¡Descarga Solucionario Sullivan y más Ejercicios en PDF de Matemáticas solo en Docsity!
Chapter 5
Exponential and Logarithmic Functions
Section 5.
1. (^) ( ) ( ) ( ) ( )
f = − + = − + = − + = −
2. (^) ( ) ( ) ( )
2 2 2
f x x x x
2 2
x f x x
( ) ( )
x x x
x ≠ − 5, x ≠ 5 Domain: (^) { x x ≠ −5, x ≠ (^5) }
4. composite function; f ( g x ( )) 5. False: f ( g x ( ) ) = ( f g )( ) x 6. False. The domain of ( f g ) ( ) x is a subset of the domain of g x ( ). 7. a. ( f g )( ) 1 = f (^) ( g ( ) (^1) ) = f ( 0 ) = − 1
b. ( f g ) ( − 1 ) = f (^) ( g ( − 1 ) (^) ) = f ( 0 ) = − 1
c. ( g f ) ( − 1 ) = g (^) ( f ( − 1 ) (^) ) = g ( − 3 ) = 8
d. ( g f )( 0 ) = g (^) ( f ( 0 )) = g ( − 1 ) = 0
e. ( g g )( − 2 ) = g (^) ( g ( − 2 )) = g ( ) 3 = 8
f. ( f f )( − 1 ) = f (^) ( f ( − 1 ) (^) ) = f ( − 3 ) = − 7
8. a. ( f g ) ( ) 1 = f (^) ( g ( ) (^1) ) = f ( 0 ) = 5
b. ( f g )( 2 ) = f (^) ( g ( 2 ) (^) ) = f ( − 3 ) = 11
c. ( g f )( 2 ) = g (^) ( f ( 2 ) (^) ) = g ( ) 1 = 0
d. ( g f )( ) 3 = g (^) ( f ( ) (^3) ) = g ( − 1 ) = 0
e. (^) ( g g (^) )( ) 1 = g (^) ( g (^) ( ) (^1) ) = g ( 0 )= 1
f. (^) ( f f (^) )( ) 3 = f (^) ( f (^) ( ) (^3) ) = f ( − (^1) ) = 7
9. a. (^) ( g f (^) ) ( − (^1) ) = g (^) ( f ( 1)− (^) ) = g ( ) 1 = 4
b. (^) ( g f (^) )( 0 ) = g (^) ( f (0) (^) ) = g ( 0 ) = 5
c. (^) ( f g (^) )( − (^1) ) = f (^) ( g ( 1)− (^) ) = f ( ) 3 = − 1
d. (^) ( f g (^) )( 4 ) = f (^) ( g (4) (^) ) = f ( 2 ) = − 2
10. a. (^) ( g f (^) )( ) 1 = g (^) ( f (1)) = g ( − (^1) ) = 3
b. (^) ( g f (^) )( 5 ) = g (^) ( f (5)) = g ( ) 1 = 4
c. (^) ( f g (^) ) ( 0 ) = f (^) ( g (0) (^) ) = f ( 5 )= 1
d. (^) ( f g (^) )( 2 ) = f (^) ( g (2) (^) ) = f ( 2 ) = − 2
11. f ( ) x = 2 x g x ( ) = 3 x^2 + 1 a.
f g f g f f
b.
2
g f g f g g
c. ( )(1) ( (1)) (2(1)) (2) 2(2) 4
f f f f f f
Section 5.1: Composite Functions
d.
2
2
g g g g g g
12. f ( ) x = 3 x + 2 g x ( ) = 2 x^2 − 1 a.
f g f g f f
b.
2
g f g f g g
c. ( )
f f f f f f
d.
2
2
g g g g g g
f x = x − g x = − x a. 2
2
f g f g f
f
= ^ −
b. 2
2
g f g f g g
c. 2
2
f f f f f f
d. ( g g )(0) = g g ( (0)) 2
2
g g
= ^ −
14. f ( ) x = 2 x^2^ g x ( ) = 1 − 3 x^2 a.
2
f g f g f f
b. 2
2
g f g f g g
c.
