Calculo purcell 9 ed solucionario

CHAPTER

0

0.1 Concepts Review
1. rational numbers

Preliminaries
1 ⎡ 2 1 ⎛ 1 1 ⎞⎤
1
8. − ⎢ − ⎜ − ⎟ ⎥ = −
3 ⎣ 5 2 ⎝ 3 5 ⎠⎦
⎡ 2 1 ⎛ 5 3 ⎞⎤
3 ⎢ − ⎜ − ⎟⎥
⎣ 5 2 ⎝ 15 15 ⎠ ⎦

2. dense

1 ⎡ 2 1 ⎛ 2 ⎞⎤
1 ⎡2 1 ⎤
= − ⎢ − ⎜ ⎟⎥ = − ⎢ − ⎥
3 ⎣ 5 2 ⎝ 15 ⎠ ⎦
3 ⎣ 5 15 ⎦
1⎛ 6 1 ⎞
1⎛ 5 ⎞
1
=− ⎜ − ⎟=− ⎜ ⎟=−
3 ⎝ 15 15 ⎠
3 ⎝ 15 ⎠
9

3. If not Q then not P.
4. theorems

2

Problem Set 0.1
1. 4 − 2(8 − 11) + 6 = 4 − 2(−3) + 6
= 4 + 6 + 6 = 16
2. 3 ⎡ 2 − 4 ( 7 − 12 ) ⎤ = 3[ 2 − 4(−5) ]
⎣
⎦
= 3[ 2 + 20] = 3(22) = 66
3.

–4[5(–3 + 12 – 4) + 2(13 – 7)]
= –4[5(5) + 2(6)] = –4[25 + 12]
= –4(37) = –148

4.

5 [ −1(7 + 12 − 16) + 4] + 2
= 5 [ −1(3) + 4] + 2 = 5 ( −3 + 4 ) + 2
= 5 (1) + 2 = 5 + 2 = 7

5.

6.

7.

5 1 65 7 58
– =
– =
7 13 91 91 91
3
3 1 3
3 1
+ − =
+ −
4 − 7 21 6 −3 21 6
42 6
7
43
=− +
−
=−
42 42 42
42
1 ⎡1 ⎛ 1 1 ⎞ 1⎤ 1 ⎡1 ⎛ 3 – 4 ⎞ 1⎤
=
⎜ – ⎟+
⎜
⎟+
3 ⎢ 2 ⎝ 4 3 ⎠ 6 ⎥ 3 ⎢ 2 ⎝ 12 ⎠ 6 ⎥
⎣
⎦
⎣
⎦
1 ⎡1 ⎛ 1 ⎞ 1⎤
= ⎢ ⎜– ⎟+ ⎥
3 ⎣ 2 ⎝ 12 ⎠ 6 ⎦
1⎡ 1
4⎤
= ⎢– + ⎥
3 ⎣ 24 24 ⎦
1⎛ 3 ⎞ 1
= ⎜ ⎟=
3 ⎝ 24 ⎠ 24

Instructor’s Resource Manual

2

2
14 ⎛ 2 ⎞
14 ⎛ 2 ⎞
14 6
⎜
⎟ = ⎜ ⎟ = ⎛ ⎞
9.
⎜ ⎟
21 ⎜ 5 − 1 ⎟
21 ⎜ 14 ⎟
21 ⎝ 14 ⎠
3⎠
⎝
⎝ 3 ⎠
2

=

14 ⎛ 3 ⎞
2⎛ 9 ⎞ 6
⎜ ⎟ = ⎜ ⎟=
21 ⎝ 7 ⎠
3 ⎝ 49 ⎠ 49

⎛2
⎞ ⎛ 2 35 ⎞ ⎛ 33 ⎞
⎜ − 5⎟ ⎜ − ⎟ ⎜ − ⎟
7
⎠ = ⎝ 7 7 ⎠ = ⎝ 7 ⎠ = − 33 = − 11
10. ⎝
6
2
⎛ 1⎞ ⎛7 1⎞
⎛6⎞
⎜1 − ⎟ ⎜ − ⎟
⎜ ⎟
⎝ 7⎠ ⎝7 7⎠
⎝7⎠
7
11 – 12 11 – 4
7 21 = 7 7 = 7 = 7
11.
11 + 12 11 + 4 15 15
7 21 7 7
7
1 3 7 4 6 7 5
− +
− +
5
12. 2 4 8 = 8 8 8 = 8 =
1 3 7 4 6 7 3 3
+ −
+ −
2 4 8 8 8 8 8

13. 1 –

1
1
2 3 2 1
=1– =1– = – =
1
3
3 3 3 3
1+ 2
2

14. 2 +

15.

(

3
5
1+
2

5+ 3

3
3
= 2+
2 5
7
−
2 2
2
6 14 6 20
= 2+ = + =
7 7 7 7

= 2+

)(

) ( 5) – ( 3)

5– 3 =

2

2

=5–3= 2

Section 0.1

1

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16.

(

5− 3

) = ( 5)
2

2

−2

( 5 )( 3 ) + ( 3 )

2

27.

= 5 − 2 15 + 3 = 8 − 2 15

17. (3x − 4)( x + 1) = 3 x 2 + 3 x − 4 x − 4

= 3x2 − x − 4
18. (2 x − 3)2 = (2 x − 3)(2 x − 3)

12

4
2
+
x + 2x x x + 2
12
4( x + 2)
2x
=
+
+
x( x + 2) x( x + 2) x( x + 2)
12 + 4 x + 8 + 2 x 6 x + 20
=
=
x( x + 2)
x( x + 2)
2(3 x + 10)
=
x( x + 2)
2

+

= 4 x2 − 6 x − 6 x + 9
= 4 x 2 − 12 x + 9

19.

28.

(3x – 9)(2 x + 1) = 6 x 2 + 3 x –18 x – 9

2
y
+
2(3 y − 1) (3 y + 1)(3 y − 1)
2(3 y + 1)
2y
=
+
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)
=

2

= 6 x –15 x – 9

20. (4 x − 11)(3x − 7) = 12 x 2 − 28 x − 33 x + 77

=

21. (3t 2 − t + 1) 2 = (3t 2 − t + 1)(3t 2 − t + 1)
3

2

3

2

2

= 9t − 3t + 3t − 3t + t − t + 3t − t + 1

6y + 2 + 2y
8y + 2
=
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)

=

= 12 x 2 − 61x + 77

4

2
y
+
6 y − 2 9 y2 −1

2(4 y + 1)
4y +1
=
2(3 y + 1)(3 y − 1) (3 y + 1)(3 y − 1)

= 9t 4 − 6t 3 + 7t 2 − 2t + 1

0⋅0 = 0

b.

0
is undefined.
0

c.

0
=0
17

d.

3
is undefined.
0

e.

05 = 0

f. 170 = 1

29. a.
22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3)
= (4t 2 + 12t + 9)(2t + 3)
= 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27
= 8t 3 + 36t 2 + 54t + 27

23.

x 2 – 4 ( x – 2)( x + 2)
=
= x+2, x ≠ 2
x–2
x–2

24.

x 2 − x − 6 ( x − 3)( x + 2)
=
= x+2, x ≠3
x−3
( x − 3)

25.

t 2 – 4t – 21 (t + 3)(t – 7)
=
= t – 7 , t ≠ −3
t +3
t +3

26.

2

2x − 2x
3

2

2

x − 2x + x

=

2 x(1 − x)
2

x( x − 2 x + 1)
−2 x( x − 1)
=
x( x − 1)( x − 1)
2
=−
x −1

Section 0.1

0
= a , then 0 = 0 ⋅ a , but this is meaningless
0
because a could be any real number. No
0
single value satisfies = a .
0

30. If

31.

.083
12 1.000
96
40
36
4



32.

.285714
7 2.000000
14
60
56
40
35
50
49
10
7
30
28
2

33.

.142857
21 3.000000
21
90
84
60
42
180
168
120
105
150
147
3

34.

.294117...
17 5.000000... → 0.2941176470588235
34
160
153
70
68
20
17
30
17
130
119
11


35.

3.6
3 11.0
9
20
18
2

36.

.846153
13 11.000000
10 4
60
52
80
78
20
13
70
65
50
39
11

37. x = 0.123123123...
1000 x = 123.123123...
x = 0.123123...
999 x = 123
123 41
x=
=
999 333
38. x = 0.217171717 …
1000 x = 217.171717...
10 x = 2.171717...
990 x = 215
215 43
x=
=
990 198
39. x = 2.56565656...
100 x = 256.565656...
x = 2.565656...
99 x = 254
254
x=
99
40. x = 3.929292…
100 x = 392.929292...
x = 3.929292...
99 x = 389
389
x=
99

Section 0.1

3


41. x = 0.199999...
100 x = 19.99999...

52.

10 x = 1.99999...
90 x = 18
18 1
x=
=
90 5

54.
55.

10 x = 3.99999...
90 x = 36
36 2
x=
=
90 5

56.

43. Those rational numbers that can be expressed
by a terminating decimal followed by zeros.
⎛1⎞
p
1
= p ⎜ ⎟ , so we only need to look at . If
q
q
⎝q⎠
q = 2n ⋅ 5m , then
n

m

1 ⎛1⎞ ⎛1⎞
= ⎜ ⎟ ⋅ ⎜ ⎟ = (0.5)n (0.2)m . The product
q ⎝ 2⎠ ⎝5⎠
of any number of terminating decimals is also a
n

m

terminating decimal, so (0.5) and (0.2) ,
and hence their product,
decimal. Thus

1
, is a terminating
q

p
has a terminating decimal
q

expansion.
45. Answers will vary. Possible answer: 0.000001,
1
≈ 0.0000010819...

π 12

46. Smallest positive integer: 1; There is no
smallest positive rational or irrational number.
47. Answers will vary. Possible answer:
3.14159101001...
48. There is no real number between 0.9999…

(repeating 9's) and 1. 0.9999… and 1 represent
the same real number.
49. Irrational
50. Answers will vary. Possible answers:
−π and π , − 2 and 2
51. ( 3 + 1)3 ≈ 20.39230485

4

Section 0.1

2− 3

)

4

≈ 0.0102051443

53. 4 1.123 – 3 1.09 ≈ 0.00028307388

42. x = 0.399999…
100 x = 39.99999...

44.

(

( 3.1415 )−1/ 2 ≈ 0.5641979034
8.9π2 + 1 – 3π ≈ 0.000691744752
4 (6π 2

− 2)π ≈ 3.661591807

57. Let a and b be real numbers with a < b . Let n
be a natural number that satisfies
1 / n < b − a . Let S = {k : k n > b} . Since
a nonempty set of integers that is bounded
below contains a least element, there is a
k 0 ∈ S such that k 0 / n > b but

(k 0 − 1) / n ≤ b . Then

k0 − 1 k0 1
1
=
− >b− > a
n
n n
n
k 0 −1
k 0 −1
Thus, a < n ≤ b . If n < b , then choose
r=

k 0 −1
n

. Otherwise, choose r =

k0 − 2
n

.

1
<r.
n
Given a < b , choose r so that a < r1 < b . Then
choose r2 , r3 so that a < r2 < r1 < r3 < b , and so
on.
Note that a < b −

58. Answers will vary. Possible answer: ≈ 120 in 3
ft
= 21,120, 000 ft
mi
equator = 2π r = 2π (21,120, 000)
≈ 132, 700,874 ft

59. r = 4000 mi × 5280

60. Answers will vary. Possible answer:
beats
min
hr
day
× 60
× 24
× 365
× 20 yr
70
min
hr
day
year
= 735,840, 000 beats
2

⎛ 16
⎞
61. V = πr 2 h = π ⎜ ⋅12 ⎟ (270 ⋅12)
⎝ 2
⎠
≈ 93,807, 453.98 in.3
volume of one board foot (in inches):
1× 12 × 12 = 144 in.3
number of board feet:
93,807, 453.98
≈ 651, 441 board ft
144



b. Every circle has area less than or equal to
9π. The original statement is true.

62. V = π (8.004) 2 (270) − π (8)2 (270) ≈ 54.3 ft.3
63. a.

If I stay home from work today then it
rains. If I do not stay home from work,
then it does not rain.

b. If the candidate will be hired then she
meets all the qualifications. If the
candidate will not be hired then she does
not meet all the qualifications.
64. a.

c.

Some real number is less than or equal to
its square. The negation is true.

71. a.

True; If x is positive, then x 2 is positive.

b. False; Take x = −2 . Then x 2 > 0 but
x<0.

If I pass the course, then I got an A on the
final exam. If I did not pass the course,
thn I did not get an A on the final exam.

c.

2

e.

2

a + b = c . If a triangle is not a right
2

2

72. a.

2

triangle, then a + b ≠ c .

c.

If angle ABC is an acute angle, then its
measure is 45o. If angle ABC is not an
acute angle, then its measure is not 45o.
2

2

2

The statement, converse, and
contrapositive are all true.

True; 1/ 2n can be made arbitrarily close
to 0.

73. a.

If n is odd, then there is an integer k such
that n = 2k + 1. Then
n 2 = (2k + 1) 2 = 4k 2 + 4k + 1
= 2(2k 2 + 2k ) + 1

The statement and contrapositive are true.
The converse is false.

Some isosceles triangles are not
equilateral. The negation is true.

b. All real numbers are integers. The original
statement is true.
c.
70. a.

True; Let x be any number. Take
1
1
y = + 1 . Then y > .
x
x

e.

b.

b. The statement, converse, and
contrapositive are all false.
69. a.

True; x + ( − x ) < x + 1 + ( − x ) : 0 < 1

2

b. The statement, converse, and
contrapositive are all true.
68. a.

True; Let y be any positive number. Take
y
x = . Then 0 < x < y .
2

d. True; 1/ n can be made arbitrarily close
to 0.

b. If a < b then a < b. If a ≥ b then
a ≥ b.
67. a.

<x

b. False; There are infinitely many prime
numbers.

b. If the measure of angle ABC is greater than
0o and less than 90o, it is acute. If the
measure of angle ABC is less than 0o or
greater than 90o, then it is not acute.
66. a.

1
4

y = x 2 + 1 . Then y > x 2 .

If a triangle is a right triangle, then
2

1
2
. Then x =
2

d. True; Let x be any number. Take

b. If I take off next week, then I finished my
research paper. If I do not take off next
week, then I did not finish my research
paper.
65. a.

False; Take x =

Some natural number is larger than its
square. The original statement is true.

Prove the contrapositive. Suppose n is
even. Then there is an integer k such that
n = 2k . Then n 2 = (2k )2 = 4k 2 = 2(2k 2 ) .

Thus n 2 is even.
Parts (a) and (b) prove that n is odd if and

74.

only if n 2 is odd.
75. a.
b.

243 = 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3
124 = 4 ⋅ 31 = 2 ⋅ 2 ⋅ 31 or 22 ⋅ 31

Some natural number is not rational. The
original statement is true.


Section 0.1

5


5100 = 2 ⋅ 2550 = 2 ⋅ 2 ⋅1275

c.

82. a.

= 2 ⋅ 2 ⋅ 3 ⋅ 425 = 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 85
= 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 5 ⋅17 or 22 ⋅ 3 ⋅ 52 ⋅17

c.

76. For example, let A = b ⋅ c 2 ⋅ d 3 ; then

A2 = b 2 ⋅ c 4 ⋅ d 6 , so the square of the number
is the product of primes which occur an even
number of times.
77.

p
p2
;2 =
; 2q 2 = p 2 ; Since the prime
2
q
q
2 must occur an even number of
factors of p
p
times, 2q2 would not be valid and = 2
q
must be irrational.
3=

p
p2
; 3=
; 3q 2 = p 2 ; Since the prime
q
q2

factors of p 2 must occur an even number of
times, 3q 2 would not be valid and

e.

f.
83. a.

p
= 3
q

x = 2.4444...;
10 x = 24.4444...
x = 2.4444...
9 x = 22
22
x=
9

2
3
n = 1: x = 0, n = 2: x = , n = 3: x = – ,
3
2
5
n = 4: x =
4
3
The upper bound is .
2
2

Answers will vary. Possible answer: An
example
is S = {x : x 2 < 5, x a rational number}.
Here the least upper bound is 5, which is
real but irrational.

must be irrational.
79. Let a, b, p, and q be natural numbers, so

b. –2

d. 1

2=

78.

–2

a
b

p
a p aq + bp
are rational. + =
This
q
b q
bq
sum is the quotient of natural numbers, so it is
also rational.

and

b. True

0.2 Concepts Review
1. [−1,5); (−∞, −2]
2. b > 0; b < 0

p
80. Assume a is irrational, ≠ 0 is rational, and
q
p r
q⋅r
is
= is rational. Then a =
q s
p⋅s
rational, which is a contradiction.
a⋅

81. a.

