2.1 Functions 121
For Thought
1. False, since {(1,2),(1,3)}is not a function.
2. False, since f(5) is not defined. 3. True
4. False, since a student’s exam grade is a function
of the student’s preparation. If two classmates
had the same IQ and only one prepared, then
the one who prepared will most likely achieve
a higher grade.
5. False, since (x+h)2=x2+ 2xh +h2
6. False, since the domain is all real numbers.
7. True 8. True 9. True
10. False, since 3
8,8and 3
8,5are two ordered
pairs with the same first coordinate and
different second coordinates.
2.1 Exercises
1. function
2. function
3. relation
4. function
5. independent, dependent
6. domain, range
7. difference quotient
8. average rate of change
9. Note, b= 2πa is equivalent to a=b
2π.
Then ais a function of b, and bis a function
of a.
10. Note, b= 2(5 + a) is equivalent to a=b−10
2.
So ais a function of b, and bis a function of
a.
11. ais a function of bsince a given denomination
has a unique length. Since a dollar bill and a
five-dollar bill have the same length, then bis
not a function of a.
12. Since different U.S. coins have different diame-
ters, then ais a function of band bis a function
of a.
13. Since an item has only one price, bis a function
of a. Since two items may have the same price,
ais not a function of b.
14. ais not a function of bsince there may be
two students with the same semester grades
but different final exams scores. bis not a
function of asince there may be identical final
exam scores with different semester grades.
15. ais not a function of bsince it is possible that
two different students can obtain the same fi-
nal exam score but the times spent on studying
are different.
bis not a function of asince it is possible that
two different students can spend the same time
studying but obtain different final exam scores.
16. ais not a function of bsince it is possible that
two adult males can have the same shoe size
but have different ages.
bis not a function of asince it is possible for
two adults with the same age to have different
shoe sizes.
17. Since 1 in ≈2.54 cm, ais a function of band
bis a function of a.
18. Since there is only one cost for mailing a first
class letter, then ais a function of b. Since
two letters with different weights each under
1/2-ounce cost 34 cents to mail first class, bis
not a function of a.
19. No 20. No 21. Yes
22. Yes 23. Yes 24. No
25. Yes 26. Yes
27. Not a function since 25 has two different sec-
ond coordinates. 28. Yes
29. Not a function since 3 has two different second
coordinates.
30. Yes 31. Yes 32. Yes
Copyright 2015 Pearson Education, Inc.
122 Chapter 2 Functions and Graphs
33. Since the ordered pairs in the graph of
y= 3x−8 are (x, 3x−8), there are no two
ordered pairs with the same first coordinate
and different second coordinates. We have a
function.
34. Since the ordered pairs in the graph of
y=x2−3x+ 7 are (x, x2−3x+ 7), there are
no two ordered pairs with the same first co-
ordinate and different second coordinates. We
have a function.
35. Since y= (x+ 9)/3, the ordered pairs are
(x, (x+ 9)/3). Thus, there are no two ordered
pairs with the same first coordinate and differ-
ent second coordinates. We have a function.
36. Since y=3
√x, the ordered pairs are (x, 3
√x).
Thus, there are no two ordered pairs with the
same first coordinate and different second co-
ordinates. We have a function.
37. Since y=±x, the ordered pairs are (x, ±x).
Thus, there are two ordered pairs with the
same first coordinate and different second
coordinates. We do not have a function.
38. Since y=±√9 + x2, the ordered pairs are
(x, ±√9 + x2). Thus, there are two ordered
pairs with the same first coordinate and
different second coordinates. We do not
have a function.
39. Since y=x2, the ordered pairs are (x, x2).
Thus, there are no two ordered pairs with the
same first coordinate and different second
coordinates. We have a function.
40. Since y=x3, the ordered pairs are (x, x3).
Thus, there are no two ordered pairs with the
same first coordinate and different second
coordinates. We have a function.
41. Since y=|x| − 2, the ordered pairs are
(x, |x| − 2). Thus, there are no two ordered
pairs with the same first coordinate and differ-
ent second coordinates. We have a function.
42. Since y= 1 + x2, the ordered pairs are
(x, 1 + x2). Thus, there are no two ordered
pairs with the same first coordinate and differ-
ent second coordinates. We have a function.
