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1-7 Inverse Relations and Functions Graph each function using a graphing calculator, and apply the horizontal line test to determine whether its inverse function exists. Write yes or no. 2 3. f (x) = x2 – 10x + 25 SOLUTION: 2 1. f (x) = x + 6x + 9 SOLUTION: 2 The graph of f (x) = x – 10x + 25 below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. The graph of f (x) = x + 6x + 9 below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. 4. f (x) = 3x − 8 SOLUTION: It appears from the portion of the graph of f (x) = 3x − 8 shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. 2. f (x) = x2 – 16x + 64 SOLUTION: 2 The graph of f (x) = x – 16x + 64below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. 5. f (x) = SOLUTION: It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. 2 3. f (x) = x – 10x + 25 SOLUTION: 2 The graph of f (x) by = xCognero – 10x eSolutions Manual - Powered + 25 below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. Page 1 1-7 Inverse Relations and Functions 5. f (x) = 7. f (x) = SOLUTION: SOLUTION: It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. 6. f (x) = 4 SOLUTION: The graph of f (x) = 4 below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. That horizontal line is the same as the graph of f (x) itself. Therefore, you can conclude that an inverse function does not exist. 7. f (x) = SOLUTION: It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. eSolutions Manual - Powered by Cognero 8. f (x) = −4x2 + 8 SOLUTION: 2 The graph of f (x) = −4x + 8 below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. 9. f (x) = SOLUTION: It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) morePage than2 once. Therefore, you can conclude that an inverse function does exist. 1-7 Inverse Relations and Functions 11. f (x) = x3 – 9 9. f (x) = SOLUTION: SOLUTION: 3 shown below It appears from the portion of the graph of f (x) = It appears from the portion of the graph of f (x) = x – 9 shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. 12. f (x) = x 3 SOLUTION: 10. f (x) = It appears from the portion of the graph of f (x) = SOLUTION: shown below It appears from the portion of the graph of f (x) = 3 x shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. Determine whether f has an inverse function. If it does, find the inverse function and state any restrictions on its domain. 13. g(x) = −3x4 + 6x2 – x 11. f (x) = x3 – 9 SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero 4 3 It appears from the portion of the graph of f (x) = x – 9 shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. 2 Page 3 The graph of g(x) = −3x + 6x – x below shows that it is possible to find a horizontal line that intersects the graph of g(x) more than once. Therefore, you can conclude that an inverse function does not exist. 1-7 Inverse Relations and Functions Determine whether f has an inverse function. If it does, find the inverse function and state any restrictions on its domain. 15. h(x) = x7 + 2x3 − 10x2 SOLUTION: 13. g(x) = −3x4 + 6x2 – x 7 SOLUTION: 4 2 The graph of g(x) = −3x + 6x – x below shows that it is possible to find a horizontal line that intersects the graph of g(x) more than once. Therefore, you can conclude that an inverse function does not exist. 3 2 The graph of h(x) = x + 2x − 10x below shows that it is possible to find a horizontal line that intersects the graph of h(x) more than once. Therefore, you can conclude that an inverse function does not exist. 16. f (x) = 5 14. f (x) = 4x – 8x 4 SOLUTION: SOLUTION: 5 4 The graph of f (x) = 4x – 8x below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. The function f has domain [–8, ) and range [0, ). 15. h(x) = x7 + 2x3 − 10x2 SOLUTION: 7 3 2 The graph of h(x) = x + 2x − 10x below shows that it is possible to find a horizontal that intersects the graph of h(x) more than once. eSolutions Manual - Poweredline by Cognero Therefore, you can conclude that an inverse function does not exist. −1 2 Page 4 f (x) = x − 8 2 From the graph y = x – 8 below, you can see that the inverse relation has domain (– , ) and range [8, ). 1-7 Inverse Relations and Functions −1 2 f (x) = x − 8 2 From the graph y = x – 8 below, you can see that the inverse relation has domain (– , ) and range [8, ). By restricting the domain of the inverse relation to [0, and range of f are equal to the range and domain of f −1 2 Therefore, f (x) = x − 8 for x ≥ 0. 18. f (x) = | x – 6 | SOLUTION: The graph of f (x) = | x – 6 | below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. ), the domain –1 , respectively. 17. f (x) = 19. f (x) = SOLUTION: SOLUTION: The graph of f (x) = below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist. It appears from the portion of the graph of f (x) = 18. f (x) = | x – 6 | SOLUTION: The graph of f (x) = | x – 6 | below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. eSolutions Manual - Powered by Cognero Therefore, you can conclude that an inverse function does not exist. shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. The function f has domain (– (–1, ). , 0) (0, ) and range [– , –1) Page 5 20. g(x) = 1-7 Inverse Relations and Functions The function f has domain (– , 0) (0, (–1, ) and range [– , –1) SOLUTION: ). It appears from the portion of the graph of g(x) = shown below that there is no horizontal line that intersects the graph of g(x) more than once. Therefore, you can conclude that an inverse function does exist. f –−1 (x) = From the graph y = has domain (– , –1) below, you can see that the inverse relation (–1, ) and range (– , 0) (0, The function g has domain (– (1, ). , 0) (0, ) and range [– , 1) ). –1 The domain and range of f are equal to the range and domain of f , –−1 respectively. Therefore, no further restrictions are necessary. f (x) = g−1(x) = From the graph y = for x ≠ −1. has domain [– , 1) below, you can see that the inverse relation (1, ) and range (– , 0) (0, ). 