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In what year was the rate the highest? x x x 48 x x Section 4.2 Real Zeros 259 40. During the first 150 hours of an experiment, the growth rate of a bacteria population at time t 0.0003t3 0.04t2 0.3t 0.2 g hours is bacteria per hour. a. What is the growth rate at 50 hours? at 100 t 2 1 hours? b. What is the growth rate at 145 hours? What does this mean? c. At what time is the growth rate 0? d. At what time is the growth rate 50 bacteria per hour? e. Approximately at what time does the highest growth rate occur? 41. An open-top reinforced box is to be made from a 12-by-36-inch piece of cardboard by cutting along the marked lines, discarding the shaded pieces, and folding as shown in the figure. If the box must be less than 2.5 inches high, what size squares should be cut from the corners in order for the box to have a volume of 448 cubic inches? 36 x x x x x x 12 cut along fold along 42. A box with a lid is to be made from a 48-by-24- inch piece of cardboard by cutting and folding, as shown in the figure. If the box must be at least 6 inches high, what size squares should be cut from the two corners in order for the box to have a volume of 1000 cubic inches? 24 43. In a sealed chamber where the temperature varies, the instantaneous rate of change of temperature with respect to time over an 11-day period is given by time is measured in days and temperature in degrees Fahrenheit (so that rate of change is in degrees per day). a. At what rate is the temperature changing at the 0.0035t4 0.4t2 0.2t 6, where F t 1 2 beginning of the period the period t 11? 1 1 2 t 0? at the end of 2 b. When is the temperature increasing at a rate of 4°F per day? c. When is the temperature decreasing at a rate of 3°F per day? d. When is the temperature decreasing at the fastest rate? 44. Critical Thinking a. If c is a zero of 5x4 4x3 3x2 4x 5, f x 2 1 show that 1 c is also a zero. b. Do part a with f x 2 1 replaced by g x. 2
1 2x6 3x5 4x4 5x3 4x2 3x 2 x 2 g 1 c. Let a12x12 a11x11 p a2x2 a1x a0. f x 1 2 If c is a zero of f, what conditions must the coefficients ai satisfy so that 1 c is also a zero? 260 Chapter 4 Polynomial and Rational Functions 4.3 Graphs of Polynomial Functions Objectives • Recognize the shape of basic polynomial functions • Describe the graph of a polynomial function • • Identify properties of general polynomial functions: Continuity, End Behavior, Intercepts, Local Extrema, Points of Inflection Identify complete graphs of polynomial functions The graph of a first-degree polynomial function is a straight line, as discussed in Section 1.4. The graph of a second-degree, or quadratic, polynomial function is a parabola, as discussed in Section 3.3. The emphasis in this section is on higher degree polynomial functions. Basic Polynomial Shapes where The simplest polynomial functions are those of the form a is a constant and n is a nonnegative integer. The graphs of polynomial axn, functions of the form are of two basic types. The with different types are determined by whether n is even or odd. n 2, x x f f 2 1 2 1 axn, Polynomial Functions of Odd Degree is odd, When the degree of a polynomial function in the form its graph has the basic form shown is Figures 4.3-1a and 4.3-1b. Notice that the graph shown in Figure 4.3-1b has the same shape as the graph shown in Figure 4.3-1a, but it is the reflection of the Figure 4.3-1a across either the x-axis or the y-axis. x f 2 1 axn, f(x) axn, n odd a 7 0 y a 6 0 900,000 x 0 40,000 21 Figure 4.3-1a Figure 4.3-1b Graphing Exploration Graph each of the following functions of odd degree in the window, and compare each shape with and with those shown in Figure 4.3-1a and 4.3-1b. 30 y 30 5 x 5 • • 2x3 0.01x5 x3 2x
7 Section 4.3 Graphs of Polynomial Functions 261 Polynomial Functions of Even Degree When the degree of a polynomial function in the form is even, its graph has the form shown in Figures 4.3-2a or 4.3-2b. Again, the graph x when a is negative, is the reflection of Figure 4.3-2a across of 2 1 the x-axis. axn, x f f 1 2 axn f(x) axn, n even Figure 4.3-2a Figure 4.3-2b Graphing Exploration Graph each of the following functions of even degree in the window, and compare each shape with and with those shown in Figure 4.3-2a or 4.3-2b. 30 y 30 2x4 6x6 • h • k x x 1 1 2 2 2x2 3x4 Properties of General Polynomial Functions The graphs of other polynomial functions can vary considerably in shape. Understanding the properties that follow should assist you in interpreting graphs correctly and in determining when a graph of a polynomial function is complete. Continuity Every graph of a polynomial function is continuous, that is, it is an unbroken curve, with no jumps, gaps, or holes. Furthermore, graphs of polynomial functions have no sharp corners. Thus, neither of the graphs shown in Figure 4.3-3 represents a polynomial function. Note: some calculator graphs of polynomial functions may appear to have sharp corners; however, zooming in on the area in question will show a smooth curve. 262 Chapter 4 Polynomial and Rational Functions Jump Hole Sharp corner Gap Sharp corner Figure 4.3-3 End Behavior The shape of a polynomial graph at the far left and far right of the cooris large, is called the end behavior of the x dinate plane, that is, when 0 graph. End behavior of graphs of functions of the form have common characteristics when n is odd and when n is even. The Graphing Exploration below asks you to find a generalization about the end behavior of polynomial functions of odd degree. axn x f 1 2 0 Graphing Exploration x Consider the function 2 mined by its leading term f 1 2x3 x2 6x 2x3. g x 1 2 and the function deter- • In a standard viewing window, graph f and g. Describe how the graphs look different and how they
look the same. • In the viewing window 20 x 20 and 10,000 y 10,000, graph f and g. Do the graphs look almost the same? • In the viewing window 100 x 100 and 1,000,000 y 1,000,000, graph f and g. Do the graphs look virtually identical? The reason that the answer to the last question is “yes” can be understood x by observing which term contributes the most to the output value when x is large, as shown in the following chart. f 2 1 Values of Specific Terms of f(x) 2x3 x2 6x x 6x x2 100 600 10,000 50 300 2,500 70 420 4,900 100 600 10,000 g(x) 2x3 2,000,000 250,000 686,000 2,000,000 f(x) 2x3 x2 6x 1,989,400 247,200 690,480 2,009,400 Section 4.3 Graphs of Polynomial Functions 263 The chart shows that when icant compared with f the end behavior of close for large values of x. 2x3, x. 1 2 0 0 x is large, the terms x2 are insignifand they play a very minor role in determining x are relatively Hence, the values of and and x g f 6x 1 2 1 2 End Behavior of Polynomial Functions x When resembles the graph of its highest degree term. is large, the graph of a polynomial function closely 00 00 When a polynomial function has odd degree, one end of its graph shoots upward and the other end downward. When a polynomial function has even degree, both ends of its graph shoot upward or both ends shoot downward. Following are some illustrations of the facts listed in the preceding box. In Figures 4.3-4a–d, the graph of a polynomial function is shown on the left and the graph of its leading term is shown on the right. The end behavior of the graph of the polynomial is the same as the end behavior of the graph of the leading term. Note the degree of each set of graphs and whether the leading coefficient is positive or negative. 2x4 5x2 2 f x 1 2 y 4 2 2x4 g x 1 2 y 4 2 −4 −2 0 2 4 −4 −2 0 2 4 x Figure 4.3-4a 3x6 5x2
2 f x 1 2 3x6 2 −4 −2 4 2 0 −2 2 4 x 2 4 −4 −2 Figure 4.3-4b x x 264 Chapter 4 Polynomial and Rational Functions 0.4x3 x2 2x 3 f x 1 2 y 4 2 0.4x3 g x 1 2 y 4 2 −4 −2 0 2 4 −4 −2 0 2 4 x −2 −4 Figure 4.3-4c 0.6x5 4x2 2 −4 0.6x5 g x 1 2 y 4 2 −4 −2 0 2 4 −4 −2 0 2 4 x −2 −4 −2 −4 x x Figure 4.3-4d Intercepts Consider a polynomial function written in polynomial form. f x anxn an1xn1 p a1x a0 • The y-intercept of the graph of f is the constant term, a0. • The x-intercepts of the graph of f are the real zeros of f. 1 2 The graph of every polynomial function has exactly one y-intercept, and because a polynomial of degree n has at most n distinct zeros, the number of x-intercepts is limited. Intercepts The graph of a polynomial function of degree n • has one y-intercept, which is equal to the constant term. • has at most n x-intercepts. Section 4.3 Graphs of Polynomial Functions 265 That is, the number of x-intercepts can be no greater than the degree of the polynomial function. Multiplicity There is another connection between zeros and graphs. If is a factor that occurs m times in the complete factorization of a polynomial expression, then r is called a zero with multiplicity m of the related polynomial function. x r For example, and 1 are zeros of multiplicity of each zero is shown in the following chart, 1 x 1 3. The 2 21 Zero 3 1 Multiplicity 2 1 1 3 Observe in Figure 4.3-5 that the graph of f does not cross the x-axis at a zero of even multiplicity, but does cross the x-axis at odd multiplicity. 3, and 1, zeros of 1 Let c be a zero of multiplicity k of a polynomial f. • If k is odd,
the graph of f crosses the x-axis at c. • If k is even, the graph of f touches, but does not cross, the x-axis at c. Example 1 Multiplicity of Zeros x 2 Find all zeros of State the multiplicity of each zero, and state whether the graph of f touches or crosses the x-axis at each corresponding x-intercept. x 3 x 1 3. 21 x f 1 1 2 2 1 2 2 Solution The following chart lists the zeros of f, the multiplicity of each, and whether the graph touches or crosses the x-axis at the corresponding x-intercept. Zero Multiplicity x-axis 1 2 3 2 1 3 touches crosses crosses The graph of f, shown in Figure 4.3-6, verifies that the graph touches but does not cross the x-axis at and crosses the x-axis at 2 and 3. 1 ■ y 30 20 10 −4 −2 0 2 x −10 Figure 4.3-5 Multiplicity and Graphs y 60 40 20 −2 0 2 4 x Figure 4.3-6 266 Chapter 4 Polynomial and Rational Functions Local Extrema The term local extremum (plural, extrema) refers to either a local maximum or a local minimum, that is, a point where the graph has a peak or a valley. Local extrema occur when the output values change from increasing to decreasing, or vice versa, as discussed in Section 3.2. Graphing Exploration • Graph f x 0.5x5 1.5x4 2.5x3 7.5x2 2x 5 the standard viewing window. What is the total number of local extrema on the graph? What is the degree of f? in 2 1 • Graph x g x4 3x3 2x2 4x 5 in the standard viewing window. What is the total number of local extrema on the graph? What is the degree of x g? 1 2 1 2 The two polynomials graphed in the Exploration are illustrations of the following fact. Number of Local Extrema A polynomial function of degree n has at most extrema. n 1 local That is, that total number of peaks and valleys on the graph is at most one less than the degree of the function. Points of Inflection Recall from Section 3.2 that an inflection point occurs where the concavity of a graph of a
function changes. The number of inflection points on the graph of a polynomial is governed by the degree of the function. Number of Points of Inflection • The graph of a polynomial function of degree n, with n 2, has at most n 2 points of inflection. • The graph of a polynomial function of odd degree, with n 7 2, has at least one point of inflection. Thus, the graph of a quadratic function, which has degree 2, has no points The graph of of inflection because it can have at most a cubic has exactly one point of inflection because it has at least 1 and at most 3 2 1. n 2 2 2 0. Technology Tip Points of inflection may be found by using INFLC in the TI-86/89 GRAPH MATH menu. Section 4.3 Graphs of Polynomial Functions 267 Complete Graphs of Polynomial Function By using the facts discussed in this section, you can often determine whether or not the graph of a polynomial function is complete, that is, shows all the important features. Example 2 A Complete Graph of a Polynomial Find a complete graph of f x 2 1 x4 10x3 21x2 40x 80. 100 Solution Because the y-intercept is 100 y 100, and as shown in Figure 4.3-7. 80, graph f in the window with 10 x 10 10 10 The three peaks and valleys shown are the only local extrema because a fourth-degree polynomial graph has at most three local extrema. 100 Figure 4.3-7 There cannot be more x-intercepts than the two shown because if the graph turned toward the x-axis farther to the right or farther to the left, there would be an additional peak, which is impossible. Finally, the end behavior of the graph resembles the graph of highest degree term. y x4, the Figure 4.3-7 includes all the important features of the graph and is therefore complete. ■ Example 3 A Complete Graph of a Polynomial Find a complete graph of f x3 1.8x2 x 2. x 1 2 Solution The graph of f, shown in Figure 4.3-8a on the next page, is similar to the graph of its leading term but it does not appear to have any local extrema. However, if you use the trace feature on the flat portion of the graph to the right of the y
-axis, you should see that the y-coordinates increase, then decrease, and then increase again. y x3, Zoom in on the portion of the graph between 0 and 1, as shown in Figure 4.3-8b. Observe that the graph actually has two local extrema, one peak and one valley, which is the maximum possible number of local extrema for a cubic function. Figures 4.3-8a and 4.3-8b together provide a complete graph of f. NOTE No polynomial graph of a function of degree horizontal line segments like those shown in Figure 4.3-8a. Always investigate such segments by using trace or zoom-in to determine any hidden behavior. contains n 7 1 268 Chapter 4 Polynomial and Rational Functions 6 2.2 6 6 6 Figure 4.3-8a 0 2 Figure 4.3-8b 1 ■ Example 4 A Complete Graph of a Polynomial Determine if the graph shown in Figure 4.3-9a is a complete graph of 0.01x5 x4 x3 6x2 5x 4. f x 1 2 Solution The graph shown in Figure 4.3-9a cannot be a complete graph because, 0.01x5, when whose left end goes downward. is large, the graph of f must resemble the graph of x g x 2 1 0 0 So the graph of f must turn downward and cross the x-axis somewhere to the left of the origin. Therefore, the graph must have one more peak, where the graph turns downward, and must have another x-intercept. One additional peak and the ones shown in Figure 4.3-9a make a total of four, the maximum possible for a polynomial of degree 5. Similarly, the additional x-intercept makes a total of 5 x-intercepts. Because f has degree 5, there are no other x-intercepts. A viewing window that includes the local maximum and the x-intercept shown in Figure 4.3-9b will not display the local extrema and x-intercepts shown in Figure 4.3-9a. Consequently, a complete graph of f requires both Figure 4.3-9a and Figure 4.3-9b to illustrate the important features of the graph. 10 10,000,000 10 10 115 10 Figure 4.3-9a 5,000,000 Figure 4.3-9
b 50 ■ The graphs shown in Examples 2–4 were known to be complete because they included the maximum possible number of local extrema. Many graphs, however, may have fewer than the maximum number of possible peaks and valleys. In such cases, use any available information and try several viewing windows to obtain the most complete graph. Section 4.3 Graphs of Polynomial Functions 269 Exercises 4.3 In Exercises 1–6, decide whether the given graph could possibly be the graph of a polynomial function. 5. 1. 2. 3. 4. In Exercises 7–12, determine whether the given graph could possibly be the graph of a polynomial function of degree 3, degree 4, or degree 5. 7. 8. 9. y y y x x x 270 10. 11. 12. Chapter 4 Polynomial and Rational Functions y y y x x x 16. 4 −5 −5 −5 17. 18. 5 5 5 −14 10 −5 20 −4 In Exercises 13 and 14, find a viewing window in which the graph of the given polynomial function f appears to have the same general shape as the graph of its leading term. 13. 14. f f x x 1 1 2 2 x4 6x3 9x2 3 x3 5x2 4x 2 In Exercises 15–18, the graph of a polynomial function is shown. List each zero of the polynomial and state whether its multiplicity is even or odd. 15. −5 10 −6 5 In Exercises 19–24, use your knowledge of polynomial graphs, not a calculator, to match the given function with one of graphs a–f. a. b. y y x x c. y d. y e. f. y y x x x x 19. f 21. g 23. f 24 2x 3 x3 4x 20. g 22. f x x 1 1 2 2 x2 4x 7 x4 5x2 4 x4 6x3 9x2 2 2x2 3x 1 In Exercises 25–28, graph the function in the standard viewing window and explain why that graph cannot possibly be complete. 25. f 26. g 27.01x3 0.2x2 0.4x 7 0.01x4 0.1x3 0.8x2 0.7x 9 0.005x4
x2 5 Section 4.3 Graphs of Polynomial Functions 271 28. f x 1 2 0.001x5 0.01x4 0.2x3 x2 x 5 In Exercises 29–34, find a single viewing window that shows a complete graph of the function. 29. 30. 31. 32. 33. 34 x3 8x2 5x 14 x3 3x2 4x 5 x4 3x3 24x2 80x 15 x4 10x3 35x2 50x 24 2x5 3.5x4 10x3 5x2 12x 6 x5 8x4 20x3 9x2 27x 7 In Exercises 35–40, find a complete graph of the function and list the viewing window(s) that show(s) this graph. 35. 36. 37. 38 39. g 40. f x x 1 1 2 2 0.1x5 3x4 4x3 11x2 3x 2 x4 48x3 101x2 49x 50 0.03x3 1.5x2 200x 5 0.25x6 0.25x5 35x4 7x3 823x2 25x 2750 2x3 0.33x2 0.006x 5 0.3x5 2x4 7x3 2x2 41. a. Explain why the graph of a cubic polynomial function has either two local extrema or none at all. Hint: If it had only one, what would the x graph look like when is very large? 0 b. Explain why the general shape of the graph of a cubic polynomial function must be one of the following. 0 a. b. c. d. 42. The figure shows an incomplete graph of an even polynomial function f of fourth degree. (Even functions were defined in Excursion 3.4.A.) a. Find the zeros of f. b. Explain why f x x c where a, b, c, d are the zeros of f. x b x a k 21 21 1 2 1 x d 2 21 272 Chapter 4 Polynomial and Rational Functions c. Experiment with your calculator to find the value of k that produces the graph in the figure. d. Find all local extrema of f. e. List the approximate intervals on which f is increasing and those on which it is decreasing. 45. The figure below is
a partial view of the graph of a cubic polynomial whose leading coefficient is negative. Which of the patterns shown in Exercise 41 does this graph have? −10 20 −10 10 43. A complete graph of a polynomial function g is 46. The figure below is a partial view of the graph of a fourth-degree polynomial. Sketch the general shape of the graph and state whether the leading coefficient is positive or negative. shown below. a. Is the degree of b. Is the leading coefficient of even or odd? g x x g 2 1 positive or 1 2 negative? c. What are the real zeros of d. What is the smallest possible degree of 15 10 5 −5 −4 −2 −10 −15 x 2 4 6 44. Do Exercise 43 for the polynomial function g whose complete graph is shown here. y 8 4 −4 −8 −3 −2 −1 −12 x 1 2 3 In Exercises 47–56, sketch a complete graph of the function. Label each x-intercept and the coordinates of each local extremum; find intercepts and coordinates exactly when possible, otherwise approximate them. 47. f 49. h 50 51. g x 1 2 52. h 53. f x x 1 1 2 2 54. g x 1 2 55. h 56. f x x 2 2 1 1 x3 3x2 4 48. g x 2 1 4x 4x3 3 0.25x4 2x3 4x2 0.25x4 2x3 3 3x3 18.5x2 4.5x 45 2x3 x2 4x 2 x5 3x3 x 1 0.25x4 x2 0.5 8x4 22.8x3 50.6x2 94.8x 138.6 32x6 48x4 18x2 1 g 57. Critical Thinking a. Graph x 2 the viewing window with 0 y 6 coincide with the horizontal line 1 0.01x3 0.06x2 0.12x 3.92 3 x 3 and and verify that the graph appears to between y 4 in x 3. and x 1 every x with equation In other words, it appears that is a solution of the 1 x 3 0.01x3 0.06x2 0.12x 3.92 4. Explain why this is impossible. Conclude that the actual graph is not horizontal between and x 3. b. Use the
trace feature to verify that the graph is x 1 Find a viewing window that shows actually rising from left to right between and this. x 3. x 1 c. Show that it is not possible for the graph of a 1 2 f x to contain a horizontal polynomial segment. Hint: A horizontal line segment is part of the horizontal line for some constant k. Adapt the argument in part a, k 4. which is the case y k 58. Critical Thinking 2 1 f x x be a polynomial of odd degree. a. Let f Explain why Hint: Why must the graph of f cross the x-axis, and what does this mean? b. Let be a polynomial of even degree, with a must have at least one real zero. g x 1 2 negative leading coefficient and a positive 1 2 Section 4.3.A Excursion: Polynomial Models 273 constant term. Explain why least one positive and at least one negative zero. must have at g x 2 1 59. Critical Thinking The graph of x2 20 x 18 x 2 1 2 1 f 21 x 2 x 10 2 has x-intercepts at each of its zeros, that is, at x 18, 2, and 10. It is also true x 2. f x that a. Draw the x-axis and mark the zeros of ± 120 ± 4.472, has a relative minimum at 21 x f. 1 1 2 2 1 2 Then use the fact that to sketch the general shape of the graph, as was done for cubics in Exercise 41. has degree 6 (Why?) x f 2 1 x 1 2 b. Now graph f in the standard viewing window. Does the graph resemble your sketch? Does it even show all the x-intercepts between 10 c. Graph and 10? x f 19 x 11 window include all the x-intercepts, as it should? in the viewing window with 10 y 10. Does this and 2 1 d. List viewing windows that give a complete graph of f x. 2 1 4.3.A Excursion: Polynomial Models Objectives • Fit a polynomial model to data Linear regression was used in Section 1.5 to construct a linear function that modeled a set of data points. When the scatter plot of the data points looks more like a higher degree polynomial graph than a straight line, similar least squares regression procedures are available on most calculators for constructing quadratic, cubic, and