43. Since (2,1) and (2,−1) are two ordered pairs
with the same first coordinate and different
second coordinates, the equation does not
define a function.
44. Since (2,1) and (2,−1) are two ordered pairs
with the same first coordinate and different
second coordinates, the equation does not
define a function.
45. Domain {−3,4,5}, range {1,2,6}
46. Domain {1,2,3,4}, range {2,4,8,16}
47. Domain (−∞,∞), range {4}
48. Domain {5}, range (−∞,∞)
49. Domain (−∞,∞);
since |x| ≥ 0, the range of y=|x|+ 5 is [5,∞).
50. Domain (−∞,∞);
since x2≥0, the range of y=x2+ 8 is [8,∞).
51. Since x=|y| − 3≥ −3, the domain
of x=|y| − 3 is [−3,∞); range (−∞,∞)
52. Since √y−2≥ −2, the domain of x=√y−2
is [−2,∞); Since √yis a real number whenever
y≥0, the range is [0,∞).
53. Since √x−4 is a real number whenever x≥4,
the domain of y=√x−4 is [4,∞).
Since y=√x−4≥0 for x≥4, the range is
[0,∞).
54. Since √5−xis a real number whenever x≤5,
the domain of y=√5−xis (−∞,5].
Since y=√5−x≥0 for x≤5, the range is
[0,∞).
55. Since x=−y2≤0, the domain of x=−y2is
(−∞,0]; range is (−∞,∞).
56. Since x=−|y| ≤ 0, the domain of x=−|y|
is (−∞,0]; range is (−∞,∞).
57. 658. 5
59. g(2) = 3(2) + 5 = 11
60. g(4) = 3(4) + 5 = 17
Copyright 2015 Pearson Education, Inc.
2.1 Functions 123
61. Since (3,8) is the ordered pair, one obtains
f(3) = 8. The answer is x= 3.
62. Since (2,6) is the ordered pair, one obtains
f(2) = 6. The answer is x= 2.
63. Solving 3x+ 5 = 26, we find x= 7.
64. Solving 3x+ 5 = −4, we find x=−3.
65. f(4) + g(4) = 5 + 17 = 22
66. f(3) −g(3) = 8 −14 = −6
67. 3a2−a68. 3w2−w
69. 4(a+2)−2 = 4a+6 70. 4(a−5)−2=4a−22
71. 3(x2+ 2x+ 1) −(x+ 1) = 3x2+ 5x+ 2
72. 3(x2−6x+ 9) −(x−3) = 3x2−19x+ 30
73. 4(x+h)−2=4x+ 4h−2
74. 3(x2+2xh+h2)−x−h= 3x2+6xh+3h2−x−h
75. 3(x2+ 2x+ 1) −(x+ 1) −3x2+x= 6x+ 2
76. 4(x+ 2) −2−4x+ 2 = 8
77. 3(x2+ 2xh +h2)−(x+h)−3x2+x=
6xh + 3h2−h
78. (4x+ 4h−2) −4x+ 2 = 4h
79. The average rate of change is
8,000 −20,000
5=−$2400 per year.
80. The average rate of change as the number of
cubic yards changes from 12 to 30 and from 30
to 60 are
528 −240
30 −12 = $16 per yd3and
948 −528
60 −30 = $14 per yd3, respectively.
81. The average rate of change on [0,2] is
h(2) −h(0)
2−0=0−64
2−0=−32 ft/sec.
The average rate of change on [1,2] is
h(2) −h(1)
2−1=0−48
2−1=−48 ft/sec.
The average rate of change on [1.9,2] is
h(2) −h(1.9)
2−1.9=0−6.24
0.1=−62.4 ft/sec.
The average rate of change on [1.99,2] is
h(2) −h(1.99)
2−1.99 =0−0.6384
0.01 =−63.84 ft/sec.
The average rate of change on [1.999,2] is
h(2) −h(1.999)
2−1.999 =0−0.063984
0.001 =−63.984
ft/sec.
82. 6−70
2−0=−64
2=−32 ft/sec
83. The average rate of change is 673 −1970
24 ≈
−54.0 million hectares per year.
84. If 54.0 million hectares are lost each year and
since 1970
54.0≈36.48 years, the forest will be
eliminated in year 2025 (≈1988 + 36.48).
