise 35 both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliffs of Insanity Point? Round your angle to the nearest tenth of a degree. 37. A naturalist sets off on a hike from a lodge on a bearing of S80°W. After 1.5 miles, she changes her bearing to S17°W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 38. The HMS Sasquatch leaves port on a bearing of N23°E and travels for 5 miles. It then changes course and follows a heading of S41°E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree. 39. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32°E. A storm moves in and after 100 miles, the captain of the Bigfoot finds he has drifted off course. If his bearing to the harbor is now S70°W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree. 40. From a point 300 feet above level ground in a firetower, a ranger spots two fires in the Yeti National Forest. The angle of depression made by the line of sight from the ranger to the first fire is 2.5° and the angle of depression made by line of sight from the ranger to the second fire is 1.3°. The angle formed by the two lines of sight is 117°. Find the distance between the two fires. Round your answer to the nearest foot. (Hint: In order to use the 117° angle between the lines of sight, you will first need to use right angle Trigonometry to find the lengths of the lines of sight. This will give you a SAS case in which to apply the Law of Cosines.) 316 41. If you apply the Law of Cosines to the ambiguous ASS case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to each of the following in order to demonstrate this result. (a) (b) (c) 42. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. 18, 63, 20ab16, 63, 20ab22, 63, 20ab 317 CHAPTER 8 POLAR COORDINATES AND APPLICATIONS Chapter Outline 8.1 Polar Coordinates 8.2 Polar Equations 8.3 Graphing Polar Equations 8.4 Polar Representations for Complex Numbers 8.5 Complex Products, Powers, Quotients and Roots Introduction Chapter 8 takes us from graphing on the Cartesian coordinate plane to graphing on a polar grid, using the pole and polar axis for reference. We begin in Section 8.1 by plotting points defined by polar coordinates. The geometry and connection between the pol
e and origin, polar axis and positive x-axis, lead to the conversion of points between polar and rectangular coordinates. Section 8.2 continues this theme by converting equations back and forth between polar form and rectangular form. Graphing is the focus of Section 8.3, beginning with circles and lines in the coordinate plane, then moving on to more complicated polar graphs. Throughout Section 8.3, techniques are introduced and emphasized that enable the student to complete polar graphs by hand, without the aid of technology. The theme of Section 8.4 is complex numbers, as represented on the complex plane. A polar definition for complex numbers is introduced, and practice is provided for converting between rectangular and polar forms. This chapter ends with Section 8.5 and the introduction of ‘arithmetic‘ with complex numbers. DeMoivre’s Theorem is included as a means for determining powers and roots of complex numbers. This chapter introduces new concepts that rely on the trigonometric tools and skills developed up to this point in the course. It provides valuable insight into polar graphs and complex numbers. 318 8.1 Polar Coordinates Learning Objectives In this section you will:  Graph points in polar coordinates.  Convert points in polar coordinates to rectangular coordinates and vice versa. Up to this point, we have graphed points in the Cartesian coordinate plane by assigning ordered pairs of numbers to points in the plane. We defined the Cartesian coordinate plane using two number lines, one horizontal and one vertical, which intersect at right angles at a point called the origin. To plot a point, say , we start at the origin, travel horizontally to the left 3 units, then up 4 units. Alternatively, we could start at the origin, travel up 4 units, then to the left 3 units and arrive at the same location. For the most part, the motions of the Cartesian system (over and up) describe a rectangle, and most points can be thought of as the corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called rectangular coordinates. In this section, we introduce polar coordinates, a new system for assigning coordinates to points in the plane. Plotting Polar Coordinates We start with an origin point, called the pole, and a ray called the polar axis. We then locate a point P using two coordinates, , where r represents a directed distance from the pole2 and θ is a measure of 1 Excluding, of course, the points in which one or both coordinates are 0. 2 We will explain more about this momentarily. 3,4P,rxyP(-3,4) 319 rotation from the polar axis. Roughly speaking, the polar coordinates of a point measure how far out the point is from the pole (that’s r) and how far to rotate from the polar axis (that’s θ). Example 8.1.1. Plot the point P with polar coordinates . Solution. We start at the pole, move out along the polar axis 4 units, then rotate radians counter- clockwise. First Second The Resulting Point We may also visualize this process by thinking of the rotation first.3 To plot this way, we rotate radians counter-clockwise from the polar axis, then move outwards from the pole 4 units. Essentially, we are locating a point on the terminal side of which is 4 units away from the pole. 3 As with anything in Mathematics, the more ways you have to look at something, the better. Take some time to think about both approaches to plotting points given in polar coordinates. ,r54,65654,6P5656Polar AxisPole r r  ,PrPole 4rPole 56Pole 54,6P 320 First Second The Resulting Point If , we begin by moving, from the pole, in the opposite direction of the polar axis. Example 8.1.2. Plot . Solution. We start at the pole, moving 3.5 units in the opposite direction of the polar axis. We then rotate units counter-clockwise. First Second The Resulting Point If we interpret the angle first, we rotate radians, then move back through the pole 3.5 units. Here we are locating a point 3.5 units away from the pole on the terminal side of , not . 0r3.5,4Q44544Pole 56Pole 56Pole 54,6PPole 3.5rPole 4Pole 3.5,4Q 321 First Second The Resulting Point As you may have guessed, means the rotation away from the polar axis is clockwise instead of counter-clockwise. Example 8.1.3. Plot . Solution. To plot , we have the following. First Second The Resulting Point From an ‘angles first’ approach, we rotate then move out 3.5 units from the pole. We see that R is the point on the terminal side of which is 3.5 units from the pole. 033.5,4R33.5,4R3434Pole 4Pole 4Pole 3.5,4QPole 3.5rPole 34Pole 33.5,4R 322 First Second The Resulting Point Multiple Representations for Polar Coordinates The points Q and R in the above examples are, in fact, the same point despite the fact that their polar coordinate representations are different. Unlike Cartesian coordinates where and represent the same point if and only if and , a point can be represented by infinitely many polar coordinate pairs. We explore this notion in the following examples. Example 8.1.4. Plot the point , given in polar coordinates, and then give two additional expressions for the point, one of which has and the other with . Solution. Whether we move 2 units along the polar axis and then rotate 240° or rotate 240° then move out 2 units from the pole, we plot below. We now set about finding alternate descriptions for the point P. Since P is 2 units from the pole, . Next, we choose angles θ for each of the r values. The given representation for P is so the angle θ we choose for the case must be coterminal with 240°. (Can you see why?) One ,ab,cdacbd2,240P0r0r2,240P,r2r2,2402rPole 34Pole 34Pole 33.5,4RPole 240Pole 2,240P 323 such angle is so one answer for this case is . For the case , we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate to arrive at a location coterminal with 240°. Hence, our answer here is . We check our answers by plotting them. Example 8.1.5. Plot the point and give two additional expressions for the point, one with and the other with . Solution. We plot by first moving 4 units to the left of the pole and then rotating radians. We find our point lies 4 units from the pole on the terminal side of . To find alternate descriptions for P, we note that the distance from P to the pole is 4 units, so any representation for P must have . As we noted above, P lies on the terminal side of , so this, coupled with , gives us as one of our answers. To find a different representation for P 1202,1202r602,6074,6P0r0r74,6766,r4r64r4,6Pole 120 2,120PPole 60 2,60PPole 76Pole 74,6P 324 with , we may choose any angle coterminal with the angle in the original representation of . We pick and get as our second answer. Example 8.1.6. Plot the point and give two additional expressions for the point, one with and the other with . Solution. To plot , we move along the polar axis 117 units from the pole and rotate clockwise radians as illustrated below. Since P is 117 units from the pole, any representation for P satisfies . For the case, we can take θ to be any angle coterminal with . In this case, we choose and get as one answer. For the case, we visualize moving left 117 units from the pole and 4r74,6P5654,65117,2P0r0r5117,2P52,r117r117r52323117,2117rPole 6 4,6PPole 54,6P 56 52 117rPole 5117,2P 325 then rotating through an angle θ to reach P. We find that satisfies this requirement, so our second answer is . Example 8.1.7. Plot the point and give two additional expressions for the point, one with and the other with . Solution. We move three units to the left of the pole and follow up with a clockwise rotation of radians to plot . We see that P lies on the terminal side of . 2117,23,4P0r0r43,4P34Pole 3117,2P 32Pole 2 117,2PPole 4Pole 3,4P 326 Since P lies on the terminal side of , one alternative representation for P is . To find a different representation for P with , we may choose any angle coterminal with . We choose for our final answer of . Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that any given point expressed in polar coordinates has infinitely many other representations in polar coordinates. The following property of the polar coordinate system summarizes characteristics of different polar coordinates that determine the same point in the plane. Equivalent Representations of Points in Polar Coordinates Suppose and are polar coordinates where , and the angles are measured in radians. Then and determine the same point P if and only if one of the following is true:   and and for some integer k for some integer k All polar coordinates of the form represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that means (directed distance from pole, angle of rotation). If , then no matter how much rotation is performed, the point never leaves the pole. Thus, is the pole for all values of θ. 3433,43r47473,4,r','r0r'0r,r','r'rr'2k'rr'21k0,,r0r0,Pole 34 33,4PPole 73,4P 74 327 Now let’s assume that neither r nor r' is zero. If and determine the same point P then the (non-zero) distance from P to the pole in each case must be the same. Since this distance is controlled by the first coordinate, we have either or . 1. If then, when plotting and , the angles and have the same initial side. Hence, if and determine the same point, we must have that
is coterminal with . We know that this means for some integer k, as required. 2. If, on the other hand, , then when plotting and the initial side of is rotated radians away from the initial side of . In this case, must be coterminal with . Hence, which we rewrite as for some integer k. Conversely, 1. If and for some integer k, then the points and lie the same (directed) distance from the pole on the terminal sides of coterminal angles, and hence are the same point. 2. Suppose that and for some integer k. To plot P, we first move a directed distance r from the pole; to plot , our first step is to move the same distance from the pole as P, but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since for some integer k, we see that is coterminal to and it is this extra π radians of rotation which aligns the points and . Converting Between Rectangular and Polar Coordinates To move points from the polar coordinate system to the Cartesian (rectangular) coordinate system, or vice versa, we identify the pole and polar axis in the polar system to the origin and positive x-axis, respectively, in the rectangular system. ,r','r'rr'rr'rr,r','r',r','r''2k'rr,r','r'''2k'21k'rr'2k,Pr'','Pr'rr'21k'P'212kk'P'P 328 If we have a polar point in Quadrant I, we can form a right triangle by first dropping a perpendicular line segment from the point to the point , on the positive x-axis, to form a vertical leg. To form a horizontal leg, we sketch the line segment from the origin to the point . Finally, the hypotenuse is the line segment from the origin to the polar point . The lengths of the legs of this triangle, x and y, are the corresponding rectangular coordinates for the polar point . Using right triangle trigonometry, we can express x and y in terms of r and θ: Translating a polar point to its rectangular representation is fairly straightforward using these two formulas for x and y. Suppose, on the other hand, we want to translate a rectangular point to its polar representation . We can find r using the Pythagorean Theorem. ,r,r,0r,0r,r,xy,rcoscosxrxrsinsinyryr,xy,r222xyrpoleoriginpolar axispositivex-axis ,rpoleoriginpolar axispositivex-axisxy ,r  Then, to determine θ, we use the tangent. 329 For Quadrant II, III or IV, we can use a reference angle, , and include the appropriate signs on x and y, as determined by the quadrant in which they lie. Recall that, since for , to find an angle θ in Quadrant II or III it will be necessary to add π to obtain the correct angle. Also keep in mind that a point in polar coordinates can be expressed in many ways since where k is an integer. We get the following result. Theorem 8.1. Conversion Between Rectangular and Polar Coordinates: Suppose P is represented and in polar coordinates as in rectangular coordinates as . Then   and and (provided To verify this result, we check out the three cases: ) , and . 1. In the case , the theorem is an immediate consequence of Theorem 2.6. Recall that We apply the quotient identity to verify that . tanyxarctantan22,,2rrk,xy,rcosxrsinyr222xyrtanyx0x0r0r0r0rcoscossinsinxxrryyrrsintancostanyxxy ,r   330 2. If , then we know an alternate representation for is . Using and , we apply Theorem 2.6 as follows. Moreover, case too. and , so the theorem is true in this 3. The remaining case is , in which case is the pole. Since the pole is identified with the origin in rectangular coordinates, the theorem in this case amounts to checking ‘ ’. The following example puts Theorem 8.1 to good use. Example 8.1.8. Convert each point in rectangular coordinates given below into polar coordinates with and . Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting back to rectangular coordinates. 1. 2. 3. 4. Solution. 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking the time to plot the point before we do any calculations. Plotting shows that it lies in Quadrant IV. With and , we get 0r,r,rcoscossinsincoscoscosxrrrsinsinsinyrrr2222xyrrtantanyx0r,0,r0,0000r022,23P3,3Q0,3R3,4S2,23P2,23P2x23y2222222341216rxyxy 2,23P 53 331 So and, since we are asked for , we choose . To find θ, we have that This tells us θ has a reference angle of , and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have , so we choose . Hence, our answer is . To check, we convert back to rectangular coordinates and we find The result is the point in rectangular coordinates, as required. 2. The point is in Quadrant III. Using , we get so We find , which means θ has a reference angle of . Since Q lies in Quadrant III, we choose , which satisfies the requirement that . Our final answer is . 4r0r4rtan2323yx3025354,35,4,3rcos54cos31422xrsin54sin334223yr2,233,3Q3xy2223318r183232r since we are asked for r03tan13454025,32,4rxy 3,3Q 54 332 To check, we find The resulting point verifies our solution. 3. The point lies along the negative y-axis. While we could go through the usual computation4 to find the polar form of R, in this case we can find the polar coordinates of R using the definition. Since the pole is identified with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation of R we know . Since we require , we choose . Concerning θ, the angle satisfies with its terminal side along the negative y-axis, so our answer is . To check, we note 4 Since x = 0, we would have to determine θ geometrically. cos532cos423223xrsin532sin423223yr3,30,3R,r3r0r3r320233,2cos33cos2300xrsin33sin2313yrxy 0,3R 32 333 4. The point lies in Quadrant II. With and , we get so . As usual, we choose and proceed to determine θ. We have Since this isn’t the tangent of one of the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy , we choose radians. Hence, our answer is . To check our answers requires a bit of tenacity since we need to simplify expressions of the form and . These are good review exercises and are hence left to the reader. We find and , so that Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems, we move on to the next section where we will convert equations from one system to another. 3,4S3x4y2223425r5r50rtan4343yx024arctan34,5,arctan5,2.213r4cosarctan34sinarctan343cosarctan3544sinarctan35cos3553xrsin4554yrxy 3,4S 4arctan3 334 8.1 Exercises In Exercises 1 – 16, plot the point given in polar coordinates and then give three different expressions for the point such that (a) and (b) and (c) and 1. 5. 9. 13. 2. 6. 10. 14. 3. 7. 11. 15. 4. 8. 12. 16. In Exercises 17 – 36, convert the point from polar coordinates into rectangular coordinates. 20. 24. 28. 18. 22. 26. 17. 21. 25. 29. 31. 33. 19. 23. 27. 30. 32. 34. 0r020r00r22,375,413,3255,26712,653,422,713,2620,354,421,33,2113,62.5,445,3,75,42,3711,620,33,5254,679,295,41342,6117,1176,arctan210,arctan343,arctan345,arctan312,arctan21,arctan5231,arctan42,arctan223 35. 335 36. In Exercises 37 – 56, convert the point from rectangular coordinates into polar coordinates with and . 37. 41. 45. 49. 53. 38. 42. 46. 50. 54. 39. 43. 47. 51. 55. 40. 44. 48. 52. 56. ,arctan1213,arctan50r020,53,37,73,33,02,24,4331,44333,10105,56,85,258,1210,6105,12525,151524,712,926,4465265,55 8.2 Polar Equations In this section you will: 336 Learning Objectives  Convert an equation from rectangular coordinates into polar coordinates.  Convert an equation from polar coordinates into rectangular coordinates. Just as we’ve used equations in x and y to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in polar coordinates. We use Theorem 8.1 to convert equations between the two systems. Converting from Rectangular to Polar Coordinates One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with and every occurrence of y with , and use identities to simplify. This is the technique we employ in the following three examples. Example 8.2.1. Convert from an equation in rectangular coordinates into an equation in polar coordinates. Solution. We start by substituting and into and then simplify. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us.1 1 Study this example and see what techniques are employed, then try your best to apply these techniques in the Exercises. cosrsinr2239xycosxrsinyr2239xy222222222222222239cos3sin9cos6cos9sin9cossin6cos0cossin6cos06cos06cos0xyrrrrrrrrrrrrrr after subtracting 9 from both sides since after factoring