– 9 = –3; rational

b.

3
0.375 = ; rational
8

c.

(1 + 3)2 = 1 + 2 3 + 3 = 4 + 2 3;
irrational

4. −1 ≤ x ≤ 5

Problem Set 0.2
1. a.

b.

(3 2)(5 2) = 15 4 = 30; rational

d.

3. (b) and (c)

c.

d.

6

Section 0.2



–3 < 1 – 6 x ≤ 4
9. –4 < –6 x ≤ 3

e.

2
1 ⎡ 1 2⎞
> x ≥ – ; ⎢– , ⎟
3
2 ⎣ 2 3⎠

f.

2. a.
c.

(2, 7)
(−∞, −2]

b.
d.

[−3, 4)

[−1, 3]

10.

3. x − 7 < 2 x − 5
−2 < x;( − 2, ∞)

4 < 5 − 3x < 7
−1 < −3x < 2
1
2 ⎛ 2 1⎞
> x > − ; ⎜− , ⎟
3
3 ⎝ 3 3⎠

4. 3x − 5 < 4 x − 6

1 < x; (1, ∞ )
11. x2 + 2x – 12 < 0;
x=

5.

7 x – 2 ≤ 9x + 3
–5 ≤ 2 x

= –1 ± 13

(

7. −4 < 3 x + 2 < 5
−6 < 3 x < 3
−2 < x < 1; (−2, −1)

)

(

)

⎡ x – –1 + 13 ⎤ ⎡ x – –1 – 13 ⎤ < 0;
⎣
⎦⎣
⎦

5 ⎡ 5 ⎞
x ≥ – ; ⎢– , ∞ ⎟
2 ⎣ 2 ⎠

6. 5 x − 3 > 6 x − 4
1 > x;(−∞,1)

–2 ± (2)2 – 4(1)(–12) –2 ± 52
=
2(1)
2

( –1 –

13, – 1 + 13

)

12. x 2 − 5 x − 6 > 0
( x + 1)( x − 6) > 0;
(−∞, −1) ∪ (6, ∞)

13. 2x2 + 5x – 3 > 0; (2x – 1)(x + 3) > 0;
⎛1 ⎞
(−∞, −3) ∪ ⎜ , ∞ ⎟
⎝2 ⎠

8. −3 < 4 x − 9 < 11
6 < 4 x < 20
3
⎛3 ⎞
< x < 5; ⎜ ,5 ⎟
2
⎝2 ⎠

14.

⎛ 3 ⎞
(4 x + 3)( x − 2) < 0; ⎜ − , 2 ⎟
⎝ 4 ⎠

15.


4 x2 − 5x − 6 < 0

x+4
≤ 0; [–4, 3)
x–3

Section 0.2

7


16.

3x − 2
2⎤
⎛
≥ 0; ⎜ −∞, ⎥ ∪ (1, ∞)
x −1
3⎦
⎝

3
>2
x+5

20.

3
−2 > 0
x+5

17.

2
−5 < 0
x
2 − 5x
< 0;
x
⎛2 ⎞
(– ∞, 0) ∪ ⎜ , ∞ ⎟
⎝5 ⎠

18.

3 − 2( x + 5)
>0
x+5

2
<5
x

7
≤7
4x
7
−7 ≤ 0
4x
7 − 28 x
≤ 0;
4x
1
( −∞, 0 ) ∪ ⎡ , ∞ ⎞
⎢4 ⎟
⎣
⎠

−2 x − 7
7⎞
⎛
> 0; ⎜ −5, − ⎟
2⎠
x+5
⎝

21. ( x + 2)( x − 1)( x − 3) > 0; (−2,1) ∪ (3,8)

3⎞ ⎛1 ⎞
⎛
22. (2 x + 3)(3x − 1)( x − 2) < 0; ⎜ −∞, − ⎟ ∪ ⎜ , 2 ⎟
2⎠ ⎝3 ⎠
⎝

3⎤
⎛
23. (2 x - 3)( x -1)2 ( x - 3) ≥ 0; ⎜ – ∞, ⎥ ∪ [3, ∞ )
2⎦
⎝

24. (2 x − 3)( x − 1) 2 ( x − 3) > 0;

19.

( −∞,1) ∪ ⎛1,
⎜

3⎞
⎟ ∪ ( 3, ∞ )
⎝ 2⎠

1
≤4
3x − 2
1
−4≤ 0
3x − 2
1 − 4(3 x − 2)
≤0
3x − 2
9 − 12 x
2 ⎞ ⎡3 ⎞
⎛
≤ 0; ⎜ −∞, ⎟ ∪ ⎢ , ∞ ⎟
3x − 2
3 ⎠ ⎣4 ⎠
⎝

25.

x3 – 5 x 2 – 6 x < 0
x( x 2 – 5 x – 6) < 0
x( x + 1)( x – 6) < 0;
(−∞, −1) ∪ (0, 6)

26. x3 − x 2 − x + 1 > 0
( x 2 − 1)( x − 1) > 0
( x + 1)( x − 1) 2 > 0;
(−1,1) ∪ (1, ∞)

27. a.
c.
8

Section 0.2

False.
False.

b.

True.



28. a.

True.

c.

False.

29. a.

b.

True.

33. a.

( x + 1)( x 2 + 2 x – 7) ≥ x 2 – 1

x3 + 3 x 2 – 5 x – 7 ≥ x 2 – 1
x3 + 2 x 2 – 5 x – 6 ≥ 0
( x + 3)( x + 1)( x – 2) ≥ 0
[−3, −1] ∪ [2, ∞)

⇒ Let a < b , so ab < b 2 . Also, a 2 < ab .

Thus, a 2 < ab < b 2 and a 2 < b 2 . ⇐ Let
a 2 < b 2 , so a ≠ b Then
0 < ( a − b ) = a 2 − 2ab + b 2
2

x4 − 2 x2 ≥ 8

b.

< b 2 − 2ab + b 2 = 2b ( b − a )

x4 − 2 x2 − 8 ≥ 0

Since b > 0 , we can divide by 2b to get
b−a > 0.

( x 2 − 4)( x 2 + 2) ≥ 0
( x 2 + 2)( x + 2)( x − 2) ≥ 0

b. We can divide or multiply an inequality by
any positive number.
a
1 1
a < b ⇔ <1⇔ < .
b
b a

(−∞, −2] ∪ [2, ∞)

c.

[( x 2 + 1) − 5][( x 2 + 1) − 2] < 0

30. (b) and (c) are true.
(a) is false: Take a = −1, b = 1 .
(d) is false: if a ≤ b , then −a ≥ −b .
31. a.

3x + 7 > 1 and 2x + 1 < 3
3x > –6 and 2x < 2
x > –2 and x < 1; (–2, 1)

( x 2 − 4)( x 2 − 1) < 0
( x + 2)( x + 1)( x − 1)( x − 2) < 0
(−2, −1) ∪ (1, 2)

34. a.

32. a.

3x + 7 > 1 and 2x + 1 < –4
5
x > –2 and x < – ; ∅
2

1 ⎞
⎛ 1
,
⎟
⎜
2.01 1.99 ⎠
⎝

2 x − 7 > 1 or 2 x + 1 < 3

b.

x > 4 or x < 1
(−∞,1) ∪ (4, ∞)

2.99 <

1
< 3.01
x+2

2.99( x + 2) < 1 < 3.01( x + 2)
2.99 x + 5.98 < 1 and 1 < 3.01x + 6.02
− 4.98 and
− 5.02
x<
x>

2 x − 7 ≤ 1 or 2 x + 1 < 3
2 x ≤ 8 or 2 x < 2

2.99
5.02
4.98
−
<x<−
3.01
2.99
⎛ 5.02 4.98 ⎞
,−
⎜−
⎟
⎝ 3.01 2.99 ⎠

x ≤ 4 or x < 1
(−∞, 4]

c.

1
< 2.01
x

1
1
<x<
2.01
1.99

2 x > 8 or 2 x < 2

b.

1.99 <

1.99 x < 1 < 2.01x
1.99 x < 1 and 1 < 2.01x
1
and x > 1
x<
1.99
2.01

b. 3x + 7 > 1 and 2x + 1 > –4
3x > –6 and 2x > –5
5
x > –2 and x > – ; ( −2, ∞ )
2
c.

( x 2 + 1)2 − 7( x 2 + 1) + 10 < 0

2 x − 7 ≤ 1 or 2 x + 1 > 3

3.01

2 x ≤ 8 or 2 x > 2

x ≤ 4 or x > 1
(−∞, ∞)

35.

x − 2 ≥ 5;
x − 2 ≤ −5 or x − 2 ≥ 5
x ≤ −3 or x ≥ 7
(−∞, −3] ∪ [7, ∞)


Section 0.2

9


36.

x + 2 < 1;
–1 < x + 2 < 1

43.

–3 < x < –1
(–3, –1)

37.

4 x + 5 ≤ 10;
−10 ≤ 4 x + 5 ≤ 10
−15 ≤ 4 x ≤ 5
−

38.

15
5 ⎡ 15 5 ⎤
≤ x ≤ ; ⎢− , ⎥
4
4 ⎣ 4 4⎦

2 x – 1 > 2;

2x – 1 < –2 or 2x – 1 > 2
2x < –1 or 2x > 3;
1
3 ⎛
1⎞ ⎛3 ⎞
x < – or x > , ⎜ – ∞, – ⎟ ∪ ⎜ , ∞ ⎟
2
2 ⎝
2⎠ ⎝2 ⎠
39.

40.

2x
−5 ≥ 7
7
2x
2x
− 5 ≤ −7 or
−5 ≥ 7
7
7
2x
2x
≤ −2 or
≥ 12
7
7
x ≤ −7 or x ≥ 42;
(−∞, −7] ∪ [42, ∞)
x
+1 < 1
4
x
−1 < + 1 < 1
4
x
−2 < < 0;
4
–8 < x < 0; (–8, 0)

41. 5 x − 6 > 1;
5 x − 6 < −1 or 5 x − 6 > 1
5 x < 5 or 5 x > 7
7
⎛7 ⎞
x < 1 or x > ;(−∞,1) ∪ ⎜ , ∞ ⎟
5
⎝5 ⎠

42.

2 x – 7 > 3;

2x – 7 < –3 or 2x – 7 > 3
2x < 4 or 2x > 10
x < 2 or x > 5; (−∞, 2) ∪ (5, ∞)

44.

1
− 3 > 6;
x
1
1
− 3 < −6 or − 3 > 6
x
x
1
1
+ 3 < 0 or − 9 > 0
x
x
1 + 3x
1− 9x
< 0 or
> 0;
x
x
⎛ 1 ⎞ ⎛ 1⎞
⎜ − , 0 ⎟ ∪ ⎜ 0, ⎟
⎝ 3 ⎠ ⎝ 9⎠
5
> 1;
x
5
5
2 + < –1 or 2 + > 1
x
x
5
5
3 + < 0 or 1 + > 0
x
x
3x + 5
x+5
< 0 or
> 0;
x
x
⎛ 5 ⎞
(– ∞, – 5) ∪ ⎜ – , 0 ⎟ ∪ (0, ∞)
⎝ 3 ⎠
2+

45. x 2 − 3x − 4 ≥ 0;
x=

3 ± (–3)2 – 4(1)(–4) 3 ± 5
=
= –1, 4
2(1)
2

( x + 1)( x − 4) = 0; (−∞, −1] ∪ [4, ∞)

4 ± (−4)2 − 4(1)(4)
=2
2(1)
( x − 2)( x − 2) ≤ 0; x = 2

46. x 2 − 4 x + 4 ≤ 0; x =

47. 3x2 + 17x – 6 > 0;
x=

–17 ± (17) 2 – 4(3)(–6) –17 ± 19
1
=
= –6,
2(3)
6
3

⎛1 ⎞
(3x – 1)(x + 6) > 0; (– ∞, – 6) ∪ ⎜ , ∞ ⎟
⎝3 ⎠

48. 14 x 2 + 11x − 15 ≤ 0;
−11 ± (11) 2 − 4(14)(−15) −11 ± 31
=
2(14)
28
3 5
x=− ,
2 7
3 ⎞⎛
5⎞
⎛
⎡ 3 5⎤
⎜ x + ⎟ ⎜ x − ⎟ ≤ 0; ⎢ − , ⎥
2 ⎠⎝
7⎠
⎝
⎣ 2 7⎦

x=

49. x − 3 < 0.5 ⇒ 5 x − 3 < 5(0.5) ⇒ 5 x − 15 < 2.5
50. x + 2 < 0.3 ⇒ 4 x + 2 < 4(0.3) ⇒ 4 x + 18 < 1.2
10

Section 0.2



51.

x−2 <

52.

x+4 <

ε
6

ε
2

⇒ 6 x − 2 < ε ⇒ 6 x − 12 < ε

59.

x –1 < 2 x – 6
( x –1) 2 < (2 x – 6)2

⇒ 2 x + 4 < ε ⇒ 2x + 8 < ε

x 2 – 2 x + 1 < 4 x 2 – 24 x + 36
3x 2 – 22 x + 35 > 0

53. 3x − 15 < ε ⇒ 3( x − 5) < ε

(3x – 7)( x – 5) > 0;

⇒ 3 x−5 < ε
⇒ x−5 <

54.

ε
3

;δ =

7⎞
⎛
⎜ – ∞, ⎟ ∪ (5, ∞)
3⎠
⎝

ε
3

4 x − 8 < ε ⇒ 4( x − 2) < ε

60.

55.

ε
4

;δ =

4 x2 − 4 x + 1 ≥ x2 + 2 x + 1

4

3x2 − 6 x ≥ 0
3 x( x − 2) ≥ 0
(−∞, 0] ∪ [2, ∞)

⇒ 6 x+6 <ε

ε
6

;δ =

2

ε

6 x + 36 < ε ⇒ 6( x + 6) < ε

⇒ x+6 <

2x −1 ≥ x + 1
(2 x − 1)2 ≥ ( x + 1)

⇒ 4 x−2 <ε
⇒ x−2 <

x –1 < 2 x – 3

ε
6

61.

2 2 x − 3 < x + 10
4 x − 6 < x + 10

56. 5 x + 25 < ε ⇒ 5( x + 5) < ε

(4 x − 6) 2 < ( x + 10)2

⇒ 5 x+5 <ε

16 x 2 − 48 x + 36 < x 2 + 20 x + 100

⇒ x+5 <

ε
5

;δ =

ε

15 x 2 − 68 x − 64 < 0

5

(5 x + 4)(3 x − 16) < 0;

57. C = π d
C – 10 ≤ 0.02
πd – 10 ≤ 0.02
10 ⎞
⎛
π ⎜ d – ⎟ ≤ 0.02
π⎠
⎝
10 0.02
d–
≤
≈ 0.0064
π
π
We must measure the diameter to an accuracy
of 0.0064 in.

58. C − 50 ≤ 1.5,

5
( F − 32 ) − 50 ≤ 1.5;
9

5
( F − 32 ) − 90 ≤ 1.5
9
F − 122 ≤ 2.7

⎛ 4 16 ⎞
⎜– , ⎟
⎝ 5 3⎠
3x − 1 < 2 x + 6

62.

3x − 1 < 2 x + 12
(3x − 1) 2 < (2 x + 12)2
9 x 2 − 6 x + 1 < 4 x 2 + 48 x + 144
5 x 2 − 54 x − 143 < 0

( 5 x + 11)( x − 13) < 0
⎛ 11 ⎞
⎜ − ,13 ⎟
⎝ 5
⎠

We are allowed an error of 2.7 F.


Section 0.2

11


63.

x < y ⇒ x x ≤ x y and x y < y y
2

⇒ x < y

Order property: x < y ⇔ xz < yz when z is positive.

2

Transitivity

(x

⇒ x2 < y 2

Conversely,
2

2

(x

2

x2 < y 2 ⇒ x < y

2

= x

2

2

= x2

)

)

2

2

⇒ x – y <0
Subtract y from each side.
⇒ ( x – y )( x + y ) < 0 Factor the difference of two squares.
⇒ x – y <0
⇒ x < y

64. 0 < a < b ⇒ a =

( a) < ( b)
2

a <

2

This is the only factor that can be negative.
Add y to each side.

( a)

2

and b =

( b)

2

, so

67.

, and, by Problem 63,

x +9
x–2

a + b + c = ( a + b) + c ≤ a + b + c
≤ a+b+c

66.

1
x2 + 3

−

⎛
1
1
1 ⎞
=
+⎜−
⎜ x +2⎟
⎟
2
x +2
x +3 ⎝
⎠
≤

1
2

x +3

+−

68.

