Table Of Content1
Chapter P: Preparing for Calculus
P.1: Functions and Their Graphs
Concepts and Vocabulary
1. If f is a function defined by y = f(x), then x is called the independent variable and y is the
dependent variable.
2. True . The independent variable is sometimes referred to as the argument of the function.
3. False . If no domain is specified for a function f, then the domain of f is taken to be the largest set
of real numbers for which the value f(x) is defined and is a real number.
3(x2−1)
4. False . The domain of the function f(x)= is the set {x|x(cid:54)=1}.
x−1
5. False . A function can have at most one y-intercept; otherwise, there would be more than one y-value
corresponding to x=0.
6. A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects
the graph in at most one point.
7. If the point (5,−3) is on the graph of f, then f( 5 )= −3 .
8. Let f(x) = ax2 +4. For the point (−1,2) to be on the graph of f, we must have f(−1) = 2. But
f(−1)=a(−1)2+4=a+4, so we must have a+4=2, or a=−2 .
9. A function f is (a) increasing on an interval I if, for any choice of x and x in I, with x < x ,
1 2 1 2
then f(x )<f(x ).
1 2
10. Afunctionis (a) even ifforeverynumberxinitsdomain,thenumber−xisalsointhedomainand
f(−x)=f(x). A function f is (b) odd if for every number x in its domain, the number −x is also
in the domain and f(−x)=−f(x).
11. False . Evenfunctionshavegraphsthataresymmetricwithrespecttothey-axis. Oddfunctionshave
graphs that are symmetric with respect to the origin.
12. The average rate of change of f(x)=2x3−3 from 0 to 2 is
f(2)−f(0) 2(2)3−3−(2(0)3−3) 13−(−3) 16
= = = = 8 .
2−0 2 2 2
Practice Problems
13. Let f(x)=3x2+2x−4. Then
(a) f(0)=3(0)2+2(0)−4= −4 .
(b) f(−x)=3(−x)2+2(−x)−4= 3x2−2x−4 .
(c) −f(x)=−(3x2+2x−4)= −3x2−2x+4 .
(d) f(x+1)=3(x+1)2+2(x+1)−4=3(x2+2x+1)+2x+2−4= 3x2+8x+1 .
2
(e) f(x+h)=3(x+h)2+2(x+h)−4=3(x2+2xh+h2)+2x+2h−4= 3x2+6xh+3h2+2x+2h−4 .
x
14. Let f(x)= . Then
x2+1
0
(a) f(0)= = 0 .
02+1
−x x
(b) f(−x)= = − .
(−x)2+1 x2+1
x
(c) −f(x)= − .
x2+1
x+1 x+1
(d) f(x+1)= = .
(x+1)2+1 x2+2x+2
x+h x+h
(e) f(x+h)= = .
(x+h)2+1 x2+2xh+h2+1
15. Let f(x)=|x|+4. Then
(a) f(0)=|0|+4= 4 .
(b) f(−x)=|−x|+4= |x|+4 .
(c) −f(x)=−(|x|+4)= −|x|−4 .
(d) f(x+1)= |x+1|+4 .
(e) f(x+h)= |x+h|+4 .
√
16. Let f(x)= 3−x. Then
√ √
(a) f(0)= 3−0= 3 .
(cid:112) √
(b) f(−x)= 3−(−x)= 3+x .
√
(c) −f(x)= − 3−x .
(cid:112) √
(d) f(x+1)= 3−(x+1)= 2−x .
√
(cid:112)
(e) f(x+h)= 3−(x+h)= 3−x−h .
17. Becausef(x)=x3−1isdefinedforanyrealnumberx,thedomainoff isthesetof all real numbers ,
or in interval notation (−∞,∞) .
x
18. Because x2 +1 is never equal to zero for any real number x, f(x) = is defined for any real
x2+1
numberx. Thedomainoff isthereforethesetof all real numbers ,orinintervalnotation (−∞,∞) .
19. Because the square root of a negative number is not a real number, the value of t2 − 9 must be
nonnegative. The solution of the inequality t2−9≥0 is {t|t≤−3}∪{t|t≥3}, so the domain of v is
the set of real numbers {t|t≤−3}∪{t|t≥3} , or in interval notation (−∞,−3]∪[3,∞) .
20. Because the expression x−1 appears under thesquare rootand in thedenominator, the value of x−1
must be positive; that is, x−1>0. The solution of this inequality is {x|x>1}, so the domain of g is
the set of real numbers {x|x>1} , or in interval notation (1,∞) .
