Type:
Solution Manual and Test Bank
Resource:
Precalculus, Mathematics for Calculus
Edition:
7th Edition
Author(s):
James Stewart Lothar Redlin Saleem Watson
2
FUNCTIONS
2.1
FUNCTIONS
1. If f (x )
x 3 + 1, then
(a) the value of f at x (b) the value of f at x
—1 is f (—1) 2 is f (2)
(—1)3 + 1
23 + 1
(c) the net change in the value of f between x
0.
9. —1 and x
2 is f (2) — f (—1)
9—0
9.
2. For a function f , the set of all possible inputs is called the domain of f , and the set of all possible outputs is called the range of f . 3. (a) f (x )
x 2 — 3x and g (x )
x—5
have 5 in their domain because they are defined when x 5. However, x h (x ) x — 10 is undefined when x 5 because 5 — 55 — 10 0 —5, so 5 is not in the domain of h. (b) f (5) 52 — 3 (5) 25 — 15 10 and g (5) 0. 5
5
4. (a) Verbal: “Subtract 4, then square and add 3.” (b) Numerical: x
f (x )
0
19
2
7
4
3
6
7
5. A function f is a rule that assigns to each element x in a set A exactly one element called f (x ) in a set B. Table (i) defines y as a function of x , but table (ii) does not, because f (1) is not uniquely defined. 6. (a) Yes, it is possible that f (1)
f (2)
(b) No, it is not possible to have f (1) value of f (x ).
5. [For instance, let f (x ) 5 and f (1)
6. A function assigns each value of x in its domain exactly one
7. Multiplying x by 3 gives 3x , then subtracting 5 gives f (x) 8. Squaring x gives x2, then adding two gives f (x) 9. Subtracting 1 gives x — 1, then squaring gives f (x )
5 for all x.]
3x — 5.
x2 + 2. (x — 1)2.
10. Adding1 gives x + 1, taking the square root gives x + 1, then dividing by 6 gives f (x ) 11. f (x )
2x + 3: Multiply by 2, then add 3.
12. g (x )
13. h (x )
5 (x + 1): Add 1, then multiply by 5.
14. k (x )
x+1 6
.
x +2 : Add 2, then divide by 3. 3 x2 — 4 3
: Square, then subtract 4, then divide by 3. 141
142
CHAPTER 2 Functions
15. Machine diagram for f (x )
subtract 1, then take square root
0
2
subtract 1, then take square root
1
5
subtract 1, then take square root
2
1
17. f (x )
19. f (x ) f 12 20. f (x ) f (0)
2 (x — 1)2
18. g (x ) x
f (x )
—1
2 (—1 — 1)2
0
2 (—1)2
1
2 (1 — 1)2
2
2 (2 — 1)2
3
2 (3 — 1)2
x 2 — 6; f (—3)
(—3)2 — 6
2 1 —6 2
— 23 .
1 —6 4
x3 + 2x; f (—2) 03 + 2 (0)
2
1
x—2
3
subtract 2, take reciprocal, multiply by 3
3
_1
subtract 2, take reciprocal, multiply by 3
_1
1
subtract 2, take reciprocal, multiply by 3
_3
2x + 3 g (x )
8
—3
2 (—3) + 3
3
2
—2
2 (—2) + 3
1
0
0
2 (0) + 3
3
2
1
2 (1) + 3
5
8
3
2 (3) + 3
9
3; f (0)
02 — 6
9—6
3; f (3)
32 — 6
9—6
3
+ 22 1
—8 — 4 1 +1 2
—12; f (—1)
9. 8
(—1)3 + 2 (—1)
—1 — 2
f (a) 24. h (x )
a
2
+ 2 (a)
02 + 2 (0) a
2
1 x + ; h (—1) x
h (x — 1)
x—1+
0; f (3)
32 + 2 (3) 2
+ 2a; f (—x)
(—x )
(—1) + 1
—1 — 1
+ 2 (—x)
;h
x—1
1
15; f (—3)
2+ 1
x
+
1
1
1 x
x
+
x.
