Solutions Manual for Precalculus An Investigation of Functions David Lippman, Melonie Rasmussen 1 st Edition Solutions created at Te Evergreen State College and Soreline Community College
1.1 Solutions to Exercises 1. (a) f(40) = 13, because te input 40 (in tousands of people) gives te output 13 (in tons of garbage) (b) f(5) = 2, means tat 5000 people produce 2 tons of garbage per week. 3. (a) In 1995 (5 years after 1990) tere were 30 ducks in te lake. (b) In 2000 (10 years after 1990) tere were 40 ducks in te lake. 5. Graps (a) (b) (d) and (e) represent y as a function of x because for every value of x tere is only one value for y. Graps (c) and (f) are not functions because tey contain points tat ave more tan one output for a given input, or values for x tat ave 2 or more values for y. 7. Tables (a) and (b) represent y as a function of x because for every value of x tere is only one value for y. Table (c) is not a function because for te input x=10, tere are two different outputs for y. 9. Tables (a) (b) and (d) represent y as a function of x because for every value of x tere is only one value for y. Table (c) is not a function because for te input x=3, tere are two different outputs for y. 11. Table (b) represents y as a function of x and is one-to-one because tere is a unique output for every input, and a unique input for every output. Table (a) is not one-to-one because two different inputs give te same output, and table (c) is not a function because tere are two different outputs for te same input x=8. 13. Graps (b) (c) (e) and (f) are one-to-one functions because tere is a unique input for every output. Grap (a) is not a function, and grap (d) is not one-to-one because it contains points wic ave te same output for two different inputs. 15. (a) f(1) = 1 (b) f(3) = 1 17. (a) g(2) = 4 (b) g( 3) = 2 19. (a) f(3) = 53 (b) f(2) = 1 21. f( 2) = 4 2( 2) = 4 + 4 = 8, f( 1) = 6, f(0) = 4, f(1) = 4 2(1) = 4 2 = 2, f(2) = 0
23. f( 2) = 8( 2) 2 7( 2) + 3 = 8(4) + 14 + 3 = 32 + 14 + 3 = 49, f( 1) = 18, f(0) = 3, f(1) = 8(1) 2 7 (1) + 3 = 8 7 + 3 = 4, f(2) = 21 25. f( 2) = ( 2) 3 + 2( 2) = ( 8) 4 = 8 4 = 4, f( 1) = ( 1) 3 + 2( 1) = ( 1) 2 = 1, f(0) = 0, f(1) = (1) 3 + 2(1) = 1, f(2) = 4 27. f( 2) = 3 + ( 2) + 3 = 3 + 1 = 3 + 1 = 4, f( 1) = 2 + 3 4.41, f(0) = 3 + 3 4.73, f(1) = 3 + (1) + 3 = 3 + 4 = 3 + 2 = 5, f(2) = 5 + 3 5.23 29. f( 2) = (( 2) 2)(( 2) + 3) = ( 4)(1) = 4, f( 1) = 6, f(0) = 6, f(1) = ((1) 2)((1) + 3) = ( 1)(4) = 4, f(2) = 0 31. f( 2) = ( 2) 3 = 5 = 5, f( 1) = undefined, f(0) = 3, f(1) = 1, f(2) = 1/3 ( 2)+1 1 33. f( 2) = 2 2 = 1 = 1, f( 1) = 1, f(0) = 1, f(1) = 2, f(2) = 4 2 2 4 2 35. Using f(x) = x 2 + 8x 4: f( 1) = ( 1) 2 + 8( 1) 4 = 1 8 4 = 11; f(1) = 1 2 + 8(1) 4 = 1 + 8 4 = 5. (a) f( 1) + f(1) = 11 + 5 = 6 (b) f( 1) f(1) = 11 5 = 16 37. Using f(t) = 3t + 5: (a) f(0) = 3(0) + 5 = 5 (b) 3t + 5 = 0 t = 5 3 39. (a) y = x (iii. Linear) (b) y = x 3 (viii. Cubic) 3 (c) y = x (i. Cube Root) (e) y = x 2 (vi. Quadratic) (d) y = 1 (ii. Reciprocal) x (f) y = x (iv. Square Root) (g) y = x (v. Absolute Value) () y = 1 2 (vii. Reciprocal Squared) 41. (a) y = x 2 (iv.) (b) y = x (ii.) (c) y = x (v.) (e) y = x (vi.) x (d) y = 1 x (i.) (f) y = x 3 (iii.)
