Section P,3 Functions and Their Graphs
23
Section P.3 Functions and Their Graphs
1. (a) Domain off." -4 < x < 4 ~ [-4, 4]
4. (a) f(-4) = ~+5 = ~ = 1
Range off:-3 < y < 5 ~ [-3, 5]
(b) f(ll) = ~+5 --,fi-g = 4
Domain of g: -3 < x < 3 ~ [-3, 3]
(c) f(-8)= xi~ + 5 = ~2-~, undefined
Range of g: -4 < y < 4 ~ [-4, 4]
(d) f(x + Ax) = ",./x + kx + S
(b) 1(-2)=-1
g(3) = -4
gtO)" "= 5-o~ = 5
5.
(c) f(x) = g(x) for x = -1
(b) g(x/~)
= -5 (x/~)2=5-5=0
(d) f(x) = 2for x : 1
(c)
(e) g(x) = 0 for x = -I, 1 and 2
2. (a)
= 4 + 2t - t2
6. (a) g(4) = 42(4-4) = 0
Domain of g: -4 _< x _< 5 ~ [-4, 5]
(b) g(-~) =\~1(312(3\~- - 4) = ~.(_~.)9 5 = ___458
Rangeofg: -4 _< y _< 2 ~ [-4, 2]
(c) g(c) = c2(c - 4) = C3 -- 4c2
= -2
g(3) = 2
(c)
(d) g(t-1) = 5-(t-1)2 = 5-(12-2t+1)
Domain of f.’ -5 _< x _< 5 ~ [-5, 5]
Range off’ -4 _< y _< 4 ~ [-4, 4]
(b)
g(-2) = 5-(-2)2 = 5-4 = 1
(d) g(t+4) = (t+4)2(t+4-4)
= (t+4)2t = t3 +gfl +16t
f(x) = g(x)for x = -2 and x = 4
(d) f(x) = 2forx =-4,4
7. (a) f(0) = cos(2(0)) = cos0 = 1
/
(e) g(x) = 0for x = -1
3. (a) S(0) = 7(0) - 4 = -4
=0
2z
1
-32
(c)
COS
(b) 7(-3)= 7(-3)- 4 = -25
8. (a) f(z) = sin n- = 0
(c) f(b) = 7(b)-4 = 7b-4
(d) f(x-1) = 7(x-1)-4 = 7x-ll
(c)
9. f(x+Ax)-f(x) = (x+Ax)3-x3 = x~ +3x2Ax+3x2(Ax)2 +(Ax)3_x,
- 3x2 +3xAx+(Ax)2,Ax . 0
10. f(x) - f(1) 3x - 1 - (3 - 1) _ 3(x - 1)
x-1 =
x-1
x-1-- = 3,x :~ 1
11. f(x)- f(2) = (1/~--i"- 1)
x-2
x-2
1-~x-1
l+~x-1
-- 2-x
(x- 2)x/Tx - 1 1 +-,./Tx- 1
-1
12, f(x) - f(1) = x3 - x - 0 _ x(x + 1)(x - 1) : 2(x + 1), x 1
x-1
x-1
x-1
© 2010 Brooks/Cole, Cengage Learning
24 Chapter P Preparation for Calculus
13. f(x) = 4x2
Domain: (-~o, 00)
Range: [0, 00)
14. g(x) = x~ Domain: (-~o, 00)
Range: [-5,
15. g(x) -- ,S~
Domain: 6x >_ 0
x > 0 ~ [0,~o)
Range: [0, ~o)
16. h(x) = -x/-~ + 3
Domain: x + 3 > 0 ~ [-3,
22. f(x) = x/x2- 3x + 2
x2 - 3x + 2 > 0
(x- 2)(x- 1) _~ 0
Domain: x >_ 2orx < 1
Domain: (-00, 11 w [2,
2
gt~’I = 1 - cos x
1- cos x 0
cos x 1
Domain: all x 2nn-, n an integer
1
24. h(x) = sin x - 0/2)
1
sinx-- 0
2
1
2
sinx -
Range: (-~, O]
17. f(t) = sec--4
__n’t (2n+l)cr ~ t 4n+2
4 2
Domain: all t 4n + 2, n an integer
Range: (-00,- 1]
18. h(t) = cot t
Domain: all t = nn-, n aninteger
Domain: all x --6 + 2nor,-- 6+ 2nn-, n integer
f(x) -ix +1 31
1x+310
x+30
Domain: all x -3
Domain: (-~,-3) ~ (-3,
1
Range: (-~o, 00)
19. f(x):-x3
Domain: all x
Range: (-00, 0)w (0, 00)
2
20. g(x) = ~
(x- 2)(x + 2) 0
Domain: all x +2
Domain: (-c~,-2) ~ (-2, 2) ~ (2,
27. f(x) = {22; ++ 2,1’ ;><00
Domain: (-0% 1) t~ (1, 00)
(a) f(-1) = 2(-1) + 1 = -1
Range: (-00, 0)w (0, 00)
(b) f(0) = 2(0)+ 2 = 2
21. f(x): ~x
x_>O and 1-x>_O
x>_O and x_<l
Domain: 0 < x _< 1 ~ [0, 1]
(c) f(2) = 2(2) + 2 = 6