2
5; h 1 2
0; f (a)
(—x )2 + 4 5
(—3)2 + 2 (—3) 1 a
1 a
x 2 — 2x ; f
2
1 x
9+6
—2; h (2)
1
1
—3;
8
1 — 2 12 5 1 — 2x 1 — 2 (2) 1 — 2 (—2) ; f (2) —1; f (—2) ; f 12 3 3 3 3 3 1 — 2 (—a) 1 + 2a 1 — 2 (a — 1) 3 — 2a f (—a) ; f (a — 1) . 3 3 3 3 8 8 (—2)2 + 4 x2 + 4 22 + 4 a2 + 4 ;h 2 ;h a 22. h x ; h 2 ;h x ( ) ( ) ( ) (— ) (— ) 5 5 5 5 5 5 2 2 2 a — 4a + 8 x+4 (a — 2) + 4 x +4 ;h x . h (a — 2) 5 5 5 5 x 2 + 2x; f (0)
—6;
4
21. f (x )
23. f (x )
.
x
(—2)3 + 2 (—2)
0; f 1
3
16. Machine diagram for f (x )
x — 1.
2
1 +2 a 1 1 2
+ 1 2
9—6 2 + . 2 a a 1 2 + 2
1 — 2a 3
;
x2 + 4 ; 5
3;
1
5
;
2
25. g (x )
1 — (2)
1—x ; g (2) 1+x 1 — (a)
g (a)
1 + (2)
— ; g (—1) 3
3
1—a
1 1 — (—1) , which is undefined; g 2 1 + (—1)
1
—1
1 — (a — 1)
143
1 — 12
1 2
1
2
3
3
1 + 12
1 — x2 — 1
2—a
1—a +1
SECTION 2.1 Functions
;
2 — x2
. ; g (a — 1) a 1 + (a) 1 + (a — 1) 1+ a 1+a — 1 x2 1 + x2 — 1 t+2 —2 + 2 2+2 0+2 a+2 26. g (t ) ; g (—2) 0; g (2) , which is undefined; g (0) —1; g (a) ; t—2 —2 — 2 2—2 0—2 a—2 a2 a2 — 2 + 2 a+3 a+1+2 . g a2 — 2 ; g (a + 1) a+1—2 a—1 a2 — 2 — 2 a 2 — 4 27. k (x ) —x2 — 2x + 3; k (0) —02 — 2 (0) + 3 3; k (2) —22 — 2 (2) + 3 —5; k (—2) — (—2)2 — 2 (—2) + 3 3; k
; g x2 — 1
—
2
k (—x) 28. k (x )
2
k x3
2 x3
3
— 3 12 — 3 x3
2 12— 1
f x2 + 2
f (5)
3
—x 2 + 2x + 3; k x 2
2
2
2 1
1. Since 1
— x2
2
— 2 x2 + 3
—a2 — 6a — 5;
—x4 — 2x 2 + 3.
2 (3)3 — 3 (3)2 27; k (—3) a3 — 3a2 2 a2 3 — 3 a2 2 ; k (—x ) 4
2 (—3)3 — 3 (—3)2
—81;
2 (—x ) 3 — 3 (—x ) 2
— 2x — 3x ;
2 —2 — 1
2 (3)
6; f (0)
2 0—1
1; f (2)
2 2—1
2 (1)
2; f (x + 1)
2
2x 2 + 2 (since x2 + 1 > 0 ). —1 1 —1; f (—1) —1 —1 x2 x2
(—2)2
0, we have f (1)
2 (1)
33. Since —4 c —1, we have f (—4) 2 — 3 2 + 2 — 3 2 9 — 43
—1 < 0 c 1, we have f (0)
2 (x + 1) — 1
1/x 1/x
4. Since —1 < 0, we have f (—1)
(—1)2
1+1
2. Since 2
2 x ;
—1; f (x) is not defined at x 1 x
1 since x 2 > 0, x / 0; f
1+1
(—4)2 + 2 (—4)
0, we have f (2)
16 — 8
— 3 . Since 4 —1 c —1, we have f (—1)
0. Since 25 > 1, we have f (25)
34. Since —5 < 0, we have f (—5) f (1)
x x
.
1. Since 0
2+1
3 (—5)
2+1
x 3 39. f (x )
5. Since 3 > 2, we
(—1)2 + 2 (—1)
1—2
0+1
—1. Since
3. Since 5 > 2, we have f (5) x 2 + 1 + (2)2 + 1
1. Since 0 c 1 c 2, we have (5 — 2)2
9.
x2 + 1 + 4 + 1
f (x) 6x — 18 3 (2x — 6) — 18 2x — 18; 2x — 6. 3 3 3 3 3x — 2, so f (1) 3 (1) — 2 1 and f (5) 3 (5) — 2 13. Thus, the net change is f (5) — f (1) 6
0, we have
3.