22cos+sin=1θθ 337 We get or . Recognizing the equation as describing a circle, we exclude the first since describes only a point (namely the pole/origin). We choose for our final answer. Note that when we substitute into , we recover the point , so we aren’t losing anything by disregarding . Example 8.2.2. Convert from an equation in rectangular coordinates into an equation in polar coordinates. Solution. We substitute and into . This gives or . Solving the latter for θ, we get for integers k. As we did in the previous example, we take a step back and think geometrically. We know describes a line through the origin. As before, describes the origin but nothing else. Consider the equation . In this equation, the variable r is free, meaning it can assume any and all values including . If we imagine plotting points for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of which is none other than . 0r6cosr2239xy0r6cosr26cosr0r0ryxcosxrsinyryxsincoscossin0cossin0yxrrrrr after rearranging after factoring0rcossin04kyx0r40r,4r4yxxy 2239xyxy yx 338 Hence, we can take as our final answer .2 Example 8.2.3. Convert from an equation in rectangular coordinates into an equation in polar coordinates. Solution. We substitute and into . Either or . We can solve the latter equation for r by dividing both sides of the equation by . As a general rule we never divide through by a quantity that may be equal to 0. In this particular case, we are safe since if then then both and, for the equation to hold, would also have to be 0. Since there are no angles with and , we are not losing any information by dividing both sides of by . Doing so, we get 2 We could take it to be θ = –π ∕ 4 + πk for any integer k. 42yxcosxrsinyr2yx2222222sincossincos0cossin0cossinyxrrrrrrrr0r2cossinr2cos2cos0cos02cossinrsincos0sin02cossinr2cos2sincos1sincoscossectanrxy 2yx 339 As before, the case is recovered in the solution when . So we state our final solution as . Converting from Polar to Rectangular Coordinates As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. We will begin with the strategy of rearranging the given polar equations so that the expressions , , and/or present themselves. Example 8.2.4. Convert from an equation in polar coordinates into an equation in rectangular coordinates. Solution. Starting with , we can square both sides. We may now substitute to get the equation . As we have seen, squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation might be satisfied by more points than . On the surface, this appears to be the case since is equivalent to , not just . However, any point with polar coordinates can be represented as , which means any point whose polar coordinates satisfy the relation has an equivalent3 representation which satisfies . 3 Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3,0) and (–3,π) are different, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r2 = 9 and r = –3 represent different relations since they correspond to different sets of ordered pairs. Since polar coordinates were defined geometrically to describe the location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, in algebra when we first defined relations as sets of points in the plane. Back then, a point in the plane was identified with a unique ordered pair given by its Cartesian coordinates. 0rsectanr0sectanr222rxycosrxsinrytanyx3r3r222339rrr222rxy229xy29r3r29r3r3r3,3,,r3r3r 3r 340 Thus, we state our final solution as . Example 8.2.5. Convert from an equation in polar coordinates into an equation in rectangular coordinates. Solution. We begin by taking the tangent of both sides of the equation. Since , we get the following. Of course, we pause a moment to wonder if, geometrically, the equations and generate the same set of points.4 The same argument presented in Example 8.2.4 applies equally well here. We conclude that our answer of is correct. Example 8.2.6. Convert from an equation in polar coordinates into an equation in rectangular coordinates. Solution. Once again, we need to manipulate a bit before using the conversion formulas given in Theorem 8.1. We could square both sides of this equation like we did in Example 8.2.4 to 4 There are infinitely many solutions to , and is only one of them. Additionally, we went from , in which x cannot be 0, to in which we assume x can be 0. 229xy43434tantan3tan3tanyx33yxyx433yx3yx1cosr1cosrtan3433yx3yx 43 341 obtain an on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r and continue manipulating the equation so that we can apply the conversion formulas from Theorem 8.1. In the last step, we applied Theorem 8.1 and we now have the equation as a solution. It can be shown that this is a legitimate solution by confirming the results when . We will forego this verification for now, as well as the verification that points with coordinates which satisfy will also satisfy . To the right is a graph of the polar equation , from this example. This curve is referred to as a cardioid. In the next section, we will graph cardioids, along with other polar equations. In practice, much of the pedantic verification of the equivalence of equations is left unsaid. Indeed, in most textbooks, squaring equations like to arrive at happens without a second thought. Your instructor will ultimately decide how much, if any, justification is warranted. 2r222222222222cos1coscoscoscosrrxrrrrrrrrrrxyxxy multiplying through by r adding to both sides squaring both sides substituting =2cosryx and 22222xyxxy0r,r222cosrrr2cosrrr1cosr3r29r 1cosr 342 8.2 Exercises In Exercises 1 – 20, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #4 through #7. In Exercises 4 – 7, you need to solve for θ. 1. 5. 9. 13. 17. 19. 2. 6. 10. 14. 3. 7. 11. 4. 8. 12. 15. 16. 18. 20. In Exercises 21 – 40, convert the equation from polar coordinates into rectangular coordinates. 21. 25. 29. 33. 37. 22. 26. 30. 34. 38. 23. 27. 31. 35. 39. 24. 28. 32. 36. 40. 41. Convert the origin into polar coordinates in four different ways. 42. With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates. 6x3x7y0yyx3yx2yx2225xy22117xy419yx31xy23yx24xy2220xyy2240xxy22xyx227yyx2224xy2239xy2214412xy7r3r2r423324cosr5cosr3sinr2sinr7secr12cscr2secr5cscr2sectanr22sinr12cosr1sinrcsccotr0,0 343 8.3 Graphing Polar Equations Learning Objectives In this section you will:  Learn techniques for graphing polar equations.  Graph polar equations. In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since any given point in the plane has infinitely many different representations in polar coordinates, practice with graphing polar equations will be an essential part of the learning process. We begin with the Fundamental Graphing Principle for polar equations. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisfy the equation. That is, a , point is on the graph of an equation if and only if there is a representation of P, say such that and satisfy the equation. Graphing a Simple Polar Equation – Constant Radius or Constant Angle Our first example focuses on some of the more structurally simple polar equations. Example 8.3.1. Graph the following polar equations. 1. 2. 3. 4. Solution. In each of these equations, only one of the variables r and θ is present, resulting in the missing variable taking on all values without restriction. This makes these graphs easier to visualize that others. 1. In the equation , θ is missing. The graph of this equation is, therefore, all points which have a polar coordinate representation , for any choice of θ. Graphically, this translates into tracing out all of the points 4 units away from the origin. This is exactly the definition of circle, centered at the origin, with a radius of 4. ,Pr','r'r'4r32r54324r4,xy 4r 344 2. Once again, we have θ missing in the equation . Plotting all of the points of the form gives us a circle of radius centered at the origin. 3. In the equation , r is missing, so we plot all of the points with polar representations . What we find is that we are tracing out the line which contains the terminal side of when plotted in standard position. 4. As in the previous problem, the variable r is missing in the equation, . Plotting for various values of r shows us that we are tracing out the y-axis. Hopefully, our experience in Example 8.3.1 makes the following result clear. 32r32,32545,4r54323,2rxy 32rxy 54 54xy 32 32 345 Theorem 8.2. Graphs of Constant r and θ: Suppose a and α are constants, .  The graph of the polar equation .  The graph of the polar equation of radius on the Cartesian plane is a circle centered at the origin on the Cart
esian plane is the line containing the terminal side of α when plotted in standard position. Graphing Polar Equations Containing Variables r and θ Suppose we wish to graph . A reasonable way to start is to treat θ as the independent variable, r as the dependent variable, evaluate at some ‘friendly’ values of θ and plot the resulting points. We generate the table below, followed by a graph of the resulting points in the xy-plane. π 0araa6cosrrf6cosr,r6cosr,r066,05432532,443232,432030,2200,27432732,43432332,4266,266,xy 346 Despite having nine ordered pairs, we only get four distinct points on the graph. For this reason, we employ a slightly different strategy. We graph on the θr-plane1 and use it as a guide for graphing the equation on the xy-plane. We first see that as θ ranges from 0 to , r ranges from 6 to 0. In the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis, when , and gradually returns to the origin, at . in the θr-plane in the xy-plane The arrows drawn in the above figures are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve . In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but is markedly faster. Next, we repeat the process as θ ranges from to . Here, the r-values are all negative. This means that in the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis. 1 The graph looks exactly like in the xy-plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates on the xy-plane. 6cosr2026cosr6cosr6cosr26cosyx,rxyCOORDINATESθ θ 6,0 32,4 0,2 ,r 347 in the θr-plane in the xy-plane As θ ranges from to , the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the for these values of θ match the r values for θ in , we have that the curve begins to retrace itself at this point. Proceeding further, we find that when , we retrace the part of the curve in Quadrant IV that we first traced out as . The reader is invited to verify that plotting any range of θ outside the interval results in retracing some portion of the curve.2 We present the final graph below. 2 The graph of r = 6cos(θ) looks suspiciously like a circle, for good reason. See Example 8.2.1. 6cosr6cosr32r,232220,xyCOORDINATESθ θ 0,2 ,r 332,4 6, 348 in the θr-plane in the xy-plane Example 8.3.2. Graph the polar equation . Solution. We first plot the fundamental cycle of on the θr-axes. To help us visualize what is going on graphically, we divide up into the usual four subintervals , , and , and proceed as we did above. 1. As θ ranges from 0 to , r decreases from 4 to 2. This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in toward a point 2 units from the origin on the positive y-axis. in the θr-plane in the xy-plane 6cosr6cosr42sinr42sinr0,20,2,23,23,22242sinr42sinrrθ θ xyθ θ 3 3 6rθ θ xyθ θ 349 2. Next, as θ runs from to , we see that r increases from 2 to 4. In the xy-plane, picking up where we left off, we gradually pull the graph toward the point 4 units away from the origin on the negative x-axis. in the θr-plane in the xy-plane 3. Over the interval , we see that r increases from 4 to 6. On the xy-plane, the curve sweeps out away from the negative x-axis toward the negative y-axis. in the θr-plane in the xy-plane 4. Finally, as θ takes on values from to , r decreases from 6 back to 4. The graph on the xy-plane pulls in from the negative y-axis to finish where we started. 242sinr42sinr3,242sinr42sinr322rθ θ xyθ θ xyθ θ rθ θ  350 in the θr-plane in the xy-plane We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval results in retracing portions of the curve, so we are finished. in the θr-plane in the xy-plane Example 8.3.3. Graph the polar equation . Solution. The first thing to note when graphing on the θr-plane over the interval is that the graph crosses through the θ-axis. This corresponds to the graph of the curve passing through the origin in the xy-plane, and our first task is to determine when this happens by determining the values of θ for which . 42sinr42sinr0,242sinr42sinr24cosr24cosr0,20rrθ θ xyθ θ rθ θ xyθ θ 4 4 2 6 351 Solving for θ in gives and . Since these values of θ are important geometrically, we break the interval into six subintervals: , , , , and . 1. As θ ranges from 0 to , r decreases from 6 to 2. Plotting this on the xy-plane, we start 6 units out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. in the θr-plane in the xy-plane 2. On the interval , r decreases from 2 to 0, which means the graph is heading into (and will eventually cross through) the origin. Not only do we reach the origin when , the curve hugs the line as it approaches the origin. 024cos01cos2r0,223430,20,22,232,34,343,323,22224cosr24cosr2,232323rθ θ xyθ θ 352 in the θr-plane in the xy-plane 3. On the interval , r ranges from 0 to . Since , the curve passes through the origin in the xy-plane, following the line and continues upwards through Quadrant IV toward the positive x-axis. With increasing from 0 to 2, the curve pulls away from the origin to finish at a point on the positive x-axis. in the θr-plane in the xy-plane 4. Next, as θ progresses from to , r ranges from to 0. Since , we continue our graph in the first quadrant, heading into the origin along the line . 24cosr24cosr2,320r23r24cosr24cosr4320r43rθ θ  23xyθ θ 23rθ θ  23xyθ θ 23 353 in the θr-plane in the xy-plane 5. On the interval , r returns to positive values and increases from 0 to 2. We hug the line as we move through the origin and head toward the negative y-axis. in the θr-plane in the xy-plane 6. In the last step, we find that as θ runs through to , r increases from 2 to 6, and we end up back where we started, 6 units from the origin on the positive x-axis. 24cosr24cosr43,324324cosr24cosr322rθ θ  43xyθ θ 43rθ θ  43xyθ θ 43 354 in the θr-plane in the xy-plane Again, we invite the reader to show that plotting the curve for values of θ outside results in retracing a portion of the curve already traced. Our final graph is below. in the θr-plane in the xy-plane Example 8.3.4. Graph the polar equation . Solution. As usual, we start by graphing a fundamental cycle of in the θr-plane, which in this case occurs as θ ranges from 0 to π. We partition our interval into subintervals to help us with the graphing, namely , , and . 24cosr24cosr0,224cosr24cosr5sin2r5sin2r0,4,423,243,4rθ θ xyθ θ rθ θ xyθ θ 43 23 355 1. As θ ranges from 0 to , r increases from 0 to 5. This means that the graph of in the xy-plane starts at the origin and gradually sweeps our so it is 5 units away from the origin on the line . in the θr-plane in the xy-plane 2. Next, we see that r decreases from 5 to 0 as θ runs through and, furthermore, r is heading negative as θ crosses . Hence, we draw the curve hugging the line (the y-axis) as the curve heads to the origin. in the θr-plane in the xy-plane 45sin2r45sin2r5sin2r,42225sin2r5sin2rxyθ θ xyθ θ θ θ  4 2 34  r θ θ  4 2 34 r   356 3. As θ runs from to , r becomes negative and ranges from 0 to . Since , the curve pulls away from the negative y-axis into Quadrant IV. in the θr-plane in the xy-plane 4. For , r increases from to 0, so the curve pulls back to the origin. in the θr-plane in the xy-plane Even though we have finished with one complete cycle of , if we continue plotting beyond , we find that the curve continues into the third quadrant! Below we present a graph of a second cycle of which continues on from the first. 23450r5sin2r5sin2r3455sin2r5sin2r5sin2r5sin2rxyθ θ θ θ  4 2 34  r θ θ  4 2 34  r xyθ θ 357 in the θr-plane in the xy-plane We have the final graph below. in the θr-plane in the xy-plane Example 8.3.5. Graph . Solution. Graphing is complicated by the r2, so we solve to get How do we sketch such a curve? First off, we sketch a fundamental period of , which is in the figure below. When , is undefined, so we don’t have any values on the 5sin2r5sin2r5sin2r5sin2r216cos2r2216cosr16cos24cos2rrcos2rcos20cos2xyθ θ θ θ   54 32 74 2 5 5xyθ θ 358 interval . On the intervals which remain, ranges from 0 to 1, inclusive. Hence, ranges from 0 to 1 as well.3 From this, we know ranges continuously from 0 to ±4, respectively. Below we graph both and on the θr-plane and use them to sketch the corresponding pieces of the curve in the xy-plane. and in the θr-plane and in the xy-plane As we have seen in earlier examples, the lines and , which are the zeros of the functions , serve as guides for us to