20. g(x) = SOLUTION: It appears from the portion of the graph of g(x) = eSolutions Manual - Powered by Cognero shown below that there is no horizontal line that intersects the graph of g(x) more than once. Therefore, you can conclude that an inverse function does exist. Page 6 –1 g−1(x) = From the graph y= below, you can see that the inverse relation 1-7 Inverse Relations and Functions has domain [– , 1) (1, ) and range (– (0, , 0) ). The function f has domain (– , 8) and range (0, ). –1 The domain and range of g are equal to the range and domain of g , –−1 respectively. Therefore, no further restrictions are necessary. g (x) = for x ≠ 1. 21. f (x) = f SOLUTION: –−1 (x) = 8 − . From the graph y = shown below It appears from the portion of the graph of f (x) = has domain [– , 0) below, you can see that the inverse relation (0, ) and range (– , 8). that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. By restricting the domain of the inverse relation to (0, ∞), the domain and range of f are equal to the range and domain of f The function f has domain (– , 8) and range (0, ). Therefore, f −1 (x) = 8 − –1 , respectively. for x > 0. 22. g(x) = SOLUTION: eSolutions Manual - Powered by Cognero Page 7 It appears from the portion of the graph of g(x) = shown Therefore, f −1 (x) = 8 − for x > 0. g−1(x) = −3 + 1-7 22. Inverse g(x) = Relations and Functions From the graph y = −3 + relation has domain (– below, you can see that the inverse , 0) (0, ) and range (–3, ). SOLUTION: shown It appears from the portion of the graph of g(x) = below that there is no horizontal line that intersects the graph of g(x) more than once. Therefore, you can conclude that an inverse function does exist. By restricting the domain of the inverse relation to (0, ), the domain –1 and range of g are equal to the range and domain of g , respectively. Therefore, g –−1 (x) = −3 + for x > 0. 23. f (x) = The function g has domain (–3, ) and range (0, ). SOLUTION: It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist. The function f has domain (– (6, ). g−1(x) = −3 + From the graph y = −3 + relation has domain (– , 8) (8, ) and range (– , 6) below, you can see that the inverse , 0) eSolutions Manual - Powered by Cognero (0, ) and range (–3, ). Page 8 24. h(x) = SOLUTION: 1-7 Inverse Relations and Functions The function f has domain (– , 8) (8, (6, ) and range (– , 6) It appears from the portion of the graph of h(x) = ). shown below that there is no horizontal line that intersects the graph of h(x) more than once. Therefore, you can conclude that an inverse function does exist. f –−1 (x) = The function h has domain has domain (– , 6) . below, you can see that the inverse relation From the graph y = (6, ) and range (– , 8) (8, and range ). –1 The domain and range of f are equal to the range and domain of f , –−1 respectively. Therefore, no further restrictions are necessary. f (x) = for x −1 h (x) = 6. From the graph y = 24. h(x) = has domain below, you can see that the inverse relation and range . SOLUTION: It appears from the portion of the graph of h(x) = shown below that there is no horizontal line that intersects the graph of h(x) more than once. Therefore, you can conclude that an inverse function does exist. eSolutions Manual - Powered by Cognero Page 9 h (x) = From the graph y = below, you can see that the inverse relation 1-7 Inverse Relations and Functions has domain and range . 26. SPEED The speed of an object in kilometers per hour y is y = 1.6x, where x is the speed of the object in miles per hour. a. Find an equation for the inverse of the function. What does each variable represent? b. Graph each equation on the same coordinate plane. SOLUTION: a. –1 The domain and range of h are equal to the range and domain of h , −1 respectively. Therefore, no further restrictions are necessary. h (x) = for x ≠ . 25. g(x) = | x + 1 | + | x – 4 | y = speed in mi/h, x = speed in km/h b. SOLUTION: The graph of g(x) = | x + 1 | + | x – 4 || below shows that it is possible to find a horizontal line that intersects the graph of g(x) more than once. Therefore, you can conclude that an inverse function does not exist. Show algebraically that f and g are inverse functions. 27. f (x) = −6x + 3 g(x) = 26. SPEED The speed of an object in kilometers per hour y is y = 1.6x, where x is the speed of the object in miles per hour. a. Find an equation for the inverse of the function. What does each variable represent? b. Graph each equation on the same coordinate plane. SOLUTION: SOLUTION: a. eSolutions Manual - Powered by Cognero Page 10 1-7 Inverse Relations and Functions Show algebraically that f and g are inverse functions. 27. f (x) = −6x + 3 29. f (x) = −3x2 + 5; x ≥ 0 g(x) = g(x) = SOLUTION: SOLUTION: 28. f (x) = 4x + 9 g(x) = 30. f (x) = SOLUTION: + 8; x ≥ 0 g(x) = SOLUTION: 2 29. f (x)Manual = −3x- Powered + 5; x ≥ 0 eSolutions by Cognero g(x) = Page 11 1-7 Inverse Relations and Functions 30. f (x) = + 8; x ≥ 0 31. f (x) = 2x3 – 6 g(x) = SOLUTION: SOLUTION: 31. f (x) = 2x3 – 6 32. f (x) = g(x) = SOLUTION: eSolutions Manual - Powered by Cognero − 8; x ≥ 0 SOLUTION: Page 12 1-7 Inverse Relations and Functions 34. g(x) = 32. f (x) = + 5 2 f (x) = x – 10x + 33; x ≥ 5 − 8; x ≥ 0 g(x) = SOLUTION: SOLUTION: 35. f (x) = 33. g(x) = − 4 2 f (x) = x + 8x + 8; x ≥ −4 g(x) = SOLUTION: SOLUTION: eSolutions Manual 34. g(x) = - Powered + 5by Cognero 2 f (x) = x – 10x + 33; x ≥ 5 Page 13 1-7 Inverse Relations and Functions 35. f (x) = 36. f (x) = g(x) = g(x) = SOLUTION: SOLUTION: eSolutions 36. f (x)Manual = - Powered by Cognero g(x) = Page 14 36. f (x) = g(x) = 1-7 Inverse Relations and Functions SOLUTION: 37. PHYSICS The kinetic energy of an object in motion in joules can be 2 described by f (x) = 0.5mx , where m is the mass of the object in kilograms and x is the velocity of the object in meters per second. a. Find the inverse of the function. What does each variable represent? b. Show that f (x) and the function you found in part a are inverses. c. Graph f (x) and f –1(x) on the same graphing calculator screen if the mass of the object is 1 kilogram. SOLUTION: a. g(x) = velocity in m/s, x = kinetic energy in joules, m = mass in kg b. eSolutions Manual - Powered by Cognero Page 15 37. PHYSICS The kinetic energy of an object in motion in joules can be 2 described by f (x) = 0.5mx , where m is the mass of the object in Because = = x, the functions are inverse when the domain g(x) = velocity in m/s, xand = kinetic energy in joules, m = mass in kg 1-7 Inverse Relations Functions b. Use the graph of each function to graph its inverse function. 38. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. Because = is restricted to [0, ). = x, the functions are inverse when the domain c. Use the graph of each function to graph its inverse function. 39. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. 38. eSolutions Manual - Powered by Cognero SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. Page 16 1-7 Inverse Relations and Functions 39. 40. SOLUTION: SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. 40. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. eSolutions Manual - Powered by Cognero 41. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. Page 17 1-7 Inverse Relations and Functions 42. 41. SOLUTION: SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. 43. 42. SOLUTION: SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. eSolutions Manual - Powered by Cognero Page 18 1-7 Inverse Relations and Functions 44. JOBS Jamie sells shoes at a department store after school. Her base salary each week is $140, and she earns a 10% commission on each pair of shoes that she sells. Her total earnings f (x) for a week in which she sold x dollars worth of shoes is f (x) = 140 + 0.1x. −1 −1 a. Explain why the inverse function f (x) exists. Then find f (x). b. What do f −1(x) and x represent in the inverse function? c. What restrictions, if any, should be placed on the domains of f (x) and 43. −1 SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph. f (x)? Explain. d. Find Jamie’s total sales last week if her earnings for that week were $220. SOLUTION: a. Sample answer: The graph of the function is linear, so it passes the horizontal line test. Therefore, it is a one-to-one function and it has an inverse. 44. JOBS Jamie sells shoes at a department store after school. Her base salary each week is $140, and she earns a 10% commission on each pair of shoes that she sells. Her total earnings f (x) for a week in which she sold x dollars worth of shoes is f (x) = 140 + 0.1x. −1 −1 a. Explain why the inverse function f (x) exists. Then find f (x). b. What do f −1(x) and x represent in the inverse function? c. What restrictions, if any, should be placed on the domains of f (x) and −1 f (x) = 10x − 1400 −1 b. x represents Jamie’s earnings for a week, and f (x) represents her sales. c. x ≥ 0; Jamie cannot have negative sales. d. −1 f (x)? Explain. d. Find Jamie’s total sales last week if her earnings for that week were $220. SOLUTION: a. Sample answer: The graph of the function is linear, so it passes the horizontal line test. Therefore, it is a one-to-one function and it has an inverse. eSolutions Manual - Powered by Cognero 45. CURRENCY The average exchange rate from Euros to U.S. dollars for a recent four-month period can be described by f (x) = 0.66x, where x is the currency value in Euros. a. Explain why the inverse function f −1(x) exists. Then find f −1(x). −1 b. What do f (x) and x represent in the inverse function? Page 19 c. What restrictions, if any, should be placed on the domains of f (x) and −1 f (x)? Explain. c. x ≥ 0; You cannot exchange negative money. d. d. 1-7 Inverse Relations and Functions Determine whether each function has an inverse function. 45. CURRENCY The average exchange rate from Euros to U.S. dollars for a recent four-month period can be described by f (x) = 0.66x, where x is the currency value in Euros. a. Explain why the inverse function f −1(x) exists. Then find f −1(x). −1 b. What do f (x) and x represent in the inverse function? c. What restrictions, if any, should be placed on the domains of f (x) and −1 f (x)? Explain. d. What is the value in Euros of 100 U.S. dollars? 46. SOLUTION: SOLUTION: a. Sample answer: The graph of the function is linear, so it passes the horizontal line test. Therefore, it is a one-to-one function and it has an inverse. The graph does not pass the Horizontal Line Test. b. x represents the value of the currency in U.S. dollars, and f −1 (x) represents the value of the currency in Euros. c. x ≥ 0; You cannot exchange negative money. d. Determine whether each function has an inverse function. 47. SOLUTION: The graph passes the Horizontal Line Test. 46. SOLUTION: The graph does not pass the Horizontal Line Test. eSolutions Manual - Powered by Cognero Page 20 48. 49. 47. SOLUTION: SOLUTION: The graph passes the Horizontal Line Test. The graphRelations passes the Horizontal Line Test. 1-7 Inverse and Functions Determine if f −1 exists. If so, complete a table for f −1. 50. SOLUTION: No output value corresponds with more than one input value, so an inverse exists. 48. SOLUTION: The graph does not pass the Horizontal Line Test. 51. SOLUTION: An output value corresponds with more than one input value, so an inverse does not exist. 52. SOLUTION: An output value corresponds with more than one input value, so an inverse does not exist. 49. SOLUTION: The graph passes the Horizontal Line Test. Determine if f −1 exists. If so, complete a table for f −1. 53. SOLUTION: No output value corresponds with more than one input value, so an inverse exists. 50. SOLUTION: NoManual output- Powered value corresponds eSolutions by Cognero inverse exists. with more than one input value, so an 54. TEMPERATURE The formula f (x) = Page 21 x + 32 is used to convert x degrees Celsius to degrees Fahrenheit. To convert x degrees No output value corresponds with more than one input value, so an inverse exists. 