Module 5 Lecture Notes Contents 5.1 5.1 An Introductory Example . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5.2 Symbolic Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 5.3 One-to-One and Invertibility . . . . . . . . . . . . . . . . . . . . . . . . . 7 5.4 Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 5.5 Graphing Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . 9 5.6 Exponential Function Introduction . . . . . . . . . . . . . . . . . . . . . 10 5.7 Modeling with Exponential Functions . . . . . . . . . . . . . . . . . . . 12 5.8 Transformations of Exponential Functions . . . . . . . . . . . . . . . . 15 5.9 What’s “e”? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 5.10 Finding the Formula for an Exponential Function (REQUIRED) . . . 22 An Introductory Example Example 1: Temperature in degrees Fahrenheit, F , can be written as a function of temperature in degrees Celsius, C. This relationship is given by F = g(C) = 59 C + 32. (a) Find and interpret g(100). 9 g(100) = (100) + 32 5 = 180 + 32 = 212 The temperature 100o C is equivalent to the temperature 212o F. 1 Math 111 Module 5 Lecture Notes (b) Solve and interpret the solution to g(C) = 32. g(C) = 32 9 C + 32 = 32 5 9 C=0 5 C=0 The temperature 32o F is equivalent to the temperature 0o C. (c) Solve the equation F = 59 C + 32 for C. 9 F = C + 32 5 9 F − 32 = C 5 5 (F − 32) = C 9 5 C = (F − 32) 9 A function f is said to be one-to-one if for every y-value in the range of f there is exactly one x-value in the domain of f . A function must be one-to-one in order to have an inverse. The inverse function of f reverses the process of the original function. In other words, the input and output switch roles. The original function is given by y = f (x). The inverse function is given by x = f −1 (y). If we want to graph both of these functions in the (x, y)-plane, then we use y = f −1 (x). The inverse function of f is denoted by f −1 . It is important to note that this 1 notation is not denoting a reciprocal. That is, f −1 (x) 6= . f (x) Instructor: A.E.Cary Page 2 of 25 Math 111 Module 5 Lecture Notes 5.2 Symbolic Examples Example 2: The function f defined by f (x) = 3x + 2 is one-to-one. Find its inverse. Then graph y = f (x) and y = f −1 (x) in Figure 5.1. Include the graph of y = x also. Original Function: y = 3x + 2 To find the inverse, we switch the variables x and y and solve for y: Figure 5.1 y y=x y = f (x) y = f −1 (x) 4 x = 3y + 2 x − 2 = 3y 1 (x − 2) = y 3 1 y = (x − 2) 3 1 −1 f (x) = (x − 2) 3 2 x −4 −2 2 4 −2 −4 In the example above, the variables x and y were switched. Notice how the graph of y = f −1 (x) is a reflection of the graph of y = f (x) about the line y = x. This is because all of the (x, y) pairs switched – the point (0, 2) is on the graph of y = f (x) while the point (2, 0) is on the graph of y = f −1 (x). Similarly, the point (1, 5) is on the graph of y = f (x) whereas the point (5, 1) is on the graph of y = f −1(x). Example 3: To verify that two functions are inverses, we show that f (f −1 (x)) = x and that f −1 (f (x)) = x. Do this for the previous example. f f −1 1 (x − 2) 3 1 =3 (x − 2) + 2 3 (x) = f Instructor: A.E.Cary f −1 (f (x)) = f −1 (3x + 2) = (x − 2) + 2 1 ((3x + 2) − 2) 3 1 = (3x) 3 = xX = xX = Page 3 of 25 Math 111 Module 5 Lecture Notes 2x