85.
f(x+h)−f(x)
h=4(x+h)−4x
h
=4h
h
= 4
86.
f(x+h)−f(x)
h=
1
2(x+h)−1
2x
h
=
1
2h
h
=1
2
87.
f(x+h)−f(x)
h=3(x+h)+5−3x−5
h
=3h
h
= 3
Copyright 2015 Pearson Education, Inc.
124 Chapter 2 Functions and Graphs
88.
f(x+h)−f(x)
h=−2(x+h)+3+2x−3
h
=−2h
h
=−2
89. Let g(x) = x2+x. Then we obtain
g(x+h)−g(x)
h=
(x+h)2+ (x+h)−x2−x
h=
2xh +h2+h
h=
2x+h+ 1.
90. Let g(x) = x2−2x. Then we get
g(x+h)−g(x)
h=
(x+h)2−2(x+h)−x2+ 2x
h=
2xh +h2−2h
h=
2x+h−2.
91. Difference quotient is
=−(x+h)2+ (x+h)−2 + x2−x+ 2
h
=−2xh −h2+h
h
=−2x−h+ 1
92. Difference quotient is
=(x+h)2−(x+h)+3−x2+x−3
h
=2xh +h2−h
h
= 2x+h−1
93. Difference quotient is
=3√x+h−3√x
h·3√x+h+ 3√x
3√x+h+ 3√x
=9(x+h)−9x
h(3√x+h+ 3√x)
=9h
h(3√x+h+ 3√x)
=3
√x+h+√x
94. Difference quotient is
=−2√x+h+ 2√x
h·−2√x+h−2√x
−2√x+h−2√x
=4(x+h)−4x
h(−2√x+h−2√x)
=4h
h(−2√x+h−2√x)
=−2
√x+h+√x
95. Difference quotient is
=√x+h+ 2 −√x+ 2
h·√x+h+2+√x+ 2
√x+h+2+√x+ 2
=(x+h+ 2) −(x+ 2)
h(√x+h+2+√x+ 2)
=h
h(√x+h+2+√x+ 2)
=1
√x+h+2+√x+ 2
96. Difference quotient is
=rx+h
2−rx
2
h·rx+h
2+rx
2
rx+h
2+rx
2
=
x+h
2−x
2
h rx+h
2+rx
2!
=
h
2
h rx+h
2+rx
2!
=1
2 rx+h
2+rx
2!
=1
√2√x+h+√x
Copyright 2015 Pearson Education, Inc.
2.1 Functions 125
97. Difference quotient is
=
1
x+h−1
x
h·x(x+h)
x(x+h)
=x−(x+h)
xh(x+h)
=−h
xh(x+h)
=−1
x(x+h)
98. Difference quotient is
=
3
x+h−3
x
h·x(x+h)
x(x+h)
=3x−3(x+h)
xh(x+h)
=−3h
xh(x+h)
=−3
x(x+h)
99. Difference quotient is
=
3
x+h+ 2 −3
x+ 2
h·(x+h+ 2)(x+ 2)
(x+h+ 2)(x+ 2)
=3(x+ 2) −3(x+h+ 2)
h(x+h+ 2)(x+ 2)
=−3h
h(x+h+ 2)(x+ 2)
=−3
(x+h+ 2)(x+ 2)
100. Difference quotient is
=
2
x+h−1−2
x−1
h·(x+h−1)(x−1)
(x+h−1)(x−1)
=2(x−1) −2(x+h−1)
h(x+h−1)(x−1)
=−2h
h(x+h−1)(x−1)
=−2
(x+h−1)(x−1)
101. a) A=s2b) s=√Ac) s=d√2
2
d) d=s√2e) P= 4sf) s=P/4
g) A=P2/16 h) d=√2A
102. a) A=πr2b) r=rA
πc) C= 2πr
d) d= 2re) d=C
πf) A=πd2
4
g) d= 2rA
π
103. C= 500 + 100n
104. a) When d= 100 ft, the atmospheric pres-
sure is A(100) = .03(100) + 1 = 4 atm.
b) When A= 4.9 atm, the depth is found by
solving 4.9=0.03d+ 1; the depth is
d=3.9
0.03 = 130 ft.
105.
a) The quantity C(4) = (0.95)(4)+5.8 = $9.6
billion represents the amount spent on
computers in year 2004.
b) By solving 0.95n+ 5.8 = 20, we obtain
n=14.2
0.95 ≈14.9.