1
x +2

1
=
+
2
x +2
x +3
1
2

+

x 2 + 3 ≥ 3 and x + 2 ≥ 2, so
1
2

x +3
1
2

x +3

12

≤

1
1
1
≤ , thus,
and
x +2 2
3

+

1
1 1
≤ +
x +2 3 2

Section 0.2

x2 + 9
x
2

x +9
x

+

–2
2

x ≤ 2 ⇒ x2 + 2 x + 7 ≤ x2 + 2 x + 7

Thus,

x2 + 2 x + 7
2

x +1

= x2 + 2 x + 7

1
2

x +1

≤ 15 ⋅1 = 15

1
x +2

x +3
by the Triangular Inequality, and since
1
1
x 2 + 3 > 0, x + 2 > 0 ⇒
> 0,
> 0.
2
x +2
x +3

≤

x + (–2)

≤ 4 + 4 + 7 = 15
1
and x 2 + 1 ≥ 1 so
≤ 1.
2
x +1

1

=

=

x +9
x +2
2
≤
+
=
2
2
2
x + 9 x + 9 x + 9 x2 + 9
1
1
Since x 2 + 9 ≥ 9,
≤
2
x +9 9
x +2 x +2
≤
9
x2 + 9
x +2
x–2
≤
9
x2 + 9

b ⇒ a< b.

of absolute values.
c.

x +9
2

a – b ≥ a – b ≥ a – b Use Property 4

b.

2

x–2

a – b = a + (–b) ≤ a + –b = a + b

65. a.

x–2

69.

1 3 1 2 1
1
x + x + x+
2
4
8
16
1 3 1 2 1
1
≤ x4 + x + x + x +
2
4
8
16
1 1 1 1
≤ 1+ + + +
since x ≤ 1.
2 4 8 16
1
1
1
1
≤ 1.9375 < 2.
So x 4 + x3 + x 2 + x +
2
4
8
16
x4 +



x < x2

70. a.

77.

x − x2 < 0
x(1 − x) < 0
x < 0 or x > 1

1 11
≤
R 60

2

x <x

b.

1 1 1
1
≤ +
+
R 10 20 30
1 6+3+ 2
≤
R
60

2

x −x<0
x( x − 1) < 0
0 < x <1

R≥

60
11

1
1
1
1
≥
+
+
R 20 30 40
1 6+4+3
≥
R
120
120
R≤
13

71. a ≠ 0 ⇒
2

1⎞
1
⎛
0 ≤ ⎜ a – ⎟ = a2 – 2 +
a⎠
⎝
a2
1
1
or a 2 +
≥2.
so, 2 ≤ a 2 +
2
a
a2

Thus,

72. a < b
a + a < a + b and a + b < b + b
2a < a + b < 2b
a+b
a<
<b
2

60
120
≤R≤
11
13

78. A = 4π r 2 ; A = 4π (10)2 = 400π
4π r 2 − 400π < 0.01
4π r 2 − 100 < 0.01

73. 0 < a < b

r 2 − 100 <

a 2 < ab and ab < b 2

0.01
4π

0.01 2
0.01
< r − 100 <
4π
4π

a 2 < ab < b 2
a < ab < b

74.

−

0.01
0.01
< r < 100 +
4π
4π
δ ≈ 0.00004 in
100 −

(

)

1
1
( a + b ) ⇔ ab ≤ a 2 + 2ab + b2
2
4
1 2 1
1 2 1 2
⇔ 0 ≤ a − ab + b = a − 2ab + b 2
4
2
4
4
1
2
⇔ 0 ≤ (a − b) which is always true.
4

ab ≤

(

)
0.3 Concepts Review
1.

75. For a rectangle the area is ab, while for a
2

⎛ a+b⎞
square the area is a 2 = ⎜
⎟ . From
⎝ 2 ⎠
1
⎛ a+b⎞
(a + b) ⇔ ab ≤ ⎜
⎟
2
⎝ 2 ⎠
so the square has the largest area.

Problem 74,

ab ≤

76. 1 + x + x 2 + x3 + … + x99 ≤ 0;
(−∞, −1]


( x + 2)2 + ( y − 3)2

2. (x + 4)2 + (y – 2)2 = 25
2

⎛ −2 + 5 3 + 7 ⎞
3. ⎜
,
⎟ = (1.5,5)
2 ⎠
⎝ 2

4.

d −b
c−a

Section 0.3

13


Problem Set 0.3

5. d1 = (5 + 2) 2 + (3 – 4)2 = 49 + 1 = 50
d 2 = (5 − 10)2 + (3 − 8)2 = 25 + 25 = 50

1.

d3 = (−2 − 10)2 + (4 − 8)2
= 144 + 16 = 160

d1 = d 2 so the triangle is isosceles.

6. a = (2 − 4)2 + (−4 − 0) 2 = 4 + 16 = 20
b = (4 − 8)2 + (0 + 2)2 = 16 + 4 = 20
c = (2 − 8)2 + (−4 + 2) 2 = 36 + 4 = 40

d = (3 – 1)2 + (1 – 1)2 = 4 = 2

a 2 + b 2 = c 2 , so the triangle is a right triangle.

2.
7. (–1, –1), (–1, 3); (7, –1), (7, 3); (1, 1), (5, 1)
8.

( x − 3) 2 + (0 − 1) 2 = ( x − 6)2 + (0 − 4) 2 ;
x 2 − 6 x + 10 = x 2 − 12 x + 52

6 x = 42
x = 7 ⇒ ( 7, 0 )

d = (−3 − 2)2 + (5 + 2)2 = 74 ≈ 8.60

3.

⎛ –2 + 4 –2 + 3 ⎞ ⎛ 1 ⎞
9. ⎜
,
⎟ = ⎜1, ⎟ ;
2 ⎠ ⎝ 2⎠
⎝ 2

2
25
⎛1
⎞
d = (1 + 2)2 + ⎜ – 3 ⎟ = 9 +
≈ 3.91
2
4
⎝
⎠
⎛1+ 2 3 + 6 ⎞ ⎛ 3 9 ⎞
10. midpoint of AB = ⎜
,
⎟=⎜ , ⎟
2 ⎠ ⎝2 2⎠
⎝ 2
⎛ 4 + 3 7 + 4 ⎞ ⎛ 7 11 ⎞
midpoint of CD = ⎜
,
⎟=⎜ , ⎟
2 ⎠ ⎝2 2 ⎠
⎝ 2
2

⎛ 3 7 ⎞ ⎛ 9 11 ⎞
d = ⎜ − ⎟ +⎜ − ⎟
⎝2 2⎠ ⎝2 2 ⎠

2

= 4 + 1 = 5 ≈ 2.24

d = (4 – 5)2 + (5 + 8) 2 = 170 ≈ 13.04

4.

11 (x – 1)2 + (y – 1)2 = 1
12. ( x + 2)2 + ( y − 3)2 = 42
( x + 2)2 + ( y − 3)2 = 16

13. ( x − 2) 2 + ( y + 1) 2 = r 2
(5 − 2)2 + (3 + 1) 2 = r 2
r 2 = 9 + 16 = 25

( x − 2) 2 + ( y + 1) 2 = 25

d = (−1 − 6)2 + (5 − 3) 2 = 49 + 4 = 53
≈ 7.28

14

Section 0.3



14. ( x − 4) 2 + ( y − 3) 2 = r 2

21. 4 x 2 + 16 x + 15 + 4 y 2 + 6 y = 0

(6 − 4) 2 + (2 − 3) 2 = r 2

3
9⎞
9
⎛
4( x 2 + 4 x + 4) + 4 ⎜ y 2 + y + ⎟ = −15 + 16 +
2
16 ⎠
4
⎝

2

r = 4 +1 = 5

2

3⎞
13
⎛
4( x + 2)2 + 4 ⎜ y + ⎟ =
4⎠
4
⎝

( x − 4)2 + ( y − 3)2 = 5
⎛ 1+ 3 3 + 7 ⎞
15. center = ⎜
,
⎟ = (2, 5)
2 ⎠
⎝ 2
1
1
radius =
(1 – 3)2 + (3 – 7)2 =
4 + 16
2
2
1
=
20 = 5
2
2

2

( x – 2) + ( y – 5) = 5

2

3⎞
13
⎛
( x + 2)2 + ⎜ y + ⎟ =
4⎠
16
⎝
3⎞
⎛
center = ⎜ −2, − ⎟ ; radius =
4⎠
⎝

105
+ 4 y2 + 3 y = 0
16
3
9 ⎞
⎛
2
4( x + 4 x + 4) + 4 ⎜ y 2 + y + ⎟
4
64 ⎠
⎝
105
9
=−
+ 16 +
16
16

22. 4 x 2 + 16 x +

16. Since the circle is tangent to the x-axis, r = 4.
( x − 3)2 + ( y − 4) 2 = 16

17. x 2 + 2 x + 10 + y 2 – 6 y –10 = 0

2

3⎞
⎛
4( x + 2)2 + 4 ⎜ y + ⎟ = 10
8⎠
⎝

x2 + 2 x + y 2 – 6 y = 0

2

( x 2 + 2 x + 1) + ( y 2 – 6 y + 9) = 1 + 9

3⎞
5
⎛
( x + 2)2 + ⎜ y + ⎟ =
8⎠
2
⎝

( x + 1) 2 + ( y – 3) 2 = 10

3⎞
5
10
⎛
center = ⎜ −2, − ⎟ ; radius =
=
8⎠
2
2
⎝

center = (–1, 3); radius = 10
x 2 + y 2 − 6 y = 16

18.

x 2 + ( y 2 − 6 y + 9) = 16 + 9

23.

2 –1
=1
2 –1

24.

7−5
=2
4−3

25.

–6 – 3 9
=
–5 – 2 7

26.

−6 + 4
=1
0−2

27.

5–0
5
=–
0–3
3

28.

6−0
=1
0+6

x 2 + ( y − 3) 2 = 25

center = (0, 3); radius = 5

19. x 2 + y 2 –12 x + 35 = 0
x 2 –12 x + y 2 = –35

( x 2 –12 x + 36) + y 2 = –35 + 36
( x – 6) 2 + y 2 = 1
center = (6, 0); radius = 1
x 2 + y 2 − 10 x + 10 y = 0

20.
2

29.

y − 2 = −1( x − 2)
y − 2 = −x + 2
x+ y−4 = 0

30.

y − 4 = −1( x − 3)
y − 4 = −x + 3

2

( x − 10 x + 25) + ( y + 10 y + 25) = 25 + 25

x+ y−7 = 0

( x − 5) 2 + ( y + 5)2 = 50
center = ( 5, −5 ) ; radius = 50 = 5 2

13
4

31.

y = 2x + 3
2x – y + 3 = 0

32.


y = 0x + 5
0x + y − 5 = 0

Section 0.3

15


33. m =

8–3 5
= ;
4–2 2
5
y – 3 = ( x – 2)
2
2 y – 6 = 5 x – 10

c.

3 y = –2 x + 6
y=–

x − 4y + 0 = 0

2
x + 2;
3

2
m=– ;
3

5x – 2 y – 4 = 0
2 −1 1
34. m =
= ;
8−4 4
1
y − 1 = ( x − 4)
4
4y − 4 = x − 4

2x + 3 y = 6

2
y + 3 = – ( x – 3)
3
2
y = – x –1
3

d.

3
m= ;
2
3
( x – 3)
2
3
15
y= x–
2
2

y+3=

2
1
2
35. 3y = –2x + 1; y = – x + ; slope = – ;
3
3
3
1
y -intercept =
3

–1 – 2
3
=– ;
3 +1
4
3
y + 3 = – ( x – 3)
4
3
3
y=– x–
4
4

36. −4 y = 5 x − 6
5
3
y = − x+
4
2
5
3
slope = − ; y -intercept =
4
2

e.

m=

37. 6 – 2 y = 10 x – 2
–2 y = 10 x – 8
y = –5 x + 4;
slope = –5; y-intercept = 4

f.

x=3

38. 4 x + 5 y = −20
5 y = −4 x − 20
4
y = − x−4
5
4
slope = − ; y -intercept = − 4
5
39. a.

b.

m = 2;
y + 3 = 2( x – 3)
y = 2x – 9
1
m=– ;
2

1
y + 3 = – ( x – 3)
2
1
3
y=– x–
2
2

16

Section 0.3

40. a.

g. y = –3

3 x + cy = 5
3(3) + c(1) = 5

c = −4

b.

c=0

c.

2 x + y = −1

y = −2 x − 1
m = −2;
3x + cy = 5
cy = −3x + 5
3
5
y = − x+
c
c
3
−2 = −
c
3
c=
2

d. c must be the same as the coefficient of x,
so c = 3.



e.

y − 2 = 3( x + 3);

1
perpendicular slope = − ;
3
1
3
− =−
3
c
c=9

3
( x + 2)
2
3
y = x+2
2

y +1 =

b.

c.

2x + 3 y = 4
9 x – 3 y = –15
= –11

11x

x = –1
–3(–1) + y = 5

3
41. m = ;
2

42. a.

45. 2 x + 3 y = 4
–3x + y = 5

m = 2;
kx − 3 y = 10
−3 y = − kx + 10
10
k
y = x−
3
3
k
= 2; k = 6
3
1
m=− ;
2
k
1
=−
3
2
3
k=−
2
2x + 3 y = 6
3 y = −2 x + 6
2
y = − x + 2;
3
3 k 3
9
m= ; = ; k=
2 3 2
2

y=2

Point of intersection: (–1, 2)
3 y = –2 x + 4
2
4
y = – x+
3
3
3
m=
2
3
y − 2 = ( x + 1)
2
3
7
y = x+
2
2
46. 4 x − 5 y = 8
2 x + y = −10
4x − 5 y = 8
−4 x − 2 y = 20
− 7 y = 28
y = −4
4 x − 5(−4) = 8
4 x = −12
x = −3
Point of intersection: ( −3, −4 ) ;
4x − 5 y = 8
−5 y = −4 x + 8
y=

43. y = 3(3) – 1 = 8; (3, 9) is above the line.
b−0
b
=−
0−a
a
b
bx
x y
y = − x + b;
+ y = b; + = 1
a
a
a b

44. (a, 0), (0, b); m =


m=−

4
8
x−
5
5

5
4

5
y + 4 = − ( x + 3)
4
5
31
y = − x−
4
4

Section 0.3

17


47. 3x – 4 y = 5
2x + 3y = 9
9 x – 12 y = 15
8 x + 12 y = 36
17 x

= 51

x=3
3(3) – 4 y = 5
–4 y = –4
y =1

Point of intersection: (3, 1); 3x – 4y = 5;
–4 y = –3x + 5
3
5
y= x–
4
4
4
m=–
3
4
y – 1 = – ( x – 3)
3
4
y = – x+5
3
48. 5 x – 2 y = 5

2x + 3y = 6
15 x – 6 y = 15
4 x + 6 y = 12
19 x

= 27
27
x=
19

⎛ 27 ⎞
2⎜ ⎟ + 3y = 6
⎝ 19 ⎠
60
3y =
19
20
y=
19
⎛ 27 20 ⎞
Point of intersection: ⎜ , ⎟ ;
⎝ 19 19 ⎠
5x − 2 y = 5
–2 y = –5 x + 5
5
5
y= x–
2
2
2
m=–
5
20
2⎛
27 ⎞
y–
= – ⎜x− ⎟
19
5⎝
19 ⎠
2
54 20
y = – x+ +
5
95 19
2
154
y = − x+
5
95

18

Section 0.3

⎛ 2 + 6 –1 + 3 ⎞
,
49. center: ⎜
⎟ = (4, 1)
2 ⎠
⎝ 2
⎛ 2+ 6 3+3⎞
midpoint = ⎜
,
⎟ = (4, 3)
2 ⎠
⎝ 2
inscribed circle: radius = (4 – 4)2 + (1 – 3)2
= 4=2
2

( x – 4) + ( y – 1)2 = 4
circumscribed circle:
radius = (4 – 2)2 + (1 – 3)2 = 8
( x – 4)2 + ( y –1)2 = 8

50. The radius of each circle is 16 = 4. The centers
are (1, −2 ) and ( −9,10 ) . The length of the belt is

the sum of half the circumference of the first
circle, half the circumference of the second circle,
and twice the distance between their centers.
1
1
L = ⋅ 2π (4) + ⋅ 2π (4) + 2 (1 + 9)2 + (−2 − 10)2
2
2
= 8π + 2 100 + 144
≈ 56.37
51. Put the vertex of the right angle at the origin
with the other vertices at (a, 0) and (0, b). The
⎛a b⎞
midpoint of the hypotenuse is ⎜ , ⎟ . The
⎝ 2 2⎠
distances from the vertices are
2

2

a⎞ ⎛
b⎞
⎛
⎜a – ⎟ +⎜0 – ⎟ =
2⎠ ⎝
2⎠
⎝
=
2

2

a⎞ ⎛
b⎞
⎛
⎜0 – ⎟ + ⎜b – ⎟ =
2⎠ ⎝
2⎠
⎝
=
2

2

a⎞ ⎛
b⎞
⎛
⎜0 – ⎟ + ⎜0 – ⎟ =
2⎠ ⎝
2⎠
⎝
=

a 2 b2
+
4
4
1 2
a + b2 ,
2
a 2 b2
+
4
4
1 2
a + b 2 , and
2
a 2 b2
+
4
4
1 2
a + b2 ,
2

which are all the same.
52. From Problem 51, the midpoint of the
hypotenuse, ( 4,3, ) , is equidistant from the

vertices. This is the center of the circle. The
radius is 16 + 9 = 5. The equation of the
circle is
( x − 4) 2 + ( y − 3) 2 = 25.