is one-to-one. Find the inverse funcx−1 tion. Confirm that the inverse function you found is correct by showing f (f −1 (x)) = x and f −1 (f (x)) = x. Example 4: The function f defined by f (x) = − Original Function: y = − Finding the Inverse: 2x x−1 x=− 2y y−1 x(y − 1) = −2y xy − x = −2y −x = −2y − xy −x = y(−2 − x) −x =y −2 − x −x −2 − x −x y= −1(x + 2) x y= x+2 x f −1 (x) = x+2 y= Confirming the Inverse: Instructor: A.E.Cary Page 4 of 25 Math 111 Module 5 Lecture Notes f (f −1(x)) = f =− =− =− x x+2 x x+2 2 x x+2 2 f −1 x x+2 x − x+2 x+2 1 x+2 • −1 2x − (f (x)) = f x−1 2x − x−1 = 2x − x−1 +2 2x − x−1 x−1 = • 2x − x−1 + 2 x − 1 2x x − 1(x + 2) −1 = −2x −2x + 2(x − 1) −2x −2x + 2x − 2 −2x = −2 2x x−x−2 2x =− −2 =− = =x =x To state the domain and range of each f and f −1 , we need to recognize that the inputs of f are the outputs of f −1 . Likewise, the outputs of f are the inputs of f −1 . The domain of f is the range of f −1 . Similarly, the range of f is the domain of f −1 . Table 5.1 Function Domain Range f f −1 {x | x 6= 1} {x | x 6= −2} {x | x 6= −2} {x | x 6= 1} Instructor: A.E.Cary Page 5 of 25 Math 111 Module 5 Lecture Notes √ Example 5: The function g defined by g(x) = 3 x + 8 is one-to-one. Find the inverse function and confirm that it is the inverse by showing g (g −1 (x)) = x and g −1 (g(x)) = x. In Figure 5.2, use transformations to sketch y = g(x), y = g −1 (x) and y = x. √ Original Function y = 3 x + 8 Finding the Inverse: p x= 3 y+8 x3 = y + 8 x3 − 8 = y y = x3 − 8 g −1 (x) = x3 − 8 Confirming the Inverse: g −1 (g(x)) = g −1 −1 g g (x) = g(x3 − 8) p = 3 (x3 − 8) + 8 √ 3 = x3 = √ 3 √ 3 x+8 x+8 3 −8 =x+8−8 = xX = xX Figure 5.2 y y=x 10 y = g(x) y = g −1 (x) 8 6 4 2 x −10 −8 −6 −4 −2 −2 2 4 6 8 10 −4 −6 −8 −10 √ √ To graph y = 3 x + 8, shift the graph of y = 3 x left 8 units. To graph y = x3 − 8, shift the graph of y = x3 down 8 units. Instructor: A.E.Cary Page 6 of 25 Math 111 Module 5 Lecture Notes 5.3 One-to-One and Invertibility The horizontal line test is a way of determining if a function is one-to-one. It states that if every horizontal line passes through a graph at most once, then the function is one-to-one. In the same way that the vertical line test verifies if a graph represents a function, the horizontal line test verifies if the graph of a function is one-to-one (and thus invertible). Example 6: Graph y = f (x) for f (x) = x2 in Figure 5.3. Then graph y = g(x) for g(x) = x2 , x ≥ 0 in Figure 5.4. Is either function invertible? Why or why not? Figure 5.4: Graph of y = g(x) Figure 5.3: Graph of y = f (x) y y 4 4 2 2 x −4 −2 2 4 x −4 −2 −2 y=x y = x2 x=y 2 −4 2 4 −2 y=x −4 y = g(x) y=g −1 (x) We can see from the graph of y = x2 that this function cannot be inverted. Since this function fails the horizontal line test, when we reflect it across the line y = x, the resulting graph is not a function as it fails the vertical line test. However, if we restict the domain on y = x2 to x ≥ 0, the resulting function is indeed invertible. This is shown in Figure ??. Instructor: A.E.Cary Page 7 of 25 Math 111 Module 5 Lecture Notes 5.4 Numerical Examples Example 7: Use the functions f and g given in Table 5.2 to determine the following. Table 5.2 x -2 -1 0 1 2 f (x) 5 4 2 -1 1 g(x) 7 2 0 -2 9 Recall that if y = g(x) and g is invertible, then