draw the curve as it passes through the origin. As we plot points corresponding to values of θ outside of the interval parts of the curve,4 so our final answer is below. , we find ourselves retracing 3 Owing to the relationship between and over , we also know wherever the former is defined. 4 In this case, we could have generated the entire graph by using just the plot , but graphed over the interval in the θr-plane. We leave the details to the reader. 3,44cos2cos24cos(2)r4cos(2)r4cos(2)r216cos2r4cos2r=-4cos2rθ4cos2r=-4cos2rθ4344cos2r0,yxyx0,1cos(2)cos(2)4cos2r0,2xy1324θ θ 34 41234θ θ  r 4 2 34   cos2r 359 in the θr-plane in the xy-plane A few remarks are in order. 1. There is no relation, in general, between the period of the function and the length of the interval required to sketch the complete graph of in the xy-plane. As we saw at the beginning of this section, despite the fact that the period of is , we sketched the complete graph of in the xy-plane just using the values of θ as θ ranged from 0 to π. In Example 8.3.4, the period of is π, but in order to obtain the complete graph of we needed to run θ from 0 to 2π. While many of the ‘common’ polar graphs can be grouped into families,5 taking the time to work through each graph in the manner presented here is the best way to not only understand the polar coordinate system but also prepare you for what is needed in Calculus. 2. The symmetry seen in the examples is a common occurrence when graphing polar equations. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin), it is possible to talk about rotational symmetry. Keep rotational symmetry in mind as you work through the Exercises. 5 Example 8.3.2 and Example 8.3.3 are examples of limacons. Example 8.3.4 is an example of a polar rose, and Example 8.3.5 is the famous Lemniscate of Bernoulli. 4cos2r216cos2rfrf6cosf26cosr5sin2f5sin2rθ θ  r 4 2 34  xyθ θ 34 4 360 8.3 Exercises In Exercises 1 – 20, plot the graph of the polar equation by hand, without the aid of a calculator. Carefully label your graphs. 1. Circle: 3. Rose: 5. Rose: 7. Rose: 9. Cardioid: 11. Cardioid: 13. Limacon: 15. Limacon: 17. Limacon: 2. Circle: 4. Rose: 6. Rose: 8. Rose: 10. Cardioid: 12. Cardioid: 14. Limacon: 16. Limacon: 18. Limacon: 19. Lemniscate: 20. Lemniscate: Exercises 21 – 30 give you some curves to graph using a graphing calculator or other form of technology. Notice that some of the curves have explicit bounds on θ and others do not. 21. , 23. , 25. 27. , 29. , , 22. 24. 26. 28. 30. 6sinr2cosr2sin2r4cos2r5sin3rcos5rsin4r3cos4r33cosr55sinr22cosr1sinr12cosr12sinr234cosr35cosr35sinr27sinr2sin2r24cos2rr012lnr1120.1re0123r1.21.2sin53cosr32sincos23rarctanr11cosr12cosr123cosr 361 31. How many petals does the polar rose have? What about , and ? With the help of your classmates, make a conjecture as to how many petals the polar rose has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does have for each natural number n? 32. In this exercise, we have you and your classmates explore transformations of polar graphs. For both parts (a) and (b), let and . (a) Using a graphing calculator or other form of technology, compare the graph of to each of the graphs of , , and . Repeat this process for . In general, how do you think the graph of compares with the graph of ? (b) Using a graphing calculator or other form of technology, compare the graph of to each of the graphs of , , and . Repeat this process for . In general, how do you think the graph of compares with the graph of ? (Does it matter if or ?) 33. With the help of your classmates, research cardioid microphones. sin2rsin3rsin4rsin5rsinrncosrncosf2singrf4rf34rf4rf34rfgrfrfrf2rf12rfrf3rfgrkfrf0k0k 362 8.4 Polar Representations for Complex Numbers In this section you will: Learning Objectives  Find the real part, the imaginary part, and the modulus of a complex number.  Graph complex numbers.  Learn the properties of the modulus and the argument of a complex number and be able to apply them. Complex Numbers and the Complex Plane A complex number is a number of the form where a and b are real numbers and i is the imaginary unit defined by . The number a is called the real part of z, denoted , while the number b is called the imaginary part of z, denoted . If for real numbers a, b, c and d, then and , verifying that and are well-defined.1 To start off this section, we associate each complex number with the point on the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line, and the y-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. The Complex Plane 1 ‘Well-defined’ means that no matter how we express z, the number Re(z) is always the same, and the number Im(z) is always the same. In other words, Re and Im are functions of complex numbers. zabi1iRezImzzabicdiacbdRezImzzabi,abReal AxisImaginary Axis i 2i 3i 4i i 2i 3i 4i 4,242zi 0,33zi 1,01z 363 Since the ordered pair gives the rectangular coordinates associated with the complex number , the expression is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates . Although it is not as straightforward as the definitions of and , we can still give r and θ special names in relation to z. The Modulus and Argument of Complex Numbers Definition. The Modulus and Argument of Complex Numbers: Let number with be a complex be a polar representation of the point with . Let and rectangular coordinates where .  The modulus of z, denoted , is defined by .  The angle θ is an argument of z. The set of all arguments of z is denoted .  If and , then θ is the principal argument of z, written . Some remarks are in order. We know from Section 8.1 that every point in the plane has infinitely many polar coordinate representations , which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-defined.  Concerning the modulus, if then the point associated with z is the origin. In this case, the only possible r-value is . Hence, for , is well-defined.  If , then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated in our definition so this pins down the value of to one and only one number. Thus, the modulus is well-defined in this case too.2  Even with the requirement , there are infinitely many angles θ which can be used in a polar representation of a point . If then the point in question is not the origin, so all of these angles θ are coterminal. Since the coterminal angles are exactly 2π radians apart, we are guaranteed that only one of them lies in the interval , and this angle is what we call the principal argument of z, . 2 In case you’re wondering, the use of the absolute value notation \|z\| for modulus will be explained shortly. ,abzabizabi,rRezImzzabiReazImbz,r,ab0rzzrargz0zArgz,r0z0r0z0z0z0rz0r,r0z,Argz 364 In fact, the set of all arguments of z can be described using set-builder notation as . Note that since is a set, we will write ‘ ’ to mean ‘θ is in3 the set of arguments of z’.  If then the point in question is the origin, which we know can be represented in polar coordinates as for any angle θ. In this case, we have and since there is no one value of θ which lies in , we leave undefined. It is high time for an example. Example 8.4.1. For each of the following complex numbers find , , , and . Plot z in the complex plane. 1. 2. 3. 4. Solution. 1. For , we have and . To find , and , we need to find a polar representation with for the point associated with z. We first determine a value for r. We require , so we choose , and have . Next, we find a corresponding angle θ. Since and P lies in Quadrant IV, θ is a Quadrant IV angle. We have 3 Recall the symbol being used here, , is the mathematical symbol which denotes membership in a set. argzargArg2\| is an integerzzkkargzargz0z0,arg0,,Arg0RezImzzargzArgz3zi24zi3zi117z331ziiRe3zIm1zzargzArgz,r0r3,1P22222223142rxyrrr from 0r2r2z0r13tan332 for integerst an6kkyx from since is a Quadrant IV angle 365 Thus, . Of these values, only satisfies the requirement that , hence 2. The complex number has . , , and is associated with the point . Our next task is to find a polar representation for P where . Running through the usual calculations gives , so . To find θ, we get . Since and P lies in Quadrant II, we know θ is a Quadrant II angle. Thus, Hence, the requirement , so . Only . satisfies 3. We rewrite as to find and . The point in the plane which corresponds to z is and while we could go through the usual calculations to find the required polar form of this point, we can almost ‘see’ the answer. The point lies 3 units away from the origin on the positive y-axis. Hence, and for integers k. We get and . 4. As in the previous problem, we write , so and . The number corresponds to the point , and this is another instance where we can determine the polar form ‘by eye’. The point is 117 units away from the origin along the negative x-axis. Hence, and for integers k. We have . Only one of the
se values, , lies in the interval which means that . arg2 is an integer6zkk6Arg6z24ziRe2zIm4z2,4P,r0r25r25ztan20rarctan22 for integers or arctan22 for integers kkkk since is a Quadrant II angle from odd property of arctangentargarctan22\| is an integerzkkarctan2Argarctan2z3zi03ziRe0zIm3z0,30,33rz22karg2 is an integer2zkkArg2z1171170ziRe117zIm0z117z117,0117,0117rz2karg2\| is an integerzkk,Argz We plot the four numbers from this example below. 366 Properties of the Modulus and Argument Now that we’ve had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem 8.3. Properties of the Modulus: Let z and w be complex numbers. is the distance from z to 0 in the complex plane , and if and only if     Product Rule:  Power Rule: for all natural numbers n  Quotient Rule: , provided To prove the first three properties in Theorem 8.3, suppose where a and b are real numbers. To determine , we find a polar representation with for the point . From Section 8.1, we know so that . Since we require , then it must be that , which means . z0z0z0z22ReImzzzzwzwnnzzzzww0wzabiz,r0r,ab222rab22rab0r22rab22zabReal AxisImaginary Axis i 2i 3i 4i i 2i 3i 4i 117 2 1 1 2 3 \| \| \| \| \| 3zi 24zi 3zi 117z 367  Using the distance formula, we find the distance from to is also , establishing the first property.4  For the second property, note that since is a distance, . Furthermore, if and only if the distance from z to 0 is 0, and the latter happens if and only if , which is what we are asked to show.5  For the third property, we note that since and ,  To prove the product rule, suppose and for real numbers a, b, c and d. Then Therefore, 4 Since the absolute value \|x\| of a real number x can be viewed as the distance from x to 0 on the number line, this first property justifies the notation \|z\| for modulus. We leave it to the reader to show that if z is real, then the definition of modulus coincides with absolute value so the notation \|z\| is unambiguous. 5 This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is true. We know if and only if if and only if , which is true if and only if . The latter happens if and only if . There. 0,0,ab22abz0z0z0zReazImbz2222ReImzabzzzabiwcdi2211izwabicdiacadibcibdiacadibcibdacbdadbci from 2222222222222222222222222222222222zwacbdadbcacabcdbdadabcdbcacadbcbdacdbcdabcdabcd after expanding terms rearranged zwzw product rule for radicals definition of and 0z220ab220ab0ab0zabi 368 Hence as required.  Now that the product rule has been established, we use it and the Principle of Mathematical Induction to prove the power rule. Let be the statement . Then is true since . Next, assume is true. That is, assume for some . Our job is to show that is true, namely . As is customary with induction proofs, we first try to reduce the problem in such a way as to use the induction hypothesis: . Hence, is true, which means is true for all natural numbers n.  Like the power rule, the quotient rule can also be established with the help of the product rule. We assume , so that , and get Hence, the proof really boils down to showing . This is left as an exercise. Next, we characterize the argument of a complex number in terms of its real and imaginary parts. zwzwPnnnzz1P11zzzPkkkzz1k1Pk11kkzzkkzz11kkkkkzzzzzzzz property of exponents product rule of modulus induction hypothesis property of exponents1Pknnzz0w0w11zzwwzw product rule of modulus11ww 369 Theorem 8.4. Properties of the Argument: Let z be a complex number. and and and  If  If   If If , then . , then , then , then and . . . To prove Theorem 8.4, suppose for real numbers a and b. By definition, and , so the point associated with z is .  From Section 8.1, we know that if is a polar representation for , then , provided .  If and , then z lies on the positive imaginary axis. Since we take , we have that θ is coterminal with , and the result follows.  If and , then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with .  The last property was already discussed following the definition at the beginning of this section. Polar Form of Complex Numbers Our next goal is to link the geometry and algebra of the complex numbers. To that end, consider the figure below. Re0zargzImtanRezzRe0zIm0zarg2 is an integer2zkkRe0zIm0zarg2 is an integer2zkkReIm0zz0zarg,zzabiReazImbz,Re,Imabzz,rRe,ImzzImtanRezzRe0zRe0zIm0z0r2Re0zIm0z2 370 Polar coordinate associated with , with We know from Theorem 8.1 that and . Making these substitutions for a and b gives The expression is abbreviated so we can write . Since and , we get Definition. A Polar Form of a Complex Number: Suppose z is a complex number and The expression . is called a polar form for z. Since there are infinitely many choices for , there are infinitely many polar forms for z, so we used the indefinite article ‘a’ in the preceding definition. It is time for an example. Example 8.4.2. Find the rectangular form of the following complex numbers. Find and . 1. 2. 3. Solution. The key to this problem is to write out as 4. . ,rzabi0rcosarsinbrcossincossinzabirriricossinicisciszrrzargzargzciscossinzziargzRezImz24cis3z32cis4z3cis0zcis2zciscossiniImaginary AxisReal Axis bi a 0 argz 22zabr ,,abzabir 371 1. By definition, After some simplifying, we get , so that and . 2. Expanding, we get From this, we find , so . 3. We get Writing number. 4. Lastly, we have , we get and , which makes sense seeing that 3 is a real Since , we get and . Since i is called the ‘imaginary unit’, these answers make sense. Example 8.4.3. Use the results from Example 8.4.1 to find a polar form of the following complex numbers. 1. 2. 3. 4. 24cis3224cossin33zi223ziRe2zIm23z32cis4332cossin44zi22ziRe2Imzz3cis03cos0sin03zi330iRe3zIm0zcis2cossin22zii01iiRe0zIm1z3zi24zi3zi117z 372 Solution. To write a polar form of a complex number z, we need two pieces of information: the modulus and an argument (not necessarily the principal argument) of z. We shamelessly mine our solution to Example 8.4.1 to find what we need. 1. For , and , so . We can check our answer by converting it back to rectangular form to see that it simplifies to 2. For , and . Hence, a good exercise to actually show that this polar form reduces to . . . It is 3. For , and . In this case, . This can be checked geometrically. Head out 3 units from 0 along the positive real axis. Rotating radians counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at . 4. Last but not least, for , and . We get . As with the last problem, our answer is easily checked geometrically. z3zi2z62cis6z3zi24zi25zarctan225cisarctan2z24zi3zi3z23cis2z23zi117z117z117cisz 373 8.4 Exercises In Exercises 1 – 20, find a polar representation for the complex number z and then identify , , , and . These exercises should be worked without the aid of a calculator. 1. 5. 9. 13. 17. 2. 6. 10. 14. 18. 3. 7. 11. 15. 19. 4. 8. 12. 16. 20. In Exercises 21 – 40, find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values. These exercises should be worked without the aid of a calculator. 24. 28. 32. 21. 25. 29. 33. 35. 37. 39. 22. 26. 30. 23. 27. 31. 34. 36. 38. 40. RezImzzargzArgz99zi553zi6zi3232zi636zi2z3122zi33zi5zi2222zi6z37zi34zi2zi724zi26zi125zi52zi42zi13zi6cis0z2cis6z72cis4z3cis2z24cis3z36cis4z9cisz43cis3z37cis4z313cis2z17cis24z12cis3z8cis12z72cis8z45cisarctan3z110cisarctan3z15cisarctan2z3cisarctan2z750cisarctan24z15cisarctan212z 374 41. Complete the proof of Theorem 8.4, Properties of the Modulus, by showing that if then . 42. Recall that the complex conjugate of a complex number is denoted and is given by . (a) Prove that . (b) Prove that . (c) Show that and . (d) Show that if then . Interpret this result geometrically. (e) Is it always true that ? 0w11wwzabizzabizzzzzRe2zzzIm2zzziargzargzArgArgzz 375 8.5 Products, Powers, Quotients and Roots of Complex Numbers In this section you will: Learning Objectives  Find the product, power, quotient and roots of complex number(s).  Learn and apply DeMoivre’s Theorem. Products, Powers and Quotients of Complex Numbers The following theorem summarizes the advantages of working with complex numbers in polar form. Theorem 8.5. Products, Powers and Quotients of Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms . Then and  Product Rule:  Power Rule (DeMoivre’s Theorem): for every natural number n  Quotient Rule: , provided The proof of Theorem 8.5 requires a healthy mix of definition, arithmetic and identities.  We start with the product rule. We now focus on the quantities in brackets on the right hand side of the equation. Putting this together with our earlier work, we get .  Moving right along, we take aim at the power rule, better known as DeMoivre’s Theorem. We proceed by induction on n.