53. x 2 + y 2 – 4 x – 2 y – 11 = 0
( x 2 – 4 x + 4) + ( y 2 – 2 y + 1) = 11 + 4 + 1
( x – 2)2 + ( y – 1)2 = 16
x 2 + y 2 + 20 x – 12 y + 72 = 0
( x 2 + 20 x + 100) + ( y 2 – 12 y + 36)
= –72 + 100 + 36
2

2

( x + 10) + ( y – 6) = 64
center of first circle: (2, 1)
center of second circle: (–10, 6)

d = (2 + 10)2 + (1 – 6) 2 = 144 + 25
= 169 = 13
However, the radii only sum to 4 + 8 = 12, so
the circles must not intersect if the distance
between their centers is 13.
54. x 2 + ax + y 2 + by + c = 0
⎛ 2
a2 ⎞ ⎛ 2
b2 ⎞
⎜ x + ax +
⎟ + ⎜ y + by + ⎟
⎜
4 ⎟ ⎜
4 ⎟
⎝
⎠ ⎝
⎠
= −c +

a 2 b2
+
4
4

2
2
a⎞ ⎛
b⎞
a 2 + b 2 − 4c
⎛
⎜x+ ⎟ +⎜ y+ ⎟ =
2⎠ ⎝
2⎠
4
⎝
2

2

a + b − 4c
> 0 ⇒ a 2 + b 2 > 4c
4

55. Label the points C, P, Q, and R as shown in the
figure below. Let d = OP , h = OR , and
a = PR . Triangles ΔOPR and ΔCQR are

similar because each contains a right angle and
they share angle ∠QRC . For an angle of
30 ,

a 1
d
3
and = ⇒ h = 2a . Using a
=
h 2
h
2

56. The equations of the two circles are
( x − R)2 + ( y − R)2 = R 2
( x − r )2 + ( y − r )2 = r 2

Let ( a, a ) denote the point where the two
circles touch. This point must satisfy
(a − R)2 + (a − R)2 = R 2
R2
2
⎛
2⎞
a = ⎜1 ±
⎟R
⎜
2 ⎟
⎝
⎠

(a − R)2 =

⎛
2⎞
Since a < R , a = ⎜1 −
⎟ R.
⎜
2 ⎟
⎝
⎠
At the same time, the point where the two
circles touch must satisfy
(a − r )2 + (a − r )2 = r 2
⎛
2⎞
a = ⎜1 ±
⎟r
⎜
2 ⎟
⎝
⎠
⎛
2⎞
Since a > r , a = ⎜ 1 +
⎟ r.
⎜
2 ⎟
⎝
⎠
Equating the two expressions for a yields
⎛
⎛
2⎞
2⎞
⎜1 −
⎟ R = ⎜1 +
⎟r
⎜
⎟
⎜
2 ⎠
2 ⎟
⎝
⎝
⎠
2

2
1−
2 R=
r=
2
1+
2
r=

1− 2 +

⎛
2⎞
⎜1 −
⎟
⎜
2 ⎟
⎝
⎠
R
⎛
⎞⎛
2
2⎞
⎜1 +
⎟⎜ 1 −
⎟
⎜
2 ⎟⎜
2 ⎟
⎝
⎠⎝
⎠

1
2R

1
2
r = (3 − 2 2) R ≈ 0.1716 R
1−

property of similar triangles, QC / RC = 3 / 2 ,
2
3
4
=
→ a = 2+
a−2
2
3
By the Pythagorean Theorem, we have
d = h 2 − a 2 = 3a = 2 3 + 4 ≈ 7.464


Section 0.3

19


57. Refer to figure 15 in the text. Given ine l1 with
slope m, draw ABC with vertical and
horizontal sides m, 1.
Line l2 is obtained from l1 by rotating it
around the point A by 90° counter-clockwise.
Triangle ABC is rotated into triangle AED .
We read off
1
1
slope of l2 =
=− .
m
−m

60. See the figure below. The angle at T is a right
angle, so the Pythagorean Theorem gives
( PM + r )2 = ( PT )2 + r 2
⇔ ( PM )2 + 2rPM + r 2 = ( PT )2 + r 2
⇔ PM ( PM + 2r ) = ( PT )2
PM + 2r = PN so this gives ( PM )( PN ) = ( PT ) 2

58. 2 ( x − 1)2 + ( y − 1)2 = ( x − 3) 2 + ( y − 4)2
4( x 2 − 2 x + 1 + y 2 − 2 y + 1)
= x 2 − 6 x + 9 + y 2 − 8 y + 16

3x 2 − 2 x + 3 y 2 = 9 + 16 − 4 − 4;
2
17
x + y2 = ;
3
3
1⎞
17 1
⎛ 2 2
2
⎜x − x+ ⎟+ y = +
3
9⎠
3 9
⎝
3x 2 − 2 x + 3 y 2 = 17; x 2 −

2

1⎞
52
⎛
2
⎜x− ⎟ + y =
3⎠
9
⎝

B = (6)2 + (8)2 = 100 = 10

⎛ 52 ⎞
⎛1 ⎞
center: ⎜ , 0 ⎟ ; radius: ⎜
⎜ 3 ⎟
⎟
⎝3 ⎠
⎝
⎠

59. Let a, b, and c be the lengths of the sides of the
right triangle, with c the length of the
hypotenuse. Then the Pythagorean Theorem

says that a 2 + b 2 = c 2
Thus,

πa 2 πb 2 πc 2
+
=
or
8
8
8
2

61. The lengths A, B, and C are the same as the
corresponding distances between the centers of
the circles:
A = (–2)2 + (8)2 = 68 ≈ 8.2

2

1 ⎛a⎞
1 ⎛b⎞
1 ⎛c⎞
π⎜ ⎟ + π⎜ ⎟ = π⎜ ⎟
2 ⎝2⎠
2 ⎝2⎠
2 ⎝2⎠

C = (8)2 + (0)2 = 64 = 8
Each circle has radius 2, so the part of the belt
around the wheels is
2(2π − a − π ) + 2(2π − b − π ) + 2(2π − c − π )
= 2[3π - (a + b + c)] = 2(2π ) = 4π
Since a + b + c = π , the sum of the angles of a
triangle.
The length of the belt is ≈ 8.2 + 10 + 8 + 4π
≈ 38.8 units.

2

2

1 ⎛ x⎞
π ⎜ ⎟ is the area of a semicircle with
2 ⎝2⎠
diameter x, so the circles on the legs of the
triangle have total area equal to the area of the
semicircle on the hypotenuse.

From a 2 + b 2 = c 2 ,
3 2
3 2
3 2
a +
b =
c
4
4
4
3 2
x is the area of an equilateral triangle
4
with sides of length x, so the equilateral
triangles on the legs of the right triangle have
total area equal to the area of the equilateral
triangle on the hypotenuse of the right triangle.

20

Section 0.3

62 As in Problems 50 and 61, the curved portions
of the belt have total length 2π r. The lengths
of the straight portions will be the same as the
lengths of the sides. The belt will have length
2π r + d1 + d 2 + … + d n .



63. A = 3, B = 4, C = –6
3(–3) + 4(2) + (–6) 7
d=
=
5
(3) 2 + (4)2
64. A = 2, B = −2, C = 4
d=

2(4) − 2(−1) + 4)
2

(2) + (2)

2

=

14
8

=

7 2
2

65. A = 12, B = –5, C = 1
12(–2) – 5(–1) + 1 18
d=
=
13
(12) 2 + (–5) 2
66. A = 2, B = −1, C = −5
d=

2(3) − 1(−1) − 5
2

(2) + (−1)

2

=

2
5

=

2 5
5

67. 2 x + 4(0) = 5
5
x=
2
d=

2

( 5 ) + 4(0) – 7 =
2
(2)2 + (4) 2

2
20

=

5
5

68. 7(0) − 5 y = −1
1
y=
5
⎛1⎞
7(0) − 5 ⎜ ⎟ − 6
7
7 74
⎝5⎠
d=
=
=
2
2
74
74
(7) + (−5)
−2 − 3
5
3
= − ; m = ; passes through
1+ 2
3
5
⎛ −2 + 1 3 − 2 ⎞ ⎛ 1 1 ⎞
,
⎜
⎟ = ⎜− , ⎟
2 ⎠ ⎝ 2 2⎠
⎝ 2
1 3⎛
1⎞
y− = ⎜x+ ⎟
2 5⎝
2⎠
3
4
y = x+
5
5

69. m =


0–4
1
= –2; m = ; passes through
2–0
2
⎛0+2 4+0⎞
,
⎜
⎟ = (1, 2)
2 ⎠
⎝ 2
1
y – 2 = ( x – 1)
2
1
3
y = x+
2
2
6–0
1
m=
= 3; m = – ; passes through
4–2
3
⎛ 2+4 0+6⎞
,
⎜
⎟ = (3, 3)
2 ⎠
⎝ 2
1
y – 3 = – ( x – 3)
3
1
y = – x+4
3
1
3
1
x+ = – x+4
2
2
3
5
5
x=
6
2
x=3
1
3
y = (3) + = 3
2
2
center = (3, 3)

70. m =

71. Let the origin be at the vertex as shown in the
figure below. The center of the circle is then
( 4 − r , r ) , so it has equation
( x − (4 − r ))2 + ( y − r )2 = r 2 . Along the side of

length 5, the y-coordinate is always

3
4

times

the x-coordinate. Thus, we need to find the
value of r for which there is exactly one x2

⎛3
⎞
solution to ( x − 4 + r ) 2 + ⎜ x − r ⎟ = r 2 .
⎝4
⎠
Solving for x in this equation gives
16 ⎛
⎞
x = ⎜ 16 − r ± 24 − r 2 + 7r − 6 ⎟ . There is
25 ⎝
⎠

(

)

exactly one solution when −r 2 + 7 r − 6 = 0,
that is, when r = 1 or r = 6 . The root r = 6 is
extraneous. Thus, the largest circle that can be
inscribed in this triangle has radius r = 1.

Section 0.3

21


72. The line tangent to the circle at ( a, b ) will be

The slope of PS is
1
[ y1 + y4 − ( y1 + y2 )] y − y
2
2
= 4
. The slope of
1
x4 − x2
x1 + x4 − ( x1 + x2 ) ]
[
2
1
[ y3 + y4 − ( y2 + y3 )] y − y
2
. Thus
QR is 2
= 4
1
x4 − x2
[ x3 + x4 − ( x2 + x3 )]
2
PS and QR are parallel. The slopes of SR and
y −y
PQ are both 3 1 , so PQRS is a
x3 − x1
parallelogram.

perpendicular to the line through ( a, b ) and the
center of the circle, which is ( 0, 0 ) . The line
through ( a, b ) and ( 0, 0 ) has slope
0−b b
a
r2
= ; ax + by = r 2 ⇒ y = − x +
0−a a
b
b
a
so ax + by = r 2 has slope − and is
b
perpendicular to the line through ( a, b ) and
m=

( 0, 0 ) ,

so it is tangent to the circle at ( a, b ) .

73. 12a + 0b = 36
a=3
32 + b 2 = 36
b = ±3 3
3x – 3 3 y = 36
x – 3 y = 12
3x + 3 3 y = 36
x + 3 y = 12

74. Use the formula given for problems 63-66, for
( x, y ) = ( 0, 0 ) .

77. x 2 + ( y – 6) 2 = 25; passes through (3, 2)
tangent line: 3x – 4y = 1
The dirt hits the wall at y = 8.

A = m, B = −1, C = B − b;(0, 0)
d=

m(0) − 1(0) + B − b
m2 + (−1) 2

=

B−b
m2 + 1

75. The midpoint of the side from (0, 0) to (a, 0) is
⎛0+a 0+0⎞ ⎛ a ⎞
,
⎜
⎟ = ⎜ , 0⎟
2 ⎠ ⎝2 ⎠
⎝ 2
The midpoint of the side from (0, 0) to (b, c) is
⎛0+b 0+c⎞ ⎛b c ⎞
,
⎜
⎟=⎜ , ⎟
2 ⎠ ⎝2 2⎠
⎝ 2
c–0
c
m1 =
=
b–a b–a
c –0
c
m2 = 2
=
; m1 = m2
b–a
b–a
2

0.4 Concepts Review
1. y-axis
2.

( 4, −2 )

3. 8; –2, 1, 4
4. line; parabola

Problem Set 0.4
1. y = –x2 + 1; y-intercept = 1; y = (1 + x)(1 – x);
x-intercepts = –1, 1
Symmetric with respect to the y-axis

2

76. See the figure below. The midpoints of the
sides are
⎛ x2 + x3 y2 + y3 ⎞
⎛ x + x y + y2 ⎞
P⎜ 1 2 , 1
,
,
⎟, Q⎜
2 ⎠
2 ⎟
⎝ 2
⎝ 2
⎠
⎛ x + x y + y4 ⎞
R⎜ 3 4 , 3
, and
2 ⎟
⎝ 2
⎠
⎛ x + x y + y4 ⎞
S⎜ 1 4 , 1
.
2 ⎟
⎝ 2
⎠

22

Section 0.4



2. x = − y 2 + 1; y -intercepts = −1,1;
x-intercept = 1 .
Symmetric with respect to the x-axis.

3. x = –4y2 – 1; x-intercept = –1
Symmetric with respect to the x-axis

4.

y = 4 x 2 − 1; y -intercept = −1
1 1
y = (2 x + 1)(2 x − 1); x-intercepts = − ,
2 2
Symmetric with respect to the y-axis.

5. x2 + y = 0; y = –x2
x-intercept = 0, y-intercept = 0

6. y = x 2 − 2 x; y -intercept = 0
y = x(2 − x); x-intercepts = 0, 2

7
7. 7x2 + 3y = 0; 3y = –7x2; y = – x 2
3
x-intercept = 0, y-intercept = 0

8. y = 3x 2 − 2 x + 2; y -intercept = 2


Section 0.4

23


9. x2 + y2 = 4
x-intercepts = -2, 2; y-intercepts = -2, 2
Symmetric with respect to the x-axis, y-axis,
and origin

10. 3x 2 + 4 y 2 = 12; y-intercepts = − 3, 3
x-intercepts = −2, 2
and origin

11. y = –x2 – 2x + 2: y-intercept = 2
2± 4+8 2±2 3
x-intercepts =
=
= –1 ± 3
–2
–2

24

Section 0.4

12. 4 x 2 + 3 y 2 = 12; y -intercepts = −2, 2

x-intercepts = − 3, 3
and origin

13. x2 – y2 = 4
x-intercept = -2, 2
and origin

14. x 2 + ( y − 1)2 = 9; y -intercepts = −2, 4

x-intercepts = −2 2, 2 2



15. 4(x – 1)2 + y2 = 36;
y-intercepts = ± 32 = ±4 2

18. x 4 + y 4 = 1; y -intercepts = −1,1
x-intercepts = −1,1
and origin

16. x 2 − 4 x + 3 y 2 = −2

19. x4 + y4 = 16; y-intercepts = −2, 2
x-intercepts = −2, 2
Symmetric with respect to the y-axis, x-axis
and origin

x-intercepts = 2 ± 2

17. x2 + 9(y + 2)2 = 36; y-intercepts = –4, 0
x-intercept = 0


20. y = x3 – x; y-intercepts = 0;
y = x(x2 – 1) = x(x + 1)(x – 1);
x-intercepts = –1, 0, 1
Symmetric with respect to the origin

Section 0.4

25


21. y =

1
2

; y-intercept = 1

x +1

2

24. 4 ( x − 5 ) + 9( y + 2) 2 = 36; x-intercept = 5

25. y = (x – 1)(x – 2)(x – 3); y-intercept = –6
x-intercepts = 1, 2, 3
22. y =

x

; y -intercept = 0
x +1
x-intercept = 0
Symmetric with respect to the origin
2

26. y = x2(x – 1)(x – 2); y-intercept = 0
x-intercepts = 0, 1, 2

23. 2 x 2 – 4 x + 3 y 2 + 12 y = –2
2( x 2 – 2 x + 1) + 3( y 2 + 4 y + 4) = –2 + 2 + 12
2( x – 1)2 + 3( y + 2)2 = 12

y-intercepts = –2 ±

30
3

x-intercept = 1
27. y = x 2 ( x − 1)2 ; y-intercept = 0
x-intercepts = 0, 1

26

Section 0.4



28. y = x 4 ( x − 1)4 ( x + 1)4 ; y -intercept = 0
x-intercepts = −1, 0,1

Intersection points: (0, 1) and (–3, 4)

32. 2 x + 3 = −( x − 1) 2
29.

x + y = 1; y-intercepts = –1, 1;

Symmetric with respect to the x-axis, y-axis
and origin

2 x + 3 = − x2 + 2 x − 1
x2 + 4 = 0
No points of intersection

33. −2 x + 3 = −2( x − 4)2
30.

x + y = 4; y-intercepts = –4, 4;

Symmetric with respect to the x-axis, y-axis
and origin

−2 x + 3 = −2 x 2 + 16 x − 32
2 x 2 − 18 x + 35 = 0
x=

18 ± 324 – 280 18 ± 2 11 9 ± 11
=
=
;
4
4
2

⎛ 9 – 11
⎞
Intersection points: ⎜
⎜ 2 , – 6 + 11 ⎟ ,
⎟
⎝
⎠
⎛ 9 + 11
⎞
⎜
⎜ 2 , – 6 – 11 ⎟
⎟
⎝
⎠

31.