1-7 Inverse Relations and Functions 54. TEMPERATURE The formula f (x) = −1 f represents the formula used to convert degrees Fahrenheit to degrees Celsius. b. x + 32 is used to convert x degrees Celsius to degrees Fahrenheit. To convert x degrees Fahrenheit to Kelvin, the formula k(x) = (x + 459.67) is used. −1 a. Find f . What does this function represent? b. Show that f and f −1 are inverse functions. Graph each function on the same graphing calculator screen. c. Find (k o f )(x).What does this function represent? d. If the temperature is 60°C, what would the temperature be in Kelvin? SOLUTION: a. c. −1 f represents the formula used to convert degrees Fahrenheit to degrees Celsius. b. [k o f ](x) = x + 273.15; represents the formula used to convert degrees Celsius to degrees Kelvin. d. eSolutions Manual - Powered by Cognero Restrict the domain of each function so that the resulting function is one-to-one. Then determine the inverse of the function. Page 22 Celsius to degrees Kelvin. d. 1-7 Inverse Relations and Functions f −1 (x) = + 5 Restrict the domain of each function so that the resulting function is one-to-one. Then determine the inverse of the function. 56. SOLUTION: The domain needs to be restricted so that the graph passes the Horizontal Line Test. There is more than one way to accomplish this. Sample answer: x ≤ −9 55. SOLUTION: The domain needs to be restricted so that the graph passes the Horizontal Line Test. There is more than one way to accomplish this. Sample answer: x ≥ 5 We are selecting the negative side of the domain, so we will replace the absolute value symbols with the opposite of the expression. f −1 (x) = + 5 −1 f (x) = x – 11 Since the range of the restricted function f is y ≤ –2, we must restrict the inverse relation so that the domain is x ≤ 2. Therefore, f 11. −1 (x) = x – 56. SOLUTION: The domain needs to be restricted so that the graph passes the Horizontal Line Test. There is more than one way to accomplish this. Sample answer: x ≤ −9 57. eSolutions Manual - Powered by Cognero SOLUTION: The domain needs to be restricted so that the graph passes the Page 23 −1 f (x) = x – 11 Since the range of the restricted function f is y ≤ –2, we must restrict the inverseRelations relation so that theFunctions domain is x ≤ 2. Therefore, f 1-7 Inverse and −1 We are using the positive side of the domain, so we only need the positive value of y . (x) = x – −1 f (x) = 11. 57. 58. SOLUTION: SOLUTION: The domain needs to be restricted so that the graph passes the Horizontal Line Test. There is more than one way to accomplish this. Sample answer: x ≥ 0 The domain needs to be restricted so that the graph passes the Horizontal Line Test. There is more than one way to accomplish this. Sample answer: x ≥ −5 We are using the positive side of the domain, so we only need the positive value of y . We are selecting the positive side of the domain, so we will replace the absolute value symbols with the expression. −1 f (x) = −1 f (x) = x – 1 Since the range of the restricted function f is y ≥ –4, we must restrict the inverse relation so that the domain is x ≥ –4. Therefore, f – 1, x ≥ –4. −1 (x) = x State the domain and range of f and f −1, if f −1 exists. 58. 59. f (x) = SOLUTION: The domain needs to be restricted so that the graph passes the Horizontal Line Test. There is more than one way to accomplish this. Sample answer: x ≥ −5 We are selecting the positive side of the domain, so we will replace the absolute value symbols with the expression. eSolutions Manual - Powered by Cognero SOLUTION: For f , the square root must be at least 0, so x must be greater than or equal to 6. f: D = {x | x ≥ 6, x R}, R = {y| y ≥ 0, y R} Graph f . Page 24 −1 f (x) = x – 1 Since the range of the restricted function f is y ≥ –4, we must restrict the inverseRelations relation so that theFunctions domain is x ≥ –4. Therefore, f 1-7 Inverse and −1 (x) = x – 1, x ≥ –4. State the domain and range of f and f −1, if f −1 exists. 59. f (x) = −1 By restricting the domain of f to [0, ∞], the range becomes [6, ∞]. −1 f : D = {x| x ≥ 0, y R}, R = {y | y ≥ 6, x R} 60. f (x) = x2 + 9 SOLUTION: SOLUTION: For f , the square root must be at least 0, so x must be greater than or equal to 6. f: D = {x | x ≥ 6, x R}, R = {y| y ≥ 0, y R} Graph f . f −1 does not exist 61. f (x) = SOLUTION: The graph of f passes the Horizontal Line Test, so f −1 exists. −1 Graph f . There appears to be asymptotes at x = 4 and y = 3. f: D = {x | x 4, x R}, R = {y | y 3, x R} The graph of f passes the Horizontal Line Test, so f −1 exists. By restricting the domain of f to [0, ∞], the range becomes [6, ∞]. −1 f : D = {x| x ≥ 0, y R}, R = {y | y ≥ 6, x R} 60. f (x) = x2 + 9 SOLUTION: eSolutions Manual - Powered by Cognero Page 25 1-7 There appears to be asymptotes at x = 4 and y = 3. f: D = {x | x 4, x R}, R = {y | y 3, x R} Inverse Relations and Functions The graph of f passes the Horizontal Line Test, so f −1 exists. There appears to be asymptotes at x = 3 and y = 4. f: D = {x | x 3, x R}, R = {y | y 4, x R} The domain and range restrictions of f correspond with the range and −1 domain restrictions of f . Therefore, no further restrictions are needed. −1 f : D = {x | x 3, x R}, R = {y | y 4, x R} 62. f (x) = The domain and range restrictions of f correspond with the range and −1 domain restrictions of f . Therefore, no further restrictions are needed. −1 f : D = {x | x 4, x R}, R = {y | y 3, x R} 63. ENVIRONMENT Once an endangered species, the bald eagle was SOLUTION: Graph f . There appears to be asymptotes at x = 3 and y = 4. f: D = {x | x 3, x R}, R = {y | y 4, x R} Manual - Powered by Cognero eSolutions downlisted to threatened status in 1995. The table shows the number of nesting pairs each year. a. Use the table to approximate a linear function that relates the number of nesting pairs to the year. Let 0 represent 1984. Page 26 b. Find the inverse of the function you generated in part a. What does each variable represent? c. Using the inverse function, in approximately what year was the An inverse function is f −1 (x) = , where x represents the −1 1-7 Inverse Relations and Functions a. Use the table to approximate a linear function that relates