g −1 (y) = x. (a) g −1 (−2) g −1 (−2) = 1 as g(1) = −2 (b) f −1 (2) f −1 (2) = 0 as f (0) = 2 (c) f −1 (0) f −1 (0) is undefined as 0 is not in the range of f (d) f g −1 (0) f g −1 (0) = f (0) =2 Instructor: A.E.Cary Page 8 of 25 Math 111 Module 5 Lecture Notes 5.5 Graphing Inverse Functions Example 8: Graph the inverse function of f in Figure 5.5. To graph the inverse function, we will switch the inputs and outputs of the original function. For y = f (x), the points (−4, −4), (−3, −2), (−2, 0), (−1, 1), (0, 2), (1, 3) and (2, 4) are on the graph. Therefore the points (−4, −4), (−2, −3), (0, −2), (1, −1), (2, 0), (3, 1) and (4, 2) are on the graph of y = f −1 (x). Figure 5.5 y y=x y = f (x) y = f −1 (x) 4 2 x −4 −2 2 4 −2 −4 Instructor: A.E.Cary Page 9 of 25 Math 111 Module 5 Lecture Notes 5.6 Exponential Function Introduction In 1988, a judge in Yonkers, New York instituted an exponential fine on the city of Yonkers. Below is the background and scenario, published in the New York Times1 : Dec. 1, 1980: Justice Department sues Board of Education, City of Yonkers and Yonkers Community Development Agency, charging that the city racially discriminated in education and public housing. Nov. 20, 1985: Judge Leonard B. Sand of Federal District Court in Manhattan rules that Yonkers’s housing and schools were intentionally segregated by race. A housing remedy order directs the city to build 200 units of public housing and to plan additional subsidized housing. Jan. 28, 1988: City Council approves consent decree that sets timetable for building 200 units of public housing and commits city to an additional 800 subsidized units. July 26, 1988: Court sets Aug. 1 deadline for Council to adopt zoning amendment needed to build the 800 units. Aug. 1, 1988: Council rejects amendment in a 4-to-3 vote. Aug. 2, 1988: Judge Sand finds city and the four Councilmen who voted against the amendment in contempt of court and imposes fines. The city’s fines start at $100 and double every day. The Councilmen’s fines start at $500 a day and increase by $500 each day. Example 9: Let P be the amount fined (in dollars) t days after the fine begins. Complete the first three entries of each Table 5.3 and Table 5.4. Table 5.3: Councilmen t P Formula t P Formula 0 500 500 0 100 100 1 1000 500 + 500 1 200 100 · 2 2 1500 500 + 500(2) 2 400 (100 · 2) · 2 3 2000 500 + 500(3) 3 800 (100 · 2 · 2) · 2 4 2500 500 + 500(4) 4 1600 (100 · 2 · 2 · 2) · 2 5 .. . 3000 500 + 500(5) 5 .. . 3200 (100 · 2 · 2 · 2 · 2) · 2 .. . 500 + 500t t t 1 Table 5.4: City of Yonkers 100 · 2t http://www.nytimes.com/1988/09/10/nyregion/yonkers-legal-battle-how-it-unfolded.html Instructor: A.E.Cary Page 10 of 25 Math 111 Module 5 Lecture Notes Example 10: Figure 5.6: Councilmen: P = 500 + 500t Figure 5.7: City of Yonkers 3,600 3,000 P , fine in dollars P , fine in dollars 3,500 2,500 2,000 1,500 1,000 500 1 2 3 4 5 t, time in days • Linear function 3,200 2,800 2,400 2,000 1,600 1,200 800 400 6 1 2 3 4 5 6 t, time in days • Exponential function • Increasing by a constant amount • Increasing by a constant factor • The initial value is 500 • The initial value is 100 Example 11: On what day will will the city of Yonkers’ fine reach over $1,000,000? On day 13, the fine will be $819,200. On day 14, the fine will be $1,638,400. Example 