2
f f f f f f
Section 5.1: Composite Functions
f x x g x x
a. ( f g )(4) = f ( g (4))
2
f
f
f
= ^
= ^
= ^
b. ( )
2
g f g f g g
c. ( )
f f f f f f
d. ( g g )(0) = g g ( (0))
2
2
g
g
= ^
= ^
19. ( ) 3 ( )^3
f x g x x x
a.
3
f g f g f
b.
( ) 3
g f g f g
g
g
= ^
= ^
c. ( f f )(1) = f ( f (1)) 3 1 1 3 2 3 (^3 ) 2 3 5 2 6 5
f
f
= ^
= ^
d.
3
g g g g g g
Chapter 5: Exponential and Logarithmic Functions
20. ( ) 3/ 2^ ( ) 2
f x x g x x
a. ( f g )(4) = f ( g (4))
3/ 2
3
f
f
= ^
= ^
= ^
= ^
b.
3/ 2
3
(^2) or 4 2 2 2 2 1 7
g f g f g
g
g
c.
( )
3/ 2
3/ 2
f f f f f f
d. ( )(0) ( (0)) 2 0 1 (2) 2 2 1 2 3
g g g g g
g
= ^
21. The domain of g is { x x ≠ 0 }. The domain of f
is { x x ≠ 1 }. Thus, g x ( ) ≠ 1 , so we solve:
g x
x x
Thus, x ≠ 2 ; so the domain of f g is
{ x x^ ≠^ 0,^ x ≠^2 }.
22. The domain of g is { x x ≠ 0 }. The domain of f
is { x x ≠ − 3 }. Thus, g x ( ) ≠ − 3 , so we solve:
g x
x x
Thus, 2 3
x ≠ ; so the domain of f g is
(^) x x ≠ x ≠
23. The domain of g is { x x ≠ 0 }. The domain of f
is { x x ≠ 1 }. Thus, g x ( ) ≠ 1 , so we solve:
g x
x x
Thus, x ≠ − 4 ; so the domain of f g is
{ x x^ ≠ −^ 4,^ x ≠^0 }.
24. The domain of g is { x x ≠ 0 }. The domain of f
is { x x ≠ − 3 }. Thus, g x ( ) ≠ − 3 , so we solve:
g x
x x
Thus, 2 3
x ≠ − ; so the domain of f g is
(^2) , 0 3
(^) x x ≠ − x ≠
Chapter 5: Exponential and Logarithmic Functions
31. f ( ) x = 3 x + 1 g x ( ) = x^2 The domain of f is (^) { x x is any real number}. The domain of g is (^) { x x is any real number}. a.
2 2
f g x f g x f x x
Domain: (^) { x x is any real number}. b.
2 2
g f x g f x g x x x x
Domain: (^) { x x is any real number}. c. ( )( ) ( ( )) (3 1) 3(3 1) 1 9 3 1 9 4
f f x f f x f x x x x
Domain: (^) { x x is any real number}. d.
2
2 2 4
( g g )( ) x g g x ( ( )) g x
x x
Domain: (^) { x x is any real number}.
32. f ( ) x = x + 1 g x ( ) = x^2 + 4 The domain of f is (^) { x x is any real number}. The domain of g is (^) { x x is any real number}. a.
2 2 2
f g x f g x f x x x
Domain: (^) { x x is any real number}. b.
2 2 2
g f x g f x g x x x x x x
Domain: (^) { x x is any real number}.
c. ( )( ) ( ( )) ( 1) ( 1) 1 2
f f x f f x f x x x
Domain: (^) { x x is any real number}. d.
2
2 2 4 2 4 2
g g x g g x g x
x x x x x
Domain: (^) { x x is any real number}.
33. f ( ) x = x^2^ g x ( ) = x^2 + 4 The domain of f is (^) { x x is any real number}. The domain of g is (^) { x x is any real number}. a.
2
2 2 4 2
f g x f g x f x
x x x
Domain: (^) { x x is any real number}. b.
2
2 2 4
g f x g f x g x
x x
Domain: (^) { x x is any real number}. c.
2
22 4
( f f )( ) x f ( f ( )) x f x
x x
Domain: (^) { x x is any real number}. d.
2
2 2 4 2 4 2
g g x g g x g x
x x x x x
Domain: (^) { x x is any real number}.
Section 5.1: Composite Functions
34. f ( ) x = x^2^ + 1 g x ( ) = 2 x^2 + 3
The domain of f is (^) { x x is any real number}. The domain of g is (^) { x x is any real number}. a.