postage eigt Last edited 4/13/15 43. (x 3) 2 + (y + 9) 2 = (6) 2 or (x 3) 2 + (y + 9) 2 = 36 45. (a) Grap (a) At te beginning, as age increases, eigt increases. At some point, eigt stops increasing (as a person stops growing) and eigt stays te same as age increases. Ten, wen a person as aged, teir eigt decreases sligtly. age (b) Grap (b) As time elapses, te eigt of a person s ead wile jumping on a pogo stick as observed from a fixed point will go up and down in a periodic manner. (c) weigt of letter Grap (c) Te grap does not pass troug te origin because you cannot mail a letter wit zero postage or a letter wit zero weigt. Te grap begins at te minimum postage and weigt, and as te weigt increases, te postage increases. 47. (a) t (b) x = a (c) f(b) = 0 so z = 0. Ten f(z) = f(0) = r. (d) L = (c, t), K = (a, p) 1.2 Solutions to Exercises 1. Te domain is [ 5, 3); te range is [0, 2]
3. Te domain is 2 < x 8; te range is 6 y < 8 5. Te domain is 0 x 4; te range is 0 y 3 7. Since te function is not defined wen tere is a negative number under te square root, x cannot be less tan 2 (it can be equal to 2, because 0 is defined). So te domain is x 2. Because te inputs are limited to all numbers greater tan 2, te number under te square root will always be positive, so te outputs will be limited to positive numbers. So te range is f(x) 0. 9. Since te function is not defined wen tere is a negative number under te square root, x cannot be greater tan 3 (it can be equal to 3, because 0 is defined). So te domain is x 3. Because te inputs are limited to all numbers less tan 3, te number under te square root will always be positive, and tere is no way for 3 minus a positive number to equal more tan tree, so te outputs can be any number less tan 3. So te range is f(x) 3. 11. Since te function is not defined wen tere is division by zero, x cannot equal 6. So te domain is all real numbers except 6, or {x x R, x 6}. Te outputs are not limited, so te range is all real numbers, or {y R}. 13. Since te function is not defined wen tere is division by zero, x cannot equal 1/2. So te domain is all real numbers except 1/2, or {x x R, x 1/2}. Te outputs are not limited, so te range is all real numbers, or {y R}. 15. Since te function is not defined wen tere is a negative number under te square root, x cannot be less tan 4 (it can be equal to 4, because 0 is defined). Since te function is also not defined wen tere is division by zero, x also cannot equal 4. So te domain is all real numbers less tan 4 excluding 4, or {x x 4, x 4}. Tere are no limitations for te outputs, so te range is all real numbers, or {y R}. 17. It is easier to see were tis function is undefined after factoring te denominator. Tis gives x 3 f(x) =. It ten becomes clear tat te denominator is undefined wen x = 11 and (x+11)(x 2) wen x = 2 because tey cause division by zero. Terefore, te domain is {x x R, x 11, x 2}. Tere are no restrictions on te outputs, so te range is all real numbers, or {y R}. 19. f( 1) = 4; f(0) = 6; f(2) = 20; f(4) = 24 21. f( 1) = 1; f(0) = 2; f(2) = 7; f(4) = 5 23. f( 1) = 5; f(0) = 3; f(2) = 3; f(4) = 16 2 if 6 x 1 25. f(x) = { 2 if 1 < x 2 4 if 2 < x 4 3 if x 0 27. f(x) = { x 2 if x > 0