12: How much will the city of Yonkers be fined on day 30? What will each of the Councilmen’s fines be on that day? P = 500 + 500(30) P = 100(2)30 = 15500 = 107,374,182,400 The Councilmen’s fine would be $15,500. The city’s fine would be $107,374,182,400. Instructor: A.E.Cary Page 11 of 25 Math 111 Module 5 Lecture Notes An exponential function is of the form f (x) = C ax where • C is the initial value • a is the growth factor and a > 0 Consequently, an exponential function is a function that increases or decreases at a constant percent rate. Let’s review percent increase and decrease as we work through these examples. 5.7 Modeling with Exponential Functions Example 13: You start a new job with an initial salary of $36,000 per year. Each year thereafter, you receive a 3% raise. Let S(t) be your salary t years after you start your new job. (a) Write the formula for S(t). t S(t) 0 36000 1 3 36000 + 0.03(36000) = 36000(1.03) 36000(1.03) (1.03) = 36000(1.03)2 36000(1.03) (1.03) = 36000(1.03)3 4 36000(1.03)4 5 .. . 36000(1.03)5 .. . t 36000(1.03)t 2 The formula is S(t) = 36000(1.03)t (b) What will your salary be after 10 years? S(10) = 36000(1.03)10 ≈ 48380.99 Your salary will be about $48,380.99 is 10 years. Instructor: A.E.Cary Page 12 of 25 Math 111 Module 5 Lecture Notes (c) When will your salary reach $72,000? (Use your graphing calculator to solve this). We need to solve 72000 = 36000(1.03)t . To do so with a graphing calculator, graph: • Graph y1(x) = 36000(1.03)x • Graph y2(x) = 72000 • Find the point of intersection The point of intersection is about (23.45, 72000). Your salary will reach $72000 after about 23.45 years. If you receive annual increases, this means that your salary will reach $72000 in 24 years. Instructor: A.E.Cary Page 13 of 25 Math 111 Module 5 Lecture Notes Example 14: A compost pile has 27 cubic feet of waste and decays at a rate of 10% per month. Let Q(t) be the volume of compost (in cubic feet) t months since decay began. Write the formula for this decreasing exponential function. Decreasing by 10% means that 90% remains. Thus to find the volume after 1 month, we multiply by the factor 0.9. Q, volume of compost in ft3 Figure 5.8: Exponential Compost Decay 30 25 27 20 15 15.94 10 9.41 5 5.56 5 10 15 20 25 3.28 1.94 1.14 30 t, time in months • Exponential function • Decaying by a constant factor • Decreasing by a smaller and smaller amount Table 5.5: Exponential Compost Decay t Q Formula 0 27 27(0.9)0 1 27(0.9) = 24.3 27(0.9)1 2 (27(0.9)) (0.9) = 21.87 27(0.9)2 3 .. . (27(0.9)(0.9)) (0.9) = 19.683 .. . 27(0.9)3 10 27(0.9)10 ≈ 9.414 27(0.9)10 t Instructor: A.E.Cary 27(0.9)t Page 14 of 25 Math 111 Module 5 Lecture Notes 5.8 Transformations of Exponential Functions Example 15: Graph of y = 2x in Figure 5.9. Use this to graph the various transformations listed. Figure 5.10: y = 2x + 1 Figure 5.9: y = 2x y y y=0 x y=2 6 6 4 4 2 2 x x −6 −4 −2 2 4 −6 6 −4 −2 2 4 6 −2 −2 −4 −4 y=0 y=1 y = 2x −6 −6 y = 2x + 1 Shift the graph of y = 2x up 1 unit. Figure 5.11: y = 3 · 2x Figure 5.12: y = −2x y y 6 6 4 4 2 2 x −6 −4 −2 2 4 6 x −6 −4 −2 y=0 −4 y=1 y = 2x 2 4 6 −2 y=0 −4 y=1 −6 y = 3 · 2x Stretch the graph of y = 2x vertically by a factor of 3. Instructor: A.E.Cary −2 y = 2x −6 y = −2x Reflect the graph of y = 2x across the x-axis. Page 15 of 