(d) f(t~ +0 = 2(t2 +0+2 = 2t~ +4
(Note: t: + 1 > 0 for all t)
Domain: (-oo, oo)
Rans~’. (-~, 0 ~ [~, ~)
© 2010 Brooks/Cole, Cengage Learning
Section P.3 Functions and Their Graphs 25
~2
X2
s~. S(~) -- +2 ~’ ~ -< 1
X + 2,
32. g(x) 4
x>1
(a) f(-2) = (-2)2 + 2 = 6
Domain: (-~, 0) c.) (0, oo)
(b) f(0) = 02 +2 = 2
Range: (-0% 0) ~ (0, oo)
Y
(c) 1(1) = 12 +2 = 3
+
+
6
= 2s4 + 8sa + 10
(Note: s2 + 2 > 1 for all s)
Domain: (-m, m)
Range: [2, o~)
{1~
+ 1, x < 1
29. f(x) = _ + l, x >
33. h(x) = ~x - 6
y
3-
(a) f(-3) :
1-31 + 1 = 4
(b) f(1) =-1+1 = 0
(c) f(3) =-3+1 =-2
21
3
6
9
12
(d) f(b2 + 1)= -(b2 + 1)+1
= _b2
Domain: (-~o,
Domain: x- 6 > 0
x > 6 ~ [6,
Range: (-0% 0] ~ [1, oo)
Range: [0,
x_<5
30. s(~)=~x+4,
{(x- ~)~,
~~~
34. f(x): ¼x3 + 3
Y
(a) f(-3)= -~+ 4 = -,~ = 1
5
4
(b) f(0)= ~,/-~+ 4 = 2
(o) f(5) = ~+4 = 3
(d) f(lO) = 0o- 5): = 25
Domain: [-4, ~)
Range: [0, ~)
Domain: (-~o,
Range: (-~o,
31. f(x) = 4- x
Domain: (-0% oo)
Range: (-o% oo)
35. f(x) = ~- x~
Domain: [-3, 3]
Range: [0, 3]
y
4"
-4 -3 -2 -1
-2
-3
© 2010 Brooks/Cole, Cengage Learning
26
Chapter P Preparation for Calculus
s(.) -- x + x
40.
d
Domain: [-2, 2]
Range: I-2, 2wi~1 ~ [-2, 2.83]
y-intercept: (0, 2)
x-intercept: (-x/~, 0)
41.
4-
(-,~, o)
x- y2 = 0 ~ y = +~X
y is not a function ofx. Some vertical lines intersect the
graph twice.
(0, 2
42. ~x~-4-y=O~ y=x/~x~-4
y is a function ofx. Vertical lines intersect the graph at
most once.
-4-
37. g(t) : 3 sin ~-t
43. y is a function ofx. Vertical lines intersect the graph at
most once.
2
44. x + y2 = 4
y = +x/~- x2
y is not a function ofx. Some vertical lines intersect the
graph twice.
2
2
45. x +y2 = 16 ~ y = +x/16-x
y is not a function ofx because there are two values ofy
for some x.
Domain:
Range: [-3, 3]
46. x2 +y = 16 ~ y = 16-xa
38. h¢o~ = -s cos 0
2
Domain:
Range: [-5, 5]
y is a function ofx because there is one value ofy for
each x.
47. y2
= x2 _ 1
~ y = +~x2 -1
y is not a function ofx because there are two values ofy
for some x.
5
X2
48. x2y- x2 + 4 y = 0 ~ y - xZ +4
y is a function ofx because there is one value ofy for
each x.
2-0 1 - mi/min during the first 4
39. The student travels
4-0 2
minutes. The student is stationary for the next 2 minutes.
6-2
Finally, the student travels
- 1 mi/min during
10 - 6
the final 4 minutes.
49. y = f(x + 5) is a horizontal shift 5 units to the left.
Matches d.
50. y = f(x) - 5 is a vertical shift 5 units downward.
Matches b.
51. y = -f(-x) - 2 is a reflection in the y-axis, a
reflection in the x-axis, and a vertical shift downward 2
units. Matches c.
52. y = -f(x - 4) is a horizontal shift 4 units to the right,
followed by a reflection in the x-axis. Matches a.