the number of nesting pairs to the year. Let 0 represent 1984. b. Find the inverse of the function you generated in part a. What does each variable represent? c. Using the inverse function, in approximately what year was the number of nesting pairs 5094? SOLUTION: a. Use two points from the table to write the related equation of the function. Because x = 0 corresponds to 1984, two points that can be used are (0, 1757) and (21, 7066). First, find the slope of the line. One of the points chosen is (0, 1757), so the y-intercept is 1757. So, the related equation is y = 252.81x + 1757. Therefore, a linear function that can be used to relate the number of nesting pairs to the year is f (x) = 252.81x + 1757. b. number of nesting pairs and f (x) represents the number of years after 1984. c. Substitute 5094 for x in f ––1(x). Because x = 0 corresponds to 1984, 13.20 corresponds to 1984 + 13.2 or about 1997.2. Therefore, the number of nesting pairs was 5094 in 1997. 64. FLOWERS Bonny needs to purchase 75 flower stems for banquet decorations. She can choose between lilies and hydrangea, which cost $5.00 per stem and $3.50 per stem, respectively. a. Write a function for the total cost of the flowers. b. Find the inverse of the cost function. What does each variable represent? c. Find the domain of the cost function and its inverse. d. If the total cost for the flowers was $307.50, how many lilies did Bonny purchase? SOLUTION: a. Let x represent the number of stems of hydrangea; b. c(x) = 3.5x + 5(75 − x). An inverse function is f −1 (x) = , where x represents the −1 number of nesting pairs and f (x) represents the number of years after 1984. c. Substitute 5094 for x in f ––1(x). eSolutions Manual - Powered by Cognero c −1 (x) = 250 − ; x represents the total cost and c −1 (x) represents the number of stems of hydrangea c. Bonny is not purchasing more than 75 stems. Therefore, the domain Page 27 of c(x) is {x | 0 ≤ x ≤ 75, x ∈ R}. The range of c(x) is from c(0) to c(75). c −1 (x) = 250 − ; x represents the total cost and c −1 (x) represents Remember that x is the number of hydrangea. The number of lilies is 1-7 Inverse Relations and Functions the number of stems of hydrangea c. Bonny is not purchasing more than 75 stems. Therefore, the domain of c(x) is {x | 0 ≤ x ≤ 75, x ∈ R}. The range of c(x) is from c(0) to c(75). 75 − 45 or 30. Find an equation for the inverse of each function, if it exists. Then graph the equations on the same coordinate plane. Include any domain restrictions. 65. SOLUTION: The range of the first equation is y ≥ 16, which is equivalent to the domain of f The domain of c −1(x) is equal to the range of c(x), or {x | 262.5 ≤ x ≤ 375, x ∈R} d. −1 (x). The domain of f (x) is for negative values of x, so the inverse of f (x) = 2 −1 x is f (x) = − . The range of the second equation is y < 13, which is equivalent to the −1 domain of f (x). Remember that x is the number of hydrangea. The number of lilies is 75 − 45 or 30. Find an equation for the inverse of each function, if it exists. Then graph the equations on the same coordinate plane. Include any domain restrictions. 65. SOLUTION: The range of the first equation is y ≥ 16, which is equivalent to the domain of f −1 (x). 66. eSolutions Manual - Powered by Cognero The domain of f (x) is for negative values of x, so the inverse of f (x) = 2 x is f −1 (x) = − . Page 28 SOLUTION: The graph of f (x) does not pass the Horizontal Line Test. Therefore, f 1-7 Inverse Relations and Functions −1 does not exist. 67. FLOW RATE The flow rate of a gas is the volume of gas that passes 66. through an area during a given period of time. The velocity v of air flowing through a vent can be found using v(r) = SOLUTION: , where r is the flow rate in cubic feet per second and A is the cross-sectional area of the vent in square feet. The graph of f (x) does not pass the Horizontal Line Test. Therefore, f −1 does not exist. a. Find v−1 of the vent shown. What does this function represent? b. Determine the velocity of air flowing through the vent in feet per second if the flow rate is 15,000 67. FLOW RATE The flow rate of a gas is the volume of gas that passes through an area during a given period of time. The velocity v of air flowing through a vent can be found using v(r) = . c. Determine the gas flow rate of a circular vent that has a diameter of 5 feet with a gas stream that is moving at 1.8 . , where r is the flow rate in cubic feet per second and A is the cross-sectional area of the vent in square feet. SOLUTION: a. The cross section is a rectangle 2 ft by 1.5 ft. So, v(r) = and v −1 (x) = 3x. This represents the formula for the flow rate of the gas. b. The velocity of air flowing through the vent in feet per second a. Find v−1 of the vent shown. What does this function represent? b. Determine the velocity of air flowing through the vent in feet per second if the flow rate is 15,000 . . c. The cross sectional area is π(2.5)2 or 6.25π. So, v(r) = c. Determine the gas flow rate of a circular vent that has a diameter of −1 and −1 v (x) = 6.25πx. v (1.8) ≈ 35.3. 5 feet with a gas stream that is moving at 1.8 . 68. COMMUNICATION A cellular phone company is having a sale as shown. Assume the $50 rebate only after the 10% discount is given. SOLUTION: a. The cross section is a rectangle 2 ft by 1.5 ft. So, v(r) = eSolutions Manual - Powered by Cognero and v (x) = 3x. This represents the formula for the flow rate of the gas. −1 Page 29 c. The cross sectional area is π(2.5)2 or 6.25π. So, v(r) = −1 and −1 v (x) = 6.25πx. v (1.8) ≈ 35.3. 1-7 Inverse Relations and Functions 68. COMMUNICATION A cellular phone company is having a sale as shown. Assume the $50 rebate only after the 10% discount is given. a. Write a function r for the price of the phone as a function of the original price if only the rebate applies. b. Write a function d for the price of the phone as a function of the original price if only the discount applies. c. Find a formula for T(x) = [r ◦ d](x) if both the discount and the rebate apply. d. Find T −1 and explain what the inverse represents. e . If the total cost of the phone after the discount and the rebate was $49, what was the original price of the phone? The inverse represents the original price of the phone as a function of the price of the phone after the rebate and the discount. e. Use f (x) = 8x – 4 and g(x) = 2x + 6 to find each of the following. 