2
2 2 4 2 4 2
f g x f g x f x
x x x x x
Domain: (^) { x x is any real number}.
b.
2
2 2 4 2 4 2 4 2
g f x g f x g x
x x x x x x x
Domain: (^) { x x is any real number}.
c.
2
2 2 4 2 4 2
f f x f f x f x
x x x x x
Domain: (^) { x x is any real number}.
d.
2
2 2 4 2 4 2 4 2
g g x g g x g x
x x x x x x x
Domain: (^) { x x is any real number}.
35. ( ) 3 ( )^2
f x g x x x
The domain of f is (^) { x x ≠ (^1) }. The domain of g is { x x^ ≠^0 }. a. ( f g )( ) x = f ( g x ( )) 2 3 2 1 3 2 3 2 f x x x x x x = ^
= −
= (^) −
Domain (^) { x x ≠ 0, x ≠ (^2) }.
b. ( )( ) ( ( )) 3 1 2 3 1 2( 1) 3
g f x g f x g x
x x
= ^
Domain (^) { x x ≠ (^1) }
c. ( f f )( ) x = f ( f ( )) x 3 1 3 3 3 3 ( 1) 1 1 1 3( 1) 4
f x
x x x x x
= ^
Domain (^) { x x ≠ 1, x ≠ (^4) }.
d.
x g g x g g x g x x x
= = ^ = = =
Domain (^) { x x ≠ (^0) }.
Section 5.1: Composite Functions
38. ( ) ( )^2
f x x g x x x
The domain of f is (^) { x x ≠ − (^3) }. The domain of g is (^) { x x ≠ (^0) }. a. ( f g )( ) x = f ( g x ( )) 2
2 2
2 3 2 3
2 2 3
f x
x x x x x
x
= ^
Domain 2 , 0 3
(^) x x ≠ − x ≠
b. ( )( ) ( ( ))
g f x g f x g x x
x x x x
= ^
Domain (^) { x x ≠ −3, x ≠ (^0) }.
c. ( f f )( ) x = f ( f ( )) x
f x x x x x x x x x x x x
= ^
= +^ = +
Domain 3, 9 4
(^) x x ≠ − x ≠ −
d. ( )( ) ( ( )) 2 2 2 (^2 ) g g x g g x g x x x x
= = ^ = = =
Domain (^) { x x ≠ (^0) }.
39. f ( ) x = x g x ( ) = 2 x + 3 The domain of f is (^) { x x ≥ (^0) }. The domain of g is (^) { x x is any real number}. a. ( f g )( ) x = f ( g x ( )) = f (^) ( 2 x + (^3) )= 2 x + 3
Domain 3 2
(^) x x ≥ −
b. ( g f )( ) x = g ( f ( )) x = g ( x )= 2 x + 3
Domain (^) { x x ≥ (^0) }.
c.
1/ 21/ 2 1/ 4 4
( f f )( ) x f ( f ( )) x f x
x x x x
Domain (^) { x x ≥ (^0) }.
d. ( )
g g x g g x g x x x x
Domain (^) { x x is any real number}.
40. f ( ) x = x − 2 g x ( ) = 1 − 2 x The domain of f is (^) { x x ≥ (^2) }. The domain of g is (^) { x x is any real number}. a. ( )
f g x f g x f x x x
Domain 1 2
(^) x x ≤ −
b.
g f x g f x g x x
Domain (^) { x x ≥ (^2) }.
Chapter 5: Exponential and Logarithmic Functions
c.
f f x f f x f x
x
Now, 2 2 0 2 2 2 4 6
x x x x
Domain (^) { x x ≥ (^6) }.
d. ( )
g g x g g x g x x x x
Domain (^) { x x is any real number}.
41. f ( ) x = x^2 + 1 g x ( ) = x − 1 The domain of f is (^) { x x is any real number}. The domain of g is (^) { x x ≥ (^1) }. a.
2
f g x f g x f x
x x x
Domain (^) { x x ≥ (^1) }.
b.
2
2 2
g f x g f x g x x x x
Domain (^) { x x is any real number}.
c.