quartic polynomial functions to model the data. Example 1 A Polynomial Model The following data, which is based on statistics from the Department of Health and Human Services, gives the cumulative number of reported cases of AIDS in the United States from 1982 through 2000. Find a quadratic, a cubic, and a quartic regression equation and determine which equation best models the data. 274 Chapter 4 Polynomial and Rational Functions Technology Tip Quadratic, cubic, and quartic regression are denoted by QuadReg, CubicReg, QuartReg in the CALC submenus of the TI STAT menu and by x 3, submenu of the Casio STAT menu. x 2, in the CALC REG x 4 900,000 0 40,000 21 Figure 4.3.A-1b Year 1982 1984 1986 1988 1990 Cases 1563 10,845 41,662 105,489 188,872 Year 1991 1992 1993 1994 1995 Cases 232,383 278,189 380,601 457,789 528,421 Year 1996 1997 1998 1999 2000 Cases 595,559 652,439 698,527 743,418 784,518 Solution x 0 Let correspond to 1980 and plot the data points (2, 1563), (4, 10845), etc., to obtain the scatter plot shown in Figure 4.3.A-1a. The points are not in a straight line, but could be part of a polynomial graph of degree 2 or more. 900,000 0 40,000 Figure 4.3.A-1a 21 Using the same procedure as for linear regression, find a quadratic, a cubic, and a quartic regression equation for the data. See the Technology Tip in the margin on this page for the specific calculator procedure needed. The polynomial functions shown below have rounded coefficients, but the graph in Figure 4.3.A-1b shows the data points along with the quadratic regression equation and was produced using full coefficients. 1758.0x2 9893.3x 59,024.3 219.18x3 9111.66x2 59,991.75x 103,255.32 20.29x4 681.94x3 4318.57x2 15,550.81x 17,877.25 Quadratic Cubic Quartic The graphs of f, g, and h are virtually identical in the viewing window shown. Although any one of f, g,
or h provides a reasonable model for the given data, knowledge of polynomial graphs suggests that the cubic and quartic models, should not be used for predicting future results. As x gets y 219.2x3 larger, the graphs of g and h will resemble respectively those of and, which turn downward. However, the cumulative number of cases cannot decrease because even when there are no new cases, the cumulative total stays the same. y 20.3x4 Graphing Exploration Graph the functions f, g, and h in the window with and 0 y 1,800,000. In this window, can you distinguish the graphs of f and g? Assuming no medical breakthroughs or changes in the current social situation, does the graph of f seem to be a plausible model for the next few years? What about the graph of g? 0 x 27 ■ Section 4.3.A Excursion: Polynomial Models 275 Example 2 Estimating Data Values The population of San Francisco in selected years is given in the table. Year 1950 1960 1970 1980 1990 2000 Population 775,357 740,316 715,674 678,974 723,959 776,733 [Source: U.S. Census Bureau] x 0 Let correspond to 1950. Find a polynomial regression model that is a reasonably good fit and estimate the population of San Francisco in 1995 and in 2004. 900,000 Solution The scatter plot of the data shown in Figure 4.3.A-2a suggests a parabola. However, the data points do not climb quite so steeply in the later years, so a higher-degree polynomial graph might fit the data better. The quadratic, cubic, and quartic regression models are shown below. 5 0 Figure 4.3.A-2a 60 128.14x2 6632.39x 783,517.18 2.48x3 58.10x2 3230.49x 776,067.76 0.11x4 13.20x3 387.24x2 168.65x 774,230.65 Quadratic Cubic Quartic 900,000 900,000 900,000 5 0 60 5 0 60 5 0 60 Quadratic Cubic Quartic Figure 4.3.A-2b The quartic appears to be the best fitting function. Use h to estimate the population in 1995 and 2004 by finding h(45) and h(54). 45 h
1 2 745,843.98 and h 803,155.18 54 2 1 That is, the estimated population of San Francisco in 1995 was approximately 745,844 and the estimated population of San Francisco in 2004 was approximately 803,155. ■ In Example 2, a model may not be accurate when applied outside the 5,817,115, range of points used to construct it. For instance, suggesting that the population of San Francisco in 1776 was about 5,817,115. 174 f 1 2 276 Chapter 4 Polynomial and Rational Functions NOTE The following table lists the minimum number of data points required for polynomial regression. In each case, the minimum number of data points required is the number of coefficients in the polynomial function modeling the data. If you have exactly the required minimum number of data points, no two of them can have the same first coordinate. Model Minimum number of data points Quadratic Regression Cubic Regression Quartic Regression 3 4 5 When using the minimum number of data points, the polynomial regression function will pass through all the data points and will fit exactly. When using more than the minimum number of data points required, the fit will generally be approximate rather than exact. Exercises 4.3.A In Exercises 1–4, a scatter plot of data is shown. State the type of polynomial model that seems most appropriate for the data (linear, quadratic, cubic, or quartic). If none of them is likely to provide a reasonable model, say so. 4. y x 1. y 2. y 3. y x x x 5. The table, which is based on the United States FBI Uniform Crime Report, shows the rate of property crimes per 100,000 population. Year 1982 1984 1986 1988 1990 1992 Crimes 5032.5 4492.1 4862.6 5027.1 5088.5 4902.7 Year 1994 1996 1997 1998 1999 2000 Crimes 4660.0 4450.1 4318.7 4051.8 3743.6 3617.9 a. Use cubic regression to find a polynomial x 0 function that models this data, with corresponding to 1980. b. According to this model, what was the property crime rate in 1987 and 1995? c. The actual crime rate was about 3698 in 2001. 8. The table, which is based on data from the Section 4.3.A Excursion: Poly
nomial Models 277 What does the model predict? d. Is this model a reasonable one? 6. The table, which is based on the U.S. National Center for Educational Statistics, shows actual and projected enrollment (in millions) in public high schools in selected years. Year 1975 1980 1985 1990 Enrollment 14.3 13.2 12.4 11.3 Year 1995 2000 2005 2010 Enrollment 12.5 13.5 14.4 14.1 a. Use quartic regression to find a polynomial function that models this data, with corresponding to 1975. x 0 b. According to this model, what was the enrollment in 1998 and 1999? c. According to the model, in what year between 1975 and 2000 was enrollment at its lowest level? d. Does this estimate appear to be accurate? 7. The table shows the air temperature at various times during a spring day in Gainesville, Florida. Association of Departments of Foreign Languages, shows the fall enrollment (in thousands) in college level Spanish classes. Year 1970 1974 1977 1980 1983 Enrollment 389.2 362.2 376.7 379.4 386.2 Year 1986 1990 1995 1998 Enrollment 411.3 533.9 606.3 656.6 a. Sketch a scatter plot of the data, with x 0 corresponding to 1970. b. Find a cubic polynomial model for this data. Use the following table for Exercises 9–10. It shows the median income of U.S. households in 1999 dollars. Year 1985 1987 1989 1991 Median Income $36,568 38,220 38,836 36,850 Year 1993 1995 1997 1999 Median Income $36,019 037,251 038,411 040,816 Time 6 a.m. 7 a.m. 8 a.m. 9 a.m. 10 a.m. 11 a.m. noon (F) Temp 52 56 61 67 72 77 80 Time 1 p.m. 2 p.m. 3 p.m. 4 p.m. 5 p.m. 6 p.m. Temp 82 86 85 83 78 72 (F°) [Source: U.S. Census Bureau] 9. a. Sketch a scatter plot of the data from 1985 to 1999, with x 0 corresponding to 1985. b. Decide whether a quadratic or quartic model seems more appropriate. c. Find an appropriate polynomial model. a. Sketch a scatter plot of the data
, with x 0 corresponding to midnight. b. Find a quadratic polynomial model for the data. c. What is the predicted temperature for noon? for 9 a.m.? for 2 p.m.? 278 Chapter 4 Polynomial and Rational Functions d. Use the model to predict the median income in a. Sketch a scatter plot of the data with x 0 2002. e. Does this model seem reasonable after 2002? 10. a. Sketch a scatter plot of the data from 1989 to 1999, with x 0 corresponding to 1989. b. Find both a cubic and a quartic model for this data. c. Is there any significant difference between the models from 1989 to 1999? What about from 1999 to 2005? d. According to these models, when will the median income reach $45,000? 11. The table shows the U.S. public debt per person, in dollars, in selected years. Year 1981 1983 1985 1987 1989 1991 Debt $4,338 5,870 7,598 9,615 11,545 14,436 Year 1993 1995 1997 1999 2001 Debt $17,105 18,930 20,026 20,746 20,353 [Source: U.S. Department of Treasury, Bureau of Public Debt] corresponding to 1980. b. Find a quartic model for the data. c. Use the model to estimate the public debt per person in 1996. How does your estimate compare with the actual figure of $19,805? 12. The table shows the total advertising expenditures, in billions of dollars, in selected years. Year 1990 1992 1994 Expenditures $129.59 132.65 151.68 Year 1996 1998 2000 Expenditures $175.23 201.59 236.33 [Source: Statistical abstract of the United States 2001] a. Sketch a scatter plot of the data with x 0 corresponding to 1990. b. Find a quadratic model for the data. c. Use the model to estimate expenditures in 1995 and 2002. d. If this model remains accurate, when will expenditures reach $350 billion? 4.4 Rational Functions Objectives • Find the domain of a rational function • Find intercepts, vertical Recall that a polynomial is an algebraic expression that can be written as anxn an1xn1 p a2x2 a1x a0 where n is a nonnegative integer. asymptotes, and horizontal asymptotes A rational function is a function whose rule is the quotient of two polynomials
, such as • Identify holes • Describe end behavior • Sketch complete graphs 1 x f x 2 1 4x 3 2x 1 x t 1 2 2x3 5x 2 x2 7x 6 k x 1 2 Section 4.4 Rational Functions 279 Although a polynomial function is defined for every real number x, a rational function is defined only when its denominator is nonzero. Domain of Rational Functions The domain of a rational function is the set of all real numbers that are not zeros of its denominator. Example 1 The Domain of a Rational Function Find the domain of each rational function. a. f x 1 2 1 x2 Solution b. g x 1 2 x2 3x 1 x2 x 6 a. The domain of f x is the set of all real numbers except x 0, 1 x2 2 1 because the denominator is 0 when undefined. x 0, making the fraction x2 3x 1 x2 x 6 x2 x 6 0. b. The domain of g x 2 1 is the set of all real numbers except the solutions of into 1 x 3. except x 2 x 2 21 Therefore, the domain of g is the set of all real numbers x 2 Because x2 x 6 0 the solutions to x2 x 6 are x 3. x 3 and, 2 factors and ■ Properties of Rational Graphs Because calculators often do a poor job of graphing rational functions, the emphasis in this section is on the algebraic analysis of rational functions. Such analysis should enable you to interpret misleading screen images. Intercepts As with any function, the y-intercept of the graph of a rational function f The x-intercepts of the occurs at graph of a rational function occur when its numerator is 0 and its denominator is nonzero. provided that f is defined at x 0. 0 f, 2 1 Intercepts of Rational Functions If f has a y-intercept, it occurs at f(0). The x-intercepts of the graph of a rational function occur at the numbers that • are zeros of the numerator • are not zeros of the denominator 280 Chapter 4 Polynomial and Rational Functions Locating the intercepts can help you determine if you correctly entered the parentheses when graphing a rational function on a graphing calculator. Example 2 Intercepts of a Rational Graph Find the intercepts of f x2 x 2 x 1. x 1 2 y 16 8 (0, 2) (−1, 0
) 4 (2, 0) −8 −4 0 −8 −16 Figure 4.4-1 x 8 Solution The y-intercept is f 0 1 2 02 0 2 0 1 2. 2 1 x2 x 2 0 The x-intercepts are solutions of x 1 0. Solutions of x2 x 2 0 that are not solutions of can be found by factoring. x 1 x2 x 2 0 x 2 0 1 x 1 or x 2. x 1 0, 21 2 1 Neither graph of f, as shown in Figure 4.4-1. nor 2 is a solution of so both are x-intercepts of the ■ Continuity There are breaks in the graph of a rational function wherever the function is not defined, that is, at the zeros of the denominator. Except for breaks, the graph is a continuous unbroken curve. Additionally, the graph has no sharp corners. Vertical Asymptotes Unlike polynomial functions, a rational function has breaks in its graph at all points where the function is not defined. Vertical asymptotes occur at every number that is a zero of the denominator but not of the numerator. The key to understanding the behavior of a rational function near these asymptotes is a fact from arithmetic. The Big-Little Concept If c is a number far from 0, then 1 c is a number close to 0. If c is close to 0, then 1 c is far from 0. In less precise, but more suggestive terms 1 big little and 1 little big Section 4.4 Rational Functions 281 For example, 5000 is big and 1 5000 is little. Similarly, 1 1000 is little and 1000 is big. Note that even though 1000 is negative, it is far 1 1 1000 from zero and therefore is large in absolute value. The role played by the Big-Little Concept when graphing rational functions is illustrated in Example 3. Example 3 A Rational Function Near a Vertical Asymptote Without using a calculator, describe the graph of x 2. x 2. Then sketch the graph for values near x 1 2x 4 x f 1 2 near Solution because the denominator is 0 there. The function is not defined at When x is greater than 2 but very close to 2, x 1, • The numerator, is very close to x 2 • The denominator, 4 0. 2 2 1 2 2x 4, 2 1 3. is a positive number very close to Figure 4.4-2
a y 10 5 0 −5 −10 x 1 2 3 4 Figure 4.4-2b By the Big-Little Concept, x 1 2x 4 3 little f x 2 1 3 1 little 3 big 1 2 very big when This fact can be confirmed by a table of values for x 2.01, 2.001, 2.0001, etc., as shown in Figure 4.4-2a. In graphical terms, the points with x-coordinates slightly greater than 2 have very large x 2. y-coordinates, so the graph shoots upward just to the right of That is, near x f 2 1 x 2 f increases without bound as x approaches 2 from the right. x 1, A similar analysis when x is less than 2 but very close to 2 shows that the numerator, is very close to 3 and the denominator is negative and very close to 0. Using the Big-Little Concept, the quotient is a negative number far from 0. As x approaches 2 from values less than 2, the quotient becomes a larger and larger negative number. Therefore, the graph of f shoots downward just to left of x 2. That is, f decreases without bound as x approaches 2 from the left. x 2 is shown in Figure 4.4–2b. The portion of the graph of f near ■ The dashed vertical line in Figure 4.4-2b is included for easier visualization, but it is not part of the graph. Such a line is called a vertical asymptote of the graph. The graph approaches a vertical asymptote very closely, but never touches or crosses it because the function is not defined at that value of x. 282 Chapter 4 Polynomial and Rational Functions All rational functions have vertical asymptotes at values that are zeros of their denominators but not zeros of their numerators. Vertical Asymptotes A rational function has a vertical asymptote at x c, provided • c is a zero of the denominator • c is not a zero of the numerator Near a vertical asymptote, the graph of a rational fraction may look like the graph in Figure 4.4-2b, or like one of the graphs in Figure 4.4-3. c x c x c x vertical asymptotes at x c Figure 4.4-3 Holes When a number c is a zero of both the numerator and denominator of a x c, rational function, the
function might have a vertical asymptote at or it might behave differently. You have often cancelled factors to reduce fractions. x2 4 x 2 1 But the functions x 2 x 2 21 x 2 2 x 2 x2 4 x 2 x p 1 2 and q x 2 x 1 2 are not the same, because when x 2 22 4 0 2 2 0 2 2 4. 2 x p q 1 1 2 2, which is not defined, but For any number other than 2, the two functions have the same values, is a straight line and hence, the same graphs. The graph of that includes the point (2, 4), as shown in Figure 4.4-4a. The graph of p(x) is the same straight line, but with the point (2, 4) omitted. That is, there x 2 is a hole in the graph of p at because p is not defined there. The graph of p is shown in Figure 4.4-4b2 −1 1 2 3 −2 Figure 4.4-4a y 4 2 −2 1 2 3 −2 Figure 4.4-4b Section 4.4 Rational Functions 283 y The graph of g x 1 2 of f x 1 2 1 x. At x 0 shown in Figure 4.4-5, is the same as the graph x2 x3, neither function is defined. There is a vertical asymp- x 0 4 8 tote rather than a hole at at multiplicity 3 in the denominator. x 0, Note that the vertical asymptote occurs which is a zero of multiplicity 2 in the numerator, but of larger x 0. 9 6 3 −3 −6 −9 −8 −4 Figure 4.4-5 Holes Technology Tip To avoid erroneous vertical lines, use a window with a vertical asymptote in the center of the screen. In Figure 4.4-6b, the asymptote at x 2 8 nology Appendix for further information. is halfway between and 12. See the Tech- Let f(x) g(x) h(x) of both g and h. be a rational function and let d denote a zero • If the multiplicity of d as a zero of g is greater than or equal to its multiplicity as a zero of h, then the graph of f has a hole at x d. • Otherwise, the graph has a vertical asymptote at x d. Acc
urate Rational Function Graphs Getting an accurate graph of a rational function on a calculator often depends on choosing an appropriate viewing window. For example, the following are graphs of f x 1 2 x 1 2x 4 in different viewing windows. 10 6 10 10 8 12 10 Figure 4.4-6a 6 Figure 4.4-6b x 2, but not at The vertical segment shown in Figure 4.4-6a is not a vertical asymptote. It is a result of the calculator evaluating f just to the left of and just x 2, to the right of and then erroneously connecting these points with a near vertical segment that looks like an asymptote. In the accurate graph shown in Figure 4.4-6b, the calculator attempted to plot a point with was not defined, skipped a pixel and did not join the points on either side of the one skipped. and when it found that x 2 x 2 2 f 1 2 284 Chapter 4 Polynomial and Rational Functions A calculator graph may also fail to show holes in graphs that should have them. Even if a window is chosen so that the graph skips a pixel where the hole should be, the hole may be difficult to see. End Behavior As with polynomials, the behavior of a rational function when is large is called its end behavior. Known facts about the end behavior of polynomial functions make it easy to determine the end behavior of rational functions in which the degree of the numerator is less than or equal to the degree of the denominator. x 0 0 Example 4 End Behavior of Rational Functions List the vertical asymptotes and describe the end behavior of the following functions. Then sketch each graph. a. f x 1 2 3x 6 5 2x b. g x 1 2 x x2 4 c. h x 1 2 2x3 x x3 1 Solution a. The zero of the denominator of f 3x 6 5 2x is 5 2 x 1 2 and it is not a x zero of the numerator. So the vertical asymptote occurs at x 5 2 is large, a polynomial function behaves like its highest When degree term, as shown in Section 4.3. The highest degree term of the numerator of f is 3x and the highest degree term of the denominator is is large, the function reduces to 2x. x. 0 0 Therefore, when 3 2. 0 0 approximately 3x 6 5 2x 3x 6 2x
5 3x 2x 3 2 f x 1 2 Thus, when 0 horizontal line x 0 is large, the graph of f gets very close to the y 3 2 which is called a horizontal asymptote of, the graph. The dashed lines in Figure 4.4-7 indicate the vertical and horizontal asymptotes of the graph. b. The zeros of the denominator of g x 2 1 x x2 4 are ± 2 and neither is a zero of the numerator. So the graph has vertical asymptotes at x 2 x 2. and at When x 0 0 is large, x g 1 2 x x2 4 x x2 1 x and 1 x is very close to 0 by the Big-Little Concept. Therefore, the (the x-axis) when is large and this line is a horizontal asymptote of the graph, as graph of g approaches the horizontal line x 0 shown in Figure 4.4-8. 0 y 0 y 8 4 −8 −4 0 4 8 x −4 −8 Figure 4.4-7 Technology Tip When the vertical asymptotes of a rational function occur at numbers such as 2.1, 2, 1.9, p,2.9, 3, etc., a decimal window normally produces an accurate graph because the calculator actually evaluates the function at the asymptotes, finds that it is undefined, and skips a pixel. CAUTION Unlike a vertical asymptote that is never crossed by a graph, a graph may cross a horizontal or oblique asymptote at some values of x. 0 0 y 4 2 −8 −4 0 4 Figure 4.4-9 Section 4.4 Rational Functions 285 y 4 2 −4 −2 0 2 4 x −2 −4 Figure 4.4-8 c. The only real zero of the denominator of h 2x3 x x3 1 x 1 2 is x 1, which is not a zero of the numerator. So, the graph has a vertical asymptote at x 1. When x 0 0 is large, 2x3 x x3 1 2x3 x3 2 1 x h 1 2 2 Therefore, the graph of h has a horizontal asymptote at shown in Figure 4.4-9. y 2, as x ■ The function in Example 4b illustrates a useful fact. When the degree of the numerator is less than the degree of the denominator of a
rational function, the x-axis is the horizontal asymptote of the graph. When the numerator and denominator have the same degree, as in Examples 4a and 4c, the horizontal asymptote is determined by the leading coefficients of the numerator and denominator: Function 3x 6 2x 5 2x3 x x3 1 f x 1 2 h x 1 2 Horizontal asymptote y 3 2 y 2 1 2 Other Asymptotes When the degree of the numerator of a rational function is greater than the degree of its denominator, the graph will not have a horizontal asymptote. To determine the end behavior in this case, the Division Algorithm must be used. Example 5 A Slant Asymptote Describe the end behavior of the graph of f x2 x 2 x 5. x 1 2 286 Chapter 4 Polynomial and Rational Functions y 40 20 x 15 30 45 −45 0 −30 −15 −20 −40 Figure 4.4-10 y 90 60 30 x 6 12 18 −18 −12 0 −6 −30 Figure 4.4-11 Solution Use synthetic or long division to divide the denominator into the numerator, and rewrite the rational expression by using the Division Algorithm. Dividend Divisor Quotient Remainder x2 x 2 1 21 x 5 x2 18 2 1 1 x 5 x 4 21 x 5 x 4 2 2 18 18 x 5 x 5 21 x 5 1 x 4 18 x 5 2 0 0 x When x 5 is large, is also large, and by the Big-Little Concept 18 x 5 and the graph of f approaches is very close to 0. Therefore, y x 4 y x 4 the line is called a slant or oblique asymptote of the graph. Note that is the quotient without the remainder in the division of the numerator by the denominator. gets large (see Figure 4.4-10). The line x 4, x 4 as x x f 1 2 0 0 ■ Example 6 A Parabolic Asymptote Describe the end behavior of the graph of f x3 3x2 x 1 x 1. x 1 2 Solution Divide the denominator into the numerator and rewrite the function. x3 3x2 x 1 x 1 1 f x 1 2 x2 4x 5 6 x 1 2 Quotient Remainder Divisor When x 0 0 is large, so is x 1