3
21. Because division by zero is not defined, x3−4x=x(x2−4)=x(x−2)(x+2) cannot be zero; that is,
x(cid:54)=0,x(cid:54)=2,andx(cid:54)=−2. Thus,thedomainofhisthesetofrealnumbers {x|x(cid:54)=−2,x(cid:54)=0,x(cid:54)=2} .
22. Because the square root of a negative number is not a real number, the value of t + 1 must be
nonnegative. The solution of the inequality t+1 ≥ 0 is {t|t ≥ −1}. Moreover, because division by
zero is not defined, t−5 cannot be 0; that is, t cannot be equal to 5. The intersection of the sets
{t|t≥−1} and {t|t(cid:54)=5} is the set {t|−1≤t<5}∪{t|t>5}. The domain of s is therefore the set of
real numbers {t|−1≤t<5}∪{t|t>5} , or in interval notation [−1,5)∪(5,∞) .
23. Let f(x)=−3x+1. Then
f(x+h)−f(x) −3(x+h)+1−(−3x+1) −3x−3h+1+3x−1 −3h
= = = = −3 .
h h h h
1
24. Let f(x)= . Then
x+3
f(x+h)−f(x) 1 − 1 (x+h+3)(x+3)
= x+h+3 x+3 ·
h h (x+h+3)(x+3)
x+3−(x+h+3) −h 1
= = = − .
h(x+h+3)(x+3) h(x+h+3)(x+3) (x+h+3)(x+3)
√
25. Let f(x)= x+7. Then
√ √ √ √
f(x+h)−f(x) x+h+7− x+7 x+h+7+ x+7
= · √ √
h h x+h+7+ x+7
x+h+7−(x+7) h 1
= √ √ = √ √ = √ √ .
h( x+h+7+ x+7) h( x+h+7+ x+7) x+h+7+ x+7
2
26. Let f(x)= √ . Then
x+7
√ √
f(x+h)−f(x) √ 2 − √2 x+h+7 x+7
= x+h+7 x+7 · √ √
h h x+h+7 x+7
√ √ √ √
2 x+7−2 x+h+7 x+7+ x+h+7
= √ √ · √ √
h x+h+7 x+7 x+7+ x+h+7
2(x+7)−2(x+h+7)
= √ √ √ √
h x+h+7 x+7( x+7+ x+h+7)
−2h
= √ √ √ √
h x+h+7 x+7( x+7+ x+h+7)
2
= −√ √ √ √ .
x+h+7 x+7( x+7+ x+h+7)
27. Let f(x)=x2+2x. Then
f(x+h)−f(x) (x+h)2+2(x+h)−(x2+2x) x2+2xh+h2+2x+2h−x2−2x
= =
h h h
2xh+h2+2h h(2x+h+2)
= = = 2x+h+2 .
h h
4
28. Let f(x)=(2x+3)2. Then
f(x+h)−f(x) (2(x+h)+3)2−(2x+3)2 4(x+h)2+12(x+h)+9−(4x2+12x+9)
= =
h h h
4x2+8xh+4h2+12x+12h+9−4x2−12x−9
=
h
8xh+4h2+12h h(8x+4h+12)
= = = 8x+4h+12 .
h h
29. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two
points), so it does not represent a function .
30. This graph passes the vertical line test, and so it does represent a function .
(a) From the graph, the domain of the function is the set of all real numbers or the interval
(−∞,∞) , and the range of the function is the set of all positive real numbers or the inter-
val (0,∞) .
(b) The graph has no x-intercepts, and the y-intercept is 1. Thus, the only intercept is (0,1) .
(c) The graph is not symmetric with respect to the x-axis, the y-axis, or the origin.
31. This graph passes the vertical line test, and so it does represent a function .
(a) From the graph, the domain of the function is the set {x|−π ≤x≤π} or the closed interval
[−π,π] ,andtherangeofthefunctionistheset {y|−1≤y ≤1} ortheclosedinterval [−1,1] .
π (cid:16) π (cid:17)
(b) Thegraphhasx-interceptsof± anday-interceptof1,sotheinterceptsare ± ,0 and (0,1) .
2 2
(c) The graph is symmetric with respect to the y-axis but not symmetric with respect to the x-axis
or the origin.
32. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two
points), so it does not represent a function .