2
2 2 4 2 4 2
f f x f f x f x
x x x x x
Domain (^) { x x is any real number}.
d.
g g x g g x g x x
Now, 1 1 0 1 1 1 1 2
x x x x
Domain (^) { x x ≥ (^2) }.
42. f ( ) x = x^2 + 4 g x ( ) = x − 2 The domain of f is (^) { x x is any real number}. The domain of g is (^) { x x ≥ (^2) }. a.
2
f g x f g x f x
x x x
Domain (^) { x x ≥ (^2) }.
b.
2
2 2
g f x g f x g x x x
Domain (^) { x x is any real number}.
c.
2
2 2 4 2 4 2
f f x f f x f x
x x x x x
Domain (^) { x x is any real number}.
d.
g g x g g x g x x
Now, 2 2 0 2 2 2 4 6
x x x x
Domain (^) { x x ≥ (^6) }.
Chapter 5: Exponential and Logarithmic Functions
b.^2 2
( )( ) ( ( )) x x
g f x g f x g − −
= = ^
or 8 8
x x x x x (^) x x x (^) x x x x x x x x x x x x x x
= ^
^ −
= −^ +^ −
= −^ +^ −
Now, − x + 8 ≠ 0, so x ≠ 8. Also, from the domain of f , we know x ≠ 2. Domain of f g : (^) { x x ≠ 2, x ≠8 .}
c. 2 1 2
x x
f f x f f x f − −
= = ^
x x x x x (^) x x x (^) x x x x x x x x x x x x
^ −
= ^ ^
From the domain of f , we know x ≠ 2. Domain of f f : (^) { x x ≠2 .}
d.^4 2 5
( )( ) ( ( )) x x
g g x g g x g + −
= = ^
(^4 ) 2 5 2 4 5 2 5 (^4) 4 (2 5) 2 5 2 4 5 (2 5) 2 5 4 4(2 5) 2( 4) 5(2 5) 4 8 20 2 8 10 25 9 16 9 16 8 33 8 33 or
x x x x x (^) x x x (^) x x x x x x x x x x x x x x
- (^) + = − + (^) − (^) − + + − (^) − = + (^) − − ^ − = +^ +^ −
- − − = +^ +^ −
- − + = −^ − − − + − Now, 8 33 0, so 33. 8
x − ≠ x ≠ Also, from the
domain of g , we know 5 2
x ≠.
Domain of f g : 5 , 33. 2 8
x x (^) ≠ x ≠
f g x = f g x = f^ ^ x ^ = ^ x = x
g f x = g f x = g x = x = x
f g x = f g x = f^ ^ x ^ = ^ x = x
g f x = g f x = g x = x = x
3 3 3 ( f g )( ) x = f ( g x ( )) = f x = x = x
( g f )( ) x = g ( f ( )) x = g x^3^ = 3 x^3 = x
48. ( f g )( ) x = f ( g x ( )) = f (^) ( x − (^5) )= x − 5 + 5 = x ( g f )( ) x = g ( f ( )) x = g (^) ( x + (^5) ) = x + 5 − 5 = x
Section 5.1: Composite Functions
f g x f g x f x
x x x
= ^ +
= ^ + −
( )
( )
g f x g f x g x x
x x
f g x f g x f x
x x x
= ^ −
= − ^ −
( )
( )
g f x g f x g x x
x x
f g x f g x f x b a a x b b a x b b x
= ^ −
= ^ − +
( )
( )
g f x g f x g ax b ax b b a ax a x
f g x f g x f x x x x
= = ^ = = ⋅ =
g f x g f x g x x x x
= = ^ = = ⋅ =
53. H x ( ) = (2 x +3)^4 Answers may vary. One possibility is f ( ) x = x^4 , g x ( ) = 2 x + 3
23 H x ( ) = 1 + x Answers may vary. One possibility is f ( ) x = x^3^ , g x ( ) = 1 + x^2
55. H x ( ) = x^2 + 1 Answers may vary. One possibility is f ( ) x = x , g x ( ) = x^2 + 1 56. H x ( ) = 1 − x^2 Answers may vary. One possibility is f ( ) x = x , g x ( ) = 1 − x^2 57. H x ( ) = 2 x + 1 Answers may vary. One possibility is f ( ) x = x , g x ( ) = 2 x + 1 58. H x ( ) = 2 x^2 + 3 Answer may vary. One possibility is f ( ) x = x , g x ( ) = 2 x^2 + 3 59. f ( ) x = 2 x^3^ − 3 x^2 + 4 x − 1 g x ( ) = 2
3 2
f g x f g x f
( g f )( ) x = g ( f ( )) x = g 2 x^3^ − 3 x^2 + 4 x − 1 = 2
Section 5.1: Composite Functions
d. ( )( ) ( )( ) ( )
f g x g f x amx b m ax b cmx d cx d amx b amx bm cmx d cx d amx bm cmx d amx b cx d
Now, this equation will only be true if m = 1. Thus, f g = g f when m =1.