f x2 4x 5 close to 0. Therefore, of f approaches the parabola The curve x2 4x 5 y x2 4x 5 is the quotient in the division. x 2 1 and by the Big-Little Concept 6 x 1 x. The graph 0 y x2 4x 5, as shown in Figure 4.4-11. is called a parabolic asymptote. Note that for large values of is very 0 ■ Section 4.4 Rational Functions 287 End Behavior of Rational Functions Let f(x) axn % cxk % be a rational function whose numerator has degree n and whose denominator has degree k. • If n 66 k, then the x-axis is a horizontal asymptote. • If n k, then the line y a c is a horizontal asymptote. • If n 77 k, then the quotient polynomial when the numerator is divided by the denominator is the asymptote that describes the end behavior of the graph. Graphing Rational Functions Notice that when the degree of the numerator and the denominator are the same, the horizontal asymptote is the horizontal line determined by the quotient of the leading coefficients of the numerator and denominator. Graphs of Rational Functions The facts presented in this section can be used in conjunction with a calculator to find accurate, complete graphs of rational functions. 1. Analyze the function algebraically to determine its vertical asymptotes, holes, and intercepts. 2. Determine the end behavior of the graph. If the degree of the numerator is less than or equal to the degree of the denominator, find the horizontal asymptote by using the facts in the box above. Otherwise, divide the numerator by the denominator. The quotient is the nonvertical asymptote of the graph. 3. Use the preceding information to select an appropriate viewing window, or windows, to interpret the calculator’s version of the graph, and display a complete graph of the function. 10 Example 7 A Complete Graph of a Rational Function −10 10 Find a complete graph of f x 1 x2 x 6. x 2 1 Solution −10 Figure 4.4-12a The graph of f is shown in Figure 4.4-12a. It is hard to determine whether or not the graph is complete, so analyze the function algebraically. 288 Chapter 4 Polynomial and Rational Functions Begin by writing the
function in factored form. Then read off the relevant information. x 1 x2 21 2 1 x 3 and Vertical Asymptotes: x 2 Intercepts: y-intercept: 0 1 02 0 6 1 6 f 0 1 2 x-intercept: x 1 Horizontal Asymptote: y 0 zeros of the denominator but not the numerator zero of numerator but not of denominator degree of numerator is less than degree of denominator Interpreting the above information suggests that a complete graph of f looks similar to Figure 4.4-12b. y 6 4 2 −2 −1 1 2 3 4 x −2 −4 Figure 4.4-12b ■ Example 8 A Complete Graph of a Rational Function Find a complete graph of f x3 2x2 5x 6 x2 3x 2. x 1 2 Solution The denominator is easily factored. To factor the numerator, note that the ± 1, ± 2, ± 3, ± 6 only possible rational zeros of and 3 actually are zeros and by the Rational Zeros Test. Verify that use the Factor Theorem to write the numerator in factored form. Then reduce the fraction. x3 2x2 5x 6 2, 1 and are Section 4.4 Rational Functions 289 Holes: x3 2x2 5x 6 x2 3x 21 x 1 2, where x 2 x 1 x 3 21 x 2 21 x 1 2 2 21 1 x 2. Therefore, the graph of f is the same as the graph of 1 g x 1 2 x 1 x 3 21 x 1 2 x2 4x 3 x 1 except there is a hole when x 2. Because g 1 2 2 1 2 1 2 3 21 2 1 2 1 3 5 21 21 15, the hole occurs at 2, 15. 2 1 Intercepts: y-intercept: x-intercepts 21 0 1 2 1 1 3 21 1 2 3 The x-intercepts of f are the same as the 0 x-intercepts of g. Solving x 1 or x 3. yields x 1 x 3 21 2 1 Vertical Asymptote: The vertical asymptote is x 1. End Behavior: Dividing the numerator by the denominator prox 5. duces a quotient of Therefore, the slant asymptote that describes the end behavior of the y x 5. function is the line The graph of f is
shown in Figure 4.4-13. y 16 8 0 −8 −8 −4 x 4 8 Figure 4.4-13 ■ 290 Chapter 4 Polynomial and Rational Functions Exercises 4.4 In Exercises 1–6, find the domain of the function. 1. f x 1 2 3x 2x 5 2. g x 2 1 x3 x 1 2x2 5x 3 In Exercises 23–50, analyze the function algebraically: list its vertical asymptotes, holes, and horizontal asymptote. Then sketch a complete graph of the function. 3. h x 1 2 6x 5 x2 6x 4 4. g x 1 2 x3 x2 x 1 x5 36x 5. f x 1 2 x5 2x3 7 x3 x2 2x 2 6. h x 1 2 x5 5 x4 12x3 60x2 50x 125 In Exercises 7–12, use algebra to determine the location of the vertical asymptotes and holes in the graph of the function. 7. f x 1 2 x2 4 x2 5x 6 8. g x 1 2 x 5 x3 7x2 2x 9. f x 1 2 x x3 2x2 x 10. g x 1 2 x x3 5x 11. f x 1 2 x2 4x 4 x 2 x 2 21 1 3 2 12. h x 1 2 x 3 x2 x 6 In Exercises 13–22, find the horizontal or other asympis large, and tote of the graph of the function when find a viewing window in which the ends of the graph are within 0.1 of this asymptote. x 00 00 13. f x 1 2 3x 2 x 3 15. h x 2 1 5 x x 2 14. g x 1 2 3x2 x 2x2 2x 4 16. f x 1 2 4x2 5 2x3 3x2 x 17. g x 1 2 5x3 8x2 4 2x3 2x 18. h x 1 2 8x5 6x3 2x 1 0.5x5 x4 3x2 x 19. f x 1 2 x3 1 x2 4 20. g x 1 2 x3 4x2 6x 5 x 2 21. h x 1 2 x3 3x2 4x 1 x 4 22. f x 1
2 x3 3x2 4x 1 x2 x 23. f x 1 2 1 x 5 25. k x 1 2 3 2x 5 27. f x 1 2 3x x 1 29. f x 1 2 2 x x 3 24. q x 1 2 26 28. p x 1 2 x 2 x 30. g x 1 2 3x 2 x 3 1 x 1 2 32. g x 1 2 x 2x2 5x 3 x 31. f x 1 2 33. f x 1 2 35. h x 1 2 36. f x 1 2 2 1 x 3 x2 x 2 1 3 x2 6x 5 x 5 2 1 x2 1 x3 2x2 x 1 38. k x 1 2 x2 1 x2 1 40. F x 2 1 x2 x x2 2x 4 34. g x 1 2 x 2 x2 1 x 5 2 21 x 1 2 37. f x 1 2 4x2 1 x2 39. q x 1 2 x2 2x x2 4x 5 x 3 x 3 1 x 5 21 x 4 2 x 3 21 2 41. p x 1 2 42. p x 1 2 1 21 x3 3x2 x4 4x2 43. f x 1 2 x2 x 6 x 2 44. k x 1 2 x2 x 2 x 45. Q x 2 1 4x2 4x 3 2x 5 46. K x 1 2 3x2 12x 15 3x 6 47. f x 1 2 x3 2 x 1 48. p x 1 2 x3 8 x 1 50. f x 1 2 x4 1 x2 49. q x 1 2 x3 1 x 2 In Exercises 51–60, find a viewing window or windows that show(s) a complete graph of the function using asymptotes, intercepts, end behavior, and holes. Be alert for hidden behavior. 51. f x 1 2 x3 4x2 5x x2 9 x2 4 21 2 1 52. g x 1 2 x2 x 6 x3 19x 30 53. h x 1 2 2x2 x 6 x3 x2 6x 54. f x 1 2 x3 x 1 x4 2x3 2x2 x 1 55. f x 1 2 2x4 3x2 1 3x4 x2 x 1 56. g x 1 2 x4 2x3 x5 25x
3 57. h x 1 2 3x2 x 4 2x2 5x 58. f x 1 2 2x2 1 3x3 2x 1 59. g x 1 2 x 4 2x3 5x2 4x 12 60. h x 1 2 x2 9 x3 2x2 23x 60 In Exercises 61–66, find a viewing window or windows that show(s) a complete graph of the function—if possible, with no erroneous vertical line segments. Be alert for hidden behavior. 61. f x 1 2 2x2 5x 2 2x 7 62. g x 1 2 2x3 1 x2 1 63. h x 1 2 x3 2x2 x 2 x2 1 64. f x 1 2 3x3 11x 1 x2 4 65. g x 1 2 2x4 7x3 7x2 2x x3 x 50 66. h x 1 2 2x3 7x2 4 x2 2x 3 67. a. Graph f x 2 6 x 6 1 1 x and in the viewing window with 6 y 6. Section 4.4 Rational Functions 291. Without using a calculator, describe how the graph of p 2 x 3 x 1 2 4 can be obtained from the graph of 2 e. Show that the function x f 1 1 x. x p 1 2 of part d is a rational function by rewriting its rule as the quotient of two first-degree polynomials. If r, s, and t are constants, describe how the f. t can be obtained from r 1 2 q x graph of x s 1 x. g. Show that the function the graph of x f 2 1 rational function by rewriting its rule as the quotient of two first-degree polynomials. q x 1 2 of part f is a 68. The graph of f 2x3 2x2 x 1 3x3 3x2 2x 1 x 1 2 has a vertical asymptote. Find a viewing window that demonstrates this fact. 69. a. Find the difference quotient of f 1 x x 2 1 and express it as a single fraction in lowest terms. b. Use the difference quotient in part a to x determine the average rate of change of 1 x changes from 2 to 2.1, from 2 to 2.01, and from 2 to 2.001. Estimate the instantaneous rate x 2. of change of at as x f f 2
2 c. Use the different quotient in part a to 1 x determine the average rate of change of 1 x changes from 3 to 3.1, from 3 to 3.01, and from 3 to 3.001. Estimate the instantaneous rate x 3. of change of at as x f f 2 1 2 change of d. How are the estimated instantaneous rates of x 2 1 x2 related to the x 3? and x 2 values of x 3 and 1 x at at x g f 2 2 1 70. Do Exercise 69 for the functions f x g 1 2 2 x3. x 1 2 1 x2 and b. Without using a calculator, describe how the 71. a. When x 0, what rational function has the 2 x Hint: x 1 x 2. graph of g can be obtained from the 1 2 f graph of g 1 c. Without using a calculator, describe how the graphs of each of the following functions can be obtained from the graph of same graph as f x 1 2? Hint: Use the definition of absolute value. b. When x 6 0, what rational function has the same graph as f x 2 1 part a.? See the hint for 292 Chapter 4 Polynomial and Rational Functions c. Use parts a and b to explain why the graph of has two vertical asymptotes What are they? Confirm your answer by graphing the function. 72. The percentage c of a drug in a person’s bloodstream t hours after its injection is approximated by c 5t 4t2 5. t 2 1 a. Approximately what percentage of the drug is in the person’s bloodstream after four and a half hours? b. Graph the function c in an appropriate window for this situation. c. What is the horizontal asymptote of the graph? What does it tell you about the amount of the drug in the bloodstream? d. At what time is the percentage the highest? What is the percentage at that time? 73. A box with a square base and a volume of 1000 cubic inches is to be constructed. The material for the top and bottom of the box costs $3 per 100 square inches and the material for the sides costs $1.25 per 100 square inches. a. If x is the length of a side of the base, express the cost of constructing the box as a function of x. b. If the side of the base must be at least 6 inches long, for what value of x will
the cost of the box be $20? 74. A truck traveling at a constant speed on a reasonably straight, level road burns fuel at the rate of of the truck in miles per hour and gallons per mile, where x is the speed is given by x g x g 1 2 2 1 800 x2 200x. g x 1 2 a. If fuel costs $1.40 per gallon, find the rule of x the cost function 2 fuel for a 500-mile trip as a function of the x speed. Hint: 500 needed to go 500 miles. (Why?) gallons of fuel are that expresses the cost of 1 g c 2 1 b. What driving speed will make the cost of fuel for the trip $250? c. What driving speed will minimize the cost of fuel for the trip? 75. Pure alcohol is being added to 50 gallons of a coolant mixture that is 40% alcohol. a. Find the rule of the concentration function x that expresses the percentage of alcohol in the resulting mixture as a function of the number x of gallons of pure alcohol that are added. Hint: The final mixture contains 50 x gallons. c 2 1 2 1 c x is the amount of alcohol in the (Why?) So final mixture divided by the total amount 50 x. How much alcohol is in the original 50-gallon mixture? How much is in the final mixture? b. How many gallons of pure alcohol should be added to produce a mixture that is at least 60% alcohol and no more than 80% alcohol? c. Determine algebraically the exact amount of pure alcohol that must be added to produce a mixture that is 70% alcohol. 76. A rectangular garden with an area of 250 square meters is to be located next to a building and fenced on three sides, with the building acting as a fence on the fourth side. a. If the side of the garden parallel to the building has length x meters, express the amount of fencing needed as a function of x. b. For what values of x will less than 60 meters of fencing be needed? 77. A certain company has fixed costs of $40,000 and variable costs of $2.60 per unit. a. Let x be the number of units produced. Find the rule of the average cost function. (The average cost is the cost of the units divided by the number of units.) b. Graph the average cost function in a window with 0 x 100,000 and 0 y 20. c. Find the horizontal asym
ptote of the average cost function. Explain what the asymptote means in this situation, that is, how low can the average cost possibly be? 78. Radioactive waste is stored in a cylindrical tank, whose exterior has radius r and height h as shown in the figure. The sides, top, and bottom of the tank are one foot thick and the tank has a volume of 150 cubic feet including top, bottom, and walls. r h a. Express the interior height h1 (that is, the height of the storage area) as a function of h. b. Express the interior height as a function of r. c. Express the volume of the interior as a function of r. d. Explain why r must be greater than 1. 79. The relationship between the fixed focal length F of a camera, the distance u from the object being photographed to the lens, and the distance v from the lens to the film is given by 1 F 1 u 1 v. u F v a. If the focal length is 50 mm, express v as a function of u. b. What is the horizontal asymptote of the graph of the function in part a? c. Graph the function in part a when 50 mm 6 u 6 35,000 mm. Section 4.5 Complex Numbers 293 d. When you focus the camera on an object, the distance between the lens and the film is changed. If the distance from the lens to the camera changes by less than 0.1 millimeter, the object will remain in focus. Explain why you have more latitude in focusing on distant objects than on very close ones. 80. The formula for the gravitational acceleration in units of meters per second squared of an object relative to the earth is g r 2 1 3.987 1014 6.378 106 r 1 2 2 where r is the distance in meters above the earth’s surface. a. What is the gravitational acceleration at the earth’s surface? b. Graph the function g(r) for c. Can you ever escape the pull of gravity? Does r 0. the graph have any r-intercepts? 4.5 Complex Numbers Objectives • Write complex numbers in standard form • Perform arithmetic operations on complex numbers • Find the conjugate of a complex number • Simplify square roots of negative numbers • Find all solutions of polynomial equations If restricted to nonnegative numbers, you cannot solve the equation x 5 0. Enlarging the number system to include
negative integers makes it possible to find the solution to this equation. By enlarging the number system to include rational numbers, it is possible to solve equations that have no integer solution, such as Similarly, the equation x2 2 x 12 x 12 are real numhas no rational solution, but ber solutions. The idea of enlarging a number system to include solutions to equations that cannot be solved in a particular number system is a natural one. 3x 7. and Complex Numbers x2 1 21 have no solutions in the real numEquations such as ber system because are not real numbers. In order to solve such equations, that is, to find the square roots of negative numbers, the number system must be enlarged again. There is a number system, called the complex number system, with the desired properties. and and x2 4 24 294 Chapter 4 Polynomial and Rational Functions Properties of the Complex Number System 1. The complex number system contains all real numbers. 2. Addition, subtraction, multiplication, and division of complex numbers obey the same rules of arithmetic that hold in the real number system, with one exception: the exponent laws hold for integer exponents, but not necessarily for fractional ones (see p. 297). 3. The complex number system contains a number, denoted i, such that i 2 1. 4. Every complex number can be written in the standard form a bi, where a and b are real numbers. a bi c di if and only if a c 5. and b d. Numbers of the form bi, where b is a real number, are called imaginary numbers. Sums of real and imaginary numbers, numbers of the form a bi, are called complex numbers. For example, 5 2i 7 4i 18 3 2 i 3 12i are all complex numbers. Just as every integer is a rational number because it can be written as a fraction with denominator of 1, every real number a is a complex numSimilarly, every imaginary number ber because it can be written as 0 bi. bi is a complex number because it can be written as a 0i. Example 1 Equating Two Complex Numbers Find x and y if 2x 3i 6 4yi. Solution Property 5 of the Complex Number System states that two complex numbers b d. c di a bi and and So, are equal exactly when 2x 6 and a c 3 4y y 3 4 x 3 Direct substitution verifies the solution. 2 1 2 3 3i 6 4 a 6 3i 6 3
i 3 4b i ■ NOTE The mathematicians who invented the complex numbers in the seventeenth century were very uneasy about a number i such that i2 1. Consequently, they called numbers of the form bi, where b is a real number i 21, and numbers. imaginary The existence of imaginary numbers is as real as any of the familiar numbers that are called real numbers, such as 3, 2 3 or 22. NOTE Hereafter, in the c di, a bi complex numbers it is assumed and that a, b, c, and d are real numbers. Section 4.5 Complex Numbers 295 Arithmetic of Complex Numbers Because the usual laws of arithmetic hold, it is easy to add, subtract, and multiply complex numbers. As the following examples demonstrate, all symbols can be treated as if they were real numbers, provided that i2 is replaced by 1. Unless directed otherwise, express answers in the standard form a bi. Example 2 Adding, Subtracting, and Multiplying Complex Numbers Perform the indicated operation and write the result in the form a bi. 3 7i 2 1 a. 4i a b. d. 1 1 4 3i 8 6i 1 2 3 4i 2 2 i 21 2 Solution 1 i 2 4 3i b. a. 1 1 3 7i 2 8 6i 1 i 3 7i 4 3i 8 6i 4i 1 2 2 1 i b 2 3 4i a 2 i 1 1 21 c. d. 2 4i 2 4i 1 a 2 3 4i 2 1 2 6 5i 8i 2i2 8i 2 2 1 3 4i 1 6 8i 3i 4i2 6 4 5i 10 5i 2 2 i 4 6i 2 i 4 9i 2 2 8i ■ The familiar multiplication patterns and exponent laws for integer exponents hold in the complex number system. Example 3 Products and Powers of Complex Numbers Perform the indicated operation and write the result in the form a bi. a. 1 3 2i 3 2i 2 21 b. 4 i 2 2 1 Solution 3 2i 4 i b. a. 1 1 2 3 2i 21 2 42 2 32 i 4 2 9 4i2 9 4 9 4 13 i2 16 8i i2 16 8i 1 15 8i 1 2i 1 2 2 2 1 1 21 2 ■ Powers of i Observe that i1 i i2 1 i3 i2 i 1 i i i4 i2 i2 1 i5
i4 i 1 i i 1 21 2 1 1 Definition of i 296 Chapter 4 Polynomial and Rational Functions The powers of i form a cycle. Any power of i must be one of four values: i, 1, i, in, divide n by 4 and match the remainder to one of the powers listed above. or 1. To find higher powers of i, such as Example 4 Powers of i Find i54. Solution The remainder when 54 is divided by 4 is 2, so i54 i2 1. ■ Complex Conjugates The conjugate of the complex number a bi the conjugate of 3 4i, and the conjugate of 3 4i and each real number a, every real number is its own conjugate. 3i 0 3i are called conjugate pairs. Because a bi. 3 4i a bi 0 3i 3i. is is is the number For example, the conjugate of a bi, and 3 4i is The numbers for a 0i a 0i The product of conjugate pairs is a real number, as shown below. NOTE The result of Example 4 can also be seen i54 by rewriting exponent rules. using i54 i52 i2 i413 i2 i4 13i2 2 1 1 113 1 1. 2 be a complex number. Then the product of a bi and its con- a bi Let jugate a bi a bi is a bi 1 21 and b2 2 a2 1 1 a2 b2. are nonnegative real numbers, so is 2 a2 b2i2 a2 b2 bi 2 2 1 a2 b2 Because a2 Quotients of Complex Numbers The procedure used to find the quotient of two complex numbers uses the fact that the product of conjugate pairs is a real number. Example 5 Quotients of Two Complex Numbers Express the quotient Solution To find the quotient 3 4i 1 2i 3 4i 1 2i in standard form., multiply both the numerator and denomi- 1 2i. nator by the conjugate of the denominator, 3 4i 3 4i 1 2i 1 2i 1 2i 1 2i 1 2i 3 4i 2 2 1 2i 1 2i 2 2 3 6i 4i 8 1 1 4 2 1 1 2 12 1 2i 3 1 4i 2i 1 1 1 3 8 6i 4i 1 4 1 2i 1 2 2 1
1 1 2 2 3 6i 4i 8i2 1 4i 2 11 2i 5 11 5 2 5 i ■ Section 4.5 Complex Numbers 297 Square Roots of Negative Numbers Because i2 1, 21 is defined to be i. Similarly, because 1 2 52i2 25 25, 5i 225 1 is defined to be 5i. In general, 2 1 2 Square Roots of Negative Numbers Let b be a positive real number. 2b is defined to be i2b because i2b A B 2 i2 2b B A 2 1 b b Example 6 Square Roots of Negative Numbers Write each of the following as a complex number. b. 1 27 3 c. A 7 24 5 29 B BA a. 23 Solution a. b. c. Technology Tip Most calculators that do complex number arithmetic will return a complex number when asked for the square root of a negative number. Make sure the MODE is set to “rectangular” or a bi. “ ” by definition 23 i23 1 27 3 7 24 A 1 i27 3 5 29 BA 1 3 B i 27 3 7 i24 A 7 2i 5 i29 B BA 5 3i 2 1 21 35 21i 10i 6i2 35 11i 6 41 11i 1 2 1 ■ CAUTION 2cd 2c 2d —or equivalently in exponential The property 1 notation —which is valid for positive real numbers, does not hold when both c and d are negative. To avoid difficulty, 2 c 2 d cd 1 2 2 1 1 NOTE When i is multiplied by a radical, it is always write square roots of negative numbers in terms of i before doing any simplification. i2b to make customary to write 2b i instead of clear that i is not under the radical. For example, 220 25 i220 i25 i2 220 5 2100 10. Therefore, 220 25 220 5. 298 Chapter 4 Polynomial and Rational Functions Because every negative real number has two square roots in the complex number system, complex solutions can be found for equations that have no real solutions. For example, the solutions of x2 25 are x ± 225 ± 5i Every quadratic equation with real coefficients has solutions in the complex number system. Example 7 Complex Solutions to a Quadratic Equation Find all solutions to 2x2 x 3 0. Solution Apply the quadratic formula. x 1 ± 212 4 2 3 2 2 1 ± 223 4 223 Because solutions. However, Thus