2x + 3 if 3 x < 1 29. f(x) = { x 1 if 1 x 2 2 if 2 < x 5 31. 33. 35. 1.3 Solutions to Exercises 1. (a) (b) 249 243 = 6 2002 2001 1 249 243 = 6 2004 2001 3 = 6 million dollars per year = 2 million dollars per year 3. Te inputs x = 1 and x = 4 produce te points on te grap: (4,4) and (1,5). Te average rate of cange between tese two points is 5 4 1 4 = 1 3 = 1 3. 5. Te inputs x = 1 and x = 5 wen put into te function f(x) produce te points (1,1) and (5,25). Te average rate of cange between tese two points is 25 1 5 1 = 24 4 = 6. 7. Te inputs x = 3 and x = 3 wen put into te function g(x) produce te points (-3, -82) and (3,80). Te average rate of cange between tese two points is 80 ( 82) 3 ( 3) = 162 6 = 27. 9. Te inputs t = 1 and t = 3 wen put into te function k(t) produce te points (-1,2) and (3,54.148 ). Te average rate of cange between tese two points is 54.148 2 3 ( 1) = 52.148 4 13. 11. Te inputs x = 1 and x = b wen put into te function f(x) produce te points (1,-3) and (b, 4b 2 7). Explanation: f(1) = 4(1) 2 7 = 3, f(b) = 4(b) 2 7. Te average rate of
cange between tese two points is (4b2 7) ( 3) b 1 4(b + 1). = 4b2 7+3 b 1 = 4b2 4 = 4(b2 1) = 4(b+1)(b 1) = b 1 b 1 (b 1) 13. Te inputs x = 2 and x = 2 + wen put into te function (x) produce te points (2,10) and (2 +, 3 + 10). Explanation: (2) = 3(2) + 4 = 10, (2 + ) = 3(2 + ) + 4 = 6 + 3 + 4 = 3 + 10. Te average rate of cange between tese two points is (3+10) 10 3. (2+) 2 = 3 = 15. Te inputs t = 9 and t = 9 + wen put into te function a(t) produce te points (9, 1 13 ) and (9 +, 1 +13 1 ). Explanation: a(9) = = 1, a(9 + ) = 1 9+4 13 rate of cange between tese two points is 1 = 1 1 13 2 +13 1 13 1 2 +13 = ( 13 +1 13 13 13 2 +13 = 13 ( 1 2 +13 ) = 1 +13 1 1 13 = +13 1 13 (9+) 9 = 1 (9+)+4 +13 +1) (to make a common denominator) = 1 = = 13( 2 +13) 13(+13). Te average = ( 1 +13 1 13 ) (1 ) = 1 (+13) 13(+13). ( 13 +1 ) = 2 +13 2 +13 17. Te inputs x = 1 and x = 1 + wen put into te function j(x) produce te points (1,3) and (1 +, 3(1 + ) 3 ). Te average rate of cange between tese two points is 3(1+)3 3 (1+) 1 = 3(1+) 3 3 3). = 3(3 +3 2 +3+1) 3 = 33 +9 2 +9+3 3 = 33 +9 2 +9 = 3 2 + 9 + 9 = 3( 2 + 3 + 19. Te inputs x = x and x = x + wen put into te function f(x) produce te points (x, 2x 2 + 1) and (x +, 2(x + ) 2 + 1). Te average rate of cange between tese two points is (2(x+) 2 +1) (2x 2 +1) (x+) x 2(x 2 +2x+ 2 ) 2x 2 = (2(x+)2 +1) (2x 2 +1) = 2x2 +4x+2 2 2x 2 = 2(x+)2 +1 2x 2 1 = 4x+22 = 2(x+)2 2x 2 = 4x + 2 = 2(2x + ). = 21. Te function is increasing (as a positive slope) on te interval ( 1.5,2), and decreasing (as a negative slope) on te intervals (, 1.5) and (2, ).