33. (a) f(−1)=−1+3=2; f(0)=0+3=3; f(1)=5; f(8)=−8+2=−6
(b) The graph of f consists of three pieces corresponding to each equation in the definition. On the
interval [−2,1), the graph is the line y = x+3, and on the interval (1,∞), the graph is the line
y =−x+2. The graph also contains the point (1,5).
5
4
3
2
1
-2 -1 1 2 3
-1
5
(c) The individual components of f have domains of {x|−2 ≤ x < 1}, {x|x = 1}, and {x|x > 1}.
Thedomainoff istheunionofthesethreesets;thatis, {x|x≥−2} insetnotationor [−2,∞)
in interval notation. Moreover, the individual components of f have ranges of {y|1 ≤ y < 4},
{y|y =5},and{y|y <1}. Therangeoff istheunionofthesethreesets;thatis, {y|y =5 or y <4}
in set notation or (−∞,4)∪{5} in interval notation. The x-intercept is 2, and the y-intercept
is 3, so the intercepts are (2,0) and (0,3) .
34. (a) f(−1)=2(−1)+5=3; f(0)=−3; f(1)=−5(1)=−5; f(8)=−5(8)=−40
(b) The graph of f consists of three pieces corresponding to each equation in the definition. On the
interval [−3,0), the graph is the line y =2x+5, and on the interval (0,∞), the graph is the line
y =−5x. The graph also contains the point (0,−3).
5
-3 -2 -1 1 2 3
-5
-10
-15
(c) The individual components of f have domains of {x|−3 ≤ x < 0}, {x|x = 0}, and {x|x > 0}.
Thedomainoff istheunionofthesethreesets;thatis, {x|x≥−3} insetnotationor [−3,∞)
in interval notation. Moreover, the individual components of f have ranges of {y|−1 ≤ y < 5},
{y|y =−3}, and {y|y <0}. The range of f is the union of these three sets; that is, {y|y <5} in
5
set notation or (−∞,5) in interval notation. The x-intercept is − , and the y-intercept is −3,
2
(cid:18) (cid:19)
5
so the intercepts are − ,0 and (0,−3) .
2
35. (a) f(−1)=1+(−1)=0; f(0)=02 =0; f(1)=12 =1; f(8)=82 =64
(b) The graph of f consists of two pieces corresponding to each equation in the definition. On the
interval (−∞,0), the graph is the line y = 1+x, and on the interval [0,∞), the graph is the
parabola y =x2.
6
8
6
4
2
-5 -4 -3 -2 -1 1 2 3
-2
-4
(c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is
the union of these two sets; that is, all real numbers or the interval (−∞,∞) . Moreover, the
individual components of f have ranges of {y|y <1} and {y|y ≥0}. The range of f is the union
of these two sets; that is, all real numbers or the interval (−∞,∞) . The x-intercepts are −1
and 0, and the y-intercept is 0, so the intercepts are (−1,0) and (0,0) .
1 √ √ √
36. (a) f(−1)= =−1; f(0)= 30=0; f(1)= 31=1; f(8)= 38=2
−1
(b) The graph of f consists of two pieces corresponding to each equation in the definition. On the
√
interval (−∞,0), the graph is y = 1, and on the interval [0,∞), the graph is y = 3x.
x
4
2
-4 -2 2 4 6 8 10
-2
-4
-6
-8
(c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is
the union of these two sets; that is, all real numbers or the interval (−∞,∞) . Moreover, the
individual components of f have ranges of {y|y <0} and {y|y ≥0}. The range of f is the union
of these two sets; that is, all real numbers or the interval (−∞,∞) . The x-intercept is 0, and
the y-intercept is 0, so the only intercept is (0,0) .
37. Because the graph of f includes the point (0,3), f(0)=3 ; because the graph also includes the point
(−6,−3), f(−6)=−3 .
38. Because the graph of f lies above the x-axis at x=3, f(3) is positive .
7
39. Because the graph of f lies below the x-axis at x=−4, f(−4) is negative .
40. Because the graph of f includes the points (−3,0), (6,0), and (10,0), f(x) = 0 for
x=−3, x=6, and x=10 .
41. Because the graph of f lies above the x-axis for −3 < x < 6 and for 10 < x ≤ 11, f(x) > 0 for
−3<x<6 and for 10<x≤11 . In interval notation, this can be written as (−3,6)∪(10,11] .