Thus, spending for computers will be $20
billion in year 2015.
106.
a) The quantity E(4) + C(4) = [0.5(4) + 1] +
9.6 = $12.6 billion represents the total
amount spent on electronics and comput-
ers in year 2004.
b) By solving
(0.5n+ 1) + (0.95n+ 5.8) = 30
1.45n= 23.2
n= 16
we find that the total spending will reach
$30 billion in year 2016 (= 2000 + 16).
c) The amount spent on computers is growing
faster since the slope of C(n) [which is 1]
is greater than the slope of E(n) [which
is 0.95].
Copyright 2015 Pearson Education, Inc.
126 Chapter 2 Functions and Graphs
107. Let abe the radius of each circle. Note, trian-
gle 4ABC is an equilateral triangle with side
2aand height √3a.
A B
C
Thus, the height of the circle centered at C
from the horizontal line is √3a+ 2a. Hence,
by using a similar reasoning, we obtain that
height of the highest circle from the line is
2√3a+ 2a
or equivalently (2√3 + 2)a.
108. In the triangle below, P S bisects the 90-angle
at Pand SQ bisects the 60-angle at Q.
d
P
a
R Q
S
300
900
450
In the 45-45-90 triangle 4SP R, we find
P R =SR =√2d/2.
And, in the 30-60-90 triangle 4SQR we get
RQ =√6
2d.
Since P Q =P R +RQ, we obtain
a
2=√2
2d+√6
2d
a=√2d+√6d
a= (√6 + √2)d
d=a
√6 + √2
d=√6−√2
4a.
109. When x= 18 and h= 0.1, we have
R(18.1) −R(18)
0.1= 1950.
The revenue from the concert will increase by
approximately $1,950 if the price of a ticket is
raised from $18 to $19.
If x= 22 and h= 0.1, then
R(22.1) −R(22)
0.1=−2050.
The revenue from the concert will decrease by
approximately $2050 if the price of a ticket is
raised from $22 to $23.
110. When r= 1.4 and h= 0.1, we obtain
A(1.5) −A(1.4)
0.1≈ −16.1
The needed amount of tin decreases by approx-
imately 16.1 in.2if the radius increases from
1.4 in. to 2.4 in.
If r= 2 and h= 0.1, then
A(2.1) −A(2)
0.1≈8.6
The amount of tin needed increases by about
8.6 in.2if the radius increases from 2 in. to 3
in.
113.
3
2x−5
9x=1
3−5
6
17
18x=−1
2
x=−1
2·18
17
x=−9
17
114. If mis the number of males, then
m+1
2m= 36
3
2m= 36
m= (36)2
3
x= 24 males
Copyright 2015 Pearson Education, Inc.
2.1 Functions 127
115. p(−4 + 6)2+ (−3−3)2=√4 + 36 =
√40 = 2√10
116. The slope is 3−2
5+1 =1
6. The line is given
by y=1
6x+bfor some b. Substitute the
coordinates of (−1,2) as follows:
2 = 1
6(−1) + b
13
6=b
The line is given by
y=1
6x+13
6.
117.
x2−x−6 = 36
x2−x−42 = 0
(x−7)(x+ 6) = 0
The solution set is {−6,7}.
118. The inequality is equivalent to
−13 <2x−9<13
−4<2x < 22
−2<x<11
The solution set is (−2,11).
119. (30 + 25)2= 3025
120. Let dbe the length of the pool. Let xbe the
rate of the swimmer who after 75 feet passes
the other swimmer. Let ybe the rate of the
other swimmer. If xis the length of the pool,
then 75
x=d−75
y
and d+ 25
x=2d−25
y.
Since we may solve for the ratio y/x from both
equations, we find
y
x=d−75
75 =2d−25
d+ 25 .
Solving for d, we obtain d= 200 or d= 0.
Thus, the length of the pool is 200 feet.
2.1 Pop Quiz
1. Yes, since A=πr2where Ais the area of a
circle with radius r.
2. No, since the ordered pairs (2,4) and (2,−4)
have the same first coordinates.
3. No, since the ordered pairs (0,1) and (0,−1)
have the same first coordinates.