Let be the sentence . Then is true, since ciszzciswwciszwzwcisnnzznciszzww0wcisciscossincoscisinszwzwzwii definition of 222-1cossincossin coscoscossinsincossinsin coscossinsinsincoscossin coscossinsinsincoscossin iiiiiiiiiii rearrange terms use ; factor out  cossin cisi sum identities cis definition of ciszwzwPncisnnzzn1P11ciscis1zzzz 376 We now assume is true, that is, we assume for some . Our goal is to show that is true, or that . We have Hence, assuming is true, we have that is true, so by the Principle of Mathematical Induction, for all natural numbers n.  The last property in Theorem 8.5 to prove is the quotient rule. Assuming , we have Next, we multiply both the numerator and denominator of the right hand side by to get Finally, we have , and we are done. Pkciskkzzk1k1Pk11cis1kkzzk11ciscisciscis1kkkkkzzzzkzzzkzk property of exponents induction hypothesis product rulePk1Pkcisnnzzn0wcisciscossincossinzzwwziwicossini2222cossincossincossincossincossincossincossincossincoscoscossinsincossinsincoscossinsincossinczziiwwiiiizwiiziiiwiiizw22222oscossinsinsincoscossincossincossincossincis1iiziwzwciszzww 377 Example 8.5.1. Let and . Use Theorem 8.5 to find the following. 1. 2. 3. Write your final answers in rectangular form. Solution. In order to use Theorem 8.5, we need to write z and w in polar form. For , we find If , we know Since z lies in Quadrant I, we have for integers k. Hence, . For , we have For an argument θ of w, we have Since w lies in Quadrant II, for integers k and . We can now proceed. 232zi13wizw5wzw232zi22232164zargzImtanRe22313 or 33zz26k4cis6z13wi22132w3tan13223k22cis3w 378 1. We get After simplifying, we get . 2. We use DeMoivre’s Theorem which yields Since is coterminal with , we get 3. Last, but not least, we have Since is a quadrantal angle, we can ‘see’ the rectangular form by moving out 2 units along the positive real axis, then rotating radians clockwise to arrive at the point 2 units below 0 on the imaginary axis. The long and short of it is that . 24cis2cis6328cis6358cis6558cossin66zwi434zwi55522cis322cis531032cis3w1034354432cossin3316163wii4cis622cis342cis2632cis2zw222ziw Some remarks are in order. 379  First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Example 8.5.1 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out the old-fashioned way. However, Theorem 8.5 pays huge dividends when computing powers of complex numbers. Consider how we computed in Example 8.5.1 and compare that to accomplishing the same feat by expanding . With division being tricky in the best of times, we saved ourselves a lot of time and effort using Theorem 8.5 to find and simplify using their polar forms as opposed to starting with , rationalizing the denominator and so forth.  There is geometric reason for studying these polar forms. Take the product rule, for instance. If and , the formula can be viewed geometrically as a two-step process. The multiplication of by can be interpreted as magnifying1 the distance , from 0 to z, by the factor . Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counter-clockwise.2 Focusing on and from Example 8.5.1, we can arrive at the product by plotting z, doubling its distance from 0 (since ), and rotating radians counter- clockwise. The sequence of diagrams below attempts to describe this process geometrically. 1 Assuming \|w\| > 1. 2 Assuming β > 0. 23213zwii5w513izw23213iiciszzciswwciszwzwzwzw4cis6z22cis3wzw2w23 380 Multiplying z by Rotating counter-clockwise by radians We may visualize division similarly. Here, the formula may be interpreted as shrinking3 the distance from 0 to z by the factor , followed by a clockwise4 rotation of β radians. In the case of and from Example 8.5.1, we arrive at by first halving the distance from 0 to z, then rotating clockwise radians, as we visualize below. Dividing z by Rotating clockwise by Roots of Complex Numbers Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers. 3 Again, assuming \|w\| > 1. 4 Again, assuming β > 0. 2w2Arg3wciszzwww4cis6z22cis3wzw232w2Arg3wImaginary AxisReal Axis i 2i 3i 4i 5i 6i 4cis6z 8cis6zwImaginary AxisReal Axis i 2i 3i 12cis6zw 4cis6zImaginary AxisReal Axis i 2i 3i 4i 5i 6i 8cis6zw 28cis63pizw 23Imaginary AxisReal Axis i i 2i 12cis6zw 22cis63zw 381 Definition. Let z and w be complex numbers. If there is a natural number n such that , then w is an nth root of z. Here, we do not specify one particular principal nth root, hence the use of the indefinite article in defining w as ‘an’ nth root of z. Using this definition, both 4 and –4 are square roots of 16, while means the principal square root of 16, as in . Suppose we wish to find the complex third (cube) roots of 8. Algebraically, we are trying to solve . We know that there is only one real solution to this equation, namely , but if we take the time to rewrite this equation as and factor, we get . The quadratic factor gives two more cube roots, , for a total of three cube roots of 8. Recall from College Algebra that, since the degree of the polynomial is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express in polar form. Since lies 8 units away from the origin on the positive real axis, we get . If we let be a polar form of w, the equation becomes The complex number on the left hand side of the equation corresponds to the point with polar coordinates while the complex number on the right hand side corresponds to the point with polar coordinates . Since by definition , so is , which means and are two polar representations corresponding to the same complex number, with both representations having positive r values. Thus, and for integers k.  Since is a real number, we solve by extracting the principal cube root to get . nwz1616438w382w380w22240www13wi38Pww8z8z8cis0zcisww38w3338cis8cis0cis38cis0www DeMoivre's Theorem3,3w8,00w3w3,3w8,038w302kw38w382w 382  As for , we get for integers k. This produces three distinct points with polar coordinates corresponding to k = 0, 1, and 2: specifically , and . The corresponding complex and rectangular forms are listed in following table. Polar Coordinate Complex Number Rectangular Form The cube roots of 8 can be visualized geometrically in the complex plane, as follows. Keeping the geometric picture in mind throughout the remainder of this section will lead to an interesting observation regarding geometric properties of complex roots. While the process for finding the cube roots of 8 seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of 32. (Try using factoring techniques on that!) If we start with a generic complex number in polar form and solve in the same manner as above, we arrive at the following theorem. 23k2,022,342,32,022,342,302cis0w122cis3w242cis3w02w113wi213wiciszznwzImaginary AxisReal Axis 3i 2i i i 2i 3i 23 43 02w 113wi 213wi 383 Theorem 8.6. The nth Roots of a Complex Number: Let be a complex number with polar form . For each natural number n, z has n distinct nth roots, which we denote by , and they are given by the formula The proof of Theorem 8.6 breaks into two parts: first, showing that each is an nth root, and second, showing that the set consists of n different complex numbers.  To show is an nth root of z, we use DeMoivre’s Theorem to show . Since k is a whole number, and . Hence, it follows that , so , as required.  To show that the formula in Theorem 8.6 generates n distinct numbers, we assume (or else there is nothing to prove) and note that the modulus of each of the is the same, namely . Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments differ by an integer multiple of . Suppose k and j are whole numbers between 0 and , inclusive, with . Then . For this to be an integer multiple of , must be a multiple of n. But because of the restrictions on k and j, . (Think this through.) Hence, is a positive number less than n, so it cannot be a multiple of n. As a result, and are different complex numbers, and we are done. From College Algebra, we know there are at most n distinct solutions to , and we have just found all of them. We illustrate Theorem 8.6 in the next example. 0zciszr011, ,,nwww2cisnkwrknnkw\|0,1,,1kwknkwnkwz2cis2ciscis2nnnknnwrknnrnknnrk DeMoivre's Theorem
cos2cosksin2sinkcis2ciskcisnkwrz2nkwnr21nkj222kjkjnnnnn2kj01kjnkjkwjwnwz 384 Example 8.5.2. Find the following: 1. both square roots of 2. 3. 4. the four fourth roots of the three cube roots of the five fifth roots of Solution. 1. To find both square roots of , we start by writing . To use Theorem 8.6, we identify , and . We know that z has two square roots, and in keeping with the notation in the theorem, we’ll call them and . We get and We can check each of these roots by squaring to get . 2. To find the four fourth roots of , proceeding as above, we write z as . With , and , we get the four fourth roots of z to be Converting these to rectangular form gives , , and . 223zi16z22zi1z223zi22234cis3zi4r232n0w1w002324cis0222cis313kwi Theorem 8.6 with rectangular form112324cis12242cis313kwi Theorem 8.6 with rectangular form223zi16z1616cisz16r4n40414243216cis02cis4442316cis12cis4442516cis22cis4442716cis32cis444wwww022wi122wi222wi322wi 385 3. For finding the cube roots of , we have . With , and the usual computations yield If we were to convert these to rectangular form, we would need to use either sum and difference identities or half-angle identities to evaluate and . Since we are not explicitly told to do so, we leave this as a good, but messy, exercise. 4. To find the five fifth roots of 1, we write . We have , and . Since , the roots are The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of . At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem 8.6, we take a step back to look at things geometrically. Essentially, Theorem 8.6 says that to find the nth roots of a complex number, we first take the nth root of the modulus and divide the argument by n. This gives the first root . Each successive root is found by adding to the argument, which amounts to rotating by radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number 2 in Example 8.5.2. 22zi2cis4z2r43n30331322cis12932cis2cis124172cis12www0w2w11cis01r05n51101234cis012cis54cis56cis58cis5wwwww250w2n0w2n 386 The four fourth roots of equally spaced around the plane We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications. For now, the following exercises will have to suffice. 16z242Imaginary AxisReal Axis 2i i i 2i 0w 1w 2w 3w 387 8.5 Exercises In Exercises 1 – 12, use and to compute the quantity. Express your answers in polar form using the principal argument. These exercises should be worked without the aid of a calculator. 1. 5. 9. 2. 6. 3. 7. 4. 8. 10. 11. 12. In Exercises 13 – 24, use DeMoivre’s Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form. These exercises should be worked without the aid of a calculator. 13. 17. 21. 14. 18. 22. 15. 19. 23. 16. 20. 24. In Exercises 25 – 36, find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. These exercises should be worked without the aid of a calculator. 25. the two square roots of 26. the two square roots of 27. the two square roots of 28. the two square roots of 29. the three cube roots of 30. the three cube roots of 31. the three cube roots of 32. the three cube roots of 33322zi3232wizwzwwz4z3w52zw32zw2zw2wz32zw23wz6wz3223i33i433i43i35522i61322i33322i43133i42222i522i53i81i4zi25zi13zi55322i64z125zzi8zi 388 33. the four fourth roots of 34. the four fourth roots of 35. the six sixth roots of 36. the six sixth roots of 37. Use the four complex fourth roots of –4 to show that the factorization of over the real numbers is . You may want to refer to the Complex Factorization Theorem from College Algebra. 38. Use the 12 complex 12th roots of 4096 to factor over the real numbers, into a product of linear and irreducible quadratic factors. 39. Given any natural number , the n complex nth roots of the number are called the nth Roots of Unity. In the following exercises, assume that n is a fixed, but arbitrary, natural number such that . (a) Show that is an nth root of unity. (b) Show that if both and are nth roots of unity then so is their product . (c) Show that if is an nth root of unity then there exists another nth root of unity such that . Hint: If let . You’ll need to verify that is indeed an nth root of unity. 40. Another way to express the polar form of a complex number is to use the exponential function. For real numbers t, Euler’s Formula defines . (a) Use Theorem 8.5 (Products Powers and Quotients of Complex Numbers in Polar Form) to show that for all real numbers x and y. (b) Use Theorem 8.5 to show that for any real number x and any natural number n. (c) Use Theorem 8.5 to show that for all real numbers x and y. (d) If is the polar form of z, show that where radians. 16z81z64z729z4()4pxx222222pxxxxx124096pxx2n1z2n1wjwkwjkwwjw'jw'1jjwwcisjw'cis2jw'cis2jwcossinitetitixyixiyeeeninxixeeixixyiyeeeciszritzret 389 (e) Show that . (This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics.) (f) Show that and that for all real numbers t. 10iecos2ititeetsin2ititeeti 390 CHAPTER 9 VECTORS AND PARAMETRIC EQUATIONS Chapter Outline 9.1 Vector Properties and Operations 9.2 The Unit Vector and Vector Applications 9.3 The Dot Product 9.4 Sketching Curves Described by Parametric Equations 9.5 Finding Parametric Descriptions for Oriented Curves Introduction Chapter 9 introduces vectors, their many resulting applications, and parametric equations. We begin in Section 9.1 by learning about the geometric representation of vectors along with vector arithmetic, properties and applications involving bearings. Section 9.2 continues with applications by focusing on component forms of vectors in the solution process. The unit vector is introduced along with operations on vectors in an i and j format. In this section, vectors are used to model forces. The focus of Section 9.3 is the dot product – operations, properties and applications. In Section 9.4, parametric equations are defined and graphed. Graph behavior is explored through sketching both x and y as a function of t before displaying the resulting parametric graph in the xy-plane. Section 9.5 follows with the elimination of the parameter t in an effort to transform parametric equations to Cartesian equations. Thus, the correlation between Cartesian and parametric equations is established and further efforts are made to transform Cartesian equations into their parametric form. Throughout Chapter 9, the application of Trigonometry can be seen, providing further evidence of its importance in Mathematics, Engineering, Physics and real-life applications. This chapter is a good stepping off place to conclude the study of Trigonometry before moving on to Calculus. 391 9.1 Vector Properties and Operations Learning Objectives In this section you will:  Interpret vectors and vector operations geometrically.  Perform algebraic operations on vectors, including scalar multiplication, addition and determination of inverses.  Determine the component form of a vector.  Find the magnitude and direction of a vector. As we have seen numerous times in this book, mathematics can be used to model and solve real-world problems. For many applications, real numbers suffice; that is, real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?” There are other times though, when these kinds of quantities do not suffice. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors1. The Geometry of Vectors A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrowhead at one endpoint of the segment. A vector has an initial point, where it begins, and a terminal point, indicated by an arrowhead, where it ends. There are various symbols that distinguish vectors from other quantities:  Lower case type, boldfaced or with an arrow on top, such as , , , , , .2  Given an initial point P and a terminal point Q, a vector can be represented as . The arrow on top is what indicates that it is not just a line, but a directed line segment. Below is a typical vector with endpoints and . The point P is the initial point, or tail, of and the point Q is the terminal point, or head, of . Since we can reconstruct completely from P 1 The word ‘vector’ comes from the Latin vehere meaning to convey or to carry. 2 In this textbook, we will adopt the boldfaced type notation for vectors, without the arrow. In the classroom, your instructor will likely use arrow notation, and arrow notation should be used whenever vectors are written by hand. vuwvuwPQv1,2P4,6Qvvv 392 and Q, we write , where the order of points P (initial point) and Q (terminal point) is important. (Think ab