, the equation has solutions in the complex number system. is not a real number, this equation has no real number 223 i223. is an imaginary number, namely, 223 x 1 ± 223 4 Note that the two solutions, 1 4 conjugates. 1 ± i223 4 223 4 i and ± 223 1 4 4 223 4 1 4 i, i are complex ■ NOTE See the Algebra Review Appendix to review factoring the difference of two cubes. Example 8 Zeros of Unity Find all solutions of x3 1. Solution Rewrite the equation as tern to factor the left side. x3 1 0 and use the difference of cubes pat- x3 1 x3 1 0 x 1 0 1 x 1 0 or x2 x 1 0 x2 x 1 21 2 x 1 or x 1 ± 212 4 1 1 2 1 Quadratic formula 1 ± 23 2 1 ± i23 2 1 2 ± 23 2 i Section 4.5 Complex Numbers 299 real complex solutions, Therefore, the equation has one real solution, 23 x 1 2 2 solutions is said to be a cube root of one or a cube root of unity. Observe that the two nonreal cube roots of unity are complex conjugates. and two non- Each of the x3 1 x 1 2 x 1, 23 2 and i. i ■ Examples 7 and 8 illustrated the following useful fact. Conjugate Solutions a bi If coefficients, then its conjugate, the equation. is a solution of a polynomial equation with real a bi, is also a solution of Calculator Exploration The following exploration demonstrates how matrices can be used for complex number arithmetic. a bi is expressed in matrix notation as the 1. The complex number. For example, matrix a a. Write b a b ab 3 4i, 1 2i, in your calculator as [A], [B], [C]. 1 i and 3 6i is written as 6 3b in matrix form and enter them a. 3 6 b. We know B 4 3 is A 4 3 4 6i. that 4 6 a 3 4i 1 6 4b, 1 2 1 2i 2 4 6i. Verify that which represents the complex number c. Use matrix addition, subtraction, and multiplication to find the following. Interpret the answers as complex numbers. C, 4 3 3 4 i 1 2i 3 4 d. In Example 5 we saw that B A 4 3 11.2 0.4i. 5 1 A B 4 3
4. Use the Do this problem in matrix form by computing x key for the exponent. 1 3 e. Do each of the following calculations and interpret the answer in terms of complex numbers. A B 1, C 3 4 3 4 3 1. 3. 5. 7. 8. 9. 11. 13. 15. 29. 31. 33. 35. 300 Chapter 4 Polynomial and Rational Functions Exercises 4.5 In Exercises 1–54, perform the indicated operation and write the result in the form a bi. 2 3i 2 8i 6 i 2 4 2i 2i b 2. 4. 6. 5 7i 14 3i 1 1 3 5i 2 2 2 1 3 7i 1 2 23 i A B A 25 2i B 47. 49. 50. 51. 53. 216 236 225 2 A BA 5 23 A 2 25 A 1 1 22 249 3 BA 1 29 B 1 210 B BA 48. 264 24 B 52. 54. 3 227 23 A 1 24 3 29 B 22 2 a i b 23 2 a i b 1 2 a 23 2 i b 3 4 a 523 2 i b 2 i 21 1 3 2i 3 5i 2 4 i 2 21 1 2 5i 2 2 23 i 23 i B BA 1 A 10. 12. 14. 16. 2 i 21 4 3i 5 2i 2 4 3i 1 1 21 2 i i 2 1 i 1 21 1 2 a i ba 1 4 2i b In Exercises 55–58, find x and y. 55. 3x 4i 6 2yi 56. 8 2yi 4x 12i 57. 3 4xi 2y 3i 58. 8 xi 1 2 y 2i 2 In Exercises 59–70, solve the equation and express each solution in the form a bi. 17. i15 18. i26 19. i33 20. 53 i 2 1 21. 107 i 2 1 22. 213 i 2 1 23. 1 5 2i 24. 1 i 25. 1 3i 26. i 2 i 27. 3 4 5i 28. 2 3i i 69. x4 1 0 59. 3x2 2x 5 0 60. 5x2 2x 1 0 61. x2 x 2 0 62. 5x2 6x 2 0 63. 2x2 x 4 65. 2x2 3 6x 67. x3 8 0 64. x2 1 4
x 66. 3x2 4 5x 68. x3 125 0 x4 81 0 70. i i2 i3 p i15 i i2 i3 i4 i5 p i15 71. Simplify: 72. Simplify: 73. Critical Thinking If is a complex z a bi number, then its conjugate is usually denoted Prove that for any complex that is, is a real number exactly number when z a bi. z a bi, z z z. z, 74. Critical Thinking The real part of the complex a bi number a bi The imaginary part of real number b (not bi). See Exercise 73 for notation. is defined to be the real number a. is defined to be the a. Show that the real part of z a bi b. Show that the imaginary part of z is. is z z 2 z z 2i. 75. Critical Thinking If z a bi numbers, not both 0), express (with a and b real 1 z in standard form. 1 4 5i i 1 2 3i 30. 32. 1 1 1 1 2i 34. 1 2 i 2 i 1 21 2 i 2 2 2 3i 4 i 2 21 3 i 2 3i i 3 i 3 i 4 i 37. 236 39. 214 41. 216 36. 6 2i 3 i 38. 281 40. 250 42. 212 43. 216 249 44. 225 29 45. 215 218 46. 212 23 Section 4.5.A Excursion: The Mandelbrot Set 301 4.5.A Excursion: The Mandelbrot Set Chapter 1 introduced the recursive process of beginning with a value and adding a specific number repeatedly, and Section 3.5.A discussed iterating real-valued functions. This section will extend the iterative process to the set of complex numbers and complex-valued functions and will illustrate the Mandelbrot set, which is used to produce many fractal images. Before reading this section, review Section 3.5.A for terminology and processes, if needed. The Complex Plane a bi, Every complex number where a and b are real numbers, corresponds to the point (a, b) in the coordinate plane. Therefore, complex numbers can be plotted in a coordinate plane, where the horizontal axis represents the real axis and the vertical axis represents the imaginary axis. For example, several complex numbers are plotted in Figure 4.5.A-1. i 6i 4i 2 + 3i −1 +
2i −6 −4 0 −2 −2i 4 6 2 − i 5 Real −3 − 4i −4i −4i −6i Figure 4.5.A-1 Orbits of Complex Numbers The concepts and processes described in Section 3.5.A apply equally well to functions with complex number inputs. For instance, the process of finding the orbit of a complex number under a complex-valued function is the same as finding the orbit of a real number under a real-valued function. For example, the orbit of i under was illustrated in Section 4.5, where it was shown that the powers of i form an orbit of period 4. iz i2 1 i i 1 2 1 i2 i i 2 1 i 1 i iz NOTE The complex plane is discussed in detail in Section 10.3. When dealing with functions that have the complex numbers as their domains, the input variable is usually denoted as z, not x. i 2i i −2 −1 0 1 2 Real −i −2i Figure 4.5.A-2 The orbit of i under is shown graphically in Figure 4.5.A-2. 302 Chapter 4 Polynomial and Rational Functions Technology Tip Calculators that handle complex numbers allow z z2 c you to compute orbits of f 1. Store the complex number c in memory C. easily. 2 1 2. Press 0 STO X ENTER 3. Press to store 0 in X. X 2 C ENTER to produce f 0. STO X 1 2 4. Pressing ENTER repeatedly produces the iterated values. Orbits of complex numbers of many functions are very interesting, but the discussion that follows will be limited to the orbit of 0 under f for different values of c. z2 c z 1 2 Example 1 The Orbit of 0 for f(z) z 2 c Describe the orbit of 0 under ber c. f z 2 1 z2 c for the given complex num- a. c 0.25 0.25i b. c 1.2 0.05i Solution is as follows. a. For z f 1 2 c 0.25 0.25i, z2 0.25 0.25i The orbit of 0 under z2 0.25 0.25i. 1 f z 2 z2 0.25 0.25i 02 0.25 0.25i 0.25 0.25i 0.25 0.25i.25 0.25i 2
0.25 0.125i 0.203125 0.187500i 0.243896 0.173828i 0.220731 0.165208i.2439 0.1738i 0.2207 0.1652i 0.2286 0.1771i 2 2 2 0.25 0.125i 1 0.203125 0.187500i 2 Use iteration on your calculator to find orbit of 0 is approaching a number near c 0.25 0.25i, can be shown that for. 0 It suggests that the f 15 0.2277 0.1718i. In fact, it 2 1 0 f n 1 c 1.2 0.05i, 2 b. For The orbit of 0 under z2 1.2 0.05i. 2 S 0.2276733451 0.1717803749i as n S q f z 1.2 0.05i z2 1 2 z2 1.2 0.05i.2 0.05i 1.1485 0.0168i 1.1860 0.0527i 1.1641 0.0194i f 8 1.1761 0.0515i f 10 1.1725 0.0243i f 12 1.1697 0.0476i f 14 1.1762 0.0296i f 16 is as follows. 0.2375 0.0700i 0.1188 0.0115i 0.2039 0.0751i 0.1547 0.0049i 0.1805 0.0712i 0.1741 0.0070i 0.1660 0.0613i 0.1825 0.0197i NOTE All decimals are shown rounded to four decimal places, but calculations are done using the decimal capacity of the calculator 11 f 13 f 15 Viewing additional iterations, the orbit of 0 when appears to oscillate between two values that are close to and as illustrated in Figure 4.5.A-3. 0.17 0.04i c 1.2 0.05i 1.17 0.04i Section 4.5.A Excursion: The Mandelbrot Set 303 i 1 0.5 −1.5 −1 −0.5 0 0.5 Real 0.5 Figure 4.5.A-3 ■ Example 2 The Orbit of 0 for f(z
) z 2 c Describe the orbit of 0 under ber c. f z 1 2 z2 c for the given complex num- a. c 1 i b. c 0.25 0.625i Solution a. The first seven iterations of f z 2 1 z2 1 i 1 2 are shown in Figure 4.5.A-4. If each of these numbers is plotted in the complex plane, successive iterations get farther and farther from the origin at a very fast rate. For instance 9407 193i 88,454,401 3,631,103i 7.81 1015 6.42 1014 1 2 1 i 2 The distance formula shows that after only five iterations, the distance to the origin is about 9509 and after seven iterations, it is gigantic. In this case, the orbit of 0 is said to approach infinity, which is sometimes expressed as follows. z2 S q as n S q., then f n 1 i If 0 z f 0 through b. The iterations through are shown in Figure 4.5.A-5a and f 15 are shown in Figure 4.5.A-5b. As you can see, successive iterations stay fairly close to the origin through the 16th iteration and then quickly move farther and farther away 21 f 20 1 f 21 f 22 20 0 0 1 1 2 2 2 167.4 522.8i 254,270 175,058i 2.95 1010 8.59 1010 i 2 2 1 Therefore, the orbit of 0 approaches infinity. In other words, 0.25 0.625i S q as n S q., then f n z2 if Figure 4.5.A-4 Figure 4.5.A-5a Figure 4.5.A-5b 304 Chapter 4 Polynomial and Rational Functions The Mandelbrot Set The Mandelbrot set is defined by whether or not the orbit of 0 under the approaches infinity for each complex number c. function z2 c z f 1 2 The Mandelbrot Set The Mandelbrot set is the set of complex numbers c such that the orbit of 0 under the function approach infinity. f(z) z2 c does not 1 2 To avoid awkward repetition in the following discussion, the orbit of 0 under the function will be referred to as “the orbit of c.” z2 c z f Example 1 shows that the orbit of converges and the c 1.2 0.05
i oscillates. Neither orbit approaches infinity, so orbit of both numbers are in the Mandelbrot set. Example 2 shows that the orbits of approach infinity, so these numbers are not in the Mandelbrot set. c 0.25 0.625i c 1 i and c 0.25 0.25i Diagram of the Mandelbrot Set Although the Mandelbrot set is defined analytically, it is usually viewed geometrically by plotting the numbers in the Mandelbrot set as points in the complex plane. The Mandelbrot set is the white region in Figure 4.5.A-6, in which the tick marks on each axis are unit apart. Note that 1 2 the Mandelbrot set is symmetric with respect to the real axis, but not with respect to the imaginary axis. Figure 4.5.A-6 Section 4.5.A Excursion: The Mandelbrot Set 305 Figure 4.5.A-6 illustrates the following fact in which the complex numbers are considered as points in the complex plane. If c lies outside the circle of radius 2 with center at the origin, then c is not in the Mandelbrot set. Furthermore, it can be proved that If c lies inside the circle of radius 2 with center at the origin, but some number in its orbit lies outside the circle, then c is not in the Mandelbrot set. Determining whether a particular point c is in the Mandelbrot set can be quite difficult, particularly if the numbers in its orbit move away from the origin very slowly. Even after hundreds of iterations, it may not be clear whether or not the orbit approaches infinity. Because of round-off errors, a calculator is inadequate for such calculations, and even computers have their limitations. Border of the Mandelbrot Set The border of the Mandelbrot set, which consists of the points just outside of the set, is shown in various colors in Figure 4.5.A-6. The colors indicate how quickly the orbit of the point approaches infinity. The colors are determined by the number of iterations n needed for a number in the orbit of the point to be more than 3 units from the origin, as indicated in the following table. Medium Red Light Red Red Light Yellow Yellow Blue Black n 5 6 n 7 8 n 9 10 n 12 13 n 19 20 n 49 n 50 is in the red region because the second iterExample 2 shows
that ation produces a point more than 3 units from the origin and that 0.25 0.625i is in the yellow region. 1 i The border of the Mandelbrot set is very jagged and chaotic. The varying rates at which the orbits of these border points approach infinity produce some interesting patterns of great complexity. When specific areas are magnified, you can see shapes that resemble islands, seahorse tails, and elephant trunks. The most fascinating aspect of the set is that some of these islands have the same shape as the entire set. Consequently, the Mandelbrot set is said to be self-similar under magnification. The following figures show the region near the point increasing magnification. 0.75 0.3i under The purple points in Figure 4.5.A-7a indicate the Mandelbrot set. Each subsequent image is the magnification of the region denoted by the white square in the previous image. The image shown in Figure 4.5.A-7d is a copy of the set that is contained within the set, and even though the graphs shown do not indicate that all of the purple regions are connected, they are. 306 Chapter 4 Polynomial and Rational Functions Figure 4.5.A-7a Figure 4.5.A-7b Figure 4.5.A-7c Figure 4.5.A-7d Exercises 4.5.A f 1(0), f 2(0), Then determine the distance from In Exercises 1–6, compute where f(z) z2 c. f 3(0) to the origin in the complex plane, where the distance from a point to the origin is given by a bi f 3(0) and 2a2 b2. 1. c 0.3 2. c 0.5i 3. c 0.5 0.5i 4. c 1 0.5i 5. c 1.2 0.5i 6. c 0.75 0.25i In Exercises 7–12, show that c is not in the Mandelbrot set by finding a number in its orbit that is more than 2 units from the origin. How many iterations are needed to find the first such number? 7. c 0.4 9. c 0.7i 8. c 1.1 0.4i 10. c 0.2 0.8i 11. c 0.5 0.7i 12. c
0.4 0.6i In Exercises 13–18, determine whether or not c is in the Mandelbrot set. 13. c i 15. c 1 14. c i 16. c 0.1 0.3i 17. c 0.2 0.6i 18. 0.1 0.8i Section 4.6 The Fundamental Theorem of Algebra 307 4.6 The Fundamental Theorem of Algebra Objectives • Use the Fundamental Theorem of Algebra • Find complex conjugate zeros • Find the number of zeros of a polynomial • Give the complete factorization of polynomial expressions NOTE The graph of a polynomial with complex coefficients cannot be drawn on a coordinate plane. The complex numbers were constructed in order to obtain a solution for Every the equation quadratic equation with real coefficients has solutions in the complex number system, as discussed in Section 4.5. A natural question arises: that is, a zero of the polynomial x2 1. x2 1, x f 2 1 Does the complex number system need to be enlarged, perhaps many times, to find zeros of higher degree polynomial functions? This section explains why the answer is no. In order to give the full answer, the discussion will not be limited to polynomials with real coefficients but will consider polynomials with complex coefficients, such as x3 ix2 4 3i 2 1 x 1 or 3 2i 2 1 x6 3x 5 4i. 2 1 The discussion of polynomial division in Section 4.1 can easily be extended to include polynomials with complex coefficients. In fact, All of the results in Section 4.1 are valid for polynomials with complex coefficients. 1 f x 2 i For example, for 2 x f i is a zero of f and that. 1 checked using the same procedures as before. 1 i 1 i 1 2 is a factor of x2 x i i2 i2 i i2 2 i 1 i 1 2 i 0 2, it is easy to verify that Both statements can be Therefore, i is a zero of f. x x2 2 x2 ix 1 2i 1 2i 1 2i Therefore, x i is a factor of 1 i x f 1 x and 2 2 i x2 2i. 2 4 Because every real number is also a complex number, polynomials with real coefficients are just special cases of polyn
omials with complex coefficients. In the rest of this section, “polynomial” means “polynomial with complex, possibly real, coefficients” unless specified otherwise. Fundamental Theorem of Algebra Every nonconstant polynomial has a zero in the complex number system. 308 Chapter 4 Polynomial and Rational Functions Although this is a powerful result, neither the Fundamental Theorem nor its proof provides a practical method for finding a zero of a given polynomial. You may think it strange that you can prove a zero exists without actually finding one, but such “existence” proofs are quite common in mathematics. Factorization over the Complex Numbers Number of Zeros f(x) be a polynomial of degree Let with leading coefficient a. Then there are n, not necessarily distinct, complex numbers such that c1, c2, p, cn n 77 0 f(x) a(x c1)(x c2) p (x cn) Furthermore, c1, c2 p, cn are the only zeros of f. That is, every polynomial of degree can be written as the product of n linear factors. The statement follows from the Fundamental Theorem and the Factor Theorem. By the Fundamental Theorem, has a complex zero x 2 is a factor, so x c1 c1, f 1 n 7 0 and by the Factor Theorem x c12 x f 2 1 1 is nonconstant, then it has a complex zero 1 2 g x. x g If Theorem and a factor 2 1 x c2 x f 1 2 so that x c121 1 x c221 h x 1 22. c2 by the Fundamental This process can be continued until the final factor is a constant a, at which point you have the factorization shown in the preceding box. Because the n zeros of distinct zeros may be less than n. c1, c2, p, cn of f may not all be distinct, the number Every polynomial of degree zeros in the complex number system. n 77 0 has at most n different Suppose f has repeated zeros, meaning that some of the are Recall that a zero c is said to have the same in the factorization of x c f multiplicity k if 2 is a factor. Consequently, if every zero is counted as many times as its multiplicity, then the statement in the preceding box implies that. 2 is a factor of but no higher power
of x c c1, c2, p, cn polynomial of degree n has exactly n complex zeros. Example 1 Finding a Polynomial Given Its Zeros Find a polynomial a zero of multiplicity 3, and f x f 1 2 24. 2 1 2 of degree 5 such that 1, 2, and 5 are zeros, 1 is Section 4.6 The Fundamental Theorem of Algebra 309 Solution must be a factor of. There f x 1 2 and 5. so x 1 2 Because 1 is a zero of multiplicity are two other factors corresponding to the zeros x x 2 2 x 5 and 2 3, 1 2 x 1 3 1 2 f The product of these factors has degree 5, as does 21 2 where a is the leading coefficient. 24, Because 21 a 1 2 24 2 5 12a 24 a 2 Therefore, f x 1 2 3 x 1 2 21 2x5 12x4 4x3 40x2 54x 20 x 2 x 5 2 1 1 2 y 100 (2, −24) x 2 4 6 −4 0 −2 −100 −200 −300 Figure 4.6-1 The graph of f is shown in Figure 4.6-1. ■ NOTE Complex zeros are not shown on the graph of a polynomial. Polynomials with Real Coefficients Recall that the conjugate of the complex number is the number a bi. We usually write a complex number as a single letter, say z, and z, sometimes read “z bar.” For instance, if indicate its conjugate by z 3 7i, z 3 7i. Conjugates play a role whenever a quadratic then polynomial with real coefficients has complex zeros. a bi y 8 4 −8 −4 0 4 8 −4 −8 Figure 4.6-2 Conjugate Zero Theorem Example 2 Conjugate Zeros Find the zeros of f x 2 1 x2 6x 13. x Solution 2 1 6 ± 2 The quadratic formula shows that f has two complex zeros. 6 ± 4i 2 z 3 2i. 6 ± 216 2 The complex roots are f has no real zeros, as shown in Figure 4.6-2. 2 4 1 13 and its conjugate 6 1 2 2 1 z 3 2i 3 ± 2i Notice that ■ The preceding example is a special case of a more general theorem. f(x
) Let number z is a zero of f, then its conjugate be a polynomial with real coefficients. If the complex is also a zero of f. z 310 Chapter 4 Polynomial and Rational Functions Example 3 A Polynomial with Specific Zeros Find a polynomial with real coefficients whose zeros include the numbers 2 and 3 i. Solution 3 i Because Factor Theorem, x 2 2 is a zero, is a zero, its conjugate, 3 i x 1 is a factor. So, consider the polynomial x and 1 2 2 3 i, 3 i must also be a zero. By the are factors. Similarly, because −8 −4 0 4 8 x f 1 x y 16 8 −8 −16 Figure 4.6-3 Obviously, form shows that are zeros of f. Multiplying out the factored, 3 i, and has real coefficients. 3 i 1 x 3 i 3 i x2 1 x2 3x ix 3x ix 9 i2 x2 6x 10 x3 8x2 22x 20 2 3 21 21 21 2 4 can be factored The next-to-last line of the calculation also shows that as a product of a linear and a quadratic polynomial, each with real coefficients. The graph of f is shown in Figure 4.6-3. x f 1 2 ■ The technique used in Example 3 works because the product, x x2 6x 10 has real coefficients. The proof of the following result shows why this must always be the case. Factorization over the Real Numbers Every nonconstant polynomial with real coefficients can be factored as a product of linear and irreducible quadratic polynomials with real coefficients in such a way that the quadratic factors, if any, have no real zeros. That is, every nonconstant polynomial with real coefficients can be written ax2 bx c, where each as the product of factors in the form quadratic factor is irreducible over the set of real numbers. x k or 1 2 f x a Proof The box on page 308 shows that for any polynomial x c121 c1, c2, p, cn x cr 1 are the zeros of f. If some cr is a real number, then the where is a linear polynomial with real coefficients. If some cj is a factor nonreal complex zero, then its conj