4. [1,∞)5. [2,∞)6. 9
7. If 2a= 1, then a= 1/2.
8. 40 −20
2008 −1998 = $2 per year
9. The difference quotient is
f(x+h)−f(x)
h=(x+h)2+ 3 −x2−3
h
=x2+ 2xh +h2−x2
h
=2xh +h2
h
= 2x+h
2.1 Linking Concepts
(a) The first graph shows U.S. federal debt
in trillions of dollars versus year y
1940
1970
2000
year
15t
5t
900b
debt
and the second graph shows population P
(in hundreds of millions) versus year y.
1940
1970
2000
year
100m
200m
300m
population
Copyright 2015 Pearson Education, Inc.
128 Chapter 2 Functions and Graphs
(b) The first table shows the average rates of
change for the U.S. federal debt
10 −year period ave. rate of change
1940 −50 257−51
10 = 20.6
1950 −60 291−257
10 = 3.4
1960 −70 381−291
10 = 9.0
1970 −80 909−381
10 = 52.8
1980 −90 3207−909
10 = 229.8
1990 −2000 5666−3207
10 = 245.9
2000 −2010 13,500−5666
10 = 783.4
The second table shows the average rates of
change for the U.S. population
10 −year period ave. rate of change
1940 −50 150.7−131.7
10 ≈1.9
1950 −60 179.3−150.7
10 ≈2.9
1960 −70 203.3−179.3
10 ≈2.4
1970 −80 226.5−203.3
10 ≈2.3
1980 −90 248.7−226.5
10 ≈2.2
1990 −2000 274.8−248.7
10 ≈2.6
2000 −2010 308.7−274.8
10 ≈3.4
(c) The first table shows the difference between
consecutive average rates of change for the
U.S. federal debt.
10-year periods difference
1940-50 & 1950-60 3.4−20.6 = −17.2
1950-60 & 1960-70 9.0−3.4=5.6
1960-70 & 1970-80 52.8−9.0 = 43.8
1970-80 & 1980-90 229.8−52.8 = 177.0
1980-90 & 1990-00 245.9−229.8 = 16.1
1990-00 & 2010-00 783.4−245.9 = 537.5
The second table shows the difference between
consecutive average rates of change for the
U.S. population.
10-year periods difference
1940-50 & 1950-60 2.9−1.9=1.0
1950-60 & 1960-70 2.4−2.9 = −0.5
1960-70 & 1970-80 2.3−2.4 = −0.1
1970-80 & 1980-90 2.2−2.3 = −0.1
1980-90 & 1990-00 2.6−2.2=0.4
1990-00 & 2010-00 3.4−2.6=0.8
(d) For both the U.S. federal debt and population,
the average rates of change are all positive.
(e) In part (c), for the federal debt most of the
differences are positive and for the population
most of the differences are negative.
(f) The U.S. federal debt is growing out of control
when compared to the U.S. population. See
part (g) for an explanation.
(g) Since most of the differences for the federal
debt in part (e) are positive, the federal debts
are increasing at an increasing rate. While the
U.S. population is increasing at a decreasing
rate since most of the differences for popula-
tion in part (e) are negative.
For Thought
1. True, since the graph is a parabola opening
down with vertex at the origin.
2. False, the graph is decreasing.
3. True
4. True, since f(−4.5) = [−1.5] = −2.
5. False, since the range is {±1}.
6. True 7. True 8. True
9. False, since the range is the interval [0,4].
10. True
2.2 Exercises
1. square root
2. semicircle
3. increasing
4. constant
5. parabola
6. piecewise
7. Function y= 2xincludes the points (0,0),(1,2),
domain and range are both (−∞,∞)
1 2 x
2
4
y
Copyright 2015 Pearson Education, Inc.