23. Te function is increasing (as a positive slope) on te intervals (, 1) and (3.25,4) and decreasing (as a negative slope) on te intervals (1,2.75) and (4, ). 25. Te function is increasing because as x increases, f(x) also increases, and it is concave up because te rate at wic f(x) is canging is also increasing. 27. Te function is decreasing because as x increases, (x) decreases. It is concave down because te rate of cange is becoming more negative and tus it is decreasing. 29. Te function is decreasing because as x increases, f(x) decreases. It is concave up because te rate at wic f(x) is canging is increasing (becoming less negative). 31. Te function is increasing because as x increases, (x) also increases (becomes less negative). It is concave down because te rate at wic (x) is canging is decreasing (adding larger and larger negative numbers). 33. Te function is concave up on te interval (, 1), and concave down on te interval (1, ). Tis means tat x = 1 is a point of inflection (were te grap canges concavity). 35. Te function is concave down on all intervals except were tere is an asymptote at x 3. 37. From te grap, we can see tat te function is decreasing on te interval (, 3), and increasing on te interval (3, ). Tis means tat te function as a local minimum at x = 3. We can estimate tat te function is concave down on te interval (0,2), and concave up on te intervals (2, )and (, 0). Tis means tere are inflection points at x = 2 and x = 0.
39. From te grap, we can see tat te function is decreasing on te interval ( 3, 2), and increasing on te interval ( 2, ). Tis means tat te function as a local minimum at x = 2. Te function is always concave up on its domain, (-3, ). Tis means tere are no points of inflection. 41. From te grap, we can see tat te function is decreasing on te intervals (, 3.15) and ( 0.38, 2.04), and increasing on te intervals ( 3.15, 0.38) and (2.04, ). Tis means tat te function as local minimums at x = 3.15 and x = 2.04 and a local maximum at x = 0.38. We can estimate tat te function is concave down on te interval ( 2,1), and concave up on te intervals (, 2) and (1, ). Tis means tere are inflection points at x = 2 and x = 1. 1.4 Solutions to Exercises 1. f(g(0)) = 4(7) + 8 = 26, g(f(0)) = 7 (8) 2 = 57 3. f(g(0)) = (12) + 4 = 4, g(f(0)) = 12 (2) 3 = 4
5. f(g(8)) = 4 7. g(f(5)) = 9 9. f(f(4)) = 4 11. g(g(2)) = 7 13. f(g(3)) = 0 15. g(f(1)) = 4 17. f(f(5)) = 3 19. g(g(2)) = 2 21. f(g(x)) = 1 = x, g(f(x)) = 7 + 6 = 7x 36 ( 7 x +6) 6 7 ( 1 ) x 6 23. f(g(x)) = ( x + 2) 2 + 1 = x + 3, g(f(x)) = (x 2 + 1) + 2 = (x 2 + 3) 25. f(g(x)) = 5x + 1, g(f(x)) = 5 x + 1 27. f(g((x))) = (( x) 6) 4 + 6 29. (a) 1 p(x) = x m(x) x 2 4, wic can be written as 1 x(x 2 4). Tis function is undefined wen tere is a negative number under te square root, or wen te factor in te denominator equals zero. So te domain is all positive numbers excluding 2, or {x x > 0, x 2}. (b) p(m(x)) = 1. Tis function is undefined wen tere is a negative number under x 2 4 te square root or a zero in te denominator, wic appens wen x is between -2 and 2. So te domain is 2 > x > 2. (c) m(p(x)) = ( 1 x )2 4 = 1 4. Tis function is undefined wen te denominator is x zero, or wen x = 0. So te domain is {x x R, x 0}. 31. b 3 33. (a) r(v(t)) = 3(10+20t) 4π 3 (b) Evaluating te function in (a) wen r(v(t)) = 10 gives 10 = 3(10+20t). Solving tis 4π