25 Math 111 Module 5 Lecture Notes Figure 5.13: y = 2−x Figure 5.14: y = 1 − 2x y y y=0 y=1 6 6 y = 2x y = 1 − 2x 4 2 4 2 x −6 −4 −2 y=0 2 4 6 x −6 −4 −2 2 −2 −2 −4 −4 −6 −6 4 6 y=1 y = 2x y = 2−x Reflect the graph of y = 2x across the y-axis. Instructor: A.E.Cary Reflect the graph of y = 2x across the x-axis and then shift up 1 unit. Page 16 of 25 Math 111 Module 5 Lecture Notes If ax = ay , then x = y. Pertinent Exponent Rules: • a−n = 1 an • (am )n = amn • am an = am+n Solving Exponential Equations Example 16: Solve the following equations for x. List the solution set. (a) 5x = 5−6 5x = 5−6 x = −6 Solution Set: {−6} (b) 22x−1 = 4 22x−1 = 4 22x−1 = 22 2x − 1 = 2 2x = 3 x= Solution Set: (c) 42x−5 = 3 2 3 2 1 16 Instructor: A.E.Cary Page 17 of 25 Math 111 Module 5 Lecture Notes 1 16 1 = 2 4 42x−5 = 42x−5 42x−5 = 4−2 2x − 5 = −2 2x = 3 x= Solution Set: 3 2 3 2 (d) 53x−7 = 125 53x−7 = 125 53x−7 = 53 3x − 7 = 3 3x = 10 x= Solution Set: (e) 4x 2 −7 10 3 10 3 = 642x 4x 2 −7 4x 2 −7 4x 2 −7 = 642x 2x = 43 = 43∗2x x2 − 7 = 6x x2 − 6x − 7 = 0 (x − 7)(x + 1) = 0 x − 7 = 0 or x + 1 = 0 x = 7 or x = −1 Solution Set: {−1, 7} Instructor: A.E.Cary Page 18 of 25 Math 111 Module 5 Lecture Notes 2 (f) 92x · 27x = 3−1 2 3 · 27x = 92x x2 2x 3 · 33 = 32 2 31 · 33x = 34x 2 31+3x = 34x 1 + 3x2 = 4x 3x2 − 4x + 1 = 0 (3x − 1)(x − 1) = 03x − 1 x= Solution Set: Instructor: A.E.Cary 1 3 1 3 = 0 or x − 1 = 0 or x = 1 ,1 Page 19 of 25 Math 111 Module 5 Lecture Notes 5.9 What’s “e”? The number e is a number that occurs in nature, and is a frequent base for exponential and logarithmic expressions. It is defined by: n 1 e = lim 1 + n→∞ n It can also be expressed by the following: e= 1 1 1 1 1 1 + + + + + + ··· 0! 1! 2! 3! 4! 5! This number is irrational and is approximated by 2.718281828. The graph of the function given by y = ex looks a lot like the graphs of the functions given by y = 2x and y = 3x , as shown in Figure 5.15. In calculus, you will study that the special property of e is that the slope of the tangent line at zero is exactly 1, as shown in Figure 5.16. Figure 5.15 Figure 5.16 y y=0 y y=0 5 y = 2x 5 y = ex y = ex y =x+1 y = 3x 4 4 3 3 2 2 1 1 x −3 −2 −1 1 2 x −3 3 −1 −2 −1 1 2 3 −1 Example 17: Solve the following equation. e3x = e2−x e5x = e3+x 5x = 3 + x 4x = 3 x= Solution Set: 3 4 3 Instructor: A.E.Cary 4 Page 20 of 25 Math 111 Module 5 Lecture Notes Example 18: In 1990, the population of Oregon was 2.84 million people. In 2010, the population of Oregon was 3.83 million people. Let P (t) be the population of Oregon in millions, where t is the number of years after 2000. This can be modeled by P (t) = 3.298e0.015t . (a) According to this model, what will the population be in 2020? P (20) = 3.298e0.015∗20 ≈ 4.452 In 2020, there will be approximately 4.452 million people in Oregon. (b) According to this model, when will the population reach 4 million people? Use your graphing calculator to solve this. Graphing y1(x) = 4 and y2(x) = 3.298e0.015x , we find the point of intersection to be approximately (12.87,4). The population is expected to reach 4 million in about 12.87 years. Instructor: A.E.Cary Page 21 of 25 Math 111 Module 5 Lecture Notes 5.10 Finding the Formula for an Exponential Function (REQUIRED) Example 19: Find an algebraic rule (or formula) for an exponential