2.2 Graphs of Relations and Functions 129
8. Function x= 2yincludes the points (0,0),
(2,1),(−2,−1), domain and range are both
(−∞,∞)
2-2 x
1
-1
y
9. Function x−y= 0 includes the points (−1,−1),
(0,0),(1,1),domain and range are both
(−∞,∞)
1-2 x
1
-2
y
10. Function x−y= 2 includes the points (2,0),
(0,−2),(−2,−4), domain and range are both
(−∞,∞)
2-2 x
-2
-4
y
11. Function y= 5 includes the points (0,5),
(±2,5), domain is (−∞,∞), range is {5}
5-5 x
4
6
y
12. x= 3 is not a function and includes the points
(3,0),(3,2), domain is {3}, range is (−∞,∞)
2 4 x
3
-3
y
13. Function y= 2x2includes the points (0,0),
(±1,2), domain is (−∞,∞), range is [0,∞)
1-1 x
2
8
y
14. Function y=x2−1 goes through (0,−1),
(±1,0), domain is (−∞,∞), range is [−1,∞)
1-1 x
1
4
y
15. Function y= 1 −x2includes the points (0,1),
(±1,0), domain is (−∞,∞), range is (−∞,1]
1-1 x
2
-4
y
16. Function y=−1−x2includes the points
(0,−1), (±1,−2), domain is (−∞,∞), range
is (−∞,−1]
1–1x
–4
y
17. Function y= 1+ √xincludes the points (0,1),
(1,2),(4,3), domain is [0,∞), range is [1,∞)
41 x
1
4
y
Copyright 2015 Pearson Education, Inc.
130 Chapter 2 Functions and Graphs
18. Function y= 2−√xincludes the points (0,2),
(4,0), domain is [0,∞), range is (−∞,2]
1 4 x
2
–4
y
19. x=y2+ 1 is not a function and includes the
points (1,0),(2,±1), domain is [1,∞), range
is (−∞,∞)
21 x
1
-1
y
20. x= 1 −y2is not function and includes the
points (1,0),(0,±1), domain is (−∞,1], range
is (−∞,∞)
-1 1 x
2
-2
y
21. Function x=√ygoes through
(0,0),(2,4),(3,9), domain and range is [0,∞)
23x
4
9
y
22. Function x−1 = √ygoes through (1,0),
(3,4),(4,9), domain [1,∞), and range [0,∞),
1 4 x
2
–2
y
23. Function y=3
√x+ 1 goes through
(−1,0),(1,2),(8,3), domain (−∞,∞),
and range (−∞,∞)
–4 4 x
2
–2
y
24. Function y=3
√x−2 goes through
(−1,−3),(1,−1),(8,0), domain (−∞,∞),
and range (−∞,∞)
–4x
1
–2
y
25. Function, x=3
√ygoes through
(0,0),(1,1),(2,8), domain (−∞,∞),
and range (−∞,∞)
–8 8 x
2
y
Copyright 2015 Pearson Education, Inc.
2.2 Graphs of Relations and Functions 131
26. Function, x=3
√y−1 goes through
(0,1),(1,2),(−1,0), domain (−∞,∞),
and range (−∞,∞)
–3 3 x
2
y
27. Not a function, y2= 1 −x2goes through
(1,0),(0,1),(−1,0), domain [−1,1],
and range [−1,1]
–2 2 x
2
–2
y
28. Not a function, x2+y2= 4 goes through
(2,0),(0,2),(−2,0), domain [−2,2],
and range [−2,2]
1 3 x
1
3
y
29. Function, y=√1−x2goes through
(±1,0),(0,1), domain [−1,1],
and range [0,1]
1
–1
x
y
30. Function, y=−√25 −x2goes through
(±5,0),(0,−5), domain [−5,5],
and range [−5,0]
4 6 x
–4
–6
y
31. Function y=x3includes the points (0,0),
(1,1),(2,8), domain and range are both
(−∞,∞)
1 2 x
1
8
y
32. Function y=−x3includes the points (0,0),
(1,−1),(2,−8), domain and range are both
(−∞,∞)
2–2x
–8
8
y
Copyright 2015 Pearson Education, Inc.
132 Chapter 2 Functions and Graphs
33. Function y= 2|x|includes the points (0,0),
(±1,2),domain is (−∞,∞), range is [0,∞)
1-1 x
2
5
y
34. Function y=|x−1|includes the points
(0,1),(1,0),(2,1), domain is (−∞,∞), range
is [0,∞)
13
-2 x
1
2
6
y
37. Not a function, graph of x=|y|includes the
points (0,0),(2,2),(2,−2), domain is [0,∞),
range is (−∞,∞)
21 x
-2
2
y
38. x=|y|+ 1 is not a function and includes the
points (1,0),(2,±1), domain is [1,∞), range
is (−∞,∞)
1 2 x
2
y