for t gives t 208.93, wic means tat it takes approximately 208 seconds or 3.3 minutes to blow up a balloon to a radius of 10 inces. 35. f(x) = x 2, g(x) = x + 2 37. f(x) = 3, g(x) = x 5 x 39. f(x) = 3 + x, g(x) = x 2 41. (a) f(x) = ax + b, so f(f(x)) = a(ax + b) + b, wic simplifies to a 2 x + 2b. a and b are constants, so a 2 and 2b are also constants, so te equation still as te form of a linear function. (b) If we let g(x) be a linear function, it as te form g(x) = ax + b. Tis means tat g(g(x)) = a(ax + b) + b. Tis simplifies to g(g(x)) = a 2 x + ab + b. We want g(g(x)) to equal 6x 8, so we can set te two equations equal to eac oter: a 2 x + ab + b = 6x 8. Looking at te rigt side of tis equation, we see tat te ting in front of te x as to equal 6. Looking at te left side of te equation, tis means tat a 2 = 6. Using te same logic, ab + b = 8. We can solve for, a = 6. We can substitute tis value for a into te second equation to solve for b: ( 6)b + b = 8 b( 6 + 1) = 8 b = 8 g(x) = 6x answer. 8 6+1 6+1. So, since g(x) = ax + b,. Evaluating g(g(x)) for tis function gives us 6x-8, so tat confirms te 43. (a) A function tat converts seconds s into minutes m is m = f(s) = s 60. C(f(s)) = 70( s 60 )2 10+( s 60 )2 ; tis function calculates te speed of te car in mp after s seconds. (b) A function tat converts ours into minutes m is m = g() = 60. C(g()) = 70(60) 2 10+(60) 2 ; tis function calculates te speed of te car in mp after ours. (c) A function tat converts mp s into ft/sec z is z = v(s) = ( 5280 ) s wic can be reduced to v(s) = ( 22 ) s. v(c(m)) = (22 ) ( 70m2 15 15 10+m2) ; tis function converts te speed of te car in mp to ft/sec. 3600
1.5 Solutions to Exercises 1. Horizontal sift 49 units to te rigt 3. Horizontal sift 3 units to te left 5. Vertical sift 5 units up 7. Vertical sift 2 units down 9. Horizontal sift 2 units to te rigt and vertical sift 3 units up 11. f(x) = (x + 2) + 1 13. f(x) = 1 (x 3) 4 15. g(x) = f(x 1), (x) = f(x) + 1 17. 19. f(x) w(x) g(x) f(x) 21. f(t) = (t + 1) 2 3 as a transformation of g(t) = t 2 f(t) = (t + 1) 2 3 g(t) = t 2
23. k(x) = (x 2) 3 1 as a transformation of f(x) = x 3 f(x) = x 3 k(x) = (x 2) 3 1 25. f(x) = x 3 2 27. f(x) = x + 3 1 29. f(x) = x 31. f(x) = 2 x f(x) = 2 x + 1 33. (a) f(x) = 6 x (b) f(x) = 6 x+2 3 35. f(x) = (x + 1) 2 + 2 37. f(x) = x + 1 39. (a) even (b) neiter (c) odd 41. te function will be reflected over te x-axis 43. te function will be vertically stretced by a factor of 4 45. te function will be orizontally compressed by a factor of 1 5 47. te function will be orizontally stretced by a factor of 3 49. te function will be reflected about te y-axis and vertically stretced by a factor of 3 51. f(x) = 4x 53. f(x) = 1 3(x+2) 2 3 55. f(x) = (2[x 5]) 2 + 1 = (2x 10) 2 + 1
57. f(x) = x 2 will be sifted to te left 1 unit, vertically stretced by a factor of 4, and sifted down 5 units. f(x) = x 2 f(x) = 4(x + 1) 2 5 59. (x) = x will be sifted rigt 4 units vertically stretced by a factor of 2, reflected about te x-axis, and sifted up 3 units. (x) = x (x) = 2 x 4 + 3 61. m(x) = x 3 will be vertically compressed by a factor of 1 2. m(x) = x 3 m(x) = 1 2 x3
63. p(x) = x 2 will be stretced orizontally by a factor of 3, and sifted down 3 units. p(x) = x 3 p(x) = ( 1 3 x)3 3 65. a(x) = x will be sifted left 4 units and ten reflected about te y-axis. a(x) = x + 4 a(x) = x 67. te function is decreasing on te interval x < 1 and increasing on te interval x > 1 69. te function is decreasing on te interval x 4 71. te function is concave up on te interval x < 1 and concave down on te interval x > 1 73. te function is always concave up. 75. f( x) 77. 3f(x) 79. 2f( x) 81. 2f ( 1 2 x) 83. 2f(x) 2 85. f(x) + 2 87. f(x) = (x + 2) 2 + 3 89. f(x) = 1 2 (x + 1)3 + 2 91. f(x) = 2(x + 2) + 1 93. f(x) = 1 (x 2) 2 + 3