− x + 1 = ( x + 1)2
− x + 1 = x2 + 2 x + 1
x 2 + 3x = 0
x( x + 3) = 0
x = 0, −3


Section 0.4

27


37. y = 3x + 1

34. −2 x + 3 = 3x 2 − 3 x + 12

x 2 + 2 x + (3 x + 1) 2 = 15

3x 2 − x + 9 = 0
No points of intersection

x 2 + 2 x + 9 x 2 + 6 x + 1 = 15
10 x 2 + 8 x − 14 = 0
2(5 x 2 + 4 x − 7) = 0
−2 ± 39
≈ −1.65, 0.85
5
Intersection points:
⎛ −2 − 39 −1 − 3 39 ⎞
,
⎜
⎟ and
⎜
⎟
5
5
⎝
⎠
⎛ −2 + 39 −1 + 3 39 ⎞
,
⎜
⎟
⎜
⎟
5
5
⎝
⎠
[ or roughly (–1.65, –3.95) and (0.85, 3.55) ]
x=

35. x 2 + x 2 = 4
x2 = 2
x=± 2

(

)(

Intersection points: – 2, – 2 ,

2, 2

)

38. x 2 + (4 x + 3) 2 = 81
x 2 + 16 x 2 + 24 x + 9 = 81
17 x 2 + 24 x − 72 = 0
−12 ± 38
≈ −2.88, 1.47
17
⎛ −12 − 38 3 − 24 38 ⎞
,
⎜
⎟ and
⎜
⎟
17
17
⎝
⎠
⎛ −12 + 38 3 + 24 38 ⎞
,
⎜
⎟
⎜
⎟
17
17
⎝
⎠
[ or roughly ( −2.88, −8.52 ) , (1.47,8.88 ) ]
x=

36. 2 x 2 + 3( x − 1)2 = 12
2 x 2 + 3 x 2 − 6 x + 3 = 12
5x2 − 6 x − 9 = 0
6 ± 36 + 180 6 ± 6 6 3 ± 3 6
=
=
10
10
5
⎛ 3 − 3 6 −2 − 3 6 ⎞ ⎛ 3 + 3 6 −2 + 3 6 ⎞
⎜
⎟,⎜
⎟
⎜ 5 ,
⎟ ⎜ 5 ,
⎟
5
5
⎝
⎠ ⎝
⎠
x=

28

Section 0.4



39. a.

y = x 2 ; (2)

(

b.

ax3 + bx 2 + cx + d , with a > 0 : (1)

c.

ax3 + bx 2 + cx + d , with a < 0 : (3)

d.

y = ax3 , with a > 0 : (4)

40. x 2 + y 2 = 13;(−2, −3), (−2,3), (2, −3), (2,3)
2

2

2

2

2

d3 = (2 − 2) + (3 + 3) = 6
Three such distances.

(

)

)(

)(

)

(

)

d1 = (–2 – 2) 2 + ⎡1 + 21 – 1 + 13 ⎤
⎣
⎦
21 – 13

)

21 + 13

)

f (2u ) = 3(2u ) 2 = 12u 2 ; f ( x + h) = 3( x + h)2

)

2

=

f (0) = 1 – 02 = 1
f (k ) = 1 – k 2
f (–5) = 1 – (–5) 2 = –24

f.

1 15
⎛1⎞
⎛1⎞
f ⎜ ⎟ =1– ⎜ ⎟ =1– =
16 16
⎝4⎠
⎝4⎠

g.

f (1 + h ) = 1 − (1 + h ) = −2h − h 2

h.

f (1 + h ) − f (1) = −2h − h 2 − 0 = −2h − h 2

i.

( 2 21)

2

c.

e.

)

f (–2) = 1 – (–2)2 = –3

f ( 2 + h ) − f ( 2) = 1 − ( 2 + h ) + 3

2

2

= 2 21 ≈ 9.17
d 4 = (−2 − 2)2 + ⎡1 − 21 − (1 + 13) ⎤
⎣
⎦

(

f (1) = 1 – 12 = 0

d.

)

2

2

(

= 16 + − 21 − 13

)

2

2

2

2

= −4h − h 2

= 50 + 2 273 ≈ 9.11

(

)

d5 = (−2 − 2)2 + ⎡1 − 21 − 1 − 13 ⎤
⎣
⎦

(

2

b.

d3 = (−2 + 2)2 + ⎡1 + 21 − 1 − 21 ⎤
⎣
⎦
21 + 21

2.

1. a.

= 50 + 2 273 ≈ 9.11

= 16 +

( 2 13 )

0.5 Concepts Review

2

(

(

=

Problem Set 0.5

d 2 = (–2 – 2)2 + ⎡1 + 21 – 1 – 13 ⎤
⎣
⎦

= 0+

2

= 2 13 ≈ 7.21
Four such distances ( d 2 = d 4 and d1 = d5 ).

2

= 50 – 2 273 ≈ 4.12

(

)

4. even; odd; y-axis; origin

21 , 2, 1 + 13 , 2, 1 – 13

= 16 +

13 + 13

3. asymptote

41. x2 + 2x + y2 – 2y = 20; –2, 1 + 21 ,

(

(

1. domain; range

d 2 = (2 + 2) + (−3 − 3) = 52 = 2 13

= 16 +

= 0+

2

2

d1 = (2 + 2) + (−3 + 3) = 4

( –2, 1 –

)

d6 = (2 − 2)2 + ⎡1 + 13 − 1 − 13 ⎤
⎣
⎦

13 − 21

)

2

2

2. a.
b.

F ( 2) = ( 2)3 + 3( 2) = 2 2 + 3 2
=5 2

= 50 − 2 273 ≈ 4.12

3

c.


F (1) = 13 + 3 ⋅1 = 4

⎛1⎞ ⎛1⎞
⎛ 1 ⎞ 1 3 49
F ⎜ ⎟ = ⎜ ⎟ + 3⎜ ⎟ =
+ =
⎝4⎠ ⎝4⎠
⎝ 4 ⎠ 64 4 64

Section 0.5

29


d.

F (1 + h ) = (1 + h ) + 3 (1 + h )
3

f.

Φ ( x 2 + x) =

= 1 + 3h + 3h 2 + h3 + 3 + 3h
= 4 + 6h + 3h 2 + h3

e.

F (1 + h ) − 1 = 3 + 6h + 3h + h

f.

=

F ( 2 + h) − F ( 2)

2

3

2

= 15h + 6h + h

3. a.

b.

G (0) =

d.

e.

f.

4. a.

b.

0.25 − 3
1

f ( x) =

c.

f (3 + 2) =

3

1
= –1
0 –1

π −3

G( y ) =

G (– x) =

c.

1
1
=–
– x –1
x +1

7. a.

2

1
x
=
– 1 1 – x2

1
2

≈ 0.841
≈ −3.293

(12.26) 2 + 9
12.26 – 3

≈ 1.199

1

–t
1
2

⎛1⎞
Φ⎜ ⎟ =
⎝2⎠

d.

Φ (u + 1) =

e.

Φ( x2 ) =

+

(1)
2

=

2

1
2

=

x2

=

x2 + y2 = 1

c.

x = 2 y +1

x2 = 2 y + 1
≈ 1.06

(u + 1) + (u + 1) 2

( x2 ) + ( x2 )2

Section 0.5

–t

u +1

; undefined

b. xy + y + x = 1
y(x + 1) = 1 – x
1– x
1– x
y=
; f ( x) =
x +1
x +1

t2 – t

3
4
1
2

3– 3

y = ± 1 – x 2 ; not a function

=2

–t + (– t ) 2

( 3)2 + 9

f ( 3) =

y 2 = 1– x 2

1
x2

1 + 12

Φ (–t ) =

is not

y 2 –1

⎛ 1 ⎞
G⎜ ⎟ =
⎝ x2 ⎠

Φ (1) =

=

3+ 2 −3
−0.25

0.79 – 3

b. f(12.26) =

1

− 2.75

≈ 2.658

(0.79) 2 + 9

f(0.79) =

1
G (1.01) =
= 100
1.01 – 1
2

1

=

1

=2

1
G (0.999) =
= –1000
0.999 –1

c.

30

1

f (0.25) =

b.

6. a.
c.

x2 + x

defined

= ( 2 + h ) + 3 ( 2 + h ) − ⎡ 23 − 3 ( 2 ) ⎤
⎣
⎦
= 8 + 12h + 6h 2 + h3 + 6 + 3h − 14

x2 + x

x 4 + 2 x3 + 2 x 2 + x

3

5. a.

( x 2 + x) + ( x 2 + x) 2

=

y=

u 2 + 3u + 2

x2 + x4
x

u +1

d.

x2 – 1
x2 – 1
; f ( x) =
2
2

y
y+1
xy + x = y
x = y – xy
x = y(1 – x)
x
x
; f ( x) =
y=
1– x
1– x
x=



8. The graphs on the left are not graphs of
functions, the graphs on the right are graphs of
functions.
9.

f (a + h) – f (a) [2(a + h) 2 – 1] – (2a 2 – 1)
=
h
h
4ah + 2h 2
=
= 4a + 2h
h

x 2 – 9 ≥ 0; x 2 ≥ 9; x ≥ 3

Domain: {x ∈
d.

F (a + h) – F (a ) 4(a + h) – 4a
=
h
h

4a3 + 12a 2 h + 12ah 2 + 4h3 – 4a3
h
12a 2 h + 12ah 2 + 4h3
h

14. a.

b.

= 12a 2 + 12ah + 4h 2

f ( x) =

4 – x2

=

4 – x2
( x – 3)( x + 2)

x2 – x – 6
Domain: {x ∈ : x ≠ −2, 3}
G ( y ) = ( y + 1) –1

Domain: { y ∈

= x − 4 x + hx − 2h + 4
h
−3h
=
2
h( x − 4 x + hx − 2h + 4)
3
=–
2
x – 4 x + hx – 2h + 4

: y > −1}

c.

φ (u ) = 2u + 3
(all real numbers)
Domain:

d.

F (t ) = t 2 / 3 – 4
(all real numbers)
Domain:

2

a+h
a + h+ 4

: y ≤ 5}

1
≥ 0; y > –1
y +1

3
3
g ( x + h) – g ( x) x + h –2 – x –2
=
h
h
3x − 6 − 3x − 3h + 6

G ( a + h) – G ( a )
=
h

H ( y ) = – 625 – y 4

Domain: { y ∈

=

12.

: x ≥ 3}

3

=

11.

ψ ( x) = x 2 – 9

625 – y 4 ≥ 0; 625 ≥ y 4 ; y ≤ 5
3

10.

c.

15. f(x) = –4; f(–x) = –4; even function

a
– a+4

h

2

a + 4a + ah + 4h − a 2 − ah − 4a
a 2 + 8a + ah + 4h + 16
h
4h

=
=
=

13. a.

h(a 2 + 8a + ah + 4h + 16)
4
a 2 + 8a + ah + 4h + 16

F ( z) = 2 z + 3

2z + 3 ≥ 0; z ≥ –
⎧
Domain: ⎨ z ∈
⎩

b.

16. f(x) = 3x; f(–x) = –3x; odd function

g (v ) =

3
2

3⎫
:z≥− ⎬
2⎭

1
4v – 1

4v – 1 = 0; v =
⎧
Domain: ⎨v ∈
⎩

1
4
1⎫
:v≠ ⎬
4⎭


Section 0.5

31


17. F(x) = 2x + 1; F(–x) = –2x + 1; neither

20. g (u ) =

18. F ( x) = 3x – 2; F (– x) = –3x – 2; neither
21. g ( x) =

19. g ( x) = 3x 2 + 2 x – 1; g (– x) = 3 x 2 – 2 x – 1 ;
neither

32

Section 0.5

22. φ ( z ) =

u3
u3
; g (– u ) = – ; odd function
8
8

x
2

x –1

; g (– x) =

–x
2

x –1

; odd

2z +1
–2 z + 1
; φ (– z ) =
; neither
z –1
–z –1



23.

f ( w) = w – 1; f (– w) = – w – 1; neither

2

2

24. h( x ) = x + 4; h(– x) = x + 4; even
function

26. F (t ) = – t + 3 ; F (– t ) = – –t + 3 ; neither

27. g ( x) =

x
x
; g (− x ) = − ; neither
2
2

28. G ( x) = 2 x − 1 ; G (− x) = −2 x + 1 ; neither

25.

f ( x) = 2 x ; f (– x) = –2 x = 2 x ; even

function

⎧1
if t ≤ 0
⎪
29. g (t ) = ⎨t + 1 if 0 < t < 2
⎪2
⎩t – 1 if t ≥ 2


neither

Section 0.5

33


⎧ – x 2 + 4 if x ≤ 1
⎪
30. h( x ) = ⎨
if x > 1
⎪3x
⎩

neither

35. Let y denote the length of the other leg. Then
x2 + y 2 = h2
y 2 = h2 − x 2
y = h2 − x 2
L ( x ) = h2 − x 2

36. The area is
1
1
A ( x ) = base × height = x h 2 − x 2
2
2
37. a.

31. T(x) = 5000 + 805x
Domain: {x ∈ integers: 0 ≤ x ≤ 100}

T ( x) 5000
u ( x) =
=
+ 805
x
x
Domain: {x ∈ integers: 0 < x ≤ 100}
P ( x) = 6 x – (400 + 5 x( x – 4))

32. a.

= 6 x – 400 – 5 x( x – 4)

E(x) = 24 + 0.40x

b. 120 = 24 + 0.40x
0.40x = 96; x = 240 mi
38. The volume of the cylinder is πr 2 h, where h is
the height of the cylinder. From the figure,
2
2
2 ⎛ h⎞
2 h
2
= 3r ;
r + ⎜ ⎟ = (2r ) ;
⎝ 2⎠
4
h = 12r 2 = 2r 3.
V (r ) = πr 2 (2r 3) = 2πr 3 3

P(200) ≈ −190 ; P (1000 ) ≈ 610

b.

c. ABC breaks even when P(x) = 0;
6 x – 400 – 5 x( x – 4) = 0; x ≈ 390
33. E ( x) = x – x 2
y
0.5

39. The area of the two semicircular ends is
0.5

1

x

−0.5

1
exceeds its square by the maximum amount.
2

34. Each side has length

p
. The height of the
3

πd 2
.
4

1 – πd
.
2
2
πd 2
d – πd 2
⎛ 1 – πd ⎞ πd
A(d ) =
+d⎜
+
⎟=
4
4
2
⎝ 2 ⎠

The length of each parallel side is

2d – πd 2
4
Since the track is one mile long, π d < 1, so
1
1⎫
⎧
d < . Domain: ⎨d ∈ : 0 < d < ⎬
π
π⎭
⎩
=

3p
.
6
1 ⎛ p ⎞⎛ 3p ⎞
3 p2
A( p ) = ⎜ ⎟ ⎜
⎟=
2 ⎝ 3 ⎠⎜ 6 ⎟
36
⎝
⎠

triangle is

34

Section 0.5



40. a.