8. Since — 3 2c —1, we have
—15. Since 0 c 0 c 2, we have f (0)
2. Since 0 c 2 c 2, we have f (2)
0;
—1.
35. f (x + 2) (x + 2)2 + 1 x2 + 4x + 4 + 1 x2 + 4x + 5; f (x ) + f (2) 36. f (2x ) 3 (2x ) — 1 6x — 1; 2 f (x) 2 (3x — 1) 6x — 2. . .2 37. f x 2 x 2 + 4; f (x) [x + 4]2 x 2 + 8x + 16. 38. f
2
2;
32. Since —3 c 2, we have f (—3) 5. Since 0 c 2, we have f (0) 5. Since 2 c 2, we have f (2) have f (3) 2 (3) — 3 3. Since 5 > 2, we have f (5) 2 (5) — 3 7. f — 32
3
2x 9 — 3x 6.
2 x2 + 2 — 1 2 x2 + 1 x —2 2 ; f (—2) x —2 —2 2 x 5 5 x2 1; f 5 5 x2
0+1
— (a + 2)2 — 2 (a + 2) + 3
0; k (3)
— 21 ; k a2
31. Since —2 < 0, we have f (—2) f (0)
1 — 2 2; k (a + 2)
2 (0)3 — 3 (0)2
2 x — 1 ; f (—2)
1 2
30. f (x )
2 +3
2x3 — 3x2; k (0) 2 12
f
—2
— (—x )2 — 2 (—x ) + 3
k 12
29. f (x )
2
x2 + 6.
x
40. f (x ) 4 — 5x , so f (3) 4 — 5 (3) f (5) — f (3) —21 — (—11) —10.
—11 and f (5)
4 — 5 (5)
13 — 1
—21. Thus, the net change is
12.
144
CHAPTER 2 Functions
41. g (t )
1 — t 2, so g (—2)
g (5) — g (—2)
1 — (—2)2
—24 — (—3)
1—4
—3 and g (5)
1 — 52
—24. Thus, the net change is
—21.
42. h (t ) t 2 +5, so h (—3) (—3)2 +5 14 and h (6) 62 +5 41. Thus, the net change is h (6)—h (—3) 43. f (a) 5 — 2a; f (a + h) 5 — 2 (a + h) 5 — 2a — 2h; f (a + h) — f (a) 5 — 2a — 2h — (5 — 2a) 5 — 2a — 2h — 5 + 2a —2h —2. h h h h 2 2 2 2 44. f (a) 3a + 2; f (a + h) 3 (a + h) + 2 3a + 6ah + 3h + 2; f (a + h) — f (a) h 45. f (a) 5; f (a + h) 46. f (a)
1
3a2 + 6ah + 3h2 + 2 — 3a2 + 2 5;
h f (a + h) — f (a)
; f (a + h)
1
h
5—5
6ah
+ h
3h2
6a + 3h.
0.
h
; a+h+1 a+h +1 1 a+1 — 1 — f (a + h) — f (a) (a + 1) (a + h + 1) (a + 1) (a + h + 1) a+h+1 a+1 h h h —h —1 (a + 1) (a + h + 1) . h (a + 1) (a + h + 1) a+h a ; f (a h) ; 47. f (a) + a+1 a+h+1 (a + h) (a + 1) a (a + h + 1) a+h a — — f (a + h) — f (a) (a + h + 1) (a + 1) (a + h + 1) (a + 1) a+h+1 a+1 h h h (a + h) (a + 1) — a (a + h + 1) 2 a + a + ah + h — a2 + ah + a (a + h + 1) (a + 1) h (a + h + 1) (a + 1) h 1 a+1
(a + h + 1) (a + 1) 2 (a + h) 2a ; f (a h) ; 48. f (a) + a —1 a+h—1 2 (a + h) 2a (2a + 2h) (a — 1) — 2a (a + h — 1) — f (a + h) — f (a) (a + h — 1) (a — 1) a+h—1 a — 1 (a + h — 1) (a — 1) h h h 2 (a + h) (a — 1) — 2a (a + h — 1) 2a 2+ 2ah — 2a — 2h — 2a 2— 2ah + 2a (a + h — 1) (a — 1) h h (a + h — 1) (a — 1) —2h 2 — h (a + h — 1) (a — 1) (a + h — 1) (a — 1) 49. f (a) 3 — 5a + 4a2; f (a + h)
3 — 5 (a + h) + 4 (a + h)2
3 — 5a — 5h + 4 a2 + 2ah + h2
3 — 5a — 5h + 4a2 + 8ah + 4h2; f (a + h) — f (a) h
3 — 5a — 5h + 4a2 + 8ah + 4h2 — 3 — 5a + 4a2 h 3 — 5a — 5h + 4a2 + 8ah + 4h2 — 3 + 5a — 4a2 h h (—5 + 8a + 4h) —5 + 8a + 4h. h
—5h + 8ah + 4h2 h
41—14
27.