function f that passes through the points (−1, 8) and (2, 1). Also find the algebraic rule (or formula) for a linear function g that passes through the points (−1, 8) and (2, 1). The general equation for an exponential function is f (x) = Cax . We need to find the appropriate values for C and a. As the function passes through the points (−1, 8) and (2, 1), we have f (−1) = 8 and f (2) = 1. Since f (x) = Cax , we then have Ca−1 = 8 and Ca2 = 1. Thus we can write and solve the following ratio based on these two equations: 1 Ca2 = −1 Ca 8 12 3 4 = Ca2 Ca−2 1 Ca2 = 8 Ca−1 1 a3 = 8 1 a= 2 Now that we have the value of a, we can solve for C using one of the original function values: Ca2 = 1 2 1 C =1 2 C 1 =1 4 C=4 Therefore f (x) = 4 Note that 4 1 x 2 1 x . 2 6= 2x . We can (and should) check that this function satisfies f (−1) = 8 and f (2) = 1: Check: Instructor: A.E.Cary Page 22 of 25 Math 111 Module 5 Lecture Notes −1 1 f (−1) = 4 2 =4·2 =8X 2 1 f (2) = 4 2 =1X To find the linear function passing through the points (−1, 8) and (2, 1), we need to find the slope and the y-intercept: 1−8 m= 2 − (−1) −7 3 7 =− 3 7 Thus y = − 3 x + b. We can use one of the points to find b: = 7 1 = − (2) + b 3 14 1=− +b 3 17 =b 3 The formula of the linear function passing through (−1, 8) and (2, 1) is g(x) = − 73 x + 17 . 3 Figure 5.17 y 10 y= x 1 2 7 −3x y=4 + 17 3 8 6 4 2 x −2 Instructor: A.E.Cary 2 4 6 8 Page 23 of 25 Math 111 Module 5 Lecture Notes Example 20: Find an algebraic rule (or formula) for an exponential function f that passes 3 through the points −2, 4 and (2, 12). The general equation for an exponential function is f (x) = Cax . We needto find the appropriate values for C and a. As the function passes through the points −2, 34 and (2, 12), we have f (−2) = 43 and f (2) = 12. Since f (x) = Cax , we then have Ca−2 = 34 and Ca2 = 12. Thus we can write the following two ratios: Ca2 12 = 3 −2 Ca 4 Ca2 = 16 Ca−2 a2 = 16 a−2 a4 = 16 a = ±(16)1/4 a = ±2 Note that a = 2 is the only relevant value. Now that we have the value of a, we can solve for C using one of the original function values: 12 = Ca2 12 = C(2)2 12 = 4C 3=C Therefore f (x) = 3(2)x . We can (and should) check that this function satisfies f (−2) = Check: 3 4 and f (2) = 12: f (−2) = 3(2)−2 1 =3 4 = 3 X 4 f (2) = 3(2)2 = 12 X Instructor: A.E.Cary Page 24 of 25 Math 111 Module 5 Lecture Notes Example 21: Find an algebraic rule (or formula) for an exponential function f that passes through the points (1, 8) and (3, 128). The general equation for an exponential function is f (x) = Cax . We need to find the appropriate values for C and a. As the function passes through the points (1, 8) and (3, 128), we have f (1) = 8 and f (3) = 128. Since f (x) = Cax , we then have Ca1 = 8 and Ca3 = 128. Thus we can write and solve the following ratios: Ca3 128 = 1 Ca 8 Ca3 = 16 Ca1 a3 = 16 a1 a2 = 16 √ a = ± 16 a = ±4 Note that a = 4 is the only relevant value. Now that we have the value of a, we can solve for C using one of the original function values: 8 = Ca1 8 = C(4)1 8 = 4C 2=C Therefore f (x) = 2(4)x . We can (and should) check that this function satisfies f (1) = 8 and f (3) = 128: Check: f (1) = 2(4)1 =8X f (3) = 2(4)3 = 2 · 64 = 128 X Instructor: A.E.Cary Page 25 of 25

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