1
3
A(1) = 1(1) + (1)(2 − 1) =
2
2

42. a.

f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)

b.

f ( x + y ) = ( x + y )2 = x 2 + 2 xy + y 2
≠ f ( x) + f ( y )

c.

f(x + y) = 2(x + y) + 1 = 2x + 2y + 1
≠ f(x) + f(y)

d. f(x + y) = –3(x + y) = –3x – 3y = f(x) + f(y)

b.

1
A(2) = 2(1) + (2)(3 − 1) = 4
2

c.

A(0) = 0

d.

1
1
A(c) = c(1) + (c)(c + 1 − 1) = c 2 + c
2
2

e.

43. For any x, x + 0 = x, so
f(x) = f(x + 0) = f(x) + f(0), hence f(0) = 0.
Let m be the value of f(1). For p in N,
p = p ⋅1 = 1 + 1 + ... + 1, so
f(p) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1)
= pf(1) = pm.
⎛1⎞ 1 1
1
1 = p ⎜ ⎟ = + + ... + , so
p
⎝ p⎠ p p
⎛1 1
1⎞
m = f (1) = f ⎜ + + ... + ⎟
p p
p⎠
⎝
⎛1⎞
⎛1⎞
⎛1⎞
⎛1⎞
= f ⎜ ⎟ + f ⎜ ⎟ + ... + f ⎜ ⎟ = pf ⎜ ⎟ ,
⎝ p⎠
⎝ p⎠
⎝ p⎠
⎝ p⎠
⎛1⎞ m
hence f ⎜ ⎟ = . Any rational number can
⎝ p⎠ p

be written as

f.

Domain: {c ∈
Range: { y ∈

41. a.
b.

: c ≥ 0}
: y ≥ 0}

B (0) = 0
1 1 1
⎛1⎞ 1
B ⎜ ⎟ = B (1) = ⋅ =
2 6 12
⎝2⎠ 2

p
with p, q in N.
q

⎛1⎞ 1 1
p
1
= p ⎜ ⎟ = + + ... + ,
q
q
⎝q⎠ q q
⎛ p⎞
⎛1 1
1⎞
so f ⎜ ⎟ = f ⎜ + + ... + ⎟
q⎠
q q
q⎠
⎝
⎝
⎛1⎞
⎛1⎞
⎛1⎞
= f ⎜ ⎟ + f ⎜ ⎟ + ... + f ⎜ ⎟
⎝q⎠
⎝q⎠
⎝q⎠
⎛1⎞
⎛m⎞
⎛ p⎞
= pf ⎜ ⎟ = p ⎜ ⎟ = m ⎜ ⎟
⎝q⎠
⎝q⎠
⎝q⎠

c.


Section 0.5

35


44. The player has run 10t feet after t seconds. He
reaches first base when t = 9, second base when
t = 18, third base when t = 27, and home plate
when t = 36. The player is 10t – 90 feet from
first base when 9 ≤ t ≤ 18, hence

46. a.

x

f(x)

–4

–6.1902

–3

0.4118

–2

13.7651

–1

9.9579

0

0

1

–7.3369

2

–17.7388

if 18 < t ≤ 27

3

–0.4521

if 27 < t ≤ 36

4

4.4378

b.

902 + (10t − 90)2 feet from home plate. The
player is 10t – 180 feet from second base when
18 ≤ t ≤ 27, thus he is
90 – (10t – 180) = 270 – 10t feet from third base

and 902 + (270 − 10t ) 2 feet from home plate.
The player is 10t – 270 feet from third base
when 27 ≤ t ≤ 36, thus he is
90 – (10t – 270) = 360 – 10t feet from home
plate.

a.

b.

45. a.

f(1.38) ≈ –76.8204
f(4.12) ≈ 6.7508

⎧10t
⎪
⎪ 902 + (10t − 90)2
s=⎨
⎪ 902 + (270 − 10t ) 2
⎪
⎪360 – 10t
⎩

if 0 ≤ t ≤ 9

⎧180 − 180 − 10t
⎪
⎪
⎪
s = ⎨ 902 + (10t − 90) 2
⎪
2
2
⎪ 90 + (270 − 10t )
⎪
⎪
⎩

if 0 ≤ t ≤ 9

if 9 < t ≤ 18

or 27 < t ≤ 36

47.

if 9 < t ≤ 18
if 18 < t ≤ 27

f(1.38) ≈ 0.2994
f(4.12) ≈ 3.6852
x

f(x)

–4

–4.05

–3

–3.1538

a.

–2

–2.375

b. f(x) = 0 when x ≈ –1.1, 1.7, 4.3
f(x) ≥ 0 on [–1.1, 1.7] ∪ [4.3, 5]

–1

–1.8

0

–1.25

1

–0.2

2

1.125

3

2.3846

4

b.

3.55

48.

a.
36

Section 0.5

Range: {y ∈ R: –22 ≤ y ≤ 13}

f(x) = g(x) at x ≈ –0.6, 3.0, 4.6


b. f(x) ≥ g(x) on [-0.6, 3.0] ∪ [4.6, 5]
c.

f ( x) – g ( x)
= x3 – 5 x 2 + x + 8 – 2 x 2 + 8 x + 1

= x3 – 7 x 2 + 9 x + 9

b. On ⎡ −6, −3) , g increases from
⎣
13
g ( −6 ) = ≈ 4.3333 to ∞ . On ( 2, 6⎤ , g
⎦
3
26
≈ 2.8889 . On
decreased from ∞ to
9

( −3, 2 ) the maximum occurs around

Largest value f (–2) – g (–2) = 45

x = 0.1451 with value 0.6748 . Thus, the
range is ( −∞, 0.6748⎤ ∪ ⎡ 2.8889, ∞ ) .
⎦ ⎣

49.
c.

x 2 + x – 6 = 0; (x + 3)(x – 2) = 0
Vertical asymptotes at x = –3, x = 2

d. Horizontal asymptote at y = 3

0.6 Concepts Review
1. ( x 2 + 1)3
a.

x-intercept: 3x – 4 = 0; x =

4
3

3⋅ 0 – 4

2
=
y-intercept:
2
0 +0–6 3

2. f(g(x))
3. 2; left
4. a quotient of two polynomial functions

b.
c.

x 2 + x – 6 = 0; (x + 3)(x – 2) = 0
Vertical asymptotes at x = –3, x = 2

Problem Set 0.6
1. a.

( f + g )(2) = (2 + 3) + 22 = 9

d. Horizontal asymptote at y = 0
b.

( f g )(1) = f (12 ) = 1 + 3 = 4

e.

( g f )(1) = g (1 + 3) = 4 2 = 16

f.

( g f )(–8) = g (–8 + 3) = (–5) 2 = 25

2. a.

x-intercepts:
3x 2 – 4 = 0; x = ±

( g f )(3) =

d.

a.

( f ⋅ g )(0) = (0 + 3)(02 ) = 0

c.

50.

4
2 3
=±
3
3

32
9 3
= =
3+3 6 2

( f – g )(2) = (22 + 2) –

12 + 1

2

b.

( f g )(1) =

c.

2
y-intercept:
3

1
⎡ 2 ⎤
⎛1⎞
g 2 (3) = ⎢
=⎜ ⎟ =
3 + 3⎥
3⎠
9
⎣
⎦
⎝

2
1+ 3
2


=

2
2 28
=6– =
2+3
5 5

2
4

=4

2

Section 0.6

37


2

d.

e.

f.

3. a.

(f

⎛ 2 ⎞ ⎛1⎞ 1 3
g )(1) = f ⎜
⎟=⎜ ⎟ + =
⎝1+ 3 ⎠ ⎝ 2 ⎠ 2 4

5.

= x2 + 2 x – 3

2
2
=
2+3 5

⎛ 2 ⎞
( g g )(3) = g ⎜
⎟=
⎝ 3+3⎠

2
2 3
= 10 =
1
+3 3 5
3

6. g 3 (x) = (x 2 +1) 3 = (x 4 + 2x 2 + 1)(x 2 + 1)
6

1
t

b.

c.

(Ψ Φ )(r ) = Ψ (r + 1) =

d.

Φ 3 ( z ) = ( z 3 + 1) 3

e.

(Φ – Ψ )(5t) = [(5t) 3 +1] –

1

3

b.

2

= g[(x 2 + 1) 2 + 1] = g( x 4 + 2x 2 + 2)
= ( x 4 + 2x 2 + 2) 2 + 1
= x 8 + 4x 6 + 8x 4 + 8x 2 + 5

7. g(3.141) ≈ 1.188
8. g(2.03) ≈ 0.000205

r 3 +1

1/ 3

9. ⎡ g 2 (π ) − g (π ) ⎤
⎣
⎦
≈ 4.789

1
5t

1/ 3

2
= ⎡(11 − 7π ) − 11 − 7π ⎤
⎢
⎥
⎣
⎦

10. [ g 3 (π) – g (π)]1/ 3 = [(6π – 11)3 – (6π – 11)]1/ 3
≈ 7.807

1
5t

⎛1⎞
((Φ – Ψ ) Ψ )(t ) = (Φ – Ψ )⎜ ⎟
⎝t⎠
3
1 1
⎛ 1⎞
= ⎜ ⎟ + 1– 1 = 3 + 1 – t
⎝ t⎠
t
t

11. a.
b.
12. a.

4. a.

4

= x + 3x + 3x + 1
( g g g )( x) = ( g g )( x 2 + 1)

3

f.

⎛
⎞
f ) ( x) = g ⎜ x 2 − 4 ⎟ = 1 + x 2 − 4
⎝
⎠

= 1 + x2 – 4

1
⎛1⎞ ⎛1⎞
(Φ Ψ )(r ) = Φ⎜ ⎟ = ⎜ ⎟ + 1 = 3 + 1
r⎠ ⎝r⎠
r
⎝

= 125t3 + 1 –

2

(g

( g f )(1) = g (12 + 1) =

(Φ + Ψ )(t ) = t 3 + 1 +

g ) ( x) = f ( 1 + x ) = 1 + x − 4

(f

g ( x) = x , f ( x) = x + 7

g (x) = x15 , f (x) = x 2 + x
2

f ( x) =

x

2 x2 – 1
x
Domain: (– ∞, – 1] ∪ [1, ∞)

3

, g ( x) = x 2 + x + 1

( f ⋅ g )( x) =

b.

4
⎛2⎞
f 4 ( x) + g 4 ( x) = ⎛ x 2 – 1 ⎞ + ⎜ ⎟
⎜
⎟
⎝
⎠ ⎝x⎠
16
= ( x 2 – 1)2 +
x4
Domain: (– ∞, 0 ) ∪ (0, ∞ )
2

c.

⎛2⎞
⎛2⎞
g )( x) = f ⎜ ⎟ = ⎜ ⎟ – 1 =
⎝ x⎠
⎝ x⎠
Domain: [–2, 0) ∪ (0, 2]

d.

( g f )( x) = g ⎛ x 2 – 1 ⎞ =
⎜
⎟
⎝
⎠

(f

4

f ( x) =

13. p = f

1
, g (x) = x 3 + 3 x
x
g h if f(x) =1/ x , g ( x) = x ,

h( x ) = x 2 + 1
p= f

g h if f ( x) = 1/ x , g(x) = x + 1,

h( x) = x 2

4
–1
x2

14. p = f g h l if f ( x) = 1/ x , g ( x) = x ,
2
h(x) = x + 1, l( x) = x

2
2

x –1
Domain: (– ∞ , –1) ∪ (1, ∞ )

38

Section 0.6



15. Translate the graph of g ( x) = x to the right 2
units and down 3 units.

17. Translate the graph of y = x 2 to the right 2

18. Translate the graph of y = x 3 to the left 1 unit
and down 3 units.

16. Translate the graph of h( x) = x to the left 3

19. ( f + g )( x) =

x–3
+ x
2

20. ( f + g )( x) = x + x


Section 0.6

39


21. F (t ) =

t –t

24. a.

t

F(x) – F(–x) is odd because
F(–x) – F(x) = –[F(x) – F(–x)]

b. F(x) + F(–x) is even because
F(–x) + F(–(–x)) = F(–x) + F(x)
= F(x) + F(–x)
c.

22. G (t ) = t − t

F ( x ) – F (– x)
F ( x ) + F (– x)
is odd and
is
2
2
even.
F ( x ) − F (− x) F ( x) + F (− x) 2 F ( x)
+
=
= F ( x)
2
2
2

25. Not every polynomial of even degree is an
even function. For example f ( x) = x 2 + x is
neither even nor odd. Not every polynomial of
odd degree is an odd function. For example
g ( x) = x 3 + x 2 is neither even nor odd.
26. a.

Neither

b.
c.

b. Odd;
(f + g)(–x) = f(–x) + g(–x) = –f(x) – g(x)
= –(f + g)(x) if f and g are both odd
functions.
c.

Even;
( f ⋅ g )(− x) = [ f (− x)][ g (− x)]
= [ f ( x)][ g ( x)] = ( f ⋅ g )( x)
if f and g are both even functions.

d. Even;
( f ⋅ g )(− x) = [ f (− x)][ g (− x)]
= [− f ( x)][− g ( x)] = [ f ( x)][ g ( x)]
= ( f ⋅ g )( x)
if f and g are both odd functions.
e.

40

PF

e.

Even;
(f + g)(–x) = f(–x) + g(–x) = f(x) + g(x)
= (f + g)(x) if f and g are both even
functions.

RF

d.

23. a.

PF

RF

f.

Neither

27. a.

P = 29 – 3(2 + t ) + (2 + t )2
= t + t + 27

b. When t = 15, P = 15 + 15 + 27 ≈ 6.773
28. R(t) = (120 + 2t + 3t2 )(6000 + 700t )
= 2100 t3 + 19, 400t 2 + 96, 000t + 720, 000

⎧400t
⎪
29. D(t ) = ⎨
2
2
⎪ (400t ) + [300(t − 1)]
⎩

if 0 < t < 1
if t ≥ 1

if 0 < t < 1
⎧ 400t
⎪
D(t ) = ⎨
2
⎪ 250, 000t − 180, 000t + 90, 000 if t ≥ 1
⎩

30. D(2.5) ≈ 1097 mi

Odd;
( f ⋅ g )(− x) = [ f (− x)][ g (− x)]
= [ f ( x)][− g ( x)] = −[ f ( x)][ g ( x)]
= −( f ⋅ g )( x)
if f is an even function and g is an odd
function.

Section 0.6



31.

(ax–+ab ) + b
cx
(ax–+ab ) – a
cx

36.

⎛ ax + b ⎞ a
f ( f ( x)) = f ⎜
⎟=
⎝ cx – a ⎠ c
=

a 2 x + ab + bcx – ab

=

x(a 2 + bc)

=x
acx + bc – acx + a 2
a 2 + bc
2
If a + bc = 0 , f(f(x)) is undefined, while if
x = a , f(x) is undefined.
c
⎛ x –3 – 3 ⎞
⎛ ⎛ x – 3⎞⎞
x +1
⎟
( f ( f ( x))) = f ⎜ f ⎜
32. f
⎟ ⎟ = f ⎜ x –3
⎜
⎟
x +1 ⎠ ⎠
+1 ⎠
⎝ ⎝
⎝ x +1
⎛ x – 3 – 3x – 3 ⎞
⎛ –2 x – 6 ⎞
⎛ –x – 3 ⎞
= f⎜
⎟= f ⎜
⎟= f ⎜
⎟
⎝ x – 3 + x +1 ⎠
⎝ 2x – 2 ⎠
⎝ x –1 ⎠
– x–3
– 3 – x – 3 – 3x + 3 – 4 x
1
= –xx–– 3
=
=
=x
– x – 3+ x –1
–4
+1
x –1

If x = –1, f(x) is undefined, while if x = 1,
f(f(x)) is undefined.
33. a.

b.

⎛1⎞
f⎜ ⎟=
⎝ x⎠

34. a.
b.