69. [f –−1o g −1](x) SOLUTION: SOLUTION: a. r(x) = x − 50 b. d(x) = 0.9x c. The discount comes before the rebate, so T(x) = 0.9x − 50. d. The inverse represents the original price of the phone as a function of the price of the phone after the rebate and the discount. e. eSolutions Manual - Powered by Cognero Page 30 1-7 Inverse Relations and Functions Use f (x) = 8x – 4 and g(x) = 2x + 6 to find each of the following. 69. [f –−1 −1 o g ](x) 70. [g −1o f −1](x) SOLUTION: SOLUTION: 71. [f o g]−1(x) 70. [g −1o f −1](x) SOLUTION: eSolutions Manual - Powered by Cognero SOLUTION: Page 31 1-7 Inverse Relations and Functions 71. [f o g]−1(x) SOLUTION: 73. (f · g)−1(x) SOLUTION: Now, find the inverse. Now, find the inverse. 72. [g o f ]−1(x) SOLUTION: Now, find the inverse. must be greater than or equal to 0. Therefore, x ≥ −1.25. 74. (f −1 · g −1)(x) SOLUTION: 73. (f · g)−1(x) eSolutions Manual - Powered by Cognero SOLUTION: Page 32 1-7 Inverse must Relations and be greater thanFunctions or equal to 0. Therefore, x ≥ −1.25. 74. (f −1 · g −1)(x) Use f (x) = x 2 + 1 with domain [0, ∞) and g(x) = each of the following. SOLUTION: to find 75. [f −1o g −1](x) SOLUTION: 76. [g −1o f −1](x) Use f (x) = x 2 + 1 with domain [0, ∞) and g(x) = each of the following. to find SOLUTION: 75. [f −1o g −1](x) SOLUTION: eSolutions Manual - Powered by Cognero Page 33 1-7 Inverse Relations and Functions 76. [g −1o f −1](x) SOLUTION: 77. [f o g]−1(x) SOLUTION: Now, find the inverse. 78. [g o f ]−1(x) SOLUTION: 77. [f o g]−1(x) SOLUTION: Next, find the inverse. Now, find the inverse. 79. (f · g −1) (x) SOLUTION: eSolutions Manual - Powered by Cognero −1 78. [g o f ] (x) SOLUTION: Page 34 1-7 Inverse Relations and Functions 79. (f · g −1) (x) SOLUTION: 80. (f −1 · g)(x) SOLUTION: 81. COPIES Karen’s Copies charges users $0.40 for every minute or part of a minute to use their computer scanner. Suppose you use the scanner for x minutes, where x is any real number greater than 0. a. Sketch the graph of the function, C(x), that gives the cost of using the scanner for x minutes. b. What are the domain and range of C(x)? c. Sketch the graph of the inverse of C(x). d. What are the domain and range of the inverse? e . What real-world situation is modeled by the inverse? SOLUTION: eSolutions Manual - Powered by Cognero a. The function is C(x) = 0.4x. the scanner for x minutes. b. What are the domain and range of C(x)? c. Sketch the graph of the inverse of C(x). d. What are the domain and range of the inverse? e . What real-world situation is modeled by the inverse? SOLUTION: a. The function is C(x) = 0.4x. b. The minutes are rounded up, so the domain will consist of whole numbers. D={x | x R}, R={ y| y positive multiples of 0.4} c. To graph the inverse, interchange the axes. d. D={ x| x positive multiples of 0.4} R = {y | y R} e . The inverse gives the number of possible minutes spent using the scanner that costs x dollars. 82. MULTIPLE REPRESENTATIONS In this problem, you will investigate inverses of even and odd functions. a. GRAPHICAL Sketch the graphs of three different even functions. Do the graphs pass the horizontal line test? b. ANALYTICAL What pattern can you discern regarding the inverses of even functions? Confirm or deny the pattern algebraically. Page 35 c. GRAPHICAL Sketch the graphs of three different odd functions. Do the graphs pass the horizontal line test? d. ANALYTICAL What pattern can you discern regarding the 82. MULTIPLE REPRESENTATIONS In this problem, you will 1-7 investigate inverses of even and odd functions. a. GRAPHICAL Sketch the graphs of three different even functions. Inverse Relations and Functions Do the graphs pass the horizontal line test? b. ANALYTICAL What pattern can you discern regarding the inverses of even functions? Confirm or deny the pattern algebraically. c. GRAPHICAL Sketch the graphs of three different odd functions. Do the graphs pass the horizontal line test? d. ANALYTICAL What pattern can you discern regarding the inverses of odd functions? Confirm or deny the pattern algebraically. d. Sample answer: The pattern indicates that all odd functions have inverses. While the pattern of the three graphs presented indicates that some odd functions have inverses, this is in fact a false statement. If a function is odd, then f (−x) = −f (x) for all x. An example of an odd function is . Notice from the graph of this function that, while it is odd, it fails the horizontal line test and, therefore, does not have an inverse function. SOLUTION: a. No; sample answer: 83. REASONING If f has an inverse and a zero at 6, what can you determine about the graph of f b. Sample answer: The pattern indicates that no even functions have inverses. When a function is even, f (x) = f (−x). Two x-values share a common y-value. This violates the horizontal line test. Therefore, the statement is true, and no even functions have inverse functions. c. No; sample answer: −1 ? SOLUTION: If f has an inverse, then (x, f (x)) = (f (x), x). Therefore, f intercept at (0, 6). −1 has a y- 84. Writing in Math Explain what type of restriction on the domain is needed to determine the inverse of a quadratic function and why a restriction is needed. Provide an example. SOLUTION: Sample answer: The domain of a quadratic function needs to be restricted so that only half of the parabola is shown. The cut-off point of the restriction will be along the axis of symmetry of the parabola. This essentially cuts the parabola into two equal halves. The restriction d. Sample answer: The pattern indicates that all odd functions have inverses. While the pattern of the three graphs presented indicates that some odd functions