ugate must also be a zero. Thus, some ck is the conjugate of with a and b real numbers. Thus, x cn2 x c22 a bi, a bi f x, 2 1 and say, cj, p ck cj 1 Section 4.6 The Fundamental Theorem of Algebra 311 x cj21 x ck2 1 1 1 x 2 4 x a bi 1 a bi a bi a bi x 2 4 3 3 x2 x 2 x2 ax bix ax bix a2 x2 2ax a2 b2 1 x ck2 x cj21 1 2 1 2 1 a bi bi 1 a bi 2 21 2 2 is a quadratic expression with real Therefore, the factor coefficients because a and b are real numbers. Its zeros, and, are noncj real. By taking the real zeros of f one at a time and the nonreal ones in is obtained. conjugate pairs in this fashion, the desired factorization of ck x f 1 2 y 12 6 −4 −2 0 2 4 x −6 −12 Figure 4.6-4 Example 4 Completely Factoring a Polynomial over the Real Numbers Completely factor numbers given that Solution x f 2 1 1 i x4 2x3 x2 6x 6 over the set of real is a zero of f. 1 i, is also a zero of f, and f x 2 1 1 i is a zero of f, its conjugate, Because has the following quadratic factor. x f Dividing factors as A 3 x by 1 x 23 2 1 i x 2 4 3 1 x2 2x 2 x 23 BA x B x 23 A 1 2 f 1 i 2 4 x2 2x 2 1 shows that the other factor is. Therefore, x 23 x2 2x 2 B 1 BA x4 2x3 x2 6x6 2 is the complete factorization of bers. The graph of f is shown in Figure 4.6-4. x2 3, which over the real num- ■ Complete Factorization of Polynomials The techniques illustrated in this chapter can be used to completely factor some polynomials into linear factors. Example 5 Completely Factoring a Polynomial over the Complex Numbers Completely factor numbers and then over the set of complex numbers. x4 5x3 4x2 2x 8 x f 1 2 over the set of real Solution
Because the degree of the polynomial is 4, there are exactly 4 complex zeros. Find all rational zeros: The possible rational zeros are factors of 8. ± 1, ± 2, ± 4, ± 8 312 Chapter 4 Polynomial and Rational Functions Graph sible zeros are the rational zeros. x4 5x3 4x2 2x 8 x f 2 1 and determine which of the pos- y 20 10 −8 −4 0 4 8 x −10 −20 Figure 4.6-5 The graph suggests that they are. The graph shows that there are no other real zeros. and 4 are zeros and you can easily verify that 1 Find all rational factors: Two linear factors of f are x 1 and x 4. x 1 2 Find remaining factors: Use synthetic division twice to find another factor of ⎧⎪⎪⎨⎪⎪⎩ ⎧⎪⎨⎪⎩ ⎧⎪⎪⎨⎪⎪⎩ x3 x2 2 2 1 2 2 So, x4 5x3 4x2 2x 8 x 1 1 21 Use the quadratic formula to find the two zeros of x 2 ± 2 21 21 1 x2 2x 2 21. 21 2 ⎧⎪⎪⎨⎪⎪⎩ x2 2x 2 x2 2x 2 ± 2i 2 2 2 2 2 ± 24 2 1 i x 1 and The complex factors are Section 4.6 The Fundamental Theorem of Algebra 313 The complete factorizations of the set of complex numbers are shown below. x f 1 2 over the set of real numbers and over x4 5x3 4x2 2x 8 x 1 x 1 1 1 21 21 x 4 x 4 21 2 3 x2 2x 2 x 1 i 2 2 4 3 1 Both products can be verified by multiplication. x 1 i 1 2 4 ■ An expression that has even degree of 4 or greater may have only complex roots. Zeros and factors of such functions and expressions may be difficult to find, and the techniques illustrated in this chapter are of little use. However, functions of odd degree must have at least one real zero, and the corresponding expression must have at least one real linear factor. Therefore, a cubic expression can easily be approximately factored by
estimating one zero, which yields the real linear factor, using synthetic division to determine the quadratic factor, and then using the quadratic formula to estimate the remaining two zeros. Exercises 4.6 In Exercises 1–6, determine if without using synthetic or long division. g(x) is a factor of 1. 2. 3. 4. 5. 6 x10 x8 g x 1 x 2 1 x6 10 g x 2 x 2 1 3x4 6x3 2x 1 g x5 3x2 2x 1 g x3 2x2 5x 10x75 8x65 6x45 4x32 2x15 5 x 1 f(x) 11. 13. 15. 16. 18. 19. 21 x2 2x 5 3x2 2x 7 12. 14. f f x x 1 1 2 2 x2 4x 13 3x2 5x 2 x3 27 Hint: Factor first. x3 125 17. f x 1 2 x3 8 x f Hint: Let x6 64 u x3 1 2 and factor u2 64 first. x4 1 x4 3x2 10 f f x x 1 1 2 2 20. 22. f f x x 1 1 2 2 x4 x2 6 2x4 7x2 4 In Exercises 7–10, list the zeros of the polynomial and state the multiplicity of each zero. with real In Exercises 23–44, find a polynomial coefficients that satisfies the given conditions. Some of the problems have many correct answers. f(x) 7. f x 1 2 x54 x 4 a 5b 8. g x 2 1 3 x 1 a 6b a x 1 5b a x 1 4b 9. 10. h x 2 1 k x 1 2 2x15 x p 14 x p 1 13 1 x 27 A 7 B A 3 1 2 x 25 5 B 1 2 4 2x 1 23. degree 3; only zeros are 1, 7, 4 24. degree 3; only zeros are 1 and 1 25. degree 6; only zeros are 1, 2, p 26. degree 5; only zero is 2 2 27. degree 3; zeros 3, 0, 4; f 80 5 1 2 In Exercises 11 –22, find all the zeros of f in the comas a product of plex number system; then
write linear factors. f(x) 28. degree 3; zeros 2 29. zeros include 2 i and 2 i, 2; f 0 1 1, 1 2 2 314 Chapter 4 Polynomial and Rational Functions 30. zeros include 1 3i and 1 3i 54. x4 6x3 29x2 76x 68; zero 2 of multiplicity 2 31. zeros include 2 and 2 i 32. zeros include 3 and 4i 1 33. zeros include 3, 1 i, 1 2i 34. zeros include 1, 2 i, 3i 1 35. degree 2; zeros 1 2i and 1 2i 36. degree 4; zeros 3i and 3i, each of multiplicity 2 37. degree 4; only zeros are 4, 3 i, and 3 i 38. degree 5; zeros 2 of multiplicity 3, i, and 39. degree 6; zeros 0 of multiplicity 3 and 3, 1 i, each of multiplicity 1 i 1 i, 40. degree 6; zeros include i of multiplicity 2 and 3 41. degree 2; zeros include 42. degree 2; zeros include 1 i; f 6 0 1 2 3 i; f 3 2 1 2 43. degree 3; zeros include i and 1; f 1 2 1 8 44. degree 3; zeros include 2 3i and 2; f 3 2 1 2 In Exercises 45–48, find a polynomial with complex coefficients that satisfies the given conditions. 45. degree 2; zeros i and 1 2i 46. degree 2; zeros 2i and 1 i 47. degree 3; zeros 3, i, and 2 i 48. degree 4; zeros 22, 22, 1 i, and 1 i In Exercises 49–56, one zero of the polynomial is given; find all the zeros. 49. x3 2x2 2x 3; zero 3 50. x3 x2 x 1; zero i 51. x4 3x3 3x2 3x 2; zero i 52. x4 x3 5x2 x 6; zero i 53. x4 2x3 5x2 8x 4; zero 1 of multiplicity 2 55. x4 4x3 6x2 4x 5; zero 2 i 56. x4 5x3 10x2 20x 24; zero 2i 57. Let and
be complex z a bi w c di numbers (a, b, c, d are real numbers). Prove the given equality by computing each side and comparing the results. z w z w a. (The left side says: “First find z w and then take the conjugate.” The right side says: “First take the conjugates of z and w and then add.”) z w z w b. 58. Let 1 2 h x x g and assume that there are cn1 1 2 be polynomials of degree n and numbers c1, c2,..., cn, n 1 such that ci2 g 1 x g 2 1 g x 2 ci2 1 x Prove that. 2 1 h f a zero of is nonzero, 2 what is its largest possible degree? To avoid a contradiction, conclude that h h x 2 0. for every i. Hint: Show that each ci is If 59. Suppose f ax3 bx2 cx d has real x 1 2 coefficients and z is a complex zero of f. a. Use Exercise 57 and the fact that is a real number, to show that az3 bz2 cz d f az 3 bz2 cz d f z 0 0. b. Conclude that f f is also a zero of Note: r r, 1 2 1 2 when r 60. Let 2 1 f x be a polynomial with real coefficients and z a complex zero of f. Prove that the conjugate is also a zero of f. Hint: Exercise 59 is the case when x f has degree 3; the proof in the general case is 2 similar. z 1 61. Use the Factorization over the Real Numbers statement to show that every polynomial with real coefficients and odd degree must have at least one real zero. f x 62. Give an example of a polynomial with complex, nonreal coefficients and a complex number z such that z is a zero of f but its conjugate is not. Therefore, the conclusion of the Conjugate Roots Theorem may be false if doesn’t have real coefficients Important Concepts Section 4.1 Section 4.2 Section 4.3 Polynomial.............................. 239 Coefficient.............
................. 239 Constant polynomial...................... 240 Zero polynomial.......................... 240 Degree of a polynomial.................... 240 Leading coefficient........................ 240 Polynomial function....................... 240 Polynomial division....................... 240 Synthetic division......................... 241 Division Algorithm........................ 243 Remainders and factors.................... 243 Remainder Theorem....................... 244 Zero of a polynomial...................... 245 Real zero................................ 245 Factor Theorem........................... 245 Fundamental polynomial connections......... 246 Number of zeros of a polynomial............ 248 The Rational Zero Test..................... 251 Factoring polynomials..................... 252 Irreducible polynomials....................
253 Complete factorization..................... 253 Bounds Test............................. 256 Real zeros of polynomials.................. 257 axn 2 1 f x Graph of....................... 260 Continuity.............................. 261 End behavior............................ 263 Intercepts............................... 264 Multiplicity.............................. 265 Local extrema............................ 266 Points of inflection........................ 266 Complete polynomial graphs................ 267 Section 4.3.A Polynomial models........................ 273 Section 4.4 Rational function......................... 278 Domain of rational functions................ 279 Intercepts............................... 279 315 316 Chapter Review Section 4.5 Continuity.............................. 280 Vertical asymptotes.........
............... 280 Big-Little Concept......................... 280 Holes.................................. 283 End behavior............................ 284 Horizontal asymptotes..................... 284 Other asymptotes......................... 285 Complex numbers........................ 294 Imaginary numbers....................... 294 Arithmetic of complex numbers.............. 295 Powers of i.............................. 295 Complex conjugates....................... 296 Square roots of negative numbers............ 297 Complex solutions to quadratic equations...... 298 Section 4.5.A The Mandelbrot set..................... 301 Section 4.6 The Fundamental Theorem of Algebra........ 307 Factorization over the complex numbers....... 308 Number of zeros.......................... 308 Conjugate Zero Theorem................... 309 Factorization over the real numbers........... 310 Complete factorization of polynomials......... 311 Important Facts and Formulas When f is a
polynomial and r is a real number that satisfies any of the following statements, then r satisfies all the statements. • r is a zero of the polynomial function • r is an x-intercept of the graph of f • r is a solution, or root, of the equation • 0 f is a factor of the polynomial expression • There is a one-to-one correspondence between the linear factors of of the graph of f. x f 1 2 that have real coefficients and the x-intercepts A polynomial of degree n has at most n distinct real zeros. All rational zeros of a polynomial have the form where r is a fac- r s, tor of the constant term and s is a factor of the leading coefficient. The end behavior of the graph of a polynomial function is similar to the end behavior of the graph of the highest degree term of the polynomial. Zeros of even multiplicity touch but do not cross the x-axis. Zeros of odd multiplicity cross the x-axis. The number of local extrema of the graph of a polynomial function is at most one less than the degree of a polynomial. Chapter Review 317 The number of points of inflection of the graph of a polynomial function is at most two less than the degree of the polynomial. The graph of a rational function has a vertical asymptote at every number that is a zero of the denominator and not a zero of the numerator. The x-intercepts of the graph of a rational function occur at the numbers that are zeros of the numerator but are not zeros of the denominator. Every complex number can be written in the standard form i2 1 and i 21 a bi. a bi If jugate a bi is also a zero. is a zero of a polynomial with real coefficients, then its con- A polynomial of degree n has exactly n complex zeros counting multiplicities. Every polynomial expression with real coefficients can be factored into linear and irreducible quadratic factors with real coefficients. Every polynomial expression can be factored into linear factors with complex coefficients. Review Exercises Section 4.1 1. Which of the following are polynomials? a. 23 x2 d. g. 23 x4 2x 2x2 b. e. x 1 x p3 x c. f.
x3 1 22 22 2x2 0 2. What is the remainder when h. 3. What is the remainder when x 1? x 0 x4 3x3 1 is divided by x2 1? x112 2x8 9x5 4x4 x 5 is divided by 4. Is x 1 a factor of f x 1 2 14x87 65x56 51? Justify your answer. 5. Use synthetic division to show that x6 5x5 8x4 x3 17x2 16x 4, x 2 is a factor of and find the other factor. 6. Find a polynomial f of degree 3 such that f 1 2 1 0, f 1 1 2 0, and f 5. 0 1 2 Section 4.2 7. Find the zero(s) of 2 x 5 a 7 b 3x x 2 5 4. 8. Find the zeros of 3x2 2x 5. 9. Factor the polynomial x3 8x2 9x 6. Hint: 2 is a zero. 10. Find all real zeros of x6 4x3 4. 318 Chapter Review 11. Find all real zeros of 9x3 6x2 35x 26. Hint: Try x 2. 12. Find all real zeros of 3y3 1 13. Find the rational zeros of y4 y2 5. 2 x4 2x3 4x2 1. 14. Consider the polynomial 2x3 8x2 5x 3. a. List the only possible rational zeros. b. Find one rational zero. c. Find all the zeros of the polynomial. 15. a. Find all rational zeros of x3 2x2 2x 2. b. Find two consecutive integers such that an irrational zero of lies between them. x3 2x2 2x 2 16. How many distinct real zeros does x3 4x have? 17. How many distinct real zeros does x3 6x2 11x 6 have? 18. Find the zeros of x4 11x2 18. has x3 2x 1 19. The polynomial a. no real zeros. b. only one real zero. c. three rational zeros. d. only one rational zero. e. none of the above. 20. Show that 5 is an upper bound for the real zeros of x4 4x3 16x 16. 21. Show that 1
is a lower bound for the real zeros of x4 4x3 15. In Exercises 22 and 23, find the real zeros of the polynomial. 22. x6 2x5 x4 3x3 x2 x 1 23. x5 3x4 2x3 x2 23x 20 Section 4.3 24. List the zeros of the polynomial and the multiplicity of each zero. 3 3 5 x 4 x 2 x 17 x2 4 1 2 1 21 25. List the zeros of the polynomial and the multiplicity of each zero. 2 x 3 x 4 x2 9 21 2 2 2 1 1 26. Draw the graph of a function that could not possibly be the graph of a polynomial function, and explain why. 27. Draw a graph that could be the graph of a polynomial function of degree 5. You need not list a specific polynomial nor do any computation. 28. Which of the statements is not true about the polynomial function f whose graph is shown in the figure on the next page? Chapter Review 319 y 3 2 1 x −3 −2 −1 1 2 3 −2 f 2 could possibly be a fifth-degree polynomial a. f has three zeros between b. c. d. e. 0 21 f is positive for all x in the interval and, 0 3 4 1 1 2 2 29. Which of the statements i–v about the polynomial function f whose graph is shown in the figure below are false? y 4 −4 2 −6 −4 −2 i ii iii f has 2 zeros in the interval, 3 1 2 x iv v 2 2 0 f 1 f has degree 2 4 In Exercises 30–33, find a viewing window (or windows) that shows a complete graph of the function. Be alert for hidden behavior. 30. f x 1 2 31. g 32. h x x 2 2 1 1 33. f x 1 2 0.5x3 4x2 x 1 0.3x5 4x4 x3 4x2 5x 1 4x3 100x2 600x 32x3 99x2 100x 2 In Exercises 34–37, sketch a complete graph of the function. 34. f 36. h x x 1 1 2 2 x3 9x x4 x3 4x2 4x 2 35. g 37. f x x 2 2 1
1 x3 2x2 3 x4 3x 2 Section 4.3.A 38. HomeArt makes plastic replicas of famous statues. Their total cost to produce copies of a particular statue is shown in the table on the next page. a. Sketch a scatter plot of the data. b. Use cubic regression to find a function C(x) that models the data—that 320 Chapter Review is, the cost of making x statues. Assume C is reasonably accurate when x 100. c. Use C to estimate the cost of making the seventy-first statue. d. Use C to approximate the average cost per statue when 35 are made and when 75 are made. Recall that the average cost of x statues is C x 1 2 x. Number of statues Total cost 0 10 20 30 40 50 60 70 $2,000 2,519 2,745 2,938 3,021 3,117 3,269 3,425 39. The following table gives the estimated cost of a college education at a public institution. Costs include tuition, fees, books, and room and board for four years. a. Sketch a scatter plot of the data (with b. Use quartic regression to find a function C that models the data. Estimate the cost of a college education in 2007 and in 2015. corresponding to 1990). x 0 Enrollment Year Costs Enrollment Year Costs 1998 2000 2002 2004 2006 $46,691 52,462 58,946 66,232 74,418 2008 2010 2012 2014 $ 83,616 93,951 105,564 118,611 Source: Teachers Insurance and Annuity Association College Retirement Equities Fund Section 4.4 In Exercises 40–43, sketch a complete graph of the function. Label the x-intercepts, all local extrema, holes, and asymptotes. 40. g x 2 1 2 x 4 42. k x 1 2 4x 10 3x 9 41. h x 1 2 3 x x 2 43. f x 1 2 x 1 x2 1 Chapter Review 321 In Exercises 44 and 45, list all asymptotes of the graph of the function. 44. f x 1 2 x2 1 x3 2x2 5x 6 45. g x 2 1 x4 6x3 2x2 6x 2 x2 3 In Exercises 46–49, find a viewing window (or windows) that shows a complete graph of the function. Be alert for hidden behavior.
46. f x 1 2 x 3 x2 x 2 48. h x 1 2 x4 4 x4 99x2 100 47. g x 1 2 x2 x 6 x3 3x2 3x 1 49. k x 1 2 x3 2x2 4x 8 x 10 50. Which of these statements is true about the graph of x 1 x2 1 x 3 2 x2 1? 2 1 2 f x 1 1 21 21 a. The graph has two vertical asymptotes. x 3. b. The graph touches the x-axis at c. The graph lies above the x-axis when d. The graph has a hole at e. The graph has no horizontal asymptotes. x 1. x 6 1. Section 4.5 In Exercises 51–58, solve the equation in the complex number system. 51. x2 3x 10 0 52. x2 2x 5 0 53. 5x2 2 3x 55. 3x4 x2 2 0 57. x3 8 0 54. 3x2 4x 5 0 56. 8x4 10x2 3 0 58. x3 27 0 59. One zero of x4 x3 x2 x 2 is i. Find all zeros. 60. One zero of x4 x3 5x2 x 6 is i. Find all zeros. 61. Give an example of a fourth-degree polynomial with real coefficients whose zeros include 0 and 1 i. 62. Find a fourth-degree polynomial f whose only zeros are such that f 1 1 50. 2 2 i and 2 i Section 4.5.A 63. Find the orbit of 1 for. i b 64. Find the orbit of 0 for z2 c whether c is in the Mandelbrot set. z f 2 1 using the following values of c. State a. c 1 b. c 0.5 0.6i c. c 0.3 0.5i Section 4.6 Factor each of the following over the set of real numbers and over the set of complex numbers. 65. x3 6x2 11x 6 66. x3 3x2 3x 2 67. x4 x3 x2 x 2 68. 2x3 3x2 9x 4 69. x4 2x2 1 70. 9x5 30x4 43x3 114x2 28x 24 12 Optimization Applications ctions Many real
-world situations require you to find the largest or smallest quantity satisfying certain conditions. For instance, automotive engineers want to design engines with maximum fuel efficiency. Similarly, a cereal manufacturer who needs a box of volume 300 cubic inches might want to know the dimensions of the box that requires the least amount of cardboard, which is the cheapest to make. The exact solutions of such minimum/ maximum problems require calculus. However, graphing technology can provide very accurate approximate solutions. Example 1 Maximum of a Rational Function Find two negative numbers whose product is 50 and whose sum is as large as possible. Solution Let x and z be the two negative numbers, and let y be their sum. Then xz 50 xz 50 y x z. for z yields Solving so that and z 50 x y x z x 50 x. The desired quantity is the value of x that makes y as large as possible. Since x must be negative, graph 20 x 0. y x 50 x in a window with Each point x, y 1 2 on the graph represents the following: • x represents one of the two negative number, and 50 x represents the other negative number • y is the sum of the two negative numbers The largest y possible is the point on the graph with largest y-coordinate, that is, the highest point on this part of the graph. Either zoom-in or use the maximum finder to approximate the highest point. As shown in Figure 4.C-1, the largest value of y is approximately which occurs when Therefore, the numbers are approximately 7.071067 The exact solution, as found by 7.071069. 14.14214 and using calculus, is close to the graphical solution. —which is approximately 7.071068 and is very ■ x 7.071067. 50 7.071067 150 –20 2 0 Figure 4.C-1 –26 322 Example 2 Largest Volume of a Box A box with no top is to be made from a inch sheet of cardboard by cutting squares of equal size from each corner and bending up the flaps, as shown in Figure 4.C-2. To the nearest hundredth of an inch, what size square should be cut from each corner in order to obtain a box with the largest possible volume? What is the volume of this box? 22 30 30 x x x 22 30 − 2x x 22 − 2x Figure 4.C-2 Solution Let x denote the length of the side of the square to be cut from
each corner. Then, Volume of box Length Width Height > > 1300 0 0 11 Figure 4.C-3 ⎧⎪⎨⎪⎩ ⎧⎪⎨⎪⎩ 30 2x > 22 2x x 2 2 660x 1 4x˛ 3 104x˛ x x 2 1 3 104x˛ 2 660x gives the volume y of the box Thus, the equation square from each corner. Because the that results from cutting an shortest side of the cardboard is 22 inches, the length x of the side of the cut-out square must be less than 11. (Why?) 2 660x, On the graph of 3 104x˛ y 4x˛ 4x˛ x f 1 2 • the x-coordinate of each point is the size of the square to be cut from each corner. • the y-coordinate of each point is the volume of the resulting box. The box with the largest volume corresponds to the point with the largest y-coordinate, that is, the highest point in the viewing window. A maximum find shows that the highest point is approximately (4.182, 1233.809), as shown in Figure 4.C-3. Therefore, a square measuring approximately 4.18 4.18 inches should be cut from each corner, producing a box of approximately 1233.81 cubic inches. ■ 323 Example 3 Minimum Surface Area of a Cylinder A cylindrical can of volume 58 cubic inches (approximately 1 quart) is to be designed. For convenient handling, it must be at least 1 inch high and 2 inches in diameter. What dimensions will use the least amount of material? h Solution The cylinder can be constructed by rolling a rectangular sheet of metal into a tube and then attaching the top and bottom, as shown in Figure 4.C-4. The surface area of the can, which determines the amount of material needed, has the following formula: Surface Area Area of rectangular sheet Area of top Area of bottom Ch 2pr > Ch 2 > pr 2 > pr 2 When the sheet is rolled into a tube, the width c of the sheet is the circumference of the ends of the can, so C 2pr. Surface Area Ch 2pr 2 2prh 2pr 2 C h r Figure 4.C-4 The volume of the cylinder of radius r and height h is is to have volume 58 cubic inches, pr 2 h.