out this before moving on.) While it is true that P and Q completely determine , it is important to note that since vectors are defined in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as is considered to be the same vector as , regardless of its initial point. In the case of our vector above, any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as . The Component Form of a Vector The notation we use to capture this idea is , the component form of the vector, where the first number, 3, is called the x-component of and the second number, 4, is called the y-component of . If we wanted to reconstruct with initial point then we would find the terminal point of by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point , as seen to the right. 3 If this idea of over and up seems familiar, it should. The slope of the line segment containing v is 4/3. PQvPQvvvvvv3,4vvv3,4v'2,3Pv'1,7Q3,4vxyPQ(4,6)(1,2) vxyP'Q'(1,7)(-2,3)over 3up 4QP(4,6)(1,2)up 4over 3 v v 393 The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in the following definition. Definition. Suppose is represented by a directed line segment with initial point and terminal point . The component form of is given by Example 9.1.1. Consider the vector whose initial point is and terminal point is . Write in component form. Solution. Using the definition of component form, we get Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, if and only if and . (Again, think about this before reading on.) We now set about defining operations on vectors. Vector Addition Suppose we are given two vectors and . The sum, or resultant vector , is obtained geometrically as follows. First, plot . Next, plot so that its initial point is the terminal point of . To plot the vector we begin at the initial point of and end at the terminal point of . It is helpful to think of the vector as the ‘net result’ of moving along then moving along . , and v00,Pxy11,Qxyv1010,PQxxyyv2,3P6,4QPQv62,434,1v1212,','vvvv11'vv22'vvvwvwvwvvwvwvwvwvwvw v w vw 394 Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines.4 Example 9.1.2. A plane leaves an airport with an airspeed of 175 miles per hour at a bearing of N40°E. A 35 mile per hour wind is blowing at a bearing of S60°E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution. For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before in this textbook. We let denote the plane’s velocity and denote the wind’s velocity in the diagram below. The true speed and bearing is found by analyzing the resultant vector, . Before proceeding, it will be helpful to determine the angle , labeled in the diagram to the right. We extend the vectors , with initial point at origin, and with initial point coinciding with the terminal point of , to give us the three lines , and , respectively. From vertical angles, we have . With both and containing the directed line segment , we see that they are parallel lines. We then use corresponding angles of parallel lines to conclude that . 4 If necessary, review the discussion on bearings in the 7.2 Exercises. vwvwvwwv1l2l3l402l3lw60100EN 40 60 v w vwEN 40 60 w w 1l 2l 3l vw v   395 From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of , which is 175, the magnitude of , which is 35, and the magnitude of , which we’ll call . The Law of Cosines gives us This means the true speed of the plane is approximately 184 miles per hour. To find the true bearing of the plane we need to determine the angle . Using the Law of Cosines once more5, we have We use the inverse cosine, along with the value for c from our prior calculation, to find that . Given the geometry of the situation, we add to the given 40° and find the true bearing of the plane to be approximately N51°E. We next define the addition of vectors component-wise to match the geometry.6 Definition. Suppose and . The vector is defined by Example 9.1.3. Let and suppose for and . Find and interpret this sum geometrically. Solution. Before adding the vectors using the definition, we need to write in component form. We get , and then 5 Or, since our given angle, 100°, is obtuse, we could use the Law of Sines without any ambiguity here. 6 Adding vectors component-wise should look hauntingly familiar. Compare this with matrix addition. In fact, in more advanced courses such as Linear Algebra, vectors are defined as 1 by n or n by 1 matrices, depending on the situation. vwvwc22217535217535cos1003185012250cos100184ccc222222351752175cos35175cos2175cccc1112,vvv12,wwwvw1122,vwvwvw3,4vPQw3,7P2,5Qvww23,571,2wEN17535 40 60 100  c 396 To visualize this sum, we draw with its initial point at , for convenience, so that its terminal point is . Next, we graph with its initial point at . Moving one to the right and two down, we find the terminal point of to be . We see that the vector has initial point and terminal point so its component form is as required. In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our definition of vectors to include a zero vector, . Geometrically, represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of since, after all, vectors are supposed to have both a magnitude (length) and a direction. While it seems clear that the magnitude of should be 0, it is not clear what its direction is. As we shall see, the direction of is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap by including in our discussions. We have the following theorem. Theorem 9.1. Properties of Vector Addition.  Commutative Property: For all vectors and , .  Associative Property: For all vectors , and , .   Identity Property: The vector . all vectors , acts as the additive identity for vector addition. That is, for Inverse Property: Every vector , there is a vector every vector has a unique additive inverse, denoted so that . . That is, for The properties in Theorem 9.1 are easily verified using the definition of vector addition. For the commutative property, we note that if and then 3,41,231,424,2vwv0,03,4w3,4w4,2vw0,04,24,2=0,0000000vwvwwvuvwuvwuvw0vv00vvv-vv-vv-v-vv012,vvv12,wwwxy vw w v v 397 Geometrically, we can ‘see’ the commutative property by realizing that the sums and are the same directed diagonal determined by the parallelogram. The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams. The Additive Inverse The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers. Given a vector , suppose we wish to find a vector so that . By the definition of vector addition, we have Hence, and , from which and , with the result that . Hence, has an additive inverse, and moreover it is unique and can be obtained by the formula . Geometrically, the vectors and have the same length but opposite directions. As a result, when adding the vectors geometrically, the sum results in starting at the initial point of and ending back at the initial point of . Or, in other words, the net result of moving then is not moving at all.7 7 An interesting property of a vector and its inverse is that the two vectors are ‘parallel’. In fact, we say two nonzero vectors are parallel when they have the same or opposite directions. That is, v is parallel to w if v=kw for some real, non-zero, number k. 1212112211221212,,,,,,vvwwvwvwwvwvwwvv definition of vector addition commutative property of real number addition definition of vector additionvwwvvwwv12,vvv12,wwwvw01122,0,0vwvwvw0110vw220vw11wv22wv12,vvwv12,vv-v12,vvv12,vv-vv-vvvv-v v v w w vw wv v -v 398 Using the additive inverse of a vector, we can define vector subtraction, or the difference of two vectors, as . If and then In other words, like vector addition, vector subtraction works component-wise. Below, we observe a geometrical interpretation of vector addition and vector subtraction. To interpret the vector geometrically, we note This means that the net result of moving along , then moving along , is just itself. From the diagram, we see that may be interpreted as the vector whose initial point is the terminal point of and whose terminal point is the terminal point of . It is also worth mentioning that in the parallelogram determined by the vectors and , the vector is one of the diagonals, the other being . Scalar Multiplication Next, we discuss scalar multiplication, which is taking a real number times a vector. We define scalar multiplication for vectors in the same way we defined it for matrices. vwv-w12,vvv12,www121211221122,,,,vvwwvwvwvwvwvwv-wvw definition of vector subtraction commutativity of vector addition associativity of vector addition definition of additive inverse wvwwv-ww-wvw-wv0vv definition of additive identitywvwvvwwvvwvwvw v vw v-w -w w v vw v v w w vw 399 Definition. If k is a real number and , we define by Scalar mult
iplication of vectors by k can be understood geometrically as scaling the vector (if ) or scaling the vector and reversing its direction (if ) as demonstrated to the right. Note that, by definition, This and other properties of scalar multiplication are summarized below. Theorem 9.2. Properties of Scalar Multiplication.  Associative Property: For every vector and scalars k and r, .  Identity Property: For all vectors , .  Additive Inverse Property: For all vectors , .  Distributive Property of Scalar Multiplication over Scalar Addition: For every vector and scalars k and r, .  Distributive Property of Scalar Multiplication over Vector Addition: For all vectors . and scalars k, and  Zero Product Property: If is a vector and k is a scalar, then if and only if or . The proof of Theorem 9.2, like the proof of Theorem 9.1, ultimately boils down to the definition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let . If k and r are scalars, then 12,vvvkv1212,,kkvvkvkvv0k0k12121211,1,1,vvvvvvv-vvkrkrvvv1vvv1-vvvkrkrvvvvwkkkvwvwvkv00kv012,vvv 2v 2v v 12v 400 The remaining properties are proved similarly and are left as exercises. Our next example demonstrates how Theorem 9.2 allows us to do the same kind of algebraic manipulations with vectors as we do with variables, multiplication and division of vectors notwithstanding. Example 9.1.4. Solve for . Solution. Vectors in Standard Position A vector whose initial point is is said to be in standard position. If is plotted in standard position, then its terminal point is necessarily . (Once more, think about this before reading on.) 1212121212,,(),(),,krkrvvkrvkrvkrvkrvkrvrvkrvv definition of scalar multiplication associative property of real number multiplication definition of scalar multiplication devkrfinition of scalar multiplicationv521,2vv0v521,2521,25221,2321,232,4 distributive property over vector addition distributive property over scalar addition devv0vv0vv0v0v0++++32,42,42,430,02,432,41132,433241,33finition of scalar multiplication definition of vector addition property of additive identity associative property, scalar multv0v0vvv++24,33iplication property of multiplicative identityv0,012,vvv12,vv 401 in standard position Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector. We can convert the point in rectangular coordinates to a pair in polar coordinates where . The magnitude of , which we said earlier was the length of the directed line segment, is denoted . We can see from the right triangle corresponding to the vector that , for , and we use the Pythagorean Theorem to find . Using right triangle trigonometry, we also find that From the definition of scalar multiplication and vector equality, we get This motivates the following definition. Definition. Suppose is a vector with component form . Let be a polar representation of the point with rectangular coordinates with  The magnitude of  If , denoted , is given by , the direction angle of is given by . Taken together, we get . . . 12,vvv12,vv,r0rvv12,vvvrv0r2212rvv12coscossinsinvrvrvv12,cos,sincos,sinvvvvvvv12,vvv,r12,vv0rvv2212rvvvv0vcos,sinvvvxy 12,vv 12,vvv 0,0xy 1v 2v ,r  v A few remarks are in order. 402  We note that if then even though there are infinitely many angles θ which satisfy the preceding definition, the stipulation means that all of the angles are coterminal. Hence, if and both satisfy the conditions of the definition, then and , and as such, , making well-defined.  If , then , and we know from Section 8.1 that is a polar representation for the origin for any angle θ. For this reason, the direction of is undefined. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem 9.3. Properties of Magnitude and Direction: Suppose is a vector.  Then , and if and only if .  For all scalars k, .  If then The proof of the first property in Theorem 9.3 is a direct consequence of the definition of . If , then which is by definition greater than or equal to 0. Moreover, if and only if , if and only if . Hence, if and only if , as required. The second property is a result of the definition of magnitude and scalar multiplication along with a property of radicals. If and k is a scalar then v00r'coscos'sinsin'cos,sincos',sin'vv00,0v=0,0v0v0vv0kkvvv0cos,sinvvv12,vvv2212vvv22120vv22120vv120vv0v0,0v012,vvv21212221222221222212222122212,,kkkkvvkvkvkvkvkvkvkvvkvvkvv definition of scalar multiplication definition of magnitude product rule for radicals since =vk definition of magnitude v 403 The equation vectors, in Theorem 9.3 is a consequence of the component definition for , and was worked out prior to that definition. In words, the equation says that any given vector is the product of its magnitude and direction, an important concept to keep in mind when studying and using vectors. Example 9.1.5. Find the component form of the vector with so that when is plotted in standard position, it lies in Quadrant II and makes a 60° angle8 with the negative x-axis. Solution. We are told that and are given information about its direction, so we can use the formula to get the component form of . To determine , we are told that lies in Quadrant II and makes a 60° angle with the negative x-axis, so a polar form of the terminal point of , when plotted in standard position, is . (See the diagram below.) Thus, Example 9.1.6. For , find and θ, so that . 8 Due to the utility of vectors in real-world applications, we will usually use degree measure for the angle when giving the vector’s direction. cos,sinvvcos,sinvvvcos,sinvvv5vv5vcos,sinvvvvv5,120cos120,sin120135,22553,22vv3,33vv02cos,sinvvxy 60 120 v 404 Solution. For , we get . We can find the θ we’re after by converting the point with rectangular coordinates to polar form , where . From Section 8.1, we have Since is a point in Quadrant IV, θ is a Quadrant IV angle. Hence, we pick . We may check our answer by verifying that . Example 9.1.7. For the vectors and , find the following: 1. Solution. 