S r = π r r t = t t ≥
3
2 3
6
6
S r t S t
t
t
t
= ^
= π^ = π^
= π
Thus, ( ) 16 6 9
S t = π t.
V r = π r r t = t t ≥
3
3 3
9
9
V r t V t
t
t
t
= ^
= π^ = π^
= π
Thus, ( ) 32 9 81
V t = π t.
N t t t t C N N
2 2 2
C N t C t t t t t t
Thus, C t ( ) = 15, 000 + 800,000 t − 40, 000 t^2.
68. A r ( ) = π r^2 r t ( ) = 200 t
2 A r t ( ) = A 200 t = π 200 t = 40, 000π t Thus, A ( ) t = 40, 000π t.
69.^1 100, 0 400
p x x
x p x p
x C
p
p p
Thus,
p C p p
70.^1 200, 0 1000
p x x
x p x p
C^ x
p
p p
Thus,
p C p p
71. V = π r h 2 h = 2 r V r ( ) = π r^2 (2 ) r = 2 π r^3
72.^1
V = π r h h = r
( ) 1 2 (2 )^23 3 3
V r = π r r = π r
Chapter 5: Exponential and Logarithmic Functions
73. f (^) ( x (^) )= the number of Euros bought for x dollars; g (^) ( x ) = the number of yen bought for x Euros a. f (^) ( x (^) ) =0.7143 x b. g (^) ( x ) (^) =137.402 x
c. (^) ( )( ) ( ( )) ( ) ( )
g f x g f x g x x x
d. (^) ( )( 1000 ) 98.1462486 1000( ) 98,146.2486 yen
g f =
74. a. Given (^) ( ) 5 ( 32 ) 9
C F = F − and
K C ( (^) ) = C + 273 , we need to find K C ( (^) ( F (^) )).
( ( )) ( )
( )
or 9 9 9
K C F F
F
F
F
F
= ^ − +
b. (^) ( ( )) 5 80( ) 2297 80 299.7 kelvins 9
K C
75. a. f (^) ( p (^) ) = p − 200
b. g (^) ( p (^) ) =0.80 p
c. (^) ( ) ( ) ( ( )) ( 0.80 ) 200 0.80 200
f g p f g p p p
This represents the final price when the rebate is issued on the sale price. ( )( ) ( ( )) 0.80 (^) ( 200 ) 0.80 160
g f p g f p p p
This represents the final price when the sale price is calculated after the rebate is given. Appling the 20% first is a better deal since a larger portion will be removed up front.
76. Given that f and g are odd functions, we know that f (^) ( − x ) (^) = − f (^) ( x )and g (^) ( − x (^) ) = − g (^) ( x )for all x in the domain of f and g , respectively. The composite function ( f g )( ) x = f (^) ( g (^) ( x ))has the following property: ( ( )) ( (^ )) ( ( ))
since is odd since is odd ( )( )
f g x f g x f g x g f g x f f g x
Thus, f g is an odd function.
77. Given that f is odd and g is even, we know that f (^) ( − x (^) ) = − f (^) ( x )and g (^) ( − x (^) ) = g (^) ( x ) for all x in the domain of f and g , respectively. The composite function ( f g )( ) x = f ( g ( x ))has the following property: ( (^ )) ( ( ))
since is even ( )( )
f g x f g x f g x g f g x
Thus, f g is an even function. The composite function ( g f )( ) x = g (^) ( f (^) ( x )) has the following property: ( ( )) ( (^ )) ( ( ))
since is odd since is even ( )( )
g f x g f x g f x f g f x g g f x
Thus, g f is an even function.