Since the can pr 2 h 58, or equivalently, h 58 pr 2. Therefore, surface area 2prh 2pr 2 2pr 58 pr 2b a 2pr 2 116 r 2pr 2. Note that r must be 1 or greater because the diameter 2r must be at least 2. Furthermore, r cannot be more than 5 because if then would be at least the volume which is greater than 58. h 1, r 7 5 and pr 25 p h 1, 2 1 21 2 The situation can be represented by the graph of the equation 2px2. The x-coordinate of each point represents a possible radius, and the y-coordinate represents the surface area of the corresponding can. A graphical minimum finder shows that the coordinates of the lowest point are approximately (2.098, 82.947), as shown in Figure 4.C-5. y 116 x If the radius is 2.098, then the height is 58 2.0982 p 4.19. The dimensions of can that uses the least amount of materials are approximately a radius of 2.1 inches and a height of 4.2 inches. ■ 5 Figure 4.C-5 220 1 –55 324 Exercises 1. Find the highest point on the part of the graph of 3 3x 2 y x that is shown in the given window. The answers are not all the same. a. c. 2. Find the lowest point on the part of the graph of y x3 3x 2 window. a. c. 0 x 2 3 x 2 that is shown in the given b. 2 x 2 3. An open-top box with a square base is to be constructed from 120 square centimeters of material. What dimensions will produce a box a. of volume 100 cm3? b. with largest possible volume? 4. A 20-inch square piece of metal is to be used to make an open-top box by cutting equal-sized squares from each corner and folding up the sides (as in Example 2). The length, width, and height of the box are each to be less than 12 inches. What size squares should be cut out to produce a box with a. volume 550 in3? b. largest possible volume? 5. A cylindrical waste container with no top, a diameter of at least 2 feet, and a volume of 25 cubic feet is to be constructed. What should its radius be under the given conditions? a. 65 square feet of material will
be used to construct it b. the smallest possible amount of material will be used to construct it (how much material is needed?) 6. If x c 1 2 is the cost of producing x units, then c x 1 x 2 is x the average cost per unit. Suppose the cost of producing x units is given by 2 10,000x c 1 than 300 units can be produced per week. a. If the average cost is 0.13x3 70x $1100 units are being produced? 2 and that no more per unit, how many b. What production level should be used in order to minimize the average cost per unit? What is the minimum average cost? 7. If the cost of material to make the can in Example 3 is 5 cents per square inch for the top and bottom and 3 cents per square inch for the sides, what dimensions should be used to minimize the cost of making the can? [The answer is not the same as in Example 3.] 8. A certain type of fencing comes in rigid 10-foot segments. Four uncut segments are used to fence in a garden on the side of a building, as shown in the figure. What value of x will result in a garden of the largest possible area? What is that area? x 9. A rectangle is to be inscribed in a semicircle of radius 2, as shown in the figure. What is the largest possible area of such a rectangle? Hint: The width of the rectangle is the second coordinate of the point P (Why?), and P is on the top half of the circle x 2 4. 2 y y x2 + y2 = 4 P 2 −2 −x 0 x 2 x 10. Find the point on the graph of y 5 x closest to the point (0, 1) and has positive coordinates. Hint: The distance from the point x, y 1 express y in terms of x. on the graph to (0, 1) is x 0 2 2 1 2 2 1 2 that is y 1 2; 2 325 C H A P T E R 5 Exponential and Logarithmic Functions Doorway to the past The image above is of Pueblo Benito in Chaco Canyon, New Mexico. It was the home of the Anasazi people of the desert southwest for several centuries, and it includes timbers (shown above the doorway) that were used to date the buildings by using carbon-14 dating, which involves an exponential equation. See Exercise 55 Section 5.
6. 326 Chapter Outline 5.1 Radicals and Rational Exponents 5.2 Exponential Functions 5.3 Applications of Exponential Functions 5.4 Common and Natural Logarithmic Functions 5.5 Properties and Laws of Logarithms 5.5.A Excursion: Logarithmic Functions to Other Bases 5.6 5.7 Solving Exponential and Logarithmic Equations Exponential, Logarithmic, and Other Models Chapter Review can do calculus Tangents to Exponential Functions Interdependence of Sections 5.1 > 5.2 > 5.3 > 5.4 > 5.5 > 5.6 > 5.7 Section 5.1 contains prerequisite review material for this chapter. If students are familiar enough with the objectives of this section, it may be skipped. Exponential and logarithmic functions are essential for the mathe- matical description of a variety of phenomena in the physical sciences, engineering, and economics. Although a calculator is necessary to evaluate these functions for most values, you will not be able to use your calculator efficiently or interpret its answers unless you understand the properties of these functions. When calculations can readily be done by hand, you will be expected to do them without a calculator. 5.1 Radicals and Rational Exponents Objectives n th Roots • Define and apply rational and irrational exponents • Simplify expressions containing radicals or rational exponents NOTE All constants, variables, and solutions in this chapter are real numbers. c 0, Recall that when the equation in a similar fashion as solutions of the equation the square root of c is the nonnegative solution of Cube roots, fourth roots, and higher roots are defined x n c. x 2 c. This equation can be solved graphically by finding the x-coordinate of the. (Review finding intersection points of the graphs of y ax n solutions graphically in Section 2.1 and the shape of the graph of in Section 4.3, if needed.) y x n y c and Depending on whether n is even or odd and whether c is positive or negative, may have two, one, or no solutions, as shown in the following figures. x n c 327 328 Chapter 5 Exponential and Logarithmic Functions Solutions of xn c n odd Exactly one solution for any c y y = xn c y = c x c 77 0 One positive and one negative solution y y = xn c y = c x n even c 0 One solution x 0 y
y = xn x y = 0 c 66 0 No solution y y = xn c x y = c Figure 5.1-1 Figure 5.1-2 Figure 5.1-3 Figure 5.1-4 The figures illustrate the following definition of nth roots. nth Roots Let c be a real number and n a positive integer. The nth root of c is denoted by either of the symbols 1 n2c or c n and is defined to be • the solution of • the nonnegative solution of when n is odd; or x n c x n c when n is even and c 0. Examples of nth roots are shown below. 1 23 8 81 81 3 2 because 2 is the solution of x 3 8. 8 1 1 4 3 because 3 is the nonnegative solution of x 4 81. 2 24 1 2 Expressions involving nth roots can often be simplified or written in a variety of ways by using a basic fact of exponents. d or equivalently, cd 2n n c c˛2n 1 n ˛d 2n cd 1 n 1 1 2 Example 1 Operations on nth Roots Simplify each expression. a. c. 28˛ 212 23 8x6 ˛y4 b. d. 212 275 5 2c 5 2c A B ˛A, where c 7 0 B Section 5.1 Radicals and Rational Exponents 329 Solution a. b. c. d. 28 212 28 12 296 216 6 216 26 426 212 275 24 3 225 3 24˛23 225˛23 223 523 323 23 8x 6y4 23 8 23 x3 x3 23 y3 y 2x 2 y˛23 y 5 2c 52 2 25 c 5 2c 2c A B ˛A B When using a calculator, exponent notation for nth roots is usually preferred over radical notation. B A ■ CAUTION Example 2 Evaluating nth Roots When using exponent notation to evaluate nth roots with a calculator, be sure to use parentheses when raising to the fractional power. Example: To enter press 9^(1/3). 23 9, Use a calculator to approximate each expression to the nearest tenthousandth. 1 5 1 11 a. 40˛ b. 225 ˛ Solution a. Because 1 5 0.2, the expressions 1 40˛ 5 and 0.2 40˛ are equivalent, as shown at right. 5 2.
0913 1 40˛ b. The fraction 1 11 repeating decimal 1 11 fraction is equivalent to the 0.090909 p. The Figure 5.1-5 is not equivalent to this decimal if it is rounded off, as Figure 5.1-6 shown at left. Therefore, it is better to leave the exponent in fractional form. 225 1 11 ˛ 1.6362 ■ Rational Exponents Rational exponents of the form 1 n are called nth roots. Rational exponents Rational exponents of the form r rs m n s, 2 ˛ c˛ 1 it is reasonable to can also be of the form m n, such as 3 2. 4˛ c˛ can be defined in such a way that the laws of exponents, such as are still valid. For example, because 1 2 2 ˛ say that 3 2 3 4˛ 4 2R˛3, Q 1 2 ˛ These expressions are equivalent. 2 264 8 3 8 2 64 2 24 3 4˛ 1 4. 4 2 1 3 2 4˛ 3 2 4˛ This illustrates the definition of rational exponents. 330 Chapter 5 Exponential and Logarithmic Functions Definition of Rational Exponents Let c be a positive real number and let be a rational number t k with positive denominator. t c k is defined to be the number (c t) ct k 2k t. c t 2k c BB AA In radical notation, 1 k (c 1 k)t 3.78 13 can be expressed as Every terminating decimal is a rational number; therefore, expressions 378 100. such as Although the definition of rational exponents requires c to be positive, it remains valid when c is negative, provided that the exponent is in lowest terms with an odd denominator, such as In Exercise 89, you will explore why these restrictions are necessary when c is negative. 8 13 2 3. 2 1 CAUTION 1 2 1 2 3 23 8 8 1 8 2 23 64 4, 2 3 on some calculators may produce either an error Although entering message or a complex number. If this occurs, you can get the correct answer by entering one of the equivalent expressions below. 3 or and 4 is a real number Laws of Exponents c You have seen that the law of exponents is valid for rational exponents. In fact, all of the laws of exponents are valid for rational exponents. 2 1 rs s cr Laws of
Exponents Let c and d be nonnegative real numbers and let r and s be rational numbers. Then 1. 2. crcs crs cr cs crs (c 0) 3. (cr)s c rs 4. 5. 6. (cd)r cr dr r cr d r c db a (d 0) cr 1 c r (c 0) and d 1, If c 1 cr cs cr dr • • r s. if and only if if and only if c d. Example 3 Simplifying Expressions with Rational Exponents Write the expression 3 3 4 s 8r 2 3 2 1 using only positive exponents. Section 5.1 Radicals and Rational Exponents 331 Solution 3 4 s 8r 12 23 82 1 23 64 1 4r s2 1 2 2 2 cd r crdr 1 definition and 2 simplify s crs cr 1 2 simplify and c r 1 cr The expression 1 2 4r s 2 can also be written as 42r s 2 if it is more convenient. ■ Example 4 Simplifying Expressions with Rational Exponents Simplify the expression Solution ab ac b c a 1 c r c s c rs 2 Example 5 Simplifying Expressions with Rational Exponents Simplify the expression 5 2y4 x 7 4 xy 2. 2 2 1 1 Solution 14 4 2 7 4 xy 2 2 1 y 21 y 21 7 2 5 2 2 5 2 2y 4 x x 21 1 2 y 4 x x 21 cd c r 2 1 commutative c r c s c rs simplify ■ ■ Example 6 Simplifying Expressions with Rational Exponents Let k be a positive rational number. Write the expression without radicals, using only positive exponents. 210 c5k 2 k 1 2 2 c 1 Solution 102c5k 3 k c 1 2 1 10 c5k 5k 10 ˛ definition s c rs c r 1 simplify and c r c s c rs simplify s c rs c r 1 2 ■ 332 Chapter 5 Exponential and Logarithmic Functions Rationalizing Denominators and Numerators Transforming fractions with radicals in the denominator to equivalent fractions with no radicals in the denominator is called rationalizing the denominator. Before the common use of calculators, fractions with rational denominators were preferred because they were easier to calculate or estimate. With calculators today there is no computational advantage to rationalizing denominators. However, the skill of rationalizing numerators or denominators is useful in calculus. Example 7 Rationalizing
the Denominator Rationalize the denominator of each fraction. a. 7 25 Solution b. 2 3 26 a. Multiply the fraction by 1 using a suitable radical fraction. 7 25 7 25 1 7 25 25 25 a b 725 5 a2 b2 to determine b. Use the multiplication pattern a suitable radical fraction equivalent to 1. a b 1 21 2 2 3 26 3 26 3 26 3 26 B 3 26 1 2 3 26 2 3 26 2 A 3 26 B 1 6 226 9 6 6 226 3 A 2 ■ NOTE When rationalizing a denominator or numerator which contains a radical expression, use a suitable radical fraction, equal to one, that contains the conjugate of the expression. Example 8 Rationalizing the Numerator Assume h 0. Rationalize the numerator of 2x h 2x h. Section 5.1 Radicals and Rational Exponents 333 Solution Multiply the fraction by 1 using a suitable radical fraction. 2x h 2x h 2x h 2x 2x h 2x 1 A 2x h 2x h 2x h 2x h 2x h 2 2x B 2x h 2x h x h x 2x h 2x h 2x h 2x h h A A B A A 1 2x h 2x B 2 B B Irrational Exponents ■ at is defined when t is an The example (not proof) below illustrates how irrational number. 1022, the exponent could be replaced with the equivalent nonEach of the decimal approximations given below is a more accurate approximation than the preceding 1.414213562 p. To compute terminating decimal of one. 22 1.4, 1.41, 1.414, 1.4142, 1.41421, p. We can raise 10 to each of these rational numbers. 10 10 10 10 10 10 1.4 25.1189 1.41 25.7040 1.414 25.9418 1.4142 25.9537 1.41421 25.9543 1.414213 25.9545 22, gets closer and closer to a real number whose decimal expansion The pattern suggests that as the exponent r gets closer and closer to 10r begins is defined to be this number. 25.954 p. Similarly, for any 1022 So a 7 0, at is a well-defined positive number for each real exponent t. The fact below shall be assumed. The laws of exponents are valid for all
real exponents. 334 Chapter 5 Exponential and Logarithmic Functions 5. 20.0081 6. 20.000169 47. 2x7 x 5 2 x 3 2 48. x A 1 2 y3 2 x0 y7 A B Exercises 5.1 Note: Unless directed otherwise, assume all letters represent positive real numbers. In Exercises 1–15, evaluate each expression without using a calculator. 2. 23 64 3. 24 16 1. 2144 4. 23 27 7. 23 0.008 8. 23 0.125 9. 20.56 10. 2 3 4 2 1 13. 64 1 2 3 2 11. 4 3 27 14. 2 3 1 64b a 12. 1 4 81 15. 3 2 16 In Exercises 16–40, simplify each expression without using a calculator. 16. 2315 18. 20.0812 20. 22. 24 23 0.05 1 2 26 212 17. 23 1216 19. 21. 28 2 2 11 1 23 0.418 23. 28 296 24. 23 18 23 12 25. 23 32 23 16 26. 28. 210 28 25 23 324 23 6 23 2 27. 29. 26 214 263 23 54 23 32 23 4 30. 227 223 31. 425 220 32. 33. 34. 35. 36. 37. 2 23 A 1 23 3 22 A 4 23 B A B A B A 225 4 B A 322 426 A A 3 22 B 5 223 B 325 2 B 2 A 5220 245 2280 B 38. 23 40 223 135 523 320 39. 11 2 2 2 23 2 5 7 2 10 2 2 1 40. 1 32 1 2 1 2 27 94 1 3 2 In Exercises 41–56, simplify each expression. 41. 216a8 b 2 42. 224x6 y 4 43. 2c2 d6 4 24c3 d 45. 4x 2y 29 1 18 2 46. 23 a b 23 44. 10 b 2a 2a14 d 12 4 a b 1 2 2 23 a b 2 5 d c 2 3 c6d3 BA 4 3 B 50 15 9 5b 7b 52. 1 6a 1 2 2ab 2 a2b 3 2 1 2 1 3 2 3b 2 3b 1 2 2 2 4a 2a 1 1 3 5 1 5 2 2 54. A ax 2 1 x B 56 ab 4 bc 1 2 x1 2 3 4
b a ab 1 bx 2 x b 1 49. 51. 53. 55. A 1 1 1 1 A 7a 5a 2a 4a 2 2 2 2 In Exercises 57–66, write each expression without radicals, using only positive exponents. 57. 23 a 2 b2 59. 34 24 a3 61. 25 t 216t5 63. 65. 23 xy 2 3 5 B c c51 1 2 2 3 42 5 6 c 2 A 1 58. 24 a 3 b3 60. 323 a 3b4 62. 2x 23 x 2 24 x 3 64. Q 34 r 14 s 3 7 21 5 R 66 In Exercises 67–72, simplify each expression. 68. 70. 1 2 x x A 3x A 1 3 y 3 2 2x 2x 69. 71. 72 CA BA 2 3 B x y BD In Exercises 73–78, rationalize the denominator and simplify your answer. 73. 76. 3 28 1 23 5 210 74. 77. 2 26 2 2x 2 75. 78. 3 2 212 2x 2x 2c B 67. 1 2 x Section 5.1 Radicals and Rational Exponents 335 In Exercises 79 – 84, factor the given expression. For example, 2 2 (x 2 2)(x 2 1). x x 1 1 1 79. 2 3 x 1 3 6 x 81. x 4x 1 2 3 83. 4 5 81 x 80. 2 5 11x 1 5 30 x 82. 1 3 7x 1 6 10 x 84. 2 3 6x 1 3 9 x In Exercises 85–88, rationalize the numerator and simplify your answer. Assume h 0. 85. 86. 87. 88. 2x h 1 2x 1 h 22x h 3 22x 2x2 1 h 2 1 x h h 2 2x2 x 89. Some restrictions are necessary when defining fractional powers of a negative number. x2 4, a. Explain why the equations x4 4, etc., have no real solutions. Conclude 1 4, c cannot be defined when c 4. is the same as it should be true that 1 6 x6 4, 1 c 2, c that 1 b. Since 3 2 1 3 c c 6, is false when that is, that c 8. 2, 6 23 c 26 c 2. Show that this 90. a. Suppose r is a solution
of the equation xn c Verify that rs is a xn d. and s is a solution of solution of xn cd. b. Explain why part a shows that c 2n cd 2n 2n d. 91. Write laws 3, 4, and 5 of exponents in radical notation in the case when r 1 m and s 1 n. 92. a. Graph f x5 and explain why this x 2 1 function has an inverse function. b. Show algebraically that the inverse function is x 1 5. x g 1 2 93. If n is an odd positive integer, show that xn 1 2 x f has an inverse function and find the rule of the inverse function. Hint: Exercise 92 is the case when n 5. 94. A long pendulum swings more slowly than a short pendulum. The time it takes for a pendulum to complete one full swing, or cycle, is called its period. The relationship between the period T (in seconds) of the pendulum and its length x (in meters) is given by the function T x 1 2 2p A x 9.8. Find the period for pendulums whose lengths are 0.5 m and 1.0 m. 95. In meteorology, the wind chill C can be 3.712V 5.81 0.25V A calculated by using the formula C 0.0817 where V is the wind speed in miles per hour and t is the air temperature in degrees Fahrenheit. Find the wind chill when the wind speed is 12 miles per hour and the temperature is t 91.4 35°F. B 2 1 91.4, 96. The elevation E in meters above sea level and the boiling point of water, T, in degrees Celsius at that elevation are related by the equation 580 E 1000 Find the approximate boiling point of water at an elevation of 1600 meters. 100 T 100 T 2. 1 2 1 2 97. Accident investigators can usually estimate a motorist’s speed s in miles per hour by examining the length d in feet of the skid marks on the road. The estimate of the speed also depends on the road surface and weather conditions. If f represents the coefficient of friction between rubber and the road surface, then motorist’s speed. The coefficient of friction f between rubber and concrete under wet conditions is 0.4. Estimate, to the nearest mile per hour, a motorist’s speed under these conditions if the skid marks are 200
feet long. gives an estimate of the s 230fd 98. Using a viewing window with 0 x 4 and graph the following functions on the 0 y 2, same screen, x, 4 in order of increasing size and justify In each of the following cases, arrange and your answer by using the graphs. a. x 99. Using a viewing window with 3 x 3 and graph the following functions on 1.5 y 1.5, the same screen, in order of increasing size and justify your In each of the following cases, arrange x answer by using the graphs. a. c. d, x x and 336 Chapter 5 Exponential and Logarithmic Functions 100. Graph f 2x in the standard viewing x 2 1 window. Then, without doing any more graphing, describe the graphs of these functions. a. 2x 3 Hint: g x 3 f see x ; 1 2 1 2 x g Section 3.4. 2 1 b. c. h k x 1 x 1 2 2 2x 2 2x 3 2 101. Do Exercise 100 with 23 in place of 2. 5.2 Exponential Functions Objectives • Graph and identify transformations of exponential functions • Use exponential functions to solve application problems Graphs of Exponential Functions For each positive real number a, with base a whose domain is all real numbers and whose rule is Some examples are shown below. there is an exponential function ax. x f 1 2 a 1, f x 1 2 10x g x 1 2 2x h x 1 2 x 1 2b a k f The shape of the graph of an exponential function on the size of a, as shown in the following figures. x 1 2 x 3 2b a x 2 1 ax depends only Graph of f(x) a x a > 1 y 0 << a << 1 y 1 x 1 x • graph is above x-axis • y-intercept is 1 • is increasing f(x) • graph is above x-axis • y-intercept is 1 • is decreasing f(x) • approaches the negative f(x) x-axis as x approaches • approaches the positive f(x) x-axis as x approaches a 0 a 1, or For is a constant function, not exponential. Even roots of negative numbers are not defined in the set of real is not defined for any rational exponent that numbers, so when the function a 6 0 ax, x f 2 1 ax Section 5.2 Exponential Functions 337 has an
even number as its denominator. Because within any interval there are infinitely many rational numbers that have an even denominator, a 6 0 f. Therefore, the function is not well-behaved for, so it is not defined for those values. has an infinite number of holes in every interval when a 6 0 ax x 2 1 The following two Graphing Explorations illustrate the effect that the value of a has on the shape of the graph of an exponential function for a 7 1 0 6 a 6 1 and for. Graphing Exploration a. Using a viewing window with 3 x 7 and 2 y 18, graph each function below on the same screen, and observe the behavior of each to the right of the y-axis. 1.3x 2 As the graphs continue to the right, which graph rises least steeply? most steeply? 2x 10x How does the steepness of the graph of of the y-axis seem to be related to the size of the base a? x f 1 2 to the right ax b. Using the graphs of the same three functions in the viewing and 0.5 y 2, 4 x 2 observe the window with behavior to the left of the y-axis. As the graph continues to the left, how does the size of the base a seem to be related to how quickly the graph of falls toward the x-axis? x f 2 1 ax Graphing Exploration Using a viewing window with graph each function below on the same screen, and observe the behavior of each. and 1 y 4, 4 x 4 f x 1 2 0.2x g x 1 2 0.4x h x 2 1 0.6x k 0.8x x 1 2 Notice that the bases of the exponential functions are increasing in size: 0 6 0.2 6 0.4 6 0.6 6 0.8 6 1 As the graphs continue to the right, which graph falls least steeply? most steeply? How does the steepness of the graph of be related to the size of the base a? f x 2 1 ax seem to The graphing explorations above show that the graph of or falls less steeply as the base a gets closer to 1. f x 1 2 ax rises 338 Chapter 5 Exponential and Logarithmic Functions 10 Example 1 Translations f The graph of is shown in Figure 5.2-1. Without graphing, describe the transformation from the graph of f to the graph of each function below. Verify
by graphing. x 2 1 2x 6 a. g x 1 2 2x3 b. h x 1 2 2x3 4 Solution a. If f x 2x, then g x 2x3 f x 3. So the graph of g is the 1 2 2 1 1 graph of f shifted horizontally 3 units to the left, as shown in Figure 5.2-2. x f 2x, x h 2 1 h(x) is the graph of f 1 and vertically 4 units downward, as shown in Figure 5.2-3. shifted horizontally 3 units to the right 2x3 4 x So the graph of x 3 2x 4. then f 2 2 2 1 2 1 b. If ■ 6 9 ax x The graphs of exponential functions of the form increase at an 2 1 2x x f in Figure 5.2-3. explosive rate. To see this, consider the graph of 2 1 x 50 If the x-axis were extended to the right, then would be at the right 2x f x edge of the page. At this point, the graph of units high. The scale of the y-axis in Figure 5.2-3 is about 12 units per inch, or 144 units per foot, or 760,320 units per mile. Therefore, the height of the graph at x 50 250 is is f 1 2 250 760,320 1,480,823,741 miles, which would put that part of the graph well beyond the planet Saturn! Since most quantities that grow exponentially do not change as dramatically as the graph of exponential functions that model real-life growth or decay are usually modified by the insertion of appropriate constants. These functions are generally of the form 2x, x f 1 2 such as the functions shown below. Pakx.20.45x g 2 x 1 2 3.5 10 1 0.03x h x 1 2 2 Their graphs have the same shape as the graph of or fall more or less steeply, depending on the constants P, k, and a. x f 1 2 but may rise 1.0762x 2 6 21 1 ax, 6 6 3 2 Figure 5.2-1 10 2 Figure 5.2-2 7 5 Figure 5.2-3 5.1 4.7 4.6 1.1 Figure 5.2-4 Example 2 Horizontal Stretches f The graph of is shown in Figure 5.2-4. Without graphing, describe the transformation