2. 1. For , we have . The magnitude of is . Hence, . 2. In the expression , notice that the arithmetic on the vectors comes first, then the magnitude. Hence, our first step is to find the component form of the vector . Then 3,33v223336v3,33,r0rvtan3333yx3,3353553,336cos,sin33v3,4v1,2w2vw2vw3,4v22345v1,2w22125w2522vw2vw2vw23,421,23,42,41,8vw2221,81865vwxy v  3,33 405 9.1 Exercises In Exercises 1 – 3, sketch , and . 1. 2. 3. In Exercises 4 – 6, sketch , and . 5. 4. 6. In Exercises 7 – 8, use the given pair of vectors to compute , and . 7. , 8. In Exercises 9 – 16, use the given pair of vectors to find the following quantities. State whether the result is a vector or a scalar. • • • • • • Finally, verify that the vectors satisfy the Parallelogram Law: . 9. 11. , , 10. 12. , , v3v12v2,1v1,4v3,2vuv+uv2uuvuv23uv2,3u1,5v3,4u2,1vvw2wvvwvwvwwvwv222212vwvwvw12,5v3,4w7,24v5,12w2,1v2,4w10,4v2,5w u v u v u v 406 13. , 14. , 15. , 16. , 17. Given a vector with initial point and terminal point , find an equivalent vector whose initial point is . Write the vector in component form 18. Given a vector with initial point and terminal point initial point is . Write the vector in component form 19. Given a vector with initial point and terminal point initial point is . Write the vector in component form . . . , find an equivalent vector whose , find an equivalent vector whose In Exercises 20 – 34, find the component form of the vector using the information given about its magnitude and direction. Give exact values. 20. ; when drawn in standard position lies in Quadrant I and makes a 60° angle with the positive x-axis. 21. ; when drawn in standard position lies in Quadrant I and makes a 45° angle with the positive x-axis. 22. ; when drawn in standard position lies in Quadrant I and makes a 60° angle with the positive x-axis. 23. ; when drawn in standard position lies along the positive y-axis. 24. ; when drawn in standard position lies in Quadrant II and makes a 30° angle with the negative x-axis. 25. ; when drawn in standard position lies in Quadrant II and makes a 30° angle with the positive y-axis. 26. ; when drawn in standard position lies along the negative x-axis. 3,1v23,2w34,55v43,55w22,22v22,22w13,22v1,3w5,21,30,0,ab4,23,30,0,ab7,11,70,0,abv6vv3vv23vv12vv4vv23vv72vv 407 27. ; when drawn in standard position lies in Quadrant III and makes a 45° angle with the 28. 29. negative x-axis. ; when drawn in standard position lies along the negative y-axis. ; when drawn in standard position lies in Quadrant IV and makes a 30° angle with the positive x-axis. 30. ; when drawn in standard position lies in Quadrant IV and makes a 45° angle with the negative y-axis. 31. ; when drawn in standard position lies in Quadrant I and makes an angle measuring with the positive x-axis. 32. ; when drawn in standard position lies in Quadrant II and makes an angle measuring with the negative x-axis. 33. ; when drawn in standard position lies in Quadrant III and makes an angle measuring with the negative x-axis. 34. ; when drawn in standard position lies in Quadrant IV and makes an angle measuring with the positive x-axis. In Exercises 35 – 40, approximate the component form of the vector using the information given about its magnitude and direction. Round your approximations to two decimal places. 35. 36. 37
. 38. 39. ; when drawn in standard position makes a 117° angle with the positive x-axis. ; when drawn in standard position makes a 78.3° angle with the positive x-axis. ; when drawn in standard position makes a 12° angle with the positive x-axis. ; when drawn in standard position makes a 210.75° angle with the positive x-axis. ; when drawn in standard position makes a 252° angle with the positive x-axis. 56vv6.25vv43vv52vv25vvarctan210vvarctan35vv4arctan326vv5arctan12v392vv63.92vv5280vv450vv168.7vv 408 40. ; when drawn in standard position makes a 304.5° angle with the positive x-axis. In Exercises 41 – 58, for the given vector , find the magnitude and an angle θ with so that (See definition of magnitude and direction.) Round approximations to two decimal places. 41. 44. 47. 50. 53. 56. 42. 45. 48. 51. 54. 57. 43. 46. 49. 52. 55. 58. 59. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S68°W. The river is flowing due east at 8 miles per hour. What is the boat’s true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 60. The HMS Sasquatch leaves port with bearing S20°E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N60°E, find the HMS Sasquatch’s true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 61. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line . Let and let . Let t be any real number. Show that the vector defined by , when drawn in standard position, has its terminal point on the line . (Hint: Show that for any real number t.) Now consider the non-vertical line . Repeat the previous analysis with and let . Thus, any non-vertical line can be thought of as a collection of terminal points of the 26vvvv0360cos,sinvv1,3v5,5v23,2v2,2v22,22v13,22v6,0v2.5,0v0,7v3,4v12,5v4,3v7,24v2,1v2,6v123.4,77.05v965.15,831.6v114.1,42.3v24yx0,40v=1,2st0vvs24yx,24ttt0vs=ymxb0,b0v1,ms 409 vector sum of (the position vector of the y-intercept) and a scalar multiple of the slope vector . 62. Prove the associative and identity properties of vector addition in Theorem 9.1. 63. Prove the properties of scalar multiplication in Theorem 9.2. 0,b1,ms 410 9.2 The Unit Vector and Vector Applications Learning Objectives In this section you will:  Use vectors in component form to solve applications.  Find the unit vector in a given direction.  Perform operations on vectors in terms of i and j.  Use vectors to model forces. Using Vectors in Component Form to Solve Applications We continue our discussion of component forms of vectors from Section 9.1 and resume the process of resolving vectors into their components. This next example revisits Example 9.1.2, making use of component forms and vector algebra to solve this problem. Example 9.2.1. A plane leaves an airport with an airspeed of 175 miles per hour at a bearing of N40°E. A 35 mile per hour wind is blowing at a bearing of S60°E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution. We proceed as we did in Example 9.1.2 and let denote the plane’s velocity and denote the wind’s velocity, and set about determining . If we regard the airport as being at the origin, the positive y-axis as acting as due north and the positive x-axis acting as due east, we see that the vectors and are in standard position and their directions correspond to the angles and , respectively. Hence, the component forms: Since we have no convenient way to express the exact values of cosine and sine of 50°, we leave both vectors in terms of cosines and sines. Adding corresponding components, we find the resultant vector: To find the true speed of the plane, we compute the magnitude of the resultant vector: vwvwvw5030175cos50,sin50175cos50,175sin50v35cos30,sin3035cos30,35sin30w175cos5035cos30,175sin5035sin30vwxy(E)(N) 40 60 v w 50 30 411 Hence, the true speed of the plane is approximately 184 miles per hour. To find the true bearing, we need to find the angle which corresponds to the polar form , , of the point . Since both of these coordinates are positive1, we know is a Quadrant I angle, as depicted below. Furthermore, Using the arctangent function, we get . Since, for the purposes of bearing, we need the angle between and the positive y-axis, we take the complement of and find the true bearing of the plane to be approximately N51°E. The Unit Vector In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. 1 Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that . 22175cos5035cos30175sin5035sin30184vw,r0r,175cos5035cos30,175sin5035sin30xytan175sin5035sin30175cos5035cos30yx39vw455060xy(E)(N) v w vw  412 Definition. Let be a vector. If , we say that is a unit vector. Any nonzero vector divided by its magnitude is a unit vector. If is a nonzero vector, then is a ‘unit vector in the direction of ’. Noting that magnitude is always a scalar, and that dividing by a scalar is the same as multiplying by its reciprocal, a unit vector for any nonzero vector can be found through multiplication by . The process of multiplying a nonzero vector by the reciprocal of its magnitude is called ‘normalizing the vector’. We leave it as an exercise to show that is a unit vector for any nonzero vector . The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take time to show this.) As a result, we visualize normalizing a nonzero vector as shrinking2 its terminal point, when plotted in standard position, back to the Unit Circle. Example 9.2.2. Find a unit vector in the same direction as . Solution. We begin by finding the magnitude. 2 …if the magnitude of v is greater than 1… v1vvvvvvv1v1vvvv5,12v2251216913vxy 1 1 1 1 v vv 413 Next, we divide by . We can check that is indeed a unit vector by verifying that its magnitude is 1. Try it! Note that since a unit vector has length 1, multiplying a unit vector by the magnitude of a vector results in the vector itself, provided the unit vector has the same direction as . (Try this with the unit vector we found in Example 9.2.2.) As a rule of thumb, for any nonzero vector , The Principal Unit Vectors Of all of the unit vectors, two deserve special mention. Definition. The Principal Unit Vectors:  The vector  The vector is defined by is defined by . . 5,12v13v115,1213512,1313vvvv512,1313vvvvvvmagnitude ofunit vector in the direction ofvvv × =i1,0ij0,1jxy11 1,0i 0,1jxy 512,1313vv 5,12v 414 We can think of the vector as representing the positive x-direction while represents the positive y-direction. We have the following ‘decomposition’ theorem.3 Theorem 9.4. Principal Vector Decomposition Theorem: Let be a vector with component form . Then . The proof of Theorem 9.4 is straightforward. Since and , we have from the definition of scalar multiplication and vector addition that Geometrically, the situation looks like this: In Section 9.1, we found the component form of a vector with initial point and terminal point to be . It follows from Theorem 9.4 that may also be written in terms of and as . Example 9.2.3. Given a vector with initial point and terminal point , write the vector in terms of and . 3 We will see a generalization of Theorem 9.4 in Section 9.3. Stay tuned! ijv12,vvv12vvvij1,0i0,1j121212121,00,1,00,,vvvvvvvv definition of i and j scalar multiplication vector additionijv1212,vvvvvijPQ00,Pxy11,Qxy1010,PQxxyyPQij1010PQxxyyijv2,6P6,6Qijxy i j 1vi 2vj 12,vvv Solution. 415 Performing Operations on Vectors in Terms of i and j When vectors are written in terms of and , we carry out addition, subtraction and scalar multiplication by performing operations on corresponding components. Operations on Vectors Written in terms of and : Given and , then    for any scalar k These results can be verified using definitions of addition, subtraction and scalar multiplication from Section 9.1 along with Theorem 9.4. Their verification is left to the student. Example 9.2.4. Use vectors and to find . Solution. Using Vectors to Model Forces We conclude this section with a classic example which demonstrates how vectors are used to model forces. A force is defined as a ‘push’ or a ‘pull’. The intensity of the push or pull is the magnitude of the force, and is measured in Newtons (N) in the SI system or pounds (lbs.) in the English system. The following example should be studied in great detail. 6266812vijijijij12vvvij12wwwij1122vwvwvwij1122vwvwvwij12kkvkvvij42vij3wij3vw33423342312631236195 scalar multiplication vector additionvwijijijijijijijij 416 Example 9.2.5. A 50 pound speaker is suspended from the ceiling by two support braces. If one of them makes a 60° angle with the ceiling and the other makes a 30° angle with the ceiling, what are the tensions on each of the supports? Solution. We first represent the problem schematically. We have three forces acting on the speaker: the weight of the speaker, which we’ll call , pulling the speaker directly downward, and the forces on the support rods, which we’ll
call and (for ‘tensions’) acting upward at angles 60° and 30°, respectively. We provide the corresponding vector diagram below. Note that we have used alternate interior angles to determine the added angle measures in the above diagram. We are looking for the tensions on the supports, which are the magnitudes and . In order for the speaker to remain stationary4, we require . Viewing the common initial point of these vectors as the origin and the dashed line as the x-axis, we find component representations for the three vectors involved. 4 This is the criteria for ‘static equilibrium’. w1T2T1T2T12wTT050 lbs. 60 30 60 30 w 1T 2T 30 60 417  We can model the weight of the speaker as a vector pointing directly downward with a magnitude of 50 pounds. That is, . Since the vector is directed strictly downward, is a unit vector in the direction of . Hence,  For the force in the first support, applying Theorem 9.3, we get  For the second support, we note that the angle 30° is measured from the negative x-axis, so the angle needed to write in component form is 150°. Hence, The requirement gives us 50ww0,1jw500,10,50w11111cos60,sin6013,223,22TTTTT2T22222cos150,sin15031,223,22TTTTT12wTT0xy w 1T 2T 30 60 418 Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables and . From (E1) we get . Substituting into (E2) gives Hence, pounds and pounds. 11221212330,50,,0,0222233,500,02222TTTTTTTT1T2T12123E1 0223E2 50022TTTT123TT222223350022350225TTTTT225T123253TT 419 9.2 Exercises 1. Given initial point and terminal point , write the vector in terms of and . 2. Given initial point and terminal point , write the vector in terms of and . In Exercises 3 – 4, use the vectors , and to find the following. 3. 4. 5. Let . Find a vector that is half the length and points in the same direction as . 6. Let . Find a vector that is twice the length and points in the opposite direction as . In Exercises 7 – 10, find a unit vector in the same direction as the given vector. 7. 8. 9. 10. In Exercises 11 – 12, use the given pair of vectors to find the following quantities. State whether the result is a vector or scalar. • • • • • • 11. , 12. , In Exercises 13 – 15, for the given vector , find the magnitude and an angle θ with so that 13. . Round approximations to two decimal places. 14. 15. 16. Let be any non-zero vector. Show that has length 1. 17. A woman leaves her home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home? 18. If the captain of the HMS Sasquatch wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S20°E from port, in three hours, what speed and heading should she set to take into 13,1P25,2Pvij16,0P21,3Pvij+5uij23vij4wijuvw42vu43vijv52vijv=34aij25bij10cij1532dijvw2wvvwvwvwwvwv34vij2wj12vij12wijvv0360cossinvvij10vjvij4vij12,vvv1vv 420 account an ocean current of 5 miles per hour? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If denotes the velocity of the HMS Sasquatch and denotes the velocity of the current, what does need to be to reach Chupacabra Cove in three hours? 