1
x

–1

=

1
1– x

1
f1 ( f 2 ( x)) = ;
x
f1 ( f3 ( x)) = 1 − x;
1
;
1− x
x −1
f1 ( f5 ( x)) =
;
x
x
;
f1 ( f6 ( x)) =
x −1

f1 ( f 4 ( x)) =

f 2 ( f1 ( x)) =
f 2 ( f 2 ( x)) =

x
x –1
x
–1
x –1

= x;

1
;
1− x
1
f 2 ( f 4 ( x)) =
= 1 − x;

f 2 ( f 6 ( x)) =

x
=x
x – x +1

=

1
x
x −1

x
;
x –1

=

1
x −1
x

x –1
;
x

f3 ( f1 ( x)) = 1 − x;

⎛ 1 ⎞
⎛ x – 1⎞
f⎜
⎜ f ( x) ⎟ = f ⎜ x ⎟ =
⎟
⎝
⎠
⎝
⎠
=1–x
1/ x
1 / x −1

x –1
x
x –1
–1
x

=

x –1
x –1– x

1 x −1
;
=
x
x
f3 ( f3 ( x)) = 1 – (1 – x) = x;

f3 ( f 2 ( x)) = 1 −

1
x
=
;
1 – x x –1
x –1 1
= ;
f3 ( f5 ( x)) = 1 –
x
x
x
1
f3 ( f 6 ( x)) = 1 –
=
;
x –1 1– x
f3 ( f 4 ( x)) = 1 –

1

=

x−x

f ( f ( x)) = f ( x /( x − 1)) =

=

1
x

f 2 ( f3 ( x)) =

f 2 ( f 5 ( x )) =

f (1 / x) =

1
;
x
1

1
1− x

⎛ x ⎞
f ( f ( x)) = f ⎜
⎟=
⎝ x – 1⎠

=

c.

1
x

f1 ( f1 ( x)) = x;

x /( x − 1)

x
x( x − 1) + 1 − x

35. ( f1 ( f 2 f3 ))( x) = f1 (( f 2 f3 )( x))
= f1 ( f 2 ( f3 ( x)))
(( f1 f 2 ) f3 )( x) = ( f1 f 2 )( f3 ( x))
= f1 ( f 2 ( f3 ( x)))
= ( f1 ( f 2 f3 ))( x)

x
−1
x −1

1
;
1− x
1
x
;
f 4 ( f 2 ( x)) =
=
1
1− x x −1
f 4 ( f1 ( x)) =

f 4 ( f3 ( x)) =
f 4 ( f 4 ( x)) =
f 4 ( f5 ( x)) =

f 4 ( f 6 ( x)) =


1
1
= ;
1 – (1 – x) x
1
1
1 – 1– x

1
1–

x –1
x

=

1− x
x –1
=
;
x
1− x −1

=

x
= x;
x − ( x − 1)

1
x −1
=
= 1 – x;
x
1 – x –1 x − 1 − x
Section 0.6

41


a.

x −1
f5 ( f1 ( x)) =
;
x
1 −1
f5 ( f 2 ( x)) = x
= 1 − x;

b.

=

f 5 ( f 5 ( x)) =

x –1
–1
x
x –1
x

f 5 ( f 6 ( x)) =

x
–1
x –1
x
x –1

=

1 − (1 − x)
= x;
1

f3

f4

f5

f6

f3 )

f4 )

f5 )

f6 )

= ( f 4 f 4 ) ( f5 f 6 )
= f5 f 2 = f3
c.

x − ( x − 1) 1
= ;
x
x

If F

f 6 = f 1 , then F = f 6 .

d. If G f 3
G = f5.
e.

1
;
1– x

f 6 = f 1 , then G f 4 = f 1 so

If f 2 f 5 H = f 5 , then f 6 H = f 5 so
H = f3.

37.

f 6 ( f3 ( x)) =

x –1
1– x
;
=
1– x –1
x

f 6 ( f 4 ( x)) =

1
1– x
1
–1
1– x

=

1
1
= ;
1 − (1 − x) x

f 6 ( f 5 ( x)) =

x –1
x
x –1
–1
x

=

x −1
= 1 – x;
x −1− x

f 6 ( f 6 ( x )) =

x
x –1
x
–1
x –1

=

x
=x
x − ( x − 1)

38.

f1

f2

f3

f4

f5

f6

f1

f1

f2

f3

f4

f5

f6

f2

f2

f1

f4

f3

f6

f5

f3

f3

f5

f1

f6

f2

f4

f4

f4

f6

f2

f5

f1

f3

f5

f5

f3

f6

f1

f4

f2

f6

42

f1 f 2
= (((( f 2

x
;
x –1
=

f3 ) f3 ) f3 ) f3 )

= ((((( f 1 f 2 ) f 3 ) f 4 ) f 5 ) f 6 )

x −1− x
1
=
=
;
x −1
1– x

–1

f3

= f1 f 3 = f 3

1
–1
1– x
1
1– x

1
x

f3

= (( f3 f3 ) f3 )

f 5 ( f 4 ( x)) =

f 6 ( f 2 ( x)) =

f3

= ((( f1 f3 ) f3 ) f3 )

1 – x –1
x
f5 ( f3 ( x)) =
=
;
1– x
x –1

1
x

f3

= (((( f 3

1
x

f 6 ( f1 ( x)) =

f3

f6

f4

f5

f2

f3

f1

Section 0.6

39.



Problem Set 0.7

40.

1. a.

b.

π
⎛ π ⎞
–60 ⎜
⎟=–
3
⎝ 180 ⎠

d.

⎛ π ⎞ 4π
240 ⎜
⎟=
⎝ 180 ⎠ 3

e.

37 π
⎛ π ⎞
–370 ⎜
⎟=–
18
⎝ 180 ⎠

f.

b.

⎛ π ⎞ π
45 ⎜
⎟=
⎝ 180 ⎠ 4

c.

41. a.

⎛ π ⎞ π
30 ⎜
⎟=
⎝ 180 ⎠ 6

⎛ π ⎞ π
10 ⎜
⎟=
⎝ 180 ⎠ 18

2. a.

7 ⎛ 180 ⎞
⎟ = 210°
π⎜
6 ⎝ π ⎠
⎛ 180 ⎟
⎞
π⎜
= 135°
⎝ π ⎠

b.
c.

1 ⎛ 180 ⎞
⎟ = –60°
– π⎜
3 ⎝ π ⎠

d.

4 ⎛ 180 ⎞
⎟ = 240°
π⎜
3 ⎝ π ⎠

e.

–

f.

c.

3
4

3 ⎛ 180 ⎞
π⎜
⎟ = 30°
18 ⎝ π ⎠

42.
3. a.

35 ⎛ 180 ⎞
⎟ = –350°
π⎜
18 ⎝ π ⎠

⎛ π ⎞
33.3 ⎜
⎟ ≈ 0.5812
⎝ 180 ⎠

b.

c.

⎛ π ⎞
–66.6 ⎜
⎟ ≈ –1.1624
⎝ 180 ⎠

d.

⎛ π ⎞
240.11⎜
⎟ ≈ 4.1907
⎝ 180 ⎠

e.

0.7 Concepts Review

⎛ π ⎞
46 ⎜
⎟ ≈ 0.8029
⎝ 180 ⎠

⎛ π ⎞
–369 ⎜
⎟ ≈ –6.4403
⎝ 180 ⎠

f.

⎛ π ⎞
11⎜
⎟ ≈ 0.1920
⎝ 180 ⎠

1. (– ∞ , ∞ ); [–1, 1]
2. 2 π ; 2 π ; π
3. odd; even

4
x
4. r = (–4) + 3 = 5; cos θ = = –
5
r
2

2


Section 0.7

43


4. a.

⎛ 180 ⎟
⎞
3.141⎜
≈ 180°
⎝ π ⎠

Thus

b.

⎛ 180 ⎟
⎞
6. 28⎜
≈ 359. 8°
⎝ π ⎠

c.

⎛ 180 ⎟
⎞
≈ 286.5°
5. 00⎜
⎝ π ⎠

d.

⎛ 180 ⎟
⎞
0. 001⎜
≈ 0 .057°
⎝ π ⎠

e.

⎛ 180 ⎟
⎞
–0.1⎜
≈ –5.73°
⎝ π ⎠

f.

⎛ 180 ⎟
⎞
36. 0⎜
≈ 2062.6 °
⎝ π ⎠

5. a.

56. 4 tan34. 2°
≈ 68.37
sin 34.1°

b.

cos

= cos

π

=

and by the Pythagorean Identity, sin

3

=

3
.
2

sin (–0.361) ≈ –0.3532

6. a.
b.
7. a.

234.1sin(1.56)
≈ 248.3
cos(0.34 )
sin 2 (2.51) + cos(0.51) ≈ 1.2828
56. 3 tan34. 2°
≈ 46.097
sin 56.1°

Referring to Figure 2, it is clear that sin

3

b.

sin 35°
⎛
⎞
⎜
⎟ ≈ 0. 0789
⎝ sin 26° + cos 26° ⎠

Identity, cos 2

π

= 1 − sin 2

π

π

9. a.

=1

2

3
⎛1⎞
= 1− ⎜ ⎟ = .
6
4
⎝2⎠

1
= –1
cos(π)

b.

sec(π) =

1
⎛ 3π ⎞
sec ⎜ ⎟ =
=– 2
4 ⎠ cos 3π
⎝
4

d.
Section 0.7

2

⎛π ⎞
sin ⎜ ⎟
⎛π⎞
⎝6⎠ = 3
tan ⎜ ⎟ =
3
⎝ 6 ⎠ cos ⎛ π ⎞
⎜ ⎟
6⎠
⎝

c.

6

π

= 0 . The rest of the values are
2
obtained using the same kind of reasoning in
the second quadrant.

and cos

8. Referring to Figure 2, it is clear that
sin 0 = 0 and cos 0 = 1 . If the angle is π / 6 ,
then the triangle in the figure below is
1
1
equilateral. Thus, PQ = OP = . This
2
2
π 1
implies that sin = . By the Pythagorean
6 2

44

π

tan (0.452) ≈ 0.4855

d.

6

π

3
. The results
2

=

2
were derived in the text.
4
4
2
If the angle is π / 3 then the triangle in the
π 1
figure below is equilateral. Thus cos =
3 2
sin

5.34 tan 21.3°
≈ 0.8845
sin 3.1°+ cot 23.5°

c.

π

1
⎛π⎞
csc ⎜ ⎟ =
=1
2 ⎠ sin π
⎝

( )

(2)



e.

f.

10. a.

b.

( )
( )

b. cos 3t = cos(2t + t ) = cos 2t cos t – sin 2t sin t

π
⎛ π ⎞ cos 4
cot ⎜ ⎟ =
=1
⎝ 4 ⎠ sin π
4

= (2 cos 2 t – 1) cos t – 2sin 2 t cos t
= 2 cos3 t – cos t – 2(1 – cos 2 t ) cos t

( )
( )

π
⎛ π ⎞ sin – 4
tan ⎜ – ⎟ =
= –1
⎝ 4 ⎠ cos – π
4

( )
( )

= 2 cos3 t – cos t – 2 cos t + 2 cos3 t

= 4 cos3 t – 3cos t
c.

π
⎛ π ⎞ sin 3
tan ⎜ ⎟ =
= 3
⎝ 3 ⎠ cos π
3

= 2(2sin x cos x)(2 cos 2 x –1)
= 2(4sin x cos3 x – 2sin x cos x)
= 8sin x cos3 x – 4sin x cos x

1
⎛π⎞
sec ⎜ ⎟ =
=2
⎝ 3 ⎠ cos π
3

( )

d.

( )
( )

c.

π
3
⎛ π ⎞ cos 3
cot ⎜ ⎟ =
=
3
⎝ 3 ⎠ sin π
3

d.

1
⎛π⎞
csc ⎜ ⎟ =
= 2
⎝ 4 ⎠ sin π

e.

π
3
⎛ π ⎞ sin – 6
tan ⎜ – ⎟ =
=–
6 ⎠ cos – π
3
⎝
6

f.

⎛ π⎞ 1
cos ⎜ – ⎟ =
⎝ 3⎠ 2

13. a.
b.

(4)

( )
( )

c.

d.
11. a.

(1 + sin z )(1 – sin z ) = 1 – sin 2 z
1

= cos 2 z =

sin 4 x = sin[2(2 x)] = 2sin 2 x cos 2 x

(1 + cos θ )(1 − cos θ ) = 1 − cos 2 θ = sin 2 θ
sin u cos u
+
= sin 2 u + cos 2 u = 1
csc u sec u
(1 − cos 2 x)(1 + cot 2 x) = (sin 2 x)(csc2 x)
2 ⎛ 1 ⎞
= sin x ⎜ 2 ⎟ = 1
⎝ sin x ⎠
⎛ 1
⎞
sin t (csc t – sin t ) = sin t ⎜
– sin t ⎟
⎝ sin t
⎠
2
2
= 1– sin t = cos t
1 – csc 2 t
csc 2 t

= – cos 2 t = –

2

sec z

b.

(sec t –1)(sec t + 1) = sec 2 t –1 = tan 2 t

c.

sec t – sin t tan t =
=

d.

12. a.

=–

14. a.

cot 2 t
csc 2 t

=–

cos 2 t
sin 2 t
1
sin 2 t

1
sec 2 t

y = sin 2x

1
sin 2 t
–
cos t cos t

1 – sin 2 t cos 2 t
=
= cos t
cos t
cos t

sec2 t – 1
sec 2 t
sin 2 v +

=

tan 2 t
sec 2 t

1
2

=

sin 2 t
cos 2 t
1
cos 2 t

= sin 2 t

= sin 2 v + cos 2 v = 1

sec v


Section 0.7

45


b.

y = 2 sin t

b. y = 2 cos t

c.

π⎞
⎛
y = cos ⎜ x − ⎟
4⎠
⎝

c.

y = cos 3t

d.

y = sec t

d.

⎛ π⎞
y = cos ⎜ t + ⎟
⎝ 3⎠

15. a.

y = csc t

46

Section 0.7

x
2
Period = 4π , amplitude = 3

16. y = 3 cos



17. y = 2 sin 2x
Period = π , amplitude = 2

21. y = 21 + 7 sin( 2 x + 3)
Period = π , amplitude = 7, shift: 21 units up,
3
units left
2

π⎞
⎛
22. y = 3cos ⎜ x – ⎟ – 1
2⎠
⎝

18. y = tan x
Period = π

Period = 2 π , amplitude = 3, shifts:

π
units
2

right and 1 unit down.

19. y = 2 +

1
cot(2 x)
6

Period =

π
2

, shift: 2 units up

π⎞
⎛
23. y = tan ⎜ 2 x – ⎟
⎝
3⎠
π
π
units right
Period = , shift:
6
2

20. y = 3 + sec( x − π )
Period = 2π , shift: 3 units up, π units right


Section 0.7

47


π⎞
⎛
24. a. and g.: y = sin ⎜ x + ⎟ = cos x = – cos(π – x)
2⎠
⎝
π⎞
⎛
b. and e.: y = cos ⎜ x + ⎟ = sin( x + π)
2⎠
⎝
= − sin(π − x )
π⎞
⎛
c. and f.: y = cos ⎜ x − ⎟ = sin x
2⎠
⎝
= − sin( x + π)
π⎞
⎛
d. and h.: y = sin ⎜ x − ⎟ = cos( x + π)
2⎠
⎝
= cos( x − π)

–t sin (–t) = t sin t; even

25. a.
b.
c.

1
csc(– t ) =
= – csc t; odd
sin(– t )

( π)

=

32. a.

sin(cos(–t)) = sin(cos t); even

f.

–x + sin(–x) = –x – sin x = –(x + sin x); odd
cot(–t) + sin(–t) = –cot t – sin t
= –(cot t + sin t); odd

26. a.

b.

sin 3 (–t ) = – sin 3 t ; odd

c.

sec(– t) =

1
= sec t; even
cos(–t )

sin 4 (– t ) = sin 4 t ; even

d.
e.

cos(sin(–t)) = cos(–sin t) = cos(sin t); even

f.

(– x )2 + sin(– x ) = x 2 – sin x; neither
2

27. cos 2

π ⎛
π ⎞ ⎛1⎞
1
= ⎜ cos ⎟ = ⎜ ⎟ =
3 ⎝
3⎠
4
⎝2⎠
2

28. sin 2

2

2

π ⎛ π⎞
1
⎛1⎞
= ⎜ sin ⎟ = ⎜ ⎟ =
6 ⎝
6⎠
4
⎝2⎠
3

2

sin(x – y) = sin x cos(–y) + cos x sin(–y)
= sin x cos y – cos x sin y

b.

cos(x – y) = cos x cos(–y) – sin x sin (–y)
= cos x cos y + sin x sin y

c.

tan( x – y ) =

2

e.

π

3

2– 2
4

=

sin(−t ) = – sin t = sin t ; even

d.