95. f(x) = x + 1 + 3 97. f(x) = (x 2) 1 3 + 1 99. (a) Wit te input in factored form, we first apply te orizontal compression by a factor of ½, followed by a sift to te rigt by tree units. After applying te orizontal compression, te domain becomes 1 x 3. Ten we apply te sift, to get a domain of 2 {x 3 1 x 6}. 2 (b) Since tese are orizontal transformations, te range is uncanged. (c) Tese are vertical transformations, so te domain is uncanged. (d) We first apply te vertical stretc by a factor of 2, followed by a downward sift of tree units. After te vertical stretc, te range becomes 6 y 10. Next, we apply te sift to get te final domain {y 9 y 7}. (e) Te simplest solution uses a positive value of B. Te new domain is an interval of lengt one. Before, it was an interval of lengt 5, so tere as been a orizontal compression by a factor of 1/5. Terefore, B = 5. If we apply tis orizontal compression to te original domain, we get 1 x 6. To transform tis interval into one tat starts at 5 5 8, we must add 7 4 = 39 39. Tis is our rigtward sift, so c = 5 5 (f) Te simplest solution uses a positive value of A. Te new range is an interval of lengt one. Te original range was an interval of lengt 8, so tere as been a vertical compression by a factor of 1/8. Tus, we ave A = 1. If we apply tis vertical 8 compression to te original range we get = 3 y 5. Now, in order to get an interval tat 8 8 begins at 0, we must add 3/8. Tis is a vertical sift upward, and we ave D = 3 8. 5. 1.6 Solutions to Exercises 1. Te definition of te inverse function is te function tat reverses te input and output. So if te output is 7 wen te input is 6, te inverse function f 1 (x) gives an output of 6 wen te input is 7. So, f 1 (7) = 6. 3. Te definition of te inverse function is te function wic reverse te input and output of te original function. So if te inverse function f 1 (x) gives an output of 8 wen te input is 4,
te original function will do te opposite, giving an output of 4 wen te input is 8. So f( 8) = 4. 5. f(5) = 2, so (f(5)) 1 = (2) 1 = 1 2 1 = 1 2. 7. (a) f(0) = 3 (b) Solving f(x) = 0 asks te question: for wat input is te output 0? Te answer is x = 2. So, f(2) = 0. (c) Tis asks te same question as in part (b). Wen is te output 0? Te answer is f 1 (0) = 2. (d) Te statement from part (c) f 1 (0) = 2 can be interpreted as in te original function f(x), wen te input is 2, te output is 0 because te inverse function reverses te original function. So, te statement f 1 (x) = 0 can be interpreted as in te original function f(x), wen te input is 0, wat is te output? te answer is 3. So, f 1 (3) = 0. 9. (a) f(1) = 0 (b) f(7) = 3 11. (c) f 1 (0) = 1 (d) f 1 (3) = 7 x 1 4 7 12 16 f 1 (x) 3 6 9 13 14 13. Te inverse function takes te output from your original function and gives you back te input, or undoes wat te function did. So if f(x) adds 3 to x, to undo tat, you would subtract 3 from x. So, f 1 (x) = x 3. 15. In tis case, te function is its own inverse, in oter words, putting an output back into te function gives back te original input. So, f 1 (x) = 2 x. 17. Te inverse function takes te output from your original function and gives you back te input, or undoes wat te function did. So if f(x) multiplies 11 by x and ten adds 7, to undo tat, you would subtract 7 from x, and ten divide by 11. So, f 1 (x) = x 7 11. 19. Tis function is one-to-one and non-decreasing on te interval x > 7. Te inverse function, restricted to tat domain, is f 1 (x) = x 7.
21. Tis function is one-to-one and non-decreasing on te interval x > 0. Te inverse function, restricted to tat domain, is f 1 (x) = x + 5. 3 23. (a) f(g(x)) = (( x + 5)) 3 5, wic just simplifies to x. (b) g(f(x)) = ( (x 3 5) + 5 3 ), wic just simplifies to x. (c) Tis tells us tat f(x) and g(x) are inverses, or, tey undo eac oter.