19. In calm air, a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of N8.2°E at 96 miles an hour. (The distance between the airport and the cliffs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree. 20. The SS Bigfoot leaves Yeti Bay on a course of N37°W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S40°E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 21. A 600 pound Sasquatch statue is suspended by two cables from a gymnasium ceiling. If each cable makes a 60° angle with the ceiling, find the tension on each cable. Round your answer to the nearest pound. 22. Two cables are to support an object hanging from a ceiling. If the cables are each to make a 42° angle with the ceiling, and each cable is rated to withstand a maximum tension of 100 pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound. 23. A 300 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 72° angle with the ceiling while the right hand cable makes an 18° angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places. 24. Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N77°E and the other pulls at a heading of S68°E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound. vwvw 421 25. Emboldened by the success of their late night keg pull in Exercise 24 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie ‘Ben-Hur’ by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N80°W, the second points due west and the third points S80°W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds; otherwise the couch won’t move. Does it move? If so, is it heading due west? 422 9.3 The Dot Product Learning Objectives In this section you will:  Find the dot product of two vectors.  Learn properties of the dot product.  Determine the angle between two vectors.  Determine whether or not two vectors are orthogonal.  Solve applications using the dot product. Thus far in Chapter 9, we have learned how to add and subtract vectors and how to multiply vectors by scalars. In this section, we define a product of vectors. Definition and Algebraic Properties of the Dot Product We begin with the following definition. Definition. Suppose The dot product of and and are vectors whose component forms are is given by and . Example 9.3.1. Find the dot product of and . Solution. We have Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity is often called the scalar product of and . The dot product enjoys the following properties. vw12,vvv12,wwwvw12121122,,vvwwvwvwvw3,4v1,2w3,41,231425vwvwvw 423 Theorem 9.5. Properties of the Dot Product:  Commutative Property: For all vectors and , .  Distributive Property: For all vectors , and , .  Scalar Property: For all vectors and , and scalars k, .  Relation to Magnitude: For all vectors , . Like most of the theorems involving vectors, the proof of Theorem 9.5 amounts to using the definition of the dot product and properties of real number arithmetic. To show the commutative property, for instance, we let and . Then The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that , and k is a scalar. Then We leave the proof of as an exercise. For the last property, we note that if , then The following example puts Theorem 9.5 to good use. vwvwwvuvwuvwuvuwvwkkkvwvwvwv2vvv12,vvv12,www1212112211221212,,,,vvwwvwvwwvwvwwvv definition of dot product commutativity of real number multiplication definition of dot productvwwv12,vvv12,www12121212112211221122,,,,kkvvwwkvkvwwkvwkvwkvwkvwkvwvw definition of scalar multiplication definition of dot product associativity of real number multiplication vw1212,,kvvwwk distributive law for real numbers definition of dot productvwkkvwvw12,vvv121222122,,vvvvvv definition of magnitudevvv 424 Example 9.3.2. Prove the identity: . Solution. We begin by using Theorem 9.5 to rewrite in terms of the dot product. Hence, as required. If we take a step back from the pedantry in Example 9.3.2, we see that the bulk of the work is needed to show that . If this looks familiar, it should. Since the dot product enjoys many of the same properties as real numbers, the machinations required to expand for vectors and match those required to expand for real numbers v and w, and hence we get similar looking results. The identity verified in Example 9.3.2 plays a large role in the development of the geometric properties of the dot product, which we now explore. Geometric Properties of the Dot Product Suppose and are two nonzero vectors. If we draw and with the same initial point, we define the angle between and to be the angle determined by the rays containing the vectors and , as illustrated below. We choose to define . The following theorem gives us some insight into the geometric role the dot product plays. 2222vwvvww2vw2 relation to magnitude property distributive propertyvwvwvwv-wvv-w-wvv-w-w11112 commutative property distributive property scalar & commutative propertiesv-wvvv-w-wv-w-wvvvwvwwwvvvw222 relation to magnitude propertywwvvww2222vwvvww
2vwvwvvvwwwvwvwvwvwvwvwvwvwvw000 v w v w  v w  425 Theorem 9.6. Geometric Interpretation of the Dot Product: If and are nonzero vectors then , where is the angle between and . We prove Theorem 9.6 in cases. Case 1: If , then and have the same direction. Thus, the unit vector in the direction of is also the unit vector in the direction of , and we have Thus, we have We note that , from which we get . It follows, from Theorem 9.3, that . Hence, We get This proves the formula holds for . vwcosvwvwvw0vwvwkk unit vectors in directions of and are the same scalar multiplication associative property of scalar multiplication letting wwvwvvwwvwwvvvwvwv2kkkk scalar property of dot productrelation to magnitude propertyvwvvvvv vv0kkkkkkvvvkkk scalar property of dot productvvvvvvvwcos01cos0k since vwvvvwvw0 426 Case 2: If , we repeat the argument with the difference being that , so that . Thus, where . It follows that, , resulting in We have Thus, the formula holds for . Case 3: Next, if , the vectors , and determine a triangle with side lengths , and , respectively, as seen below. The Law of Cosines yields . From Example 9.3.2, we know . Equating these two expressions for gives Thus, , as required. Determining the Angle Between Two Vectors An immediate consequence of Theorem 9.6 is the following. wvwvkwvkwv0kkkkkkvvvwcos1cos since vwvwvw0vwvwvwvw2222cosvwvwvw2222vwvvww2vw22222cos22cos2cosvwvwvvwwvwvwvwvwcosvwvw v w vw   v w vw 427 Theorem 9.7. Let and be nonzero vectors and let be the angle between and . Then We obtain the formula in Theorem 9.7 by solving the equation given in Theorem 9.6 for . Since and are nonzero, so are and . Hence, we may divide both sides of by to get Since by definition, the values of exactly match the range of the arccosine function. Hence, We are long overdue for an example. Example 9.3.3. Find the angle between the following pairs of vectors. 1. 2. 3. and and and Solution. We use the formula from Theorem 9.7 in each case below. 1. For and , we have Then vwvwarccosvwvwvwvwcosvwvwvwcosvwvw0arccosvwvw3,33v3,1w2,2v5,5w3,4v2,1warccosvwvw3,33v3,1w3,333,1333363vw22333366v223142w 428 2. We have and , so that It follows that 3. We find, for and , Then arccos63arccos123arccos256vwvw2,2v5,5w2,25,510100vwarccos0arccosarccos0200 and and v0w0vvwwvwvw3,4v2,1w3,42,1642vw2234255v22215w2arccosarccos5525arccos25vwvw 429 Since isn’t the cosine of one of the common angles, we leave our answer as . Orthogonal Vectors The vectors and are called orthogonal, and we write , because the angle between them is radians, or . Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. In the illustration to the right, and are orthogonal, and we write . We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem 9.8. The Dot Product Detects Orthogonality: Let and be nonzero vectors. Then if and only if . To prove Theorem 9.8, we first assume and are nonzero vectors with . By definition, the angle between and is . By Theorem 9.6, . Conversely, if and are nonzero vectors and , then Theorem 9.7 gives While Theorem 9.8 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. 252525arccos252,2v5,5wvw290vwvwvwvw0vwvwvwvw2cos02vwvwvw0vwarccos0arccosarccos02 thus verifying vwvwwvvw v w 430 Orthogonal Projection Consider the two nonzero vectors and drawn with a common initial point below. For the moment, assume that the angle between and , which we’ll denote , is acute. We wish to develop a formula for the vector , indicated below, which is called the orthogonal projection of onto . The vector is obtained geometrically as follows: drop a perpendicular from the terminal point T of to the vector and call the point of intersection R. The vector is then defined as . Like any vector, is determined by its magnitude and its direction. To determine the magnitude, we observe that We determine the direction of by finding the unit vector in the direction of , which is the same as the unit vector in the direction of ; that is, . It follows that vwOvwpvwpvwpORpppcssoscoco from Theorem 9.6pvpvvwpvvwvwpvwvw,wppwww v w  OTRO  p vTRO v w  ORp 431 Thus, we have a formula for the orthogonal projection of onto for an acute angle . Suppose next that the angle between and is obtuse, and consider the diagram below. In this case we see that , from which Thus, from , we have The unit vector in the direction of is the unit vector in the direction of , which is . We note that reversing the direction of does not affect the magnitude. Finally, 2 magnitude of times unit vector in direction of vwwpwwvwwwpvwpwwwvwvw'cos'coscoscossinsin1cos0sincos difference identity for cosinecos'pvcos'coscoscos'cos since from Theorem 9.6: vwvwpvvvwvvwvwwpwwwwwwTRO w v  ' ORp 432 This formula for orthogonal projection when is obtuse matches the formula for an acute angle . If the angle between and is , then 1. Since in this case, . It follows that Finally, we have the following theorem. Theorem 9.9. If and are nonzero vectors, then the orthogonal projection of onto , denoted , is given by It is time for an example. Example 9.3.4. Let and . Find , and plot , and in standard position. Solution. We find 1 In this case, the point R coincides with the point O, so that . 2 magnitude times directionvwwpwwvwwwwvwwwvw2p0vw0vw2200p0wwwvwwwvwvwprojwv2projwvwvww1,8v1,2wprojwpvvwpOROOp0 433 Hence, . We plot , and below. Suppose we wanted to verify that our answer in Example 9.3.4 is indeed the orthogonal projection of onto . We first note that, since , is a scalar multiple of and so it has the correct direction. It remains to check the orthogonality condition. Consider the vector whose initial point is the terminal point of and whose terminal point is the terminal point of . 222222proj1,81,21,21,211821,2121161,2531,2wvwvwwproj3,6wpvvwppvw3pwpwqpvxy v w pxy v w p q 434 From the definition of vector arithmetic, , so that . In the case of Example 9.3.4, and , so . Then This shows as required. Work We close this section with an application of the dot product. In Physics, if a constant force moves an object a distance d, then the work, W, done by the force is given by the magnitude of the force times the amount of displacement, or , where the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to find the work done. Consider the scenario below where the constant force is applied to move an object from the point P to the point Q. To determine the work W done in this scenario, we find that the magnitude of the force in the direction of is , where is the angle between and . The distance the object travels is . Since work is the magnitude of the force in the direction of times the distance traveled from P to Q, we get We have proved the following. pqvqvp1,8v3,6p1,83,64,2q4,21,2440qvqwFWdFFFPQcosFFPQPQFPQcoscosWPQPQPQ from Theorem 9.6FFFPQ   F F 435 Theorem 9.10. Work as a Dot Product: Suppose a constant force is applied to move an object along the vector , from P to Q. The work W done by is given by where is the angle between and . , Example 9.3.5. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 30° angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a 30° angle for the duration of the 50 feet. Solution. There are two ways to attack this problem.  One way is to find the vectors and mentioned in Theorem 9.10 and compute . To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive x-axis lies along the dashed line in the figure above. Since the force applied is a constant 10 pounds, we have . Since it is being applied at a constant angle of with respect to the positive x-axis, Theorem 9.3 gives us Since the wagon is being pulled along 50 feet in the positive direction, the displacement vector is We get FPQFcosWPQPQFFFPQFPQWPQF10F3010cos30,sin303110,2253,5F50501,050,0PQi 436 Since force is measured in pounds and distance is measured in feet, we get foot- pounds.  Alternately, we can use the formulation to get 53,550,02503WPQF2503WcosWPQF10 pounds50 feetcos303500 foot-pounds22503 foot-pounds of workW 437 9.3 Exercises 1. Given and , calculate . 2. Given and , calculate . 3. Given and , calculate 4. Given and , calculate . . In Exercises 5 – 24, use the give pair of vectors, and , to find the following quantities. • • the angle θ (in degrees) between and 5. 7. 9. 11. 13. 15. 17. 19. and and and and and and and and 21. and 23. and • • 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. (Show that .) and and and and and and and and and and 34uij23vijuvuij5vijuv2,4u3,1vuv1,6u6,1vuvvwvwprojwvvwprojwqvv0qw2,7v5,9w6,5v10,12w1,3v1,3w3,4v6,8w2,1v3,6w33,3v3,1w1,17v1,0w3,4v5,12w4,2v1,5w5,6v4,7w8,3