have inverses, this is in fact a false statement. If a function is odd, then f (−x) = −f (x) for all x. An example of an odd function is . Notice from the graph of this function eSolutions Manual - Powered by Cognero that, while it is odd, it fails the horizontal line test and, therefore, does not have an inverse function. will be or 2 for f (x) = ax + bx + c. 85. REASONING True or False. Explain your reasoning. All linear functions have inverse functions. SOLUTION: Page 36 False; sample answer: Constant functions are linear, but they do not pass the horizontal line test. Therefore, constant functions are not oneto-one functions and do not have inverse functions. restricted so that only half of the parabola is shown. The cut-off point of the restriction will be along the axis of symmetry of the parabola. This essentially cuts the parabola into two equal halves. The restriction 2 1-7 Inverse will be Relations for f (x) = ax or and Functions + bx + c. 85. REASONING True or False. Explain your reasoning. though both limits approach 0, they do it from opposite sides of 0 and no x-values ever share a corresponding y-value. Therefore, the function passes the horizontal line test. f (x) = ? SOLUTION: SOLUTION: False; sample answer: Constant functions are linear, but they do not pass the horizontal line test. Therefore, constant functions are not oneto-one functions and do not have inverse functions. 86. CHALLENGE If f (x) = x3 − ax + 8 and f −1 (23) = 3, find the value of a. Sample answer: If the ± sign is used, then f (x) will no longer be a function because it violates the vertical line test. 89. Writing in Math Explain how an inverse of f can exist. Give an example provided that the domain of f is restricted and f does not have an inverse when the domain is unrestricted. SOLUTION: SOLUTION: If f . Even 88. REASONING Why is ± not used when finding the inverse function of All linear functions have inverse functions. −1 Yes; sample answer: One function that does this is f (x) = 2 (23) = 3, then (23, 3) is equal to (x, y) in f Sample answer: If f (x) = x , f does not have an inverse because it is not one-to-one. If the domain is restricted to x ≥ 0, then the function is now −1 . one-to-one and f −1 exists; f −1 (x) = . For each pair of functions, find f o g and g o f . Then state the domain of each composite function. 90. f (x) = x2 – 9 g(x) = x + 4 Replace (x, y ) with (23, 3). SOLUTION: 87. REASONING Can f (x) pass the horizontal line test when and f (x) = 0 f (x) = 0? Explain. SOLUTION: Yes; sample answer: One function that does this is f (x) = . Even though both limits approach 0, they do it from opposite sides of 0 and no x-values ever share a corresponding y-value. Therefore, the function passes the horizontal line test. There are no restrictions on the domains. 91. f (x) = x–7 g(x) = x + 6 88. REASONING Why is ± not eSolutions Manual - Powered by Cognero f (x) = ? used when finding the inverse function of SOLUTION: Page 37 1-7 Inverse Functions There areRelations no restrictionsand on the domains. 91. f (x) = x–7 g(x) = x + 6 SOLUTION: There are no restrictions on the domains. Use the graph of the given parent function to describe the graph of each related function. 93. f (x) = x2 2 a. g(x) = (0.2x) b. h(x) = (x – 5)2 – 2 2 c. m(x) = 3x + 6 SOLUTION: a. The graph of g(x) is the graph of f (x) expanded horizontally because 2 g(x) = (0.2x) = f (0.2x) and 0 < 0.2 < 1. b. This function is of the form h(x) = f (x – 5) – 2. So, the graph of h(x) is the graph of f (x) translated 5 units to the right and 2 units down. c. This function is of the form m(x) = 3f (x) + 6. So, the graph of m(x) is the graph of f (x) expanded vertically because 3 > 1 and translated 6 units up. 94. f (x) = x3 There are no restrictions on the domains. a. g(x) = |x3 + 3| 3 92. f (x) = x – 4 g(x) = 3x 2 SOLUTION: There are no restrictions on the domains. Use the graph of the given parent function to describe the graph of each related function. eSolutions Manual - Powered by Cognero 93. f (x) = x2 2 a. g(x) = (0.2x) b. h(x) = −(2x) c. m(x) = 0.75(x + 1)3 SOLUTION: a. This function is of the form g(x) = |f (x) + 3|. So, the graph of g(x) is the graph of f (x) translated 3 units up and the portion of the graph below the x-axis is reflected in the x-axis. b. This function is of the form h(x) = –f (2x). So, the graph of h(x) is the graph of f (x) compressed horizontally because 2 > 1 and reflected in the x-axis. c. This function is of the form m(x) = 0.75f (x + 1). So, the graph of m (x) is the graph of f (x) translated 1 unit to the left and compressed vertically because 0 < 0.75 < 1. 95. f (x) = |x| a. g(x) = |2x| b. h(x) = |x – 5| c. m(x) = |3x| − 4 SOLUTION: Page 38 1-7 the graph of f (x) compressed horizontally because 2 > 1 and reflected in the x-axis. c. This function is of the form m(x) = 0.75f (x + 1). So, the graph of m (x) is the graph of f (x) translated 1 unit to the left and compressed Inverse Relations and Functions vertically because 0 < 0.75 < 1. Solve each system of equations. 97. x + 2y + 3z = 5 3x + 2y – 2z = −13 5x + 3y – z = −11 SOLUTION: 95. f (x) = |x| a. g(x) = |2x| b. h(x) = |x – 5| c. m(x) = |3x| − 4 Subtract the 2nd equation from the first. SOLUTION: a. The graph of g(x) is the graph of f (x) compressed horizontally because g(x) = |2x| = f (2x) and 2 > 1. b. This function is of the form h(x) = f (x – 5). So, the graph of h(x) is the graph of f (x) translated 5 units to the right. c. This function is of the form m(x) = f (3x) – 4. So, the graph of m(x) is the graph of f (x) compressed horizontally because 3 > 1 and translated 4 units down. rd nd Subtract 2 times the 3 original equation from 3 times the 2 equation. original 96. ADVERTISING A newspaper surveyed companies on the annual amount of money spent on television commercials and the estimated number of people who remember seeing those commercials each week. A soft-drink manufacturer spends $40.1 million a year and estimates 78.6 million people remember the commercials. For a package-delivery service, the budget is $22.9 million for 21.9 million people. A telecommunications company reaches 88.9 million people by spending $154.9 million. Use a matrix to represent these data. Solve the new system of 2 equations. Multiply the bottom equation by 2 and add. SOLUTION: Substitute to find x. Solve each system of equations. 