π

1 – cos 4 1 – 2
π 1 – cos 2 8
31. sin
=
=
=
8
2
2
2
2

sin (– t ) = sin t ; even
2

(π)

1 + cos 6 1 + 2
π 1 + cos 2 12
=
=
=
30. cos
12
2
2
2
2+ 3
=
4
2

tan x + tan(– y )
1 – tan x tan(– y )
tan x – tan y
1 + tan x tan y

tan t + tan π
tan t + 0
=
1 – tan t tan π 1 – (tan t )(0)
= tan t

33. tan(t + π) =

34. cos( x − π ) = cos x cos(−π ) − sin x sin(−π )
= –cos x – 0 · sin x = –cos x
35. s = rt = (2.5 ft)( 2π rad) = 5π ft, so the tire
goes 5π feet per revolution, or 1 revolutions
5π
per foot.
ft ⎞
⎛ 1 rev ⎞ ⎛ mi ⎞ ⎛ 1 hr ⎞ ⎛
⎜
⎟ ⎜ 60 ⎟ ⎜
⎟ ⎜ 5280 ⎟
5π ft ⎠ ⎝ hr ⎠ ⎝ 60 min ⎠ ⎝
mi ⎠
⎝
≈ 336 rev/min
36. s = rt = (2 ft)(150 rev)( 2π rad/rev) ≈ 1885 ft
37. r1t1 = r2 t2 ; 6(2π)t1 = 8(2π)(21)
t1 = 28 rev/sec
38. Δy = sin α and Δx = cos α
Δy sin α
m=
=
= tan α
Δx cos α

3

1
π⎞ ⎛1⎞
3 π ⎛
29. sin 6 = ⎜ sin 6 ⎟ = ⎜ 2 ⎟ = 8
⎝
⎠ ⎝ ⎠

48

Section 0.7



39. a.

b.

tan α = 3
π
α=
3
3x + 3 y = 6
3 y = – 3x + 6
y=–

3
3
x + 2; m = –
3
3
3
3

tan α = –

α=

44. Divide the polygon into n isosceles triangles by
drawing lines from the center of the circle to
the corners of the polygon. If the base of each
triangle is on the perimeter of the polygon, then
2π
.
the angle opposite each base has measure
n
Bisect this angle to divide the triangle into two
right triangles (See figure).

5π
6

40. m1 = tan θ1 and m2 = tan θ 2
tan θ 2 + tan(−θ1 )
tan θ = tan(θ 2 − θ1 ) =
1 − tan θ 2 tan(−θ1 )
=

tan θ 2 − tan θ1
m − m1
= 2
1 + tan θ 2 tan θ1 1 + m1m2

π b
π
π h
=
so b = 2r sin and cos = so
n 2r
n
n r
π
h = r cos .
n
π
P = nb = 2rn sin
n
π
π
⎛1 ⎞
A = n ⎜ bh ⎟ = nr 2 cos sin
n
n
⎝2 ⎠
sin

3–2
1
=
1 + 3(2 ) 7
θ ≈ 0.1419

41. a.

tan θ =

b.

tan θ =

–1 – 1
2
1+

( 1 ) (–1)
2

= –3

θ ≈ 1.8925
c.

2x – 6y = 12
2x + y = 0
–6y = –2x + 12y = –2x
1
y= x–2
3
1
m1 = , m2 = –2
3
–2 – 1
3
= –7; θ ≈ 1.7127
tan θ =
1 + 1 (–2)
3

()

42. Recall that the area of the circle is π r 2 . The
measure of the vertex angle of the circle is 2π .
Observe that the ratios of the vertex angles
must equal the ratios of the areas. Thus,
t
A
=
, so
2π π r 2
1
A = r 2t .
2
43. A =

45. The base of the triangle is the side opposite the
t
angle t. Then the base has length 2r sin
2
(similar to Problem 44). The radius of the
t
semicircle is r sin and the height of the
2
t
triangle is r cos .
2
A=

1⎛
t ⎞⎛
t ⎞ π⎛
t⎞
⎜ 2r sin ⎟⎜ r cos ⎟ + ⎜ r sin ⎟
2⎝
2 ⎠⎝
2⎠ 2⎝
2⎠

2

t
t πr 2
t
= r 2 sin cos +
sin 2
2
2
2
2

1
(2)(5) 2 = 25cm 2
2


Section 0.7

49


x
x
x
x
46. cos cos cos cos
2
4
8
16
1⎡
3
1 ⎤1 ⎡
3
1 ⎤
= ⎢cos x + cos x ⎥ ⎢cos x + cos x ⎥
⎣
⎦2 ⎣
2
4
4
16
16 ⎦
1⎡
3
1 ⎤⎡
3
1 ⎤
= ⎢ cos x + cos x ⎥ ⎢cos x + cos x ⎥
4⎣
4
4 ⎦ ⎣ 16
16 ⎦
1⎡
3
3
3
1
= ⎢ cos x cos x + cos x cos x
4⎣
4
16
4
16
3
1
1 ⎤
1
+ cos x cos x + cos x cos x ⎥
16
4
16 ⎦
4
1 ⎡1 ⎛
15
9 ⎞ 1⎛
13
11 ⎞
= ⎢ ⎜ cos + cos x ⎟ + ⎜ cos x + cos x ⎟
4 ⎣2 ⎝
16
16 ⎠ 2 ⎝
16
16 ⎠

1⎛
7
1 ⎞ 1⎛
5
3 ⎞⎤
+ ⎜ cos x + cos x ⎟ + ⎜ cos x + cos x ⎟⎥
2⎝
16
16 ⎠ 2 ⎝
16
16 ⎠ ⎦
1⎡
15
13
11
9
= ⎢cos x + cos x + cos x + cos x
8⎣
16
16
16
16
7
5
3
1 ⎤
+ cos x + cos x + cos x + cos x ⎥
16
16
16
16 ⎦

49. As t increases, the point on the rim of the
wheel will move around the circle of radius 2.
a.

x(2) ≈ 1.902
y (2) ≈ 0.618
x(6) ≈ −1.176
y (6) ≈ −1.618
x(10) = 0
y (10) = 2
x(0) = 0
y (0) = 2

b.

⎛π ⎞
⎛π ⎞
x(t ) = −2 sin ⎜ t ⎟, y (t ) = 2 cos⎜ t ⎟
⎝5 ⎠
⎝5 ⎠

c.

The point is at (2, 0) when
is , when t =

π
5

t=

π
2

; that

5
.
2

2π
. When
10
you add functions that have the same
frequency, the sum has the same frequency.

50. Both functions have frequency
47. The temperature function is
⎛ 2π ⎛ 7 ⎞ ⎞
T (t ) = 80 + 25 sin ⎜
⎟
⎜ 12 ⎜ t − 2 ⎟ ⎟ .
⎝
⎠⎠
⎝

The normal high temperature for November
15th is then T (10.5) = 67.5 °F.

a.

y (t ) = 3sin(π t / 5) − 5cos(π t / 5)

+2sin((π t / 5) − 3)

48. The water level function is
⎛ 2π
⎞
F (t ) = 8.5 + 3.5 sin ⎜
(t − 9) ⎟ .
⎝ 12
⎠
The water level at 5:30 P.M. is then
F (17.5) ≈ 5.12 ft .

b.

50

Section 0.7

y (t ) = 3cos(π t / 5 − 2) + cos(π t / 5)
+ cos((π t / 5) − 3)



a.

C sin(ωt + φ ) = (C cos φ )sin ωt + (C sin φ ) cos ω t. Thus A = C ⋅ cos φ and B = C ⋅ sin φ .

b.

51.

A2 + B 2 = (C cos φ )2 + (C sin φ ) 2 = C 2 (cos 2 φ ) + C 2 (sin 2 φ ) = C 2

Also,
c.

B C ⋅ sin φ
=
= tan φ
A C ⋅ cos φ

A1 sin(ωt + φ1 ) + A2 sin(ωt + φ 2 ) + A3 (sin ωt + φ 3 )
= A1 (sin ωt cos φ1 + cos ωt sin φ1 )
+ A2 (sin ωt cos φ 2 + cos ωt sin φ 2 )
+ A3 (sin ωt cos φ 3 + cos ωt sin φ 3 )
= ( A1 cos φ1 + A2 cos φ 2 + A3 cos φ 3 ) sin ωt
+ ( A1 sin φ1 + A2 sin φ 2 + A3 sin φ 3 ) cos ωt
= C sin (ωt + φ )

where C and φ can be computed from
A = A1 cos φ1 + A2 cos φ2 + A3 cos φ3
B = A1 sin φ1 + A2 sin φ2 + A3 sin φ3
as in part (b).

d.

Written response. Answers will vary.

52. ( a.), (b.), and (c.) all look similar to this:

d.

53. a.

b.

c.
e.

The windows in (a)-(c) are not helpful because
the function oscillates too much over the
domain plotted. Plots in (d) or (e) show the
behavior of the function.

The plot in (a) shows the long term behavior of
the function, but not the short term behavior,
whereas the plot in (c) shows the short term
behavior, but not the long term behavior. The
plot in (b) shows a little of each.
Section 0.7

51


54. a.

h( x ) = ( f g ) ( x )
3
cos(100 x) + 2
100
=
2
⎛ 1 ⎞
2
⎜
⎟ cos (100 x) + 1
100 ⎠
⎝

j ( x) = ( g f )( x) =

56.

⎧
2
f ( x ) = ⎪( x − 2n ) ,
⎨
⎪0.0625,
⎩

1
1⎤
⎡
x ∈ ⎢ 2n − , 2n + ⎥
4
4⎦
⎣
otherwise

where n is an integer.
y

1
3x + 2 ⎞
⎛
cos ⎜100
⎟
100
x2 + 1 ⎠
⎝

0.5

b.

0.25

−2

c.

−1

1

x

2

0.8 Chapter Review
Concepts Test
1. False:
2. True:

⎧
1⎞
⎡
⎪ 4 x − x + 1 : x ∈ ⎢ n, n + ⎟
4⎠
⎪
⎣
55. f ( x ) = ⎨
⎪ − 4 x − x + 7 : x ∈ ⎡ n + 1 , n + 1⎞
⎟
⎢
⎪ 3
3
4
⎣
⎠
⎩
where n is an integer.

(

)

p and q must be integers.
p1 p2 p1q2 − p2 q1
−
=
; since
q1 q2
q1q2
p1 , q1 , p2 , and q2 are integers, so
are p1q2 − p2 q1 and q1q2 .

If the numbers are opposites
(– π and π ) then the sum is 0,
which is rational.

4. True:

Between any two distinct real
numbers there are both a rational
and an irrational number.

5. False:

0.999... is equal to 1.

6. True:

( am ) = ( an )

7. False:

(a * b) * c = abc ; a *(b * c) = ab

8. True:

(

3. False:

Since x ≤ y ≤ z and x ≥ z , x = y = z

)

y

2

1

−1

1

x

9. True:

52

Section 0.8

n

m

= a mn
c

x
would
2
be a positive number less than x .

If x was not 0, then ε =



10. True:

y − x = −( x − y ) so
( x − y )( y − x) = ( x − y )(−1)( x − y )

20. True:

If r > 1, then 1 − r < 0. Thus,
since 1 + r ≥ 1 − r ,

= (−1)( x − y ) .
2

1
1
.
≤
1− r 1+ r

( x − y ) 2 ≥ 0 for all x and y, so
−( x − y ) 2 ≤ 0.

11. True:
12. True:

If r > 1, r = r , and 1 − r = 1 − r , so
1
1
1
.
=
≤
1− r 1− r 1+ r

[ a, b] and [b, c ]

If r < −1, r = − r and 1 − r = 1 + r ,

a
1 1
> 1; <
b
b a

a < b < 0; a < b;

so
share point b in

common.
13. True:

14. True:

21. True:

If (a, b) and (c, d) share a point then
c < b so they share the infinitely
many points between b and c.

1
1
1
≤
=
.
1− r 1− r 1+ r

If x and y are the same sign, then
x – y = x– y . x– y ≤ x+ y
when x and y are the same sign, so
x – y ≤ x + y . If x and y have
opposite signs then either
x – y = x – (– y ) = x + y

x 2 = x = − x if x < 0.

16. False:

17. True:

For example, if x = −3 , then
− x = − ( −3) = 3 = 3 which does

(x > 0, y < 0) or
x – y = –x – y = x + y

not equal x.

15. False:

(x < 0, y > 0). In either case
x – y = x+ y .

For example, take x = 1 and
y = −2 .
4

x < y ⇔ x < y
4

If either x = 0 or y = 0, the
inequality is easily seen to be true.

4

4

22. True:

x = x and y = y , so x < y

18. True:

4

4

4

x + y = −( x + y )

4

x2 =

23. True:

= − x + (− y ) = x + y

19. True:

If r = 0, then
1
1
1
=
=
= 1.
1+ r 1 – r 1 – r
For any r, 1 + r ≥ 1 – r . Since
r < 1, 1 – r > 0 so

1
1
;
≤
1+ r 1 – r

If y is positive, then x =

( y)

x3 =

= y.

(3 y )

3

=y

24. True:

For example x 2 ≤ 0 has solution
[0].

25. True:

x 2 + ax + y 2 + y = 0
x 2 + ax +

If –1 < r < 0, then r = – r and

2

a2
1 a2 1
+ y2 + y + =
+
4
4 4 4
2

a⎞ ⎛
a2 + 1
1⎞
⎛
x+ ⎟ +⎜ y+ ⎟ =
⎜
2⎠ ⎝
2⎠
4
⎝
is a circle for all values of a.

1 – r = 1 + r , so

1
1
1
=
≤
.
1+ r 1 – r 1 – r
1 – r = 1 – r , so

y satisfies

For every real number y, whether it
is positive, zero, or negative, the
cube root x = 3 y satisfies

also, –1 < r < 1.

If 0 < r < 1, then r = r and

2

26. False:

If a = b = 0 and c < 0 , the equation
does not represent a circle.

1
1
1
.
≤
=
1+ r 1 – r 1 – r


Section 0.8

53


27. True;

28. True:

3
( x − a)
4
3
3a
y = x − + b;
4
4
If x = a + 4:
3
3a
y = (a + 4) –
+b
4
4
3a
3a
=
+3–
+b = b+3
4
4
y −b =

40. True:

41. True:

The equation is
(3 + 2m) x + (6m − 2) y + 4 − 2m = 0
which is the equation of a straight
line unless 3 + 2m and 6m − 2 are
both 0, and there is no real number
m such that
3 + 2m = 0 and 6m − 2 = 0.
f ( x) = –( x 2 + 4 x + 3)
= –( x + 3)( x + 1)

31. True:

32. True:

If ab > 0, a and b have the same
sign, so (a, b) is in either the first or
third quadrant.

The domain does not include
π
nπ + where n is an integer.
2

43. True:

The domain is ( − ∞, ∞) and the
range is [−6, ∞) .

44. False:

The range is ( − ∞, ∞) .

45. False:

The range ( − ∞, ∞) .

46. True:

If f(x) and g(x) are even functions,
f(x) + g(x) is even.
f(–x) + g(–x) = f(x) + g(x)

47. True:

If f(x) and g(x) are odd functions,
f(–x) + g(–x) = –f(x) – g(x)
= –[f(x) + g(x)], so f(x) + g(x) is odd

48. False:

f(–x)g(–x) = –f(x)[–g(x)] = f(x)g(x),
so
f(x)g(x) is even.
If f(x) is even and g(x) is odd,
f(–x)g(–x) = f(x)[–g(x)]
= –f(x)g(x), so f(x)g(x) is odd.

50. False:

If f(x) is even and g(x) is odd,
f(g(–x)) = f(–g(x)) = f(g(x)); while if
f(x) is odd and g(x) is even,
f(g(–x)) = f(g(x)); so f(g(x)) is even.

51. False:

30. True:

42. False:

49. True:

29. True:

If the points are on the same line,
they have equal slope. Then the
reciprocals of the slopes are also
equal.

f ( g (− x)) = f(–g(x)) = –f(g(x)), so
f(g(x)) is odd.

Let x = ε / 2. If ε > 0 , then x > 0
and x < ε .
If ab = 0, a or b is 0, so (a, b) lies
on
the x-axis or the y-axis. If a
= b = 0,
(a, b) is the origin.
y1 = y2 , so ( x1 , y1 ) and ( x2 , y2 )

are on the same horizontal line.
33. True:

−( x 2 + 4 x + 3) ≥ 0 on −3 ≤ x ≤ −1 .

d = [(a + b) – (a – b)]2 + (a – a) 2
= (2b) 2 = 2b

34. False:

The equation of a vertical line
cannot be written in point-slope
form.

35. True:

This is the general linear equation.

36. True:

Two non-vertical lines are parallel
if and only if they have the same
slope.

37. False:

The slopes of perpendicular lines
are
negative reciprocals.

38. True:

If a and b are rational and
( a, 0 ) , ( 0, b ) are the intercepts, the
slope is −

39. False:

b
which is rational.
a

f (– x) =

2(– x)3 + (– x)

ax + y = c ⇒ y = − ax + c
ax − y = c ⇒ y = ax − c
(a )(− a) ≠ −1.
(unless a = ±1 )

54

52. True:

Section 0.8

=−

(– x)2 + 1

=

–2 x3 – x
x2 + 1

2 x3 + x
x2 + 1



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