v2,6w34,91v0,1w3vij4wj247vij2wi3322vijwij512vij34wij13,22v22,22w22,22v13,22w31,22v22,22w13,22v22,22w 438 25. A force of 1500 pounds is required to tow a trailer. Find the work done towing the trailer 300 feet along a flat stretch of road Assume the force is applied in the direction of the motion. 26. Find the work done lifting a 10 pound book 3 feet straight up into the air. Assume the force of gravity is acting straight downwards. 27. Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a 15° angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places. 28. In Exercise 24 in Section 9.2, two drunken college students have filled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a 13° angle with the direction of motion. (In this case, the keg was being pulled due east and the student’s heading was N77°E.) Find the work done by this student if the keg is dragged 42 feet. 29. Find the work done pushing a 200 pound barrel 10 feet up a 12.5° incline. Ignore all forces acting on the barrel except gravity, which acts downwards. Round your answer to two decimal places. HINT: Since you are working to overcome gravity only, the force being applied acts directly upwards. This means that the angle between the applied force in this case and the motion of the object is not the 12.5° of the incline! 30. Prove the distributive property of the dot product in Theorem 9.5. 31. Finish the proof of the scalar property of the dot product in Theorem 9.5. 32. Use the identity in Example 9.3.2, , to prove the Parallelogram Law: 33. We know that for all real numbers x and y by the Triangle Inequality from College Algebra. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that for all pairs of vectors and . a) (Step 1) Show that . 2222vwvvww222212vwvwvwxyxyuvuvuv2222uvuuvv 439 b) (Step 2) Show that . This is the celebrated Cauchy-Schwarz Inequality. (Hint: To show this inequality, start with the fact that and use the fact that for all θ.) c) (Step 3) Show that d) (Step 4) Use Step 3 to show that for all pairs of vectors and . e) As an added bonus, we can now show that the Triangle Inequality holds for all complex numbers z and w as well. Identify the complex number with the vector and identify the complex number with the vector and just follow your nose! uvuv cos uvuvcos122222222222uvuuvvuuvvuuvvuvuvuvuvzwzwzabi,abuwcdi,cdv 440 9.4 Sketching Curves Described by Parametric Equations Learning Objectives In this section you will:  Graph plane curves described by parametric equations.  Analyze behavior in the graphs of parametric equations. As we have seen, most recently in Section 8.3, there are scores of interesting curves which, when plotted in the xy-plane, neither represent y as a function of x nor x as a function of y. In this section, we present a new concept which allows us to use functions to study these kinds of curves. To motivate the idea, we imagine a bug crawling across a table top starting at the point O and tracing out a curve C in the plane, as shown below. The curve C does not represent y as a function of x because it fails the Vertical Line Test and it does not represent x as a function of y because it fails the Horizontal Line Test. However, since the bug can be in only one place at any given time t, we can define the x-coordinate of P as a function of t and the y-coordinate of P as a (usually but not necessarily) different function of t. Curves Described by Parametric Equations The functions describing the curve C, traditionally, use to represent x and to represent y. The independent variable t in this case is called a parameter and the system of equations is called a system of parametric equations or a parametrization of the curve C.1 1 Note the use of the indefinite article ‘a’. As we shall see, there are infinitely many different parametric representations for any given curve. ,Pxyftgtxftygt 441 The parametrization of C endows it with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this case, our bug starts at the point O, travels upwards to the left, then loops back around to cross its path2 at the point Q and finally heads off into the first quadrant. It is important to note that the curve itself is a set of points and as such is devoid of any orientation. The parametrization determines the orientation and as we shall see, different parametrizations can determine different orientations. If all of this seems hauntingly familiar, it should. By definition, the system of equations parametrizes the Unit Circle, giving it a counter-clockwise orientation. It is time for an example. Example 9.4.1. Sketch the curve described by for . Solution. We follow the same procedure here as we have time and time again when asked to graph anything new. We choose friendly values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told , we start there and as we plot successive points, we draw an arrow to indicate the direction of the path for increasing values of t. The curve sketched out in Example 9.4.1 certainly looks like a parabola and the presence of the term in the equation reinforces this hunch. In Section 9.5, we will use the technique of substitution to eliminate the parameter t and get an equation involving just x and y. As we will see, the resulting Cartesian equation describes a parabola with vertex . 2 Here, the bug reaches the point Q at two different times. While this does not contradict our claim that f (t) and g(t) are functions of t, it shows that neither f nor g can be one-to-one. (Think about this before reading on,) cossinxtyt2321xtyt2t2t2t23xt2143yx3,1t23xtt21ytt,xtyt2151,51232,30313,11212,12131,33656,5 442 Graphing Parametric Equations Example 9.4.2. Sketch the curve described by the parametric equations for . Solution. To get a feel for the curve described by the system, we first sketch the graphs of each equation, and , over the interval . , , We note that as t takes on values on the interval , ranges between and 1, and ranges between 0 and 2. This means that all of the action is happening on a portion of the plane, namely . More generally, we see that is increasing over the entire interval whereas is decreasing over the interval and then increasing over . Next, we plot a few points to get a sense of the position and orientation of the curve. 322xtyt11t3xt22yt1,13xt11t22yt11t1,13xt122yt,\|11, 02xyxy3xt1,122yt1,00,1ty 1,2 0,0 1,2t3xtt22ytt,xtyt1121,20000,01121,2txty 443 To trace out the path described by the parametric equations:  We start at , where , then move to the right (since x is increasing) and down (since y is decreasing) to .  We continue to move to the right (since x is still increasing) but now move upwards (since y is now increasing) until we reach , where . for Example 9.4.3. Sketch the curve described by the parametric equations for . Solution. We proceed as in the previous example and graph and over the interval . , , We find that the range of x in this case is and the range of y is . Since t is ranging over the unbounded interval , we take the time to analyze the behavior of both x and y.  As , and as well. This means the graph of the resulting function approaches the point .  Since both and are always decreasing for , we know that our final graph will start at , where , and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. 1,21t0,01,21t322xtyt11t22ttxeye0t2txe2tye0,2txe0t2tye0t0,20,10,t20txe20tye0,02txe2tye0t2,10txytxty 444 Next, we plug in some friendly values of t to get a sense of the orientation of the curve. Since t lies in the exponent here, friendly values of t involve natural logarithms. for Example 9.4.4. Sketch the curve described by the parametric equations for . Solution. We start by graphing and over the interval . , , We find that the range of x is while the range of y is . Before moving on, we take a closer look at these two graphs.  Since and aren’t included in the domain for t, we analyze the behavior of the system as t approaches each of these values. We find that as , and when , we get and . Piecing this information together, we get that for t near 0, and for t near , we have points with very small positive x-values, but very large y-values. 22ttxeye0tsincscxtyt0tsinxtcscyt0,sinxt0tcscyt0t0,11,0tt0ttsin0xtcscytxyt2txte2tyte,xtytln10212,1ln211411,4ln3231921,39tytx 445  As t ranges through the interval , is increasing and is decreasing. This means that we are moving to the right and downwards.  Once , the orientation reverses, and we start to head to the left, since is now decreasing, and up, since is now increasing. We plot a few points before sketching the curve. We combine the above information to first graph the system of equations on the interval , followed by the interval , and finally on the combined interval . for We see that the parametrization given above traces out this portion of the curve twice as t runs through the interval . 0,2sinxtcscyt2tsinxtcscyttsinxttcscytt,xtyt61sin62csc261,222sin12csc121,15651sin625csc
261,220,2,20,02t2tsincscxtyt0t0,xyxyxy 446 Example 9.4.5. Sketch the curve described by the parametric equations for . Solution. Proceeding as above, we set about graphing the system of parametric equations by first graphing and on the interval . , , We see that x ranges from to 4 and y ranges from to 2. The direction of the curve is as follows.  As t ranges from 0 to , x is decreasing while y is increasing, resulting in a movement left and upwards.  For , x is decreasing as is y, so the motion is still right to left but is now downwards.  On the interval , x begins to increase while y continues to decrease. Hence, the motion becomes left to right but continues downwards. Plugging in the values , , and gives the following coordinates. 13cos2sinxtyt302t13cosxt2sinyt30,213cosxt302t2sinyt302t2222t3,20t232,xytytx 447 We put all of our information together to get the following graph. for If this graph looks suspiciously like that of an ellipse, we will find in the next section that this is indeed the case. The next section is devoted to converting between parametric and Cartesian equations. t13cosxtt2sinytt,xtyt013cos042sin004,0213cos122sin221,213cos22sin02,032313cos1232sin221,213cos2sinxtyt302txy432112 1 1 2 448 9.4 Exercises In Exercises 1 – 6, graph each set of parametric equations by making a table of values. Include the orientation on the graph. 1. 3. 5. 2. 4. 6. 21xttytt21xttytttxy3210123txy321012232xttytt223xttytttxy210123txy3210132xttytt23xttytttxy21012txy21012 449 In Exercises 7 – 30, plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. 7. for 8. for 9. for 10. for 11. for 12. for 13. for 14. for 15. for 16. for 17. for 18. for 19. for 21. for 23. for 25. for 27. for 20. 22. 24. 26. 28. for for for for for 29. for 30. for 4362xtyt01t4134xtyt01t22xtyt12t23xtyt05t2132xtytt03t2211xttyt1t2118913xtyt3t3xtytt3xtytt225xtyt05txtyt0tcossinxtyt22t3cos3sinxtyt0t2cos6sinxtyt0t13cos4sinxtyt02t3cos2sin1xtyt22t2cossecxtyt02t2tancotxtyt02tsectanxtyt22tsectanxtyt322ttan2secxtyt22ttan2secxtyt322tcosxtyt0tsinxtyt22t 450 In Exercises 31 – 34, plot the set of parametric equations with the help of a graphing utility. Be sure to indicate the orientation imparted on the curve by the parametrization. 31. for 33. for 32. 34. for for In Exercises 35 – 38, use a graphing utility to view the graph of each of the four sets of parametric equations. Although they look unusual and beautiful, they are so common they have names, as indicated in each exercise. 35. An epicycloid: 36. A hypocycloid: 37. A hypotrochoid: for for for 38. A rose: for 3234xttyt22t334cos4sinxtyt02tttttxeeyee22tcos3sin4xtyt02t14cos()cos1414sinsin14xttytt02t6sin2sin66cos2cos6xttytt02t2sin5cos65cos()2sin(6)xttytt02t5sin2sin5sin2cosxttytt02t 451 9.5 Finding Parametric Descriptions for Oriented Curves Learning Objectives In this section you will:  Eliminate the parameter in a pair of parametric equations.  Parametrize curves given in Cartesian coordinates.  Reverse orientation and shift starting point of a curve described by parametric equations. Now that we have had some good practice sketching the graphs of parametric equations, we turn to the problem of eliminating the parameter in a pair of parametric equations to get an equation involving just x and y. Eliminating the Parameter in Parametric Equations Recall that several curves in the examples from Section 9.4 resembled graphs of functions we’ve seen before. We revisit these examples, eliminating the parameter to determine a Cartesian equation. Example 9.5.1. Eliminate the parameter t in the system of equations from Example 9.4.1 to determine a Cartesian equation. Recall in this example. Solution. We use the technique of substitution to eliminate the parameter in the system of equations. The first step is to solve for t. Substituting this result into the equation yields We see that the graph of this equation is a parabola with vertex which opens to the right. 2321xtyt2t21yt211212ytytyt23xt222132134143yxyxyx3,1 452 Technically speaking, the equation describes the entire parabola, while the parametric equations describe only a portion of the parabola. In this case, we can remedy the situation by restricting the bounds on y. Since the portion of the parabola we want is exactly the part where , the equation coupled with the restriction describes the same curve as the given parametric equations. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. Example 9.5.2. Eliminate the parameter t in the system of equations from Example 9.4.3, where the restriction on t was . Solution. To eliminate the parameter, one way to proceed is to solve for t to get . Substituting this for t in gives The parametrized curve is the portion of the parabola which starts at the point and heads toward, but never reaches, . Example 9.5.3. Eliminate the parameter t in the system of equations from Example 9.4.5, where the restriction on t was . Solution. To eliminate the parameter here, we note that the trigonometric functions involved, namely and , are related by the Pythagorean identity . Hence, we solve 2143yx5y2143yx5y22ttxeye0t2txeln2xt2tye22ln22ln2ln22224xxxyeeexx24xy2,10,013cos2sinxtyt302tcostsint22cossin1ttxy 22 for 0ttxetye 453 for to get and we solve for to get . Substituting these expressions into gives This is the equation of an ellipse centered at with vertices at and , and with a minor axis of length 4. The parametric equations trace out three-quarters of this ellipse in a counter-clockwise direction. We next turn to the problem of finding parametric representations of curves. Parametrizing Curves We start with the following.  To parametrize through I.  To parametrize through I. Parametrizations of Common Curves as x runs through some interval I, let , , and let t run as y runs through the interval I, let , , and let t run  To parametrize a directed line segment with initial point and terminal point , let and for .  To parametrize where and , let and for . (This will impart a counter-clockwise orientation.) The reader is encouraged to verify the above formulas by eliminating the parameter and, when indicated, checking the orientation. We put these formulas to good use in the following examples. Example 9.5.4. Find a parametrization for the curve from to . 13cosxtcost1cos3xt2sinytsintsin2yt22cossin1tt222211321194xyxy1,02,04,0yfxxtyftxgyxgtyt00,xy11,xy010xxxxt010yyyyt01t22221xhykab0a0bcosxhatsinykbt02t2yx3x2xxy432112 1 1 2 13cos3 for 022sinxttyt 454 Solution. Since is written in the form , we let and . Since , the bounds on t match precisely the bounds on x so we get Example 9.5.5. Find a parametrization for the curve . Solution. While we could attempt to solve this equation for y, we don’t need to. We can parametrize by setting so that . Since and there are no bounds placed on y, it follows that there are no bounds placed on t. Our final answer is Example 9.5.6. Find a parametrization for the line segment which starts at and ends at . Solution. We make use of the formulas and for . These formulas can be summarized as  To find the equation for x, we have that the line segment starts at and ends at . This means that the displacement in the x-direction is . Hence, the equation for x is , or .  For y, we note that the line segment starts at and ends at . Thus, the displacement in the y-direction is , so we get . Our final answer is for 2yxyfxxt2yfttxt2 for 32xttyt521xyy521xfyyyyt521xttyt521 for xtttyt2,31,5010xxxxt010yyyyt01tstarting point displacementt2x1x12121xt2xt3y5y53838yt238xtyt01txy43xy 455 line segment line segment for Example 9.5.7. Find a parametrization for the circle . Solution. In order to use the formulas and to parametrize the circle , we first need to put it into the correct form, . The formulas and can be a challenge to memorize, but they come from the Pythagorean identity . By writing the equation as , we identify and . Rearranging these last two equations, we get and . In order to complete one revolution around the circle, we let t range through the interval . Our final answer is for 2xt01t38yt01t238xtyt01t22244xxyycosxhatsinykbt22244xxyy22221xhykab22222222244214441412912199xxyyxxyyxyxycosxhatsinykbt22cossin1tt2212199xy2212133xy1cos3xt2sin3yt13cosxt23sinyt0,213cos23sinxtyt02txy 2,3 1,5tx 0,2 1,1ty 0,3 1,5xy 0; 2tt 456 Example 9.5.8. Find a parametrization for the left half of the ellipse . Solution. In the equation , we can either use the formulas or think back to the Pythagorean identity , along with , to get The normal range on the parameter in this case is , but since we are i

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