from the graph of f to the graph of each function below. Verify by graphing. x 1 2 3x g x 2 1 30.2x h x 1 2 30.8x k x 1 2 3 x p 0.4x 3 x 1 2 Section 5.2 Exponential Functions 339 30.8x Solution The graphs of g x 1 2 izontally by a factor of 30.2x 1 0.2 h and 5 x 2 1 and 30.8x 1 0.8 are the graph of f stretched hor- 1.25, respectively. The graph of x 3 3 0.4x k 1 p x 2 x 1 2 is the graph of f reflected across the y-axis. The graph of 2.5 is the graph of f stretched horizontally by a factor of 1 0.4 and reflected across the y-axis. The graphs are identified in Figure 5.2.5. ■ −0.4x 3 −x 3 3x 5.1 30.2x 4.7 −1.1 Figure 5.2-5 10 Example 3 Vertical Stretches 5 5 p The graph of is shown in Figure 5.2-6. Without graphing, describe the transformation from the graph of p to the graph of each function below. Verify by graphing. x 2 1 0.4x 3 4 3 0.4x.4x Solution 4 3 0.4x 1 q x The graph of cally by a factor of 4. The graph of p x-axis. The graphs are identified in Figure 5.2-7. stretched vertix is the graph of 2 stretched vertically by a factor of 2 and reflected across the p is the graph of x 2 2 3 0.4x 0.4x.4x 3 3 ■ Exponential Growth and Decay In this section, you will see that exponential functions are useful for modeling situations in which a quantity increases or decreases by a fixed factor. In Section 5.3 you will learn how to construct these types of functions. −4.7 5 10 Figure 5.2-6 −0.4x 4. 3 3 −0.4x 10 −5 (−2)3 −0.4x −10 Figure 5.2-7 Example 4 Finance If you invest $5000 in a stock that is increasing in value at the rate of 3% per year, then the value of your stock is given by the function f where x is measured in years. 5000 1.03
x, x 1 2 1 2 a. Assuming that the value of your stock continues growing at this rate, how much will your investment be worth in 4 years? b. When will your investment be worth $8000? Solution a. Letting x 4, 5000 In 4 years your stock is worth about $5627.54. 4 5627.54. 1.03 4 f 2 1 2 1 b. Find the value of x for which equation 5000 1.03 1 2 x 8000. 8000. f x 1 2 In other words, solve the 340 20,000 0 7,000 30 0 5 1.25 0 0.25 Chapter 5 Exponential and Logarithmic Functions Figure 5.2-8 20 120 The point of intersection of the graphs of y 8000 is approximately (15.901, 8000). f x 1 2 5000 1.03 1 x 2 and Therefore, the stock will be worth $8000 in about 16 years. ■ Example 5 Population Growth Based on data from the past 50 years, the world population, in billions, g can be approximated by the function corresponds to 1950. x 0 2.5 where 1.0185 x, x 2 2 1 1 a. Estimate the world population in 2015. b. In what year will the population be double what it is in 2015? Solution a. Since x 0 2015 corresponds to x 65. x 1 Find g(65). 2.5 1.0185 1 65 g 1 2 65 8.23 2 corresponds to 1950, to 1951, and so on, the year The world population in 2015 will be about 8.23 billion people. b. Twice the population in 2015 is 2 16.46; number x such that g x 1 2 16.46 8.23 1 that is, solve 2 billion. Find the 2.5 1.0185 x 16.46. 1 2 x 2.5 A graphical intersection finder shows that the approximate coordinates of the point of intersection of the graphs of y 16.46 g 102.81, or 103 when rounded to the nearest year, corresponds to the year 2053. Notice that it takes only 38 years for the world population to double. are (102.81, 16.46). The x-coordinate 1.0185 and 2 1 1 2 x ■ Figure 5.2-9 Example 6 Radioactive Decay The amount from one kilogram of plutonium M years can be approximated by the function amount of plutonium remaining after 10,000 years. 239
Pu 1 x 2 1 that remains after x Estimate the 0.99997x. 2 Solution Because M is an exponential function with a base smaller than 1 but very close to 1, its graph falls very slowly from left to right. The fact that the graph falls so slowly as x gets large means that even after an extremely long time, a substantial amount of plutonium will remain. 12,000 Figure 5.2-10 x 10,000, M Therefore, almost three-fourths of the origWhen inal plutonium remains after 10,000 years! This is the reason that nuclear waste disposal is such a serious concern. x 1 2 0.74. ■ 3x ex 14 4 2 Figure 5.2-11 2x 4 800 0 −200 100 Figure 5.2-12 Section 5.2 Exponential Functions 341 The Number e and the Natural Exponential Function There is an irrational number, denoted e, that arises naturally in a variety of phenomena and plays a central role in the mathematical description of the physical universe. Its decimal expansion begins as shown below. e 2.718281828459045p Most calculators have an x exponential function the display will show the first part of the decimal expansion of e. key that can be used to evaluate the natural e1 When you evaluate using a calculator, f 1 2 ex ex. Figure 5.2-11 shows that the graph of and graphs of y 2x y 3x, and less steeply than the graph of y 2x x f y 3x. has the same shape as the but it climbs more steeply than the graph of 2 1 ex Example 7 Population Growth If the population of the United States continues to grow as it has since 1980, then the approximate population, in millions, of the United States in year t, where corresponds to the year 1980, will be given by the function t 0 227e0.0093t. P t 1 2 a. Estimate the population in 2015. b. When will the population reach half a billion? Solution a. The year 2015 corresponds to t 35. 227e0.0093 Find P(35). 2 314.3 35 1 35 P 1 2 Therefore, the population in 2015 will be approximately 314.3 million people. b. Half a billion is 500 million. Find the value of t for which P A graphical intersection finder shows that the approximate coordinates of the point of intersection of the graphs of are approximately (85, 500). A t-value of
P 85 corresponds to the year 2065. Therefore, the population will reach half a billion approximately by the year 2065. and y 500 227e0.0093t t t 1 2 2 1 500. ■ Other Exponential Functions In most real-world applications, populations cannot grow infinitely large. The population growth models shown previously do not take into account factors that may limit population growth in the future. Example 8 illustrates a function, called a logistic model, which is designed to model situations that have limited future growth due to a fixed area, food supply, or other factors. 342 25,000 0 0 Chapter 5 Exponential and Logarithmic Functions Example 8 Logistic Model The population of fish in a certain lake at time t months is given by the t 0. There is an upper limit on the fish p function, where t 20,000 t 1 24e 4 1 2 100 population due to the oxygen supply, available food, etc. Graph the function, and find the upper limit on the fish population. Figure 5.2-13 Solution p y 20,000 The graph of is a horizontal asymptote of the graph. If so, the upper limit on the fish population is 20,000. at the left suggests that the horizontal line t 1 2 You can verify this by rewriting the rule of p as shown below. p t 2 1 20,000 t 1 24e 4 20,000 1 24 e t 4 As t increases, t 4 increases and grows very large. As grows very large, t e 4 t e 4 24 t e 4 gets very close to 0. As 24 t e 4 gets closer and closer to 0, p gets closer t 2 1 and closer to 20,000 1 0, or 20,000. Because e t 4 is positive and 24 t e 4 never quite is always slightly larger than 1 and reaches 0, the denominator of is always less than 20,000. p t 2 1 p 1 t 2 ■ When a cable, such as a power line, is suspended between towers of equal height, it forms a curve called a catenary, which is the graph of a function of the form shown below for suitable constants A and k. f x A ekx e kx 1 The Gateway Arch in St. Louis, shown in Figure 5.2-14, has the shape of an inverted catenary, which was chosen because it evenly distributes the internal structural forces. 1 2 2 Graphing Exploration 10 y 80, Using the
viewing window with graph each function below on the same screen, and observe their behavior. 10 5 x 5 e0.4x e e2x e e3x e 10 10 and Y3 Y2 0.4x Y1 2x 3x 1 2 1 2 2 1 How does the coefficient of x affect the shape of the graph? Predict the shape of the graph of answer by graphing. y Y1 80. Confirm your Figure 5.2-14 Section 5.2 Exponential Functions 343 In Exercises 22–29, find a viewing window (or windows) that shows a complete graph of the function. 22. x k 1 2 x e 24. f x 1 2 x ex e 2 26. g x 2 1 2x x 23. f x 1 2 x 2 e 25. h x 2 1 x ex e 2 27. x k 1 2 2 ex e x 28. x f 1 2 5 1 e x 29. x g 1 2 10 1 9e x 2 In Exercises 30–34, determine whether the function is even, odd, or neither. (See Excursion 3.4A.) 30. x f 1 2 10x x ex e 2 x 2 e 32. 34. f f x x 1 1 2 2 31. g x 1 2 2x x 33. f x 1 2 x ex e 2 35. Use the Big-Little concept (see Section 4.4) to x is approximately equal to ex ex e explain why when x is large. In Exercises 36–39, find the average rate of change of the function. (See Section 3.7). 36. f x 1 2 37. g 38. h x x 2 2 1 1 39. f x 1 2 x 2x 2 1 3x 2x as x goes from 1 to 3 as x goes from 1 to 1 x 2 5 as x goes from 1 to 0 ex e x as x goes from 3 to 1 Exercises 5.2 In Exercises 1–6, list the transformations needed to transform the graph of into the graph of the given function. (Section 3.4 may be helpful.) h(x) 2x 1. 3. 5 2x 5 3 2x 1 2 2x2 5 2. 4. 6 2x 1 2x1 2 5 2x1 1 2 7 In Exercises 7–13, list the transformations needed to h(x) 3x transform the graph of
into the graph of the given function. (Section 3.4 may be helpful.) 7. x f 1 2 3x 4 9. x k 1 2 1 4 1 3x 2 11. f 13. g x x 1 1 2 2 32x 4 1 0.15x 3 2 8. x g 1 2 x 3 10. g x 1 2 30.4x 12. x f 1 2 8 5 3x 1 2 In Exercises 14–19, sketch a complete graph of the function. 14. 16. 18. x 4 23x 25x 15. 17. 19. x f x 1 2 5 2b a 3 x 2 2x5 g g x x 1 1 2 2 In Exercises 20–21, match the functions to the graphs. Assume c 77 1. a 77 1 and 20. 21. ax ax 3 ax5 cx 3cx cx5 3cx In Exercises 40–43, find the difference quotient of the function. (See Section 3.7.) A B C D 40. 42. f f x x 1 1 2 2 10x 2x 2 x 41. g 43. f x x 1 1 2 2 5x 2 ex e x In Exercises 44–49, list all asymptotes of the graph of the function and the approximate coordinates of each local extremum. (See Section 4.3.) 44. 46. 48 2x 1 x 2 2 e x 2 e 2 45. 47. g x 1 2 k x 1 2 49 2x 2 6x2 xe x 2 20 344 Chapter 5 Exponential and Logarithmic Functions 50. If you deposit at $750 2.2% interest, compounded annually and paid from the day of deposit to the day of withdrawal, your balance at time t is given t. by after 2 years? after 3 years and 9 months? How much will you have 750 1.022 B t 2 1 1 2 51. The population of a colony of fruit flies t days 100 3 from now is given by the function a. What will the population be in 15 days? in 25 p t 1 2 t 10. days? b. When will the population reach 2500? 52. A certain type of bacteria grows according to the 2 x 5000e0.4055x, f function 1 measured in hours. a. What will the population be in 8 hours? b. When will the population reach 1 million? where the time
x is 53. According to data from the National Center for Health Statistics, the life expectancy at birth for a person born in year x is approximated by the function below. D x 1 2 79.257 1 9.7135 1024 e 1900 x 2050 1 2 0.0304x a. What is the life expectancy of someone born in 1980? in 2000? b. In what year was life expectancy at birth 60 years? 54. The number of subscribers, in millions, to basic cable TV can be approximated by the function g x 1 2 76.7 1 16 0.8444x x 0 where corresponds to 1970. (Source: The Cable TV Financial Datebook and The Pay TV Newsletter) a. Estimate the number of subscribers in 1995 and in 2005. b. When does the number of subscribers reach 70 million? c. According to this model, will the number of subscribers ever reach 90 million? 55. The estimated number of units that will be sold t N 100,000e by a certain company t months from now is given by a. What are the current sales? sales be in 2 months? in 6 months? What will t 0 0.09t. 1 2 1 2 b. From examining the graph, do you think that sales will ever start to increase again? Explain. 56. a. The function 2 1 t g 0.0479t 1 e percentage of the population (expressed as a decimal) that has seen a new TV show t weeks after it goes on the air. What percentage of people have seen the show after 24 weeks? gives the b. Approximately when will 90% of the people have seen it? 57. a. The beaver population near a certain lake in year t is approximated by the function p t 2 1 now 2000 1 199e when t 0 1 2 0.5544t. What is the population and what will it be in 5 years? b. Approximately when will there be 1000 beavers? f 58. Critical Thinking Look back at Section 4.3, where the basic properties of graphs of polynomial functions were discussed. Then review the basic discussed in properties of the graph of this section. Using these various properties, give an argument to show that for any fixed positive number a, where polynomial function such that words, no exponential function is a polynomial function. a 1, it is not possible to find a cnxn p c1x c0 g x for all numbers x. In
other ax g ax x x 2 1 1 1 2 2 59. Critical Thinking For each positive integer n, let fn be the polynomial function below. 1 x x2 2! x3 3! x4 4! fn1 x 2 p xn n! 5 y 55, a. Using the viewing window with ex 4 x 4 f41 x and and graph the same screen. Do the graphs appear to coincide? g x 1 2 2 on 2 2 x f41, then by that of b. Replace the graph of x f51 2 x, f7 1 x f6 1, and so on until you find a by 2 fn1 polynomial whose graph appears to x coincide with the graph of viewing window. Use the trace feature to move from graph to graph at the same value of x to see how accurate this approximation is. in this ex x g 2 2 1 c. Change the viewing window so that and 10 y 400. 6 x 6 Is the polynomial you found in part b a good x g approximation for in this viewing 2 window? If not, what polynomial is a good approximation? 1 Section 5.3 Applications of Exponential Functions 345 5.3 Applications of Exponential Functions Objective • Create and use exponential models for a variety of exponential growth or decay application problems In Section 5.2, you used several exponential functions that modeled exponential growth and decay. In this section you will learn how to construct such exponential models in a variety of real-life situations. Compound Interest When interest is paid on a balance that includes interest accumulated from the previous time periods it is called compound interest. Example 1 Compounding Annually If you invest in the account at the end of 10 years? $6000 at 8% interest, compounded annually, how much is Solution After one year the account balance is Principal Interest 6000 0.08 6000 1 6000 1 2 1 0.08 2 6000 1.08. 2 1 The account balance has changed by a factor of 1.08. If this amount is left in the account, the balance will again change by a factor of 1.08 after the second year. 6000 1.08 1.08 2 2 4 1 1 3, or 6000 1.08 1 2 2 Because the balance will change by a factor of 1.08 every year, the balance in the account at the end of year x is given by B x 1 2 6000 1.08 1 x. 2 Therefore, the balance (to the nearest penny) in the account after
10 years is 10 B 1 2 6000 1.08 2 1 10 $12,953.55. ■ The pattern illustrated in Example 1 can be generalized as shown below. Compound Interest If P dollars is invested at interest rate r (expressed as a decimal) per time period t, then A is the amount after t periods. A P(1 r)t Notice that in Example 1, or years, is t 10. P 6000, r 0.08, and the number of periods, 346 Chapter 5 Exponential and Logarithmic Functions Example 2 Different Compounding Periods Determine the amount that a annual interest rate of 6.4% $4000 investment over three years at an is worth for each compounding period. a. annually b. quarterly c. monthly d. daily Solution a. Use P 4000, r 0.064, and t 3 in the compound interest formula. 1 b. Quarterly compounding means that interest is compounded every 2 A 4000 1.064 3 $4818.20 one-fourth of a year or 4 times a year. Therefore, • the interest rate per period is r 0.064, and • the number of periods in 3 years is 4839.32 A 4000 1 0.064 4 b a c. Monthly compounding means that interest is compounded every 1 12 of a year or 12 times a year. Therefore, • the interest rate per period is • the number of periods in 3 years is A 4000 1 0.064 12 b a and, r 0.064 12 t 12 12 3 3. 2 1 2 $4844.21 1 d. Daily compounding means that interest is compounded every 1 365 of a year, or 365 times a year. Therefore, • the interest rate per period is • the number of periods in 3 years is A 4000 1 0.064 365 b a and, r 0.064 365 t 365 365 3 3. 2 1 2 $4846.60 1 ■ Notice in Example 2 that the more often interest is compounded, the larger the final amount will be. Example 3 shows you how to write and solve an exponential equation to determine how long it will take for an investment to be worth a given amount. Section 5.3 Applications of Exponential Functions 347 Example 3 Solving for the Time Period $5000 7% If the investment be worth is invested at $6800? annual interest, compounded daily, when will Solution Use the compound interest formula with the final amount P 5000
. Because the interest is compounded every 1 365 A 6800 and of a year, the interest rate per period is. r 0.07 365 A P 1 6800 5000 t 1 r 2 1 0.07 365 b a t 8,000 The point of intersection of the graphs of 5000 y1 t 1 0.07 365 b a and 6800 y2 be worth is approximately (1603.5, 6800). Therefore, the investment will $6800 after about 1603 days, or about 4.4 years. ■ 0 0 3,000 Figure 5.3-1 Continuous Compounding and the Number e As you have seen in previous examples, the more often interest is compounded, the larger the final amount will be. However, there is a limit that is reached, as you will see in Example 4. Example 4 The Number e Suppose you invest for one year at 100% annual interest, compounded n times per year. Find the maximum value of the investment in one year. $1 Solution Use the compound interest formula. The annual interest rate is 1.00, so the interest rate per period is 1 n, and the number of periods is n nb a n 1 1 nb a 1 Observe what happens to the final amount as n grows larger and larger. 348 Chapter 5 Exponential and Logarithmic Functions Compounding period n Annually Semiannually Quarterly Monthly Daily Hourly 1 2 4 12 365 8760 n 1 1 nb a 1 1 1 a 1b 2 2 1 1 a 2b 2.25 4 1 1 a 4b 2.4414 12 1 1 12b a 2.6130 365 1 1 a 365b 2.71457 8760 1 1 8760b a 2.718127 Every minute 525,600 Every second 31,536,000 1 a 1 525,600b 525,600 2.7182792 1 a 1 31,536,000b 31,536,000 2.7182825 The maximum amount of the mately no matter how large n is. $2.72, $1 investment after one year is approxi- ■ When the number of compounding periods increases without bound, the process is called continuous compounding. Note that the last entry in the preceding table is the same as the number e to five decimal places. Example 4 is the case when. A similar result occurs, in the general case and leads to the following formula. r 100% P 1 t 1, and Continuous Compounding If P
dollars is invested at an annual interest rate of r, compounded continuously, then A is the amount after t years. A Pert Example 5 Continuous Compounding 5% If you invest much is in the account at the end of 3 years? $4000 at annual interest compounded continuously, how Section 5.3 Applications of Exponential Functions 349 Solution Use the continuous compounding formula with t 3. P 4000, r 0.05, and A Pert 4000e0.05 4647.34 3 1 2 After 3 years the investment will be worth $4647.34. ■ Exponential Growth Compound interest is one type of exponential growth; other exponential growth functions are very similar to the compound interest formula, as you will see in Example 6. Example 6 Population Growth The world population in 1950 was about 2.5 billion people and has been increasing at approximately 1.85% per year. Write the function that gives the world population in year x, where corresponds to 1950. x 0 Solution If the population increases each year by then it increases each year by a factor of 1.0185. Notice that this pattern of population growth is the same as that of compound interest. 1.85%, Year 1950 1951 1952 1953 Population (in billions) 2.5 2.5(1.0185) 1.0185 2.5 1 2 2 1.0185 2.5 1 3 2 …... 1950 x 1.0185 2.5 1 x 2 So, the function that gives the world population, in billions, in year x, f where corresponds to 1950 is x 0 2.5 1.0185 x. x 1 2 1 2 ■ Exponential Growth Exponential growth can be described by a function of the form f(x) Pax, f(x) x 0 is the quantity at time x, P is the initial quantity and where when changes when x increases by 1. If the quantity at rate r per time period, then is the factor by which the quantity a 1 r a 77 1 f(x) is growing and f(x) Pax P(1 r)x. 350 Chapter 5 Exponential and Logarithmic Functions Example 7 Bacteria Growth At the beginning of an experiment, a culture contains 1000 bacteria. Five hours later, there are 7600 bacteria. Assuming that the bacteria grow exponentially, how many bacteria will there be after 24 hours? Solution Use the exponential growth formula with 1000ax x f 1 2 P 1000. Because there are 7600 bacteria after 5 hours
, 1000a5 7600. lation grows. and Solve for a to find the factor by which the bacteria popu- 5 1 2 f 7600 1000a5 7600 a5 7.6 a 25 7.6 a 7.6 1 5 7.60.2 7.60.2. 1000 7.60.2x f x Therefore, the function’s growth factor is Find f 24, 2 1 1 2 the bacteria population after 24 hours. 1000 7.60.2 24 2 16,900,721 24 f 1 1 2 After 24 hours, the bacteria population will be approximately 16,900,721. ■ Exponential Decay Sometimes a quantity decreases by a fixed factor as time goes on, as shown in Example 8. Example 8 Filtering 30% When tap water is filtered through a layer of charcoal and other purifyof the chemical impurities in the water are removed. If ing agents, the water is filtered through a second purifying layer, of the remaining impurities in the water are removed. How many layers are needed to ensure that of the impurities are removed from the water? 95% 30% Solution of the With the first layer, of the impuimpurities remain. With the second layer, rities remain. The table below shows this pattern of exponential decay. of the impurities are removed and of the 70% 70% 30% 70% 0.5 0 0.1 12 Figure 5.3-2 Exponential Decay Section 5.3 Applications of Exponential Functions 351 Layer Impurities remaining 1 2 3 o x 0.7 70% 0.72 0.49, or 49% 0.73 0.343, or 34.3% o 0.7x Therefore, the percentage of impurities remaining in the water after it passes through x layers of purifying material is given by the function 0.7x. f x 1 2 of the impurities are removed, 5% remain. So, find the value 95% When of x for which 0.05. f x 1 2 The point of intersection of the graphs of is approximately (8.4, 0.05). Because you cannot have a fractional part of a filter, 9 layers are needed to ensure that 95% of the impurities are removed from the water. and y2 y1 0.7x 0.05 ■ Example 8 illustrates exponential decay. Notice that the impurities were and that the amount of impurities remainremoved at a rate