SECTION 2.1 Functions
50. f (a)
a3; f (a + h)
f (a + h) — f (a) h
51. f (x )
(a + h)3
a3 + 3a2h + 3ah2 + h3;
a3 + 3a2h + 3ah2 + h3 — a3
3a2h + 3ah2 + h3 h
h h 3a2 + 3ah + h2
3a2 + 3ah + h2. h 3x. Since there is no restriction, the domain is all real numbers, (—
times the real number 31 y, the range is all real numbers (—
). Since every real number y is three
).
52. f (x ) 5x2 + 4. Since there is no restriction, the domain is all real numbers, (— + 4 4 for all x, so the range is [4 ). 53. f (x )
145
3x , —2 c x c 6. The domain is [—2 6], f (—2)
3 (—2)
—6, and f (6)
). Since 5x2 3 (6)
0 for all x , 5x2
18, so the range is [—6 18].
54. f (x ) 5x2 + 4, 0 c x c 2. The domain is [0 2], f (0) 5 (0)2 + 4 4, and f (2) 5 (2)2 + 4 24, so the range is [4 24]. 1 55. f (x ) . Since the denominator cannot equal 0 we have x — 3 / 0 e x / 3. Thus the domain is x x / 3 . In x 3 interval notation, the domain is (— 3) (3 ). 1 56. f (x ) . Since the denominator cannot equal 0, we have 3x — 6 / 0 e 3x / 6 e x / 2. In interval notation, the 3x — 6 domain is (— 2) (2 ). x+2 57. f (x ) . Since the denominator cannot equal 0 we have x 2 — 1 / 0 e x 2 / 1 → x / 1. Thus the domain is x2 — 1 x x/ 1 . In interval notation, the domain is (— —1) (—1 1) (1 ). x4 . Since the denominator cannot equal 0, x 2 x 6 0 58. f x x 3 x 2 0 x 3 or x 2. () + — / e ( + )( — ) / → / — / 2 x +x —6 In interval notation, the domain is (— —3) (—3 2) (2 ). 59. f (x )
x + 1. We must have x + 1
60. g (x )
x 2 + 9. The argument of the square root is positive for all x , so the domain is (—
61. f (t ) 62. g (x ) — 63. f (x )
0e x
—1. Thus, the domain is [—1
). ).
t — 1. Since the odd root is defined for all real numbers, the domain is the set of real numbers, (—
3
7 — 3x . For the square root to be defined, we must have 7 — 3x
0e7
3x e 7
3
).
x . Thus the domain is
7 . 3
1 — 2x. Since the square root is defined as a real number only for nonnegative numbers, we require that 1
— 2x
0 e x c 1 . So 2 the domain is x
x c 1 . 2In interval notation, the domain is —
64. g (x )
x 2 — 4. We must have x 2 — 4
0 e (x — 2) (x + 2)
1 . 2
0. We makea table:
(—
—2)
(—2 2)
(2
)
Sign of x — 2
—
—
+
Sign of x + 2
—
+
+
Sign of (x — 2) (x + 2)
+
—
+
Thus the domain is (— —2] [2 ). 2 x + 65. g (x ) . We require 2 + x 0, and the denominator cannot equal 0. Now 2 + x 0 e x —2, and 3 — x / 0 3—x e x / 3. Thus the domain is x x —2 and x / 3 , which can be expressed in interval notation as [—2 3) (3 ).