97. x + 2y + 3z = 5 3x + 2y – 2z = −13 5x + 3y – z = −11 Substitute into one of the original equations to find y . SOLUTION: eSolutions Manual - Powered by Cognero Page 39 Add the 1 and 3 equations together. 1-7 Inverse Relations and Functions Add 2 times the 3rd original equation to the 2nd original equation. Substitute into one of the original equations to find y . Substitute all three values into the other two original equations to confirm your answer. Solve the new system of 2 equations. Solve for y in the 1s t, then substitute into the 2nd. 98. 7x + 5y + z = 0 −x + 3y + 2z = 16 x – 6y – z = −18 SOLUTION: Substitute for x and solve for y . Add the 1s t and 3rd equations together. Substitute into one of the original equations to find z. Add 2 times the 3rd original equation to the 2nd original equation. eSolutions Manual - Powered by Cognero Substitute all three values into the other two original equations to confirm your answer. Page 40 st Solve the new system of 2 equations. Solve for y in the 1 , then substitute into the 2nd. 1-7 Inverse and SubstituteRelations all three values intoFunctions the other two original equations to confirm your answer. Solve the new system of 2 equations and 2 variables. Solve for x in the 1s t, then substitute into the 2nd. 99. x – 3z = 7 2x + y – 2z = 11 −x – 2y + 9z = 13 SOLUTION: Substitute into one of the original equations to find y . Add the 3rd equation to twice the 2nd equation to eliminate the y . Substitute all three values into the last original equation to confirm your answer. Solve the new system of 2 equations and 2 variables. Solve for x in the 1s t, then substitute into the 2nd. 100. BASEBALL A batter pops up the ball. Suppose the ball was 3.5 feet above the ground when he hit it straight up with an initial velocity of 80 2 feet per second. The function d(t) = 80t – 16t + 3.5 gives the ball’s height above the ground in feet as a function of time t in seconds. How long did the catcher have to get into position to catch the ball after it was hit? SOLUTION: eSolutions Manual - Powered by Cognero Page 41 The ball will hit the ground after about 5.04 seconds, so the catcher will have about 5 seconds. 1-7 Inverse Relations and Functions 100. BASEBALL A batter pops up the ball. Suppose the ball was 3.5 feet above the ground when he hit it straight up with an initial velocity of 80 101. SAT/ACT What is the probability that the spinner will land on a number that is either even or greater than 5? 2 feet per second. The function d(t) = 80t – 16t + 3.5 gives the ball’s height above the ground in feet as a function of time t in seconds. How long did the catcher have to get into position to catch the ball after it was hit? SOLUTION: A B C D E The ball will hit the ground after about 5.04 seconds, so the catcher will have about 5 seconds. 101. SAT/ACT What is the probability that the spinner will land on a SOLUTION: The sections of the spinner are all equal and four of the 6 numbers are either even or greater than 5. Therefore, the probability is = . number that is either even or greater than 5? 102. REVIEW If m and n are both odd natural numbers, which of the following must be true? 2 2 I. m + n is even. 2 A B C eSolutions D Manual - Powered by Cognero E 2 II. m + n is divisible by 4. 2 III. (m + n) is divisible by 4. F none G I only H I and II only J I and III only SOLUTION: 2 Page 42 If m is odd, then m equals an odd number multiplied by an odd number. 2 The sum of an odd number of odd numbers is always odd, so m will 2 SOLUTION: The sections of the spinner are all equal and four of the 6 numbers are 1-7 Inverse and Functions either evenRelations or greater than 5. Therefore, the probability is = . 102. REVIEW If m and n are both odd natural numbers, which of the following must be true? 2 2 I. m + n is even. 2 2 2 2 where a is an integer, thus (m + n) = (2a) = 4a , a ∈ Z. Since 4a is divisible by 4, Part III is true. Part II is not true when m = 1 and n = 3. Therefore, the correct choice is J. 103. Which of the following is the inverse of f (x) = ? A g(x) = 2 II. m + n is divisible by 4. 2 III. (m + n) is divisible by 4. F none G I only H I and II only J I and III only B g(x) = C g(x) = 2x + 5 D g(x) = SOLUTION: SOLUTION: 2 If m is odd, then m equals an odd number multiplied by an odd number. 2 The sum of an odd number of odd numbers is always odd, so m will 2 2 2 always be odd. The same is true for n . Since m and n are odd, the sum of these two numbers will always be even. Thus, part I is true. The sum of two odd numbers is always an even number of the form 2a 2 2 2 2 where a is an integer, thus (m + n) = (2a) = 4a , a ∈ Z. Since 4a is divisible by 4, Part III is true. Part II is not true when m = 1 and n = 3. Therefore, the correct choice is J. 103. Which of the following is the inverse of f (x) = A g(x) = B g(x) = ? 104. REVIEW A train travels d miles in t hours and arrives at its destination 3 hours late. At what average speed, in miles per hour, should the train have gone in order to have arrived on time? Ft–3 G H J − 3 C g(x) = 2x + 5 D g(x) = SOLUTION: d = rt SOLUTION: eSolutions Manual - Powered by Cognero The train’s current rate is r = . If the train needs to arrive 3 hours earlier, then replace t with t − 3. Therefore, r = Page 43 . 1-7 Inverse Relations and Functions 104. REVIEW A train travels d miles in t hours and arrives at its destination 3 hours late. At what average speed, in miles per hour, should the train have gone in order to have arrived on time? Ft–3 G H J − 3 SOLUTION: d = rt The train’s current rate is r = . If the train needs to arrive 3 hours earlier, then replace t with t − 3. Therefore, r = eSolutions Manual - Powered by Cognero . Page 44