42. The points on the graph of f have x-coordinates between −6 and 11 inclusive. The domain of f is
therefore the set of real numbers {x|−6≤x≤11} or the closed interval [−6,11] .
43. Thepointsonthegraphoff havey-coordinatesbetween−3and3inclusive. Therangeoff istherefore
the set of real numbers {y|−3≤y ≤3} or the closed interval [−3,3] .
44. Becausethegraphoff includesthepoints(−3,0),(6,0),and(10,0),thex-interceptsare −3, 6, and 10 .
45. Because the graph of f includes the point (0,3), the y-intercept is 3 .
1
46. The graph of the horizontal line y = will intersect the graph of f three times .
2
47. The graph of the vertical line x=5 will intersect the graph of f once .
48. Because the graph of f includes the point (x,3) for 0≤x≤4, f(x)=3 for 0≤x≤4 .
49. Because the graph of f includes the points (−5,−2) and (8,−2), f(x) = −2 for x=−5 and for
x=8 .
50. The function f is increasing on the intervals (−6,0) and (8,11) .
51. The function f is decreasing on the interval (4,8) .
52. The function f is constant on the interval (0,4) .
53. Thefunctionf isnonincreasing(thatis, thefunctionisconstantordecreasing)ontheinterval (0,8) .
54. The function f is nondecreasing (that is, the function is constant or increasing) on the intervals
(−6,4) and (8,11) .
55. Because division by zero is not defined, x−6 cannot be equal to 0. The domain of g is therefore the
set {x|x(cid:54)=6} .
3+2 5
56. Note that g(3)= =− . Because g(3)(cid:54)=14, the point (3,14) is not on the graph of g.
3−6 3
4+2
57. If x=4, then g(x)=g(4)= =−3 ; therefore, the point (4,−3) is on the graph of g.
4−6
x+2
58. If g(x)=2, then =2. Multiplying both sides of this equation by x−6 yields
x−6
x+2=2(x−6)=2x−12,
so that x=14 . Accordingly, the point (14,2) is on the graph of g.
8
59. The x-intercepts occur when g(x) = 0. The value of g can only be zero when its numerator is zero;
therefore, g(x) = 0 only when x+2 = 0. Thus, the graph of g has only one x-intercept, and this
intercept is −2 .
0+2 1 1
60. Because g(0)= =− , the y-intercept is − .
0−6 3 3
61. The domain of h is {x|x (cid:54)= ±1}, so for every number x in the domain of h, the number −x is also in
the domain. Moreover,
−x x
h(−x)= =− =−h(x),
(−x)2−1 x2−1
so that the function h is odd . Because h is an odd function, its graph is symmetric with respect to
the origin .
62. The domain of f is all real numbers, so for every number x in the domain of f, the number −x is also
in the domain. Moreover,
(cid:112) (cid:112)
f(−x)= 3 3(−x)2+1= 3 3x2+1=f(x),
so that the function f is even . Because f is an even function, its graph is symmetric with respect to
the y-axis .
63. The domain of G is {x|x≥0}. The number x=1 is in the domain of G, but x=−1 is not; therefore,
the function G is neither even nor odd . Because G is neither an even nor an odd function, its graph
is not symmetric with respect to either the y-axis or the origin .
64. The domain of F is {x|x (cid:54)= 0}, so for every number x in the domain of F, the number −x is also in
the domain. Moreover,
2(−x) 2x
F(−x)= =− =−F(x),
|−x| |x|
so that the function F is odd . Because F is an odd function, its graph is symmetric with respect to
the origin .
65. Let f(x)=−2x2+4.
(a) The average rate of change of f from 1 to 2 is
f(2)−f(1) −4−2 −6
= = = −6 .
2−1 2−1 1
(b) The average rate of change of f from 1 to 3 is
f(3)−f(1) −14−2 −16
= = = −8 .
3−1 3−1 2
(c) The average rate of change of f from 1 to 4 is
f(4)−f(1) −28−2 −30
= = = −10 .
4−1 4−1 3
(d) The average rate of change of f from 1 to x for x(cid:54)=1 is
f(x)−f(1) −2x2+4−2 2(1−x2) 2(1−x)(1+x)
= = = = −2(1+x) .
x−1 x−1 x−1 x−1
9
66. Let s(t)=20−0.8t2.
(a) The average rate of change of s from 1 to 4 is
s(4)−s(1) 7.2−19.2 −12
= = = −4 .