© 2010 Brooks/Cole, Cengage Learning
Section P.3 Functions and Their Graphs 27
53. y = f(x + 6) + 2 is a horizontal shift to the left 6
units, and a vertical shift upward 2 units. Matches e.
1
(f) The graph is stretched vertically by a factor of -~.
Y
54. y = f(x - 1) + 3 is a horizontal shift to the right 1 unit,
and a vertical shift upward 3 units. Matches g.
6
55. (a) the graph is shifted 3 units to the left.
Y
4
56. (a) g(x) = f(x - 4)
24
g(6) = f(2)= 1
g(O) = f(-4)=-3
Shift f right 4 units
(b) The graph is shifted 1 unit to the right.
4321"
4-
I
6
I ~-x
8
-11 .
-2-
-4
~
567
t
(o, -3)
(b) g(x) = f(x + 2)
(c) The graph is shifted 2 units upward.
Y
Shift f left 2 units
6-
4~
4-
O, 1)
2-4
-2
24
l
-i
(d) The graph is shifted 4 units downward.
(c) g(x) = f(x) + 4
Y
Vertical shift upwards 4 units
Y
6
(2, 5)
4
2
-5 -4 -3 -2 -1 _
(e) The graph is stretched vertically by a factor of 3.
Y
x
123
-2-
(d) x(x/= S(x/- 1
Vertical shift down 1 unit
3’
(2, 0)
23
© 2010 Brooks/Cole, Cengage Learning
Chapter P Preparation for Calculus
(e) g(x) = 2f(x)
g(2) = 2f(2)= 2
g(-4) = 2f(-4)=-6
Y
(2, 2)
58. (a) h(x) = sin(x + (st/2)) + l is a horizontal shift
a’/2 units to the left, followed by a vertical shift 1
unit upwards.
(b) h(x) = -sin(x - 1)is a horizontal shift 1 unit to the
right followed by a reflection about the x-axis.
1
-5 -4 -3 -2 -1
123
59. (a) Y(gO)) = ~(0) = 0
(b) g(f(1))= g(1)= 0
-3
(~) g(y(0))= g(o)=-~
(f)
g(x) = ½S(x)
(d) f(g(-4))= f(15)= ~
g(2) = -}f(2)= ½
(e) f(g(x)) = f(x2 -1) = ~x2 -1
g(-4) = -}f(-4)=--}
(f) g(f(x))= g(~x) : (.,,/-~)2 -l=x-1, (x > O)
2-
60. f(x) = sin x, g(x) = ax
(a) f(g(2))= f(2a’) = sin(2rc)= 0
-3-4-5"
-6-
(b) flg(~))= f/~)= sin/~) = 1
(c) g(f(O)) = g(O) = 0
57. (a) y = ~x +-2
Y
4
3
2
(e) y(g(x))= y(~x)= s~.(~x)
1
1
2
3
4
Vertical shift 2 units upward
(b) y =-~’-~
(~ g(f(x)) = g(sin x) = x sin x
61, f(x) = x2, g(x) =
(f o g)(x) = f(g(x))
=
=
=X,x>_O
Domain: [0, ~o)
(go f)(x) = g(f(x)) = g(x2) = x/~-x2 = x
Domain:
Reflection about the x-axis
(c) y = ,/Tx- 2
y
No. Their domains are different. (fo g) = (go f) for
x>0.
43
2
1
__
-1
123456
-2
Horizontal shift 2 units to the right
© 2010 Brooks/Cole, Cengage Learning
Section P.3 Functions and Their Graphs 29
~z. s(x) -- x~ -~, g(.) = cos
(
64. (fog)(x) : S ~ - ~+ 2
)
1
(fo g)(x) = f(g(x)) = f(cos x) = cosZx- 1
Domain: (-2, oo)
Domain: (-0% oo)
(go f)(x) : g(x2 --1) =
cos(xa -1)
You can find the domain of g o f by determining the
intervals where (1 + 2x) and x are both positive, or both
negative.
Domain: (-0%
No, fog , go f.
+ + +--+ + + +
3, g(x) = x~ -1
63, f(x)= 7
-i _½ 0
i
~
3
(f o g)(x) = f(g(x)) = f(x~- 1) - x~- 1
Domain: all x:~ +1 => (-co,-1) u (-1, 1)~ (1,
(go f)(x) =
g(S(.))
g
=
65. (a) (fo g)(3) = f(g(3)) = f(-1)= 4
(b) g(f(2))= g(1)= -2
-1 = -~5-- 1 -
(c) g(f(5)) : g(-5), which is undefined
x2
Domain: all x + 0 ~ (-oo, O)~ (0, oo)
(d) (fo g)(-3)= f(g(-3))= f(-2)= 3
(e) (go f)(-1) = g(f(-1))= g(4)= 2
No, fog ~ go f.