of 1 0.3 0.7. ing in the water was changing by a factor of This pattern is true in general for exponential decay. 30% 0.3 Exponential decay can be described by a function of the form f(x) Pax, x 0 where f(x) is the quantity at time x, P is the initial quantity when changes when x increases by 1. If the quantity decaying at rate r per time period then is the factor by which the quantity 0 66 a 66 1 a 1 r is and and f(x) f(x) Pax P(1 r)x. The half-life of a radioactive substance is the time it takes a given quantity of the substance to decay to one-half of its original mass. The half-life depends only on the substance, not on the size of the sample. Because radioactive substances decay exponentially, their decay can be described where x is measured in the same time by a function of the form units as the half-life. The constant a can be determined from the half-life of the substance. Pax, x f 2 1 352 Chapter 5 Exponential and Logarithmic Functions For example, suppose that the half-life of a substance is 25 years. Then after 25 years, the initial amount P decays to 0.5P, or 0.5P. 25 f 1 2 f 2 0.5P 25 1 Pa25 0.5P a25 0.5 a 0.5 1 25 Radioactive Decay The function for this radioactive decay is x f 1 2 Pax P 1 1 25 0.5 x P 1 2 0.5 x 25. 2 The amount of a radioactive substance that remains is given by the function f(x) P(0.5) x h, where P is the initial amount of the substance, corresponds to the time when the radioactive decay began, and h is the half-life of the substance. x 0 Example 9 Radioactive Decay When a living organism dies, its carbon-14 decays exponentially. An archeologist determines that the skeleton of a mastodon has lost of its carbon-14. The half-life of carbon-14 is 5730 years. Estimate how long ago the mastodon died. 64% Solution Use the exponential decay formula for radioactive decay, with h 5730. 1 Because the mastodon has lost 14, or 0.36P, remains. So, find the value of x for which 1 of its carbon-14, 2 64% 2
f x P 0.5 x 5730 0.36P P 1 0.36 0.5 0.5 x 5730 x 5730 x 5730 and 0.5 The point of intersection of the graphs of y1 is approximately (8445.6, 0.36) as shown in Figure 5.3-3. Therefore, the mastodon died about 8445.6 years ago. 0.36 y2 ■ 36% f x 1 2 of its carbon 0.36P. 2 1.5 0 0.5 15,000 Figure 5.3-3 Section 5.3 Applications of Exponential Functions 353 Exercises 5.3 1. If $1,000 is invested at 8% interest, find the value of the investment after 5 years for each compounding period. a. annually c. monthly b. quarterly d. weekly 17. A $8900 years 18. A $9500 years at 11.3% compounded monthly for 3 at 9.4% compounded monthly for 6 2. If $2500 is invested at 11.5% interest, what is the value of the investment after 10 years for each compounding period? a. annually b. monthly c. daily In Exercises 3–12, determine how much money will be in a savings account with an initial deposit of $500 and the interest rate indicated below. 3. 2% compounded annually for 8 years 4. 2% compounded annually for 10 years 5. 2% compounded quarterly for 10 years 6. 2.3% compounded monthly for 9 years 7. 2.9% compounded daily for 8.5 years 8. compounded weekly for 7 years and 7 3.5% months 9. 3% compounded continuously for 4 years 10. 3.5% compounded continuously for 10 years 11. 2.45% compounded continuously for 6.2 years 12. 3.25% compounded continuously for 11.6 years A sum of money P that can be deposited today to yield some larger amount A in the future is called the present value of A. In Exercises 13–18, find the present value of the given amount A. Hint: Substitute the given amount A, the interest rate r per period, and the number of periods t into the compound interest formula, and solve for P. 13. A $5000 years 14. A $3500 years 15. A $4800 years 16. A $7400 years at 6% compounded annually for 7 at 5.5% compounded annually
for 4 at 7.2% compounded quarterly for 5 at 5.9% compounded quarterly for 8 In Exercises 19–26, use the compound interest formula. Given three of the quantities, A, P, r, and t, find the remaining one. 19. A typical credit card company charges 18% annual interest, compounded monthly, on the unpaid balance. If your current balance is and you do not make any payments for 6 months, how much will you owe? $520 20. When his first child was born, a father put $3000 in a savings account that pays compounded quarterly. How much will be in the account on the child’s 18th birthday? annual interest, 4% 13.2% $10,000 to invest for 2 years. Fund A 21. You have pays B pays and Fund C pays monthly. Which fund will return the most money? interest, compounded annually. Fund interest, compounded quarterly, interest, compounded 12.7% 12.6% 22. If you invest $7400 more money: an interest rate of quarterly or an interest rate of continuously? for 5 years, which will return 5% compounded 4.8% compounded 23. If you borrow 14% $1200 at monthly, and pay off the loan (principle and interest) at the end of 2 years, how much interest will you have paid? interest, compounded 24. A developer borrows $150,000 at 6.5% interest, compounded quarterly, and agrees to pay off the loan in 4 years. How much interest will she owe? 25. A manufacturer has settled a lawsuit out of court by agreeing to pay 1.5 million dollars 4 years from now. How much should the company put in an account paying $1.5 compounded monthly, in order to have million in 4 years? Hint: See Exercises 13–18. annual interest, 6.4% 26. Ellen wants to have $30,000 available in 5 years for a down payment on a house. She has inherited $25,000. invested at in order for the investment to reach a value of $30,000? How much of the inheritance should be interest, compounded quarterly, 5.7% 354 Chapter 5 Exponential and Logarithmic Functions 27. Suppose you win a contest and have a choice of $3000 prizes. You can take in 4 years. If money can be invested at $4000 interest, compounded annually, which prize is more valuable in the long run? now
or you can receive 6% 28. If money can be invested at 7% interest, compounded quarterly, which is worth more: $9000 in 5 years? now or $12,500 29. If an investment of $1000 grows to $1407.10 in seven years with interest compounded annually, what is the interest rate? 30. If an investment of $2000 $2700 years, with an annual interest rate that is compounded quarterly, what is the annual interest rate? grows to in 31 2 31. If you put $3000 in a savings account today, what interest rate (compounded annually) must you receive in order to have after 5 years? $4000 32. If interest is compounded continuously, what annual rate must you receive if your investment of $2100 is to grow to in 6 years? $1500 33. a. At an interest rate of 8%, compounded annually, how long will it take to double an of investment of $1200? $500? $100? of b. What conclusion does part a suggest about doubling time? 34. At an interest rate of 6%, compounded annually, how long will it take to double an investment of P dollars? 35. How long will it take to double an investment of interest, compounded continuously? $500 7% at 36. How long will it take to triple an investment of $5000 at 8% interest, compounded continuously? 37. a. Suppose P dollars is invested for 1 year at 12% interest, compounded quarterly. What interest rate r would yield the same amount in 1 year with annual compounding? r is called the effective rate of interest. Hint: Solve the equation P 1 0.12 a 4 b 4 P 1 1 r 2 for r. The left side of the equation is the yield after 1 year at 12% interest, compounded quarterly, and the right side is the yield after 1 year at interest, compounded annually. r% b. Complete the following table: Effective rate 12% 12% interest compounding period annually quarterly monthly daily 38. This exercise investigates the continuous compounding formula, using a realistic interest rate. Consider the value of years at year, for increasing values of n. In this case, the $4000 interest, compounded n times per deposited for 3 5% interest rate per period is 0.05 n, and the number of periods in 3 years is 3n. So, the value at the end of 3 years is given by: A 4000 3n 1 0.05 n b a 4000 1
0.05 n b c a n 3 d a. Complete the following table: n 1 0.05 n b a n 1000 10,000 500,000 1,000,000 5,000,000 10,000,000 b. Compare the entries in the second column of e0.05, and the table in part a to the number complete the following sentence: As n gets larger and larger, the value of 1 0.05 n b a number gets closer and closer to the?. n c. Use your answer to part b to complete the following sentence: As n gets larger and larger, the value of A 4000 to the number 1 0.05 n b? c a d. n 3 gets closer and closer d. Compare your answer to part c with the value given by the continuous compounding formula. Section 5.3 Applications of Exponential Functions 355 39. A weekly census of the tree-frog population in a state park is given below. corresponds to the 1989–1990 school year. b. According to this model, what are the expenditures per pupil in 1999–2000? c. In what year did expenditures first exceed Week 1 2 3 4 5 6 $7000 per pupil? Population 18 54 162 486 1458 4374 44. There are now 3.2 million people who play bridge a. Find a function of the form f Pax that x 2 1 describes the frog population at time x weeks. b. What is the growth factor in this situation (that is, by what number must this week’s population be multiplied to obtain next week’s population)? c. Each tree frog requires 10 square feet of space and the park has an area of 6.2 square miles. Will the space required by the frog population exceed the size of the park in 12 weeks? in 14 weeks? ( 1 square mile 52802 square feet) 40. The fruit fly population in a certain laboratory triples every day. Today there are 200 fruit flies. a. Make a table showing the number of fruit flies present for the first 4 days (today is day 0, tomorrow is day 1, etc.). b. Find a function of the form Pax that x f describes the fruit fly population at time x days. c. What is the growth factor here (that is, by what 1 2 number must each day’s population be multiplied to obtain the next day’s population)? d. How many fruit flies will there be a week from now? 41. The population
of Mexico was 100.4 million in 2000 and is expected to grow by approximately 1.4% each year. a. If x g is the population, in millions, of Mexico corresponds to the year x 0 in year x, where 2000, find the rule of the function g. (See Example 6.) 2 1 b. Estimate the population of Mexico in the year 2010. 42. The number of dandelions in your lawn increases a week, and there are 75 dandelions now. x is the number of dandelions in week x, by a. If 5% f 1 2 find the rule of the function f. and the number increases by a. Write the rule of a function that gives the a year. 3.5% number of bridge players x years from now. b. How many people will be playing bridge 15 years from now? c. When will there be 10 million bridge players? 45. At the beginning of an experiment a culture contains 200 h-pylori bacteria. An hour later there are 205 bacteria. Assuming that the h-pylori bacteria grow exponentially, how many will there be after 10 hours? after 2 days? (See Example 7.) 46. The population of India was approximately 1030 million in 2001 and was 865 million a decade earlier. What will the population be in 2006 if it continues to grow exponentially at the same rate? 47. Use graphical methods to estimate the following values. 313 a. b. 413 c. 513 48. Kerosene is passed through a pipe filled with clay in order to remove various pollutants. Each foot of pipe removes a. Write the rule of a function that gives the percentage of pollutants remaining in the kerosene after it has passed through x feet of pipe. (See Example 8.) of the pollutants. 25% b. How many feet of pipe are needed to ensure 90% that from the kerosene? of the pollutants have been removed 49. If inflation runs at a steady per year, then the amount that a dollar is worth today decreases by 3% a. Write the function rule that gives the value of each year. 3% a dollar x years from today. b. How much will the dollar be worth in 5 years? in 10 years? c. How many years will it take before today’s b. How many dandelions will there be in 16 weeks? dollar is worth only a dime? 43. Average annual expenditure per pupil in 50. a. The half
-life of radium is 1620 years. Find the $5550 elementary and secondary schools was 1989–1990 and has been increasing at about each year. a. Write the rule of a function that gives the in 3.68% expenditure per pupil in year x, where x 0 rule of the function that gives the amount remaining from an initial quantity of 100 milligrams of radium after x years. b. How much radium is left after 800 years? after 1600 years? after 3200 years? 356 Chapter 5 Exponential and Logarithmic Functions 51. a. The half-life of polonium-210 is 140 days. Find the rule of the function that gives the amount of polonium-210 remaining from an initial 20 milligrams after t days. b. How much polonium-210 is left after 15 weeks? after 52 weeks? c. How long will it take for the 20 milligrams to decay to 4 milligrams? 52. How old is a piece of ivory that has lost 58% of its carbon-14? (See Example 9.) 53. How old is a mummy that has lost 49% of its carbon-14? 5.4 Common and Natural Logarithmic Functions Objectives • Evaluate common and natural logarithms with and without a calculator • Solve common and natural exponential and logarithmic equations by using an equivalent equation • Graph and identify transformations of common and natural logarithmic functions Technology Tip 10x The graph of can be obtained in para- x f 1 2 metric mode by letting x t and y 10t, where t is any real number. The graph of the inverse function g can then be obtained by letting x 10t and y t, where t is any real number. From their invention in the seventeenth century until the development of computers and calculators, logarithms were the only effective tools for numerical computation in astronomy, chemistry, physics, and engineering. Although they are no longer needed for computation, logarithmic functions still play an important role in the sciences and engineering. In this section you will examine the two most important types of logarithms, those to base 10 and those to base e. Logarithms to other bases are considered in Excursion 5.5A. Common Logarithms The graph of the exponential function is shown in Figure 5.4-1. Because it is an increasing function, it is a one-to-one function
, as explained in Section 3.6. Recall that the graphs of inverse functions are reflections of one another across the line The exponential function f and its inverse function are graphed in Figure 5.4-2. y x. 10x x x f 1 2 10x 1 2 y 1 f(x) = 10x x y y = x f(x) 1 x 1 g(x) Figure 5.4-1 Figure 5.4-2 The inverse function of the exponential function is called the common logarithmic function. The value of this function at the number x is denoted as log x and called the common logarithm of the number x. x f 1 2 10x The functions f 1 10x and x 2 log v u if and only if 10u v x g 1 2 are inverse functions. log x Section 5.4 Common and Natural Logarithmic Functions 357 Because logarithms are a special kind of exponent, every statement about logarithms is equivalent to a statement about exponents. Logarithmic statement log v u Equivalent exponential statement 10u v log 29 1.4624 log 378 2.5775 101.4624 29 102.5775 378 Example 1 Evaluating Common Logarithms Without using a calculator, find each value. a. log 1000 b. log 1 c. log210 d. log 3 1 2 Solution a. If b. If log 1000 x, log 1 x, then 10x 1000. Because then 10x 1. Because 103 1000, log 1000 3.. 100 1, log 1 0 c. If log210 x, then 10x 210. Because 10 1 2 210, log210 1 2. d. If 1 log 3 x, 10x 3. exponent of 10 that produces numbers. then 2 Because there is no real number is not defined for real 3, log 3 1 2 ■ Every scientific and graphing calculator has a LOG key for evaluating logarithms. For example, log 0.6 0.2218 and log 327 2.5145 A calculator is necessary to evaluate most logarithms, but you can get a rough estimate mentally. For example, because log 795 is greater than log 100 2 you can estimate that log 795 is between 2 and 3 and closer to 3. and less than log 1000 3, Example 2 Using Equivalent Statements Solve each equation by using an equivalent statement. a. log x 2 b. 10x 29 Solution
a. If log b. If x 2, 10x 29, Figure 5.4-3. then 102 x. Therefore, x 100. then log 29 x. Therefore, x 1.4624, as shown in ■ NOTE Logarithms are rounded to four decimal places and an equal sign is used rather than the “approximately equal” sign. The word “common” will be omitted except when it is necessary to distinguish the common logarithm from another type of logarithm. Figure 5.4-3 358 Chapter 5 Exponential and Logarithmic Functions Natural Logarithms The exponential function is very useful in science and engineering. Consequently, another type of logarithm exists, based on the number e instead of 10. x f 1 2 ex The graph of the exponential function Because it is an increasing function, it is one-to-one. The function and its inverse function are graphed in Figure 5.4-5. is shown in Figure 5.4-4. ex x x f f 1 1 2 2 ex y 1 f(x) = ex x y y = x f(x) 1 x 1 g(x) Figure 5.4-4 Figure 5.4-5 This inverse function of the exponential function is called the natural logarithmic function. The value of this function at the number x is denoted as ln x and called the natural logarithm of the number x. x f 1 2 e x The functions f x 1 and ex ln x 2 ln v u if and only if eu v are inverse functions. x g 2 1 Again, as with common logarithms, every statement about natural logarithms is equivalent to a statement about exponents. Logarithmic statement v u ln Equivalent exponential statement eu v ln 14 2.6391 ln 0.2 1.6094 e2.6391 14 1.6094 0.2 e Example 3 Evaluating Natural Logarithms Use a calculator to find each value. c. ln a. ln 0.15 b. ln 186 1 5 2 Solution Figure 5.4-6 ln 0.15 1.8971, 186 5.2257, 5 a. b. ln c. ln e that produces 2 1 5. which means that 1.8971 0.15. e which means that e5.2257
186. is undefined for real numbers because there is no exponent of ■ Section 5.4 Common and Natural Logarithmic Functions 359 In a few cases you can evaluate ln x without a calculator. ln e 1 because e1 e ln 1 0 because e0 1 Example 4 Solving by Using an Equivalent Statement Solve each equation by using an equivalent statement. a. ln x 4 b. e x 5 Solution x 4, then a. If ln b. If e x 5, e4 x 5 x then ln. Therefore,. Therefore, x 54.5982. x 1.6094. ■ Graphs of Logarithmic Functions Because the graphs of exponential functions have the same basic shape and each logarithmic function is the inverse of an exponential function, the graphs of logarithmic functions have common characteristics. The following table compares the graphs of exponential and logarithmic functions. Exponential functions Logarithmic functions Examples x f 1 2 10x; f ex x 1 2 Domain all real numbers g x 2 1 log x; g ln x x 2 1 all positive real numbers Range all positive real numbers all real numbers x f 1 2 increases as x increases g x 2 1 increases as x increases approaches the x-axis x f 1 as x decreases 2 approaches the y-axis as x g x approaches 0 1 2 Reference points f(x) 10x g(x) log x 1, a 1 10b, 1 0, 1, 1 2 1, 10 2 1 10 a, 1 1, 0, 1 b, 1 2 10, 1 2 f(x) ex 1, a 1 e b, 1 0, 1 1, e, 2 1 2 g(x) ln x 1 e, 1 a 1, 0, 1 b e, 1, 2 1 2 Example 5 Transforming Logarithmic Functions Figure 5.4-7 y exponential y = x (0, 1) (1, 0) x logarithmic Figure 5.4-8 Describe the transformation from the graph of of Give the domain and range of h. 2 log log x to the graph 360 Chapter 5 Exponential and Logarithmic Functions Solution The graph of after a horizontal translation of 3 units right and a vertical stretch by a factor of 2. is the graph of log x Domain of h: The domain of log x The horizontal translation of 3 units to the right changes the domain to all real
numbers greater than 3. is all positive real numbers. g x 2 1 Range of h: log x The range of vertical stretch has no effect on the range. is all real numbers, so the x g 2 1 The graphs of g and h are shown in Figure 5.4-9. The points a 9 1, 0 1, and 2 1, and 2 2 log 10, 1 2 13, 2 2 1 x 3 2 4, 0 1 h x continues to approach the asymptote at 2 1 1 on the graph of g are translated to the points b on the graph of h. Although the graph of you know that it appears to stop abruptly at x 3, 3 a, x 3. 1 10 1 10, 1, b, 2 ■ 3 –1 –3 Figure 5.4-9 3 5 5 5 Figure 5.4-10 Example 6 Transforming Logarithmic Functions Describe the transformation from Give the domain and range of h. g x 1 2 ln x to h x 2 1 ln 1 2 x 2 3. Solution g x 2 3, x h Because after a horizontal reflection across the y-axis followed by a horizontal translation of 2 units to the right and a vertical translation of 3 units downward. its graph is that of 22 g x 1 1 1 2 2 1 ln x Domain of h: The domain of g x ln x is all positive real numbers. 1 2 The reflection across the y-axis first changes the domain to all negative real numbers. Then the translation of 2 units to the right changes the domain from all negative real numbers to all real numbers less than 2. ln x The range of vertical translation does not affect the range. is all real numbers, so the g x 2 1 Range of h:, The graphs of g and h are shown in Figure 5.4-10. The points on the graph of g are translated to points on the graph of h. 2 e, 2 1, 3 and and 1, 0 e ■ Example 7 Solving Logarithmic Equations Graphically If you invest money at an interest rate r, compounded annually, then gives the time in years that it would take to double. D r 2 1 D r 1 2 ln 2 ln 1 1 r 2 Section 5.4 Common and Natural Logarithmic Functions 361 a. How long will it take to double an investment of $2500 at 6.5% annual interest? b. What annual interest rate is
needed in order for the investment in part a to double in 6 years? Solution a. The annual interest rate r is 0.065. Find D(0.065). 0.065 D 1 2 ln 2 1 0.065 2 ln 1 11.0067 Therefore, it will take approximately 11 years to double an investment of $2500 6.5% at b. If the investment doubles in 6 years, then 6 annual interest rate r, solve 6. To find the r 2 1 by graphing. The point of intersection of the graphs of and Y2 6 is approximately (0.1225, 6). Therefore, an annual interest rate of 12.25% 5.4-11. is needed for the investment to double in 6 years. See Figure ■ annual interest. D ln 1 Y1 ln 2 1 r 2 ln 2 ln 1 1 r 2 7 0 1 1 Figure 5.4-11 Exercises 5.4 Unless stated otherwise, all letters represent positive numbers. In Exercises 1–4, find the value of each logarithm. 1. log 10,000 3. log 210 1000 2. log 0.001 4. log23 0.01 In Exercises 5–14, translate the given logarithmic statement into an equivalent exponential statement. 5. log 1000 3 6. log 0.001 3 7. log 750 2.8751 8. log 0.8 0.0969 9. ln 3 1.0986 10. ln 10 2.3026 11. ln 0.01 4.6052 12. ln s r 13. ln 1 x2 2y 2 z w 14. log a c 2 1 d In Exercises 15 – 24, translate the given exponential statement into an equivalent logarithmic statement. 15. 10 2 0.01 16. 103 1000 17. 100.4771 3 18. 107k r 19. e3.25 25.79 21. 12 7 5.5527 e 23. 2 r w e 20. 4 0.0183 e 22. ek t 24. e4uv m In Exercises 25–36, evaluate the given expression without using a calculator. 25. log 10243 26. log 102x2y2 27. ln e15 28. ln e3.78 31. eln 931 34. ln ex22y 29. l

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