4−1 4−1 3
(b) The average rate of change of s from 1 to 3 is
s(3)−s(1) 12.8−19.2 −6.4
= = = −3.2 .
3−1 3−1 2
(c) The average rate of change of s from 1 to 2 is
s(2)−f(1) 16.8−19.2 −2.4
= = = −2.4 .
2−1 2−1 1
(d) The average rate of change of s from 1 to t for t(cid:54)=1 is
s(t)−s(1) 20−0.8t2−19.2 0.8(1−t2) 0.8(1−t)(1+t)
= = = = −0.8(1+t) .
t−1 t−1 t−1 t−1
67. For −1 ≤ x < 0, the graph is a line through the points (−1,1) and (0,0). The slope of this line
is m = −1, and the y-intercept is 0; therefore, the equation for this component of the function is
y =−x. For0≤x≤2, thegraphisalinethroughthepoints(0,0)and(2,1). Theslopeofthislineis
m= 1, and the y-intercept is 0; therefore, the equation for this component of the function is y = 1x.
2 2
Combining these two equations, the definition for this piecewise function is
(cid:26)
−x, if −1≤x<0
f(x)=
1x, if 0≤x≤2.
2
The domain of f is the set {x|−1≤x≤2} or the closed interval [−1,2] , and the range is the set
{y|0≤y ≤1} or the closed interval [0,1] .
68. For x ≤ 0, the graph is a line with slope 1 through the origin, and for 0 < x ≤ 2, the graph is a
horizontal line at height y =1. The definition for this piecewise function is therefore
(cid:26)
x, if x≤0
f(x)=
1, if 0<x≤2.
The domain of f is {x|x≤2} in set notation or (−∞,2] in interval notation, and the range is
{y|y =1 or y ≤0} in set notation or (−∞,0]∪{1} in interval notation.
69. For x<1, the graph is a horizontal line at height y =−1; at x=1, the value of the function is 0; and
for 1 < x ≤ 2, the graph is a line with slope −1 and x-intercept 2. The definition for this piecewise
function is therefore
−1, if x<1
f(x)= 0, if x=1
2−x, if 1<x≤2.
The domain of f is {x|x≤2} in set notation or (−∞,2] in interval notation, and the range is
{y|y =−1 or 0≤y <1} in set notation or {−1}∪[0,1) in interval notation.
10
70. For x < −1, the graph is a horizontal line at height y = 2; at x = −1, the value of the function is
0; and for −1 < x ≤ 1, the graph is a line with slope −2 and y-intercept −1. The definition for this
piecewise function is therefore
2, if x<−1
f(x)= 0, if x=−1
−2x−1, if −1<x≤1.
The domain of f is {x|x≤1} in set notation or (−∞,1] in interval notation, and the range is
{y|y =2 or −3≤y <1} in set notation of [−3,1)∪{2} in interval notation.
71. The definition for this piecewise function is
−x3, if −2<x<1
f(x)= 0, if x=1
x2, if 1<x≤3.
The domain of f is the set {x|−2<x≤3} or the interval (−2,3] , and the range is the set
{y|−1<y ≤9} or the interval (−1,9] .
72. The definition for this piecewise function is
(cid:26) x2−1, if −3≤x≤0
f(x)=
x3, if 0<x≤2.
The domain of f is the set {x|−3≤x≤2} or the closed interval [−3,2] , and the range is the set
{y|−1≤y ≤8} or the closed interval [−1,8] .
73. (a) The cost of manufacturing 100 road bikes is
C(100)=0.004(100)3−0.6(100)2+250(100)+100,500=123,500;
the cost of manufacturing 101 road bikes is
C(101)=0.004(101)3−0.6(101)2+250(101)+100,500≈123,750.60.
The average rate of change of C from 100 to 101 road bikes is therefore
C(101)−C(100) 123,750.60−123,500
≈ = $250.60 per bike .
101−100 1
(b) The cost of manufacturing 500 road bikes is
C(500)=0.004(500)3−0.6(500)2+250(500)+100,500=575,500;
the cost of manufacturing 501 road bikes is
C(501)=0.004(501)3−0.6(501)2+250(501)+100,500≈578,155.40.
The average rate of change of C from 500 to 501 road bikes is therefore
C(501)−C(500) 578,155.40−575,500
≈ = $2655.40 per bike .
501−500 1
(c) Answers will vary. One possible interpretation is that the unit cost per road bike increases as the
number of road bikes manufactured increases.