(f) f(g(-1)) = f(-4), which is undefined
66. (Ao r)(t): A(r(t))= A(0.61)= sr(0.6t)2 = 0.36n-12
(Ao r)(t)represents the area of the circle at time t.
67. F(x) = x/~x- 2
Let h(x) = 2x, g(x) = x- 2 and f(x) = x/--~x.
Then, (fo go h)(x) = f(g(2x)) = f((2x)- 2) = 4(2x)- 2 = x/~Tx- 2 = F(x).
[Other answers possible]
68. F(x) = -4 sin(1- x)
Let f(x) = -4x, g(x) = sin x and h(x) = 1- x. Then,
(fogo h)(x): f(g(1- x)): f(sin(1- x)):-4 sin(1- x): F(x).
[Other answers possible]
f(--X) = (--X)2 (4 -- (--X)2) : X2(4--X2) : f(x)
Even
70. f(-x) = ~ =-3~x =-f(x)
Odd
71.
S(-x) = (-x)cos(-x) = -x cos x = -S(x)
Odd
72. S(-x) = sin2(-x)= sin(-x)sin(-x)= (-sin x)(-sin x)= sin2 x
Even
~
© 2010 Brooks/Cole, Cengage Learning
30 Chapter P Preparation for Calculus
73. (a) If f is even, then (-~, 4)is on the graph.
(b) If f is odd, then (-~, -4)is on the graph.
74. (a) If f is even, then (-4, 9) is on the graph.
(b) If f is odd, then (-4, -9) is on the graph.
75. f is even because the graph is symmetric about the
y-axis, g is neither even nor odd. h is odd because the
graph is symmetric about the origin.
78. Slope-
8-1
7
=5-3 2
7 3)
y-1-- 7(x-
7 21
y - 1 = --x - -2
2
7
19
2
2
For the line segment, you must restrict the domain.
76. (a) If f is even, then the graph is symmetric about the
y-axis.
(5, 8)
Y
13, 1)
246
246
-6 -4 -2
-2-
8
-4-
79. x + y2 = 0
-6-
(b) If f is odd, then the graph is symmetric about the
origin.
Y
f
y2 =--X
y = _~-~
f(x) : _.~fS-~, x <_ 0
4
x
-6 -4 -2
-2-4-6-
77. Slope- 4 - (-6)
-2 - 0
10 -2
5
y - 4 = -~(x -(-~/)
y-4 =-5x-10
y = -5x - 6
80. X2 + y2
= 36
y2 = 36 - x2
y =-x/36-x2, -6_< x_< 6
y
For the line segment, you must restrict the domain.
f(x) =-5x-6,-2 <_ x < 0
Y
81. Answers will vary. Sample answer: Speed begins and
ends at 0. The speed might be constant in the middle:
~~:~x
Time (in hours)
© 2010 Brooks/Cole, Cengage Learning
Section P. 3 Functions and Their Graphs 31
82. Answers will vary. Sample answer: Height begins a few
feet above 0, and ends at 0.
88. (a) For each time t, there corresponds a depth d.
(b) Domain: 0 < t _< 5
Range: 0 < d < 30
(c)
d
252015-10.
Distance
5.
83. Answers will vary. Sample answer: In general, as the
price decreases, the store will sell more.
123456
(d) d(4) ~ 18. At time 4 seconds, the depth is
approximately 18 cm.
Y
A
89. (a)
~r~ ~ 200
Price (in dollars)
84. Answers will vary. Sample answer: As time goes on, the
value of the car will decrease
5 15 25 35 45 55
Year (5 ~ 1955)
(b) A(20) ,~ 384 acres/farm
0
85.
y = ~C--X2
(b) ~H= 0.002 0.005~-f-x
+ k,l.6J ) - 0.029
y2 = c - x2
x2 + y2 = c, acircle.
For the domain to be [-5, 5], c = 25.
86. For the domain to be the set of all real numbers, you
must require that x2 + 3cx + 6 ~: 0. So, the
discriminant must be less than zero:
(3c)2- 4(6) < 0
9c2 < 24
C2 <-~ 8
-~ff-~ < c < x/-~
= 0.00078125x2 + 0.003125x - 0.029
re(x) =l l+lxIfx < O, then f(x) :-x-(x-2) =-2x+2.
IfO < x < 2, then f(x) = x-(x-2) = 2.
If x_> 2, then f(x) = x+(x-2) = 2x-2.
So,
-2x+2, x_< 0
0 < x < 2.
f(x) = ~2,
2x- 2, x > 2
87. (a) T(4) = 16°,T(15) ~ 23°
(b) If H(t) = T(t - 1), then the changes in temperature
will occur 1 hour later.
(c) If H(t) = T(t) - 1, then the overall temperature
would be 1 degree lower.
© 2010 Brooks/Cole, Cengage Learning