Section 5.
1. The set of ordered pairs is a function because there are no ordered pairs with the same first element and different second elements. 2. The function f ( ) x = x^2 is increasing on the interval (^) ( 0, ∞ (^) ). It is decreasing on the interval ( −∞, 0^ ). 3. The function is not defined when x^2 + 3 x − 18 = 0. Solve: 2 3 18 0 ( 6)( 3) 0
x x x x
x = − 6 or x = 3 The domain is { x | x ≠ −6, x ≠ 3}.
Chapter 5: Exponential and Logarithmic Functions
Range: { Star Wars , The Phantom Menace , E.T. the Extra Terrestrial , Jurassic Park , Forrest Gump }
27. To find the inverse, interchange the elements in the domain with the elements in the range:
Monthly Cost of Life Insurance Age
Domain: {$7.09, $8.40, $11.29} Range: {30, 40, 45}
28. To find the inverse, interchange the elements in the domain with the elements in the range:
Virginia Nevada Tennessee Texas
Unemployment Rate State
Domain: {11%, 5.5%, 5.1%, 6.3%} Range: {Virginia, Nevada, Tennessee, Texas}
29. Interchange the entries in each ordered pair: {(5, −3), (9, −2), (2, −1), (11, 0), ( 5,1)} − Domain: {5, 9, 2, 11, −5} Range: { 3,− −2, −1, 0, 1} 30. Interchange the entries in each ordered pair: {(2, −2), (6, −1), (8, 0), ( 3,1), (9, 2)} − Domain: {2, 6, 8, −3, 9} Range: { 2,− −1, 0, 1, 2} 31. Interchange the entries in each ordered pair: {(1, −2), (2, −3), (0, −10), (9,1), (4, 2)} Domain: {1, 2, 0, 9, 4} Range: { 2,− −3, −10, 1, 2} 32. Interchange the entries in each ordered pair: {( 8,− −2), ( 1, − −1), (0, 0), (1,1), (8, 2)} Domain: { 8,− −1, 0, 1, 8} Range: { 2,− −1, 0, 1, 2}
f x = x + g x = x −
( )
f g x f x
x
x x
= ^ −
= ^ − +
( ) ( )
g f x g x x
x x
Thus, f and g are inverses of each other.
f x = − x g x = − x −
( )
f g x f x
x x x
= ^ − −
= − ^ − −
( )
( )
g f x g x x
x x
Thus, f and g are inverses of each other.
f x = x − g x =^ x +
( ( ) ) 2 4 4 2 8 4 8 8
x f g x f
x
x x
= ^ +
= ^ + −
Section 5.2: One-to-One Functions; Inverse Functions
( ( )) (4^ 8) (^4 8 ) 4 2 2
g f x^ g^ x x
x x
Thus, f and g are inverses of each other.
f x = x + g x = x −
( )
f g x f x
x x
x
= ^ −
= ^ − + = − +
( ( )^ ) (^2 6 ) (^1) (2 6) 3 3 3 2
g f x g x x x x
Thus, f and g are inverses of each other.
37. f ( ) x = x^3 − 8; g x ( ) = 3 x + 8
3
3 3
f g x f x
x x x
( ) 3 3 3 (^3 )
g f x g x x x x
Thus, f and g are inverses of each other.
38. f ( ) x = ( x − 2) ,^2 x ≥ 2; g x ( ) = x + 2
2
2
f g x f x
x
x x
2
g f x g x x x x
Thus, f and g are inverses of each other.
39. f ( ) x 1 ; g x ( )^1 x x
( )
x f g x f x x x
= ^ = = ⋅ =
( )
g f x g x x x x
= ^ = = ⋅ =
Thus, f and g are inverses of each other.
40. f ( ) x = x ; g x ( ) = x f (^) ( g x ( ) (^) ) = f (^) ( x ) = x g (^) ( f ( ) x (^) ) = g (^) ( x (^) )= x Thus, f and g are inverses of each other. 41. ( ) 2 3 ; ( )^4 4 2
f x x^ g x x x x
= +^ = −
( )
x f g x f x x x x x x x x x
x (^) x x x x x x x x x x x
x
= ^
+^ −
^ −
= −^ +^ −
= −^ +^ −
