Calculus 11e sol george thomas (1)

CHAPTER 1 PRELIMINARIES
1.1 REAL NUMBERS AND THE REAL LINE
1. Executing long division,

"
9

2. Executing long division,

"
11

œ 0.1,

2
9

œ 0.2,

œ 0.09,

2
11

3
9

œ 0.3,

œ 0.18,

3
11

8
9

œ 0.8,

œ 0.27,

9
11

9
9

œ 0.9

œ 0.81,

11
11

œ 0.99

3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.
a) NNT. 5 is a counter example.
b) NT. 2 < x < 6 Ê 2 c 2 < x c 2 < 6 c 2 Ê 0 < x c 2 < 2.
c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3.
d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2.

e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3.
f) NT. 2 < x < 6 Ê x < 6 Ê (x c 4) < 2 and 2 < x < 6 Ê x > 2 Ê cx < c2 Ê cx + 4 < 2 Ê c(x c 4) < 2.
The pair of inequalities (x c 4) < 2 and c(x c 4) < 2 Ê | x c 4 | < 2.
g) NT. 2 < x < 6 Ê c2 > cx > c6 Ê c6 < cx < c2. But c2 < 2. So c6 < cx < c2 < 2 or c6 < cx < 2.
h) NT. 2 < x < 6 Ê c1(2) > c1(x) < c1(6) Ê c6 < cx < c2
4. NT = necessarily true, NNT = Not necessarily true. Given: c1 < y c 5 < 1.
a) NT. c1 < y c 5 < 1 Ê c1 + 5 < y c 5 + 5 < 1 + 5 Ê 4 < y < 6.
b) NNT. y = 5 is a counter example. (Actually, never true given that 4  y  6)
c) NT. From a), c1 < y c 5 < 1, Ê 4 < y < 6 Ê y > 4.
d) NT. From a), c1 < y c 5 < 1, Ê 4 < y < 6 Ê y < 6.
e) NT. c1 < y c 5 < 1 Ê c1 + 1 < y c 5 + 1 < 1 + 1 Ê 0 < y c 4 < 2.
f) NT. c1 < y c 5 < 1 Ê (1/2)(c1 + 5) < (1/2)(y c 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3.
g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4.
h) NT. c1 < y c 5 < 1 Ê y c 5 > c1 Ê y > 4 Ê cy < c4 Ê cy + 5 < 1 Ê c(y c 5) < 1.
Also, c1 < y c 5 < 1 Ê y c 5 < 1. The pair of inequalities c(y c 5) < 1 and (y c 5) < 1 Ê | y c 5 | < 1.
5. c2x € 4 Ê x  c2
6. 8 c 3x   5 Ê c3x   c3 Ê x Ÿ 1
7. 5x c $ Ÿ ( c 3x Ê 8x Ÿ 10 Ê x Ÿ

ïïïïïïïïïñqqqqqqqqp x
1
5
4

8. 3(2 c x) € 2(3 b x) Ê 6 c 3x € 6 b 2x
Ê 0 € 5x Ê 0 € x
9. 2x c

10.

"
#

Ê

"
5

6 cx
4



  7x b

ˆc

10 ‰
6

3xc4
2

7
6

Ê c "# c

  x or c

"
3

7
6

ïïïïïïïïïðqqqqqqqqp x
0

  5x

 x

Ê 12 c 2x  12x c 16

Ê 28  14x Ê 2  x

qqqqqqqqqðïïïïïïïïỵ x
2


2
11.

Chapter 1 Preliminaries
4
5

"
3

(x c 2) 

(x c 6) Ê 12(x c 2)  5(x c 6)

Ê 12x c 24  5x c 30 Ê 7x  c6 or x  c 67
12. c xb2 5 Ÿ

12b3x
4

Ê c(4x b 20) Ÿ 24 b 6x

Ê c44 Ÿ 10x Ê c 22
5 Ÿ x

qqqqqqqqqđïïïïïïïïỵ x
c22/5

13. y œ 3 or y œ c3
14. y c 3 œ 7 or y c 3 œ c7 Ê y œ 10 or y œ c4
15. 2t b 5 œ 4 or 2t b & œ c4 Ê 2t œ c1 or 2t œ c9 Ê t œ c "# or t œ c 9#
16. 1 c t œ 1 or 1 c t œ c1 Ê ct œ ! or ct œ c2 Ê t œ 0 or t œ 2
17. 8 c 3s œ
18.

s
#

9

2

or 8 c 3s œ c #9 Ê c3s œ c 7# or c3s œ c 25
# Ê sœ

c 1 œ 1 or

s
#

c 1 œ c1 Ê

s
#

œ 2 or

s
#

7
6

or s œ

25
6

œ ! Ê s œ 4 or s œ 0

19. c2  x  2; solution interval (c2ß 2)
20. c2 Ÿ x Ÿ 2; solution interval [c2ß 2]

qqqqđïïïïïïïïđqqqqp x
c2
2

21. c3 Ÿ t c 1 Ÿ 3 Ê c2 Ÿ t Ÿ 4; solution interval [c2ß 4]
22. c1  t b 2  1 Ê c3  t  c1;
solution interval (c3ß c1)

qqqqðïïïïïïïïðqqqqp t
c3
c1

23. c%  3y c 7  4 Ê 3  3y  11 Ê 1  y 
solution interval ˆ1ß

11
3

;

11 ‰
3

24. c1  2y b 5  " Ê c6  2y  c4 Ê c3  y  c2;
solution interval (c3ß c2)
25. c1 Ÿ

z
5

c1Ÿ1 Ê 0Ÿ

z
5

qqqqðïïïïïïïïðqqqqp y
c3
c2

Ÿ 2 Ê 0 Ÿ z Ÿ 10;

solution interval [0ß 10]
26. c2 Ÿ

c 1 Ÿ 2 Ê c1 Ÿ
solution interval 3z
#

27. c "#  3 c
Ê

2
7

28. c3 

"
x



x
2
x

2
5

"
#

2
7

Ÿ 3 Ê c 32 Ÿ z Ÿ 2;
qqqqñïïïïïïïïñqqqqp z
2
c2/3

Ê c 7#  c x"  c 5# Ê

7
#

€

"
x

€

5
#

; solution interval ˆ 27 ß 25 ‰

c43 Ê 1

Ê 2€x€

3z
#

Ê

2
7

2
x

( Ê 1€

x
#

€

"
7

 x  2; solution interval ˆ 27 ß 2‰

qqqqðïïïïïïïïðqqqqp x
2
2/7

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Section 1.1 Real Numbers and the Real Line
29. 2s   4 or c2s   4 Ê s   2 or s Ÿ c2;
solution intervals (c_ß c2] r [2ß _)
30. s b 3  

"
#

or c(s b 3)  

"
#

Ê s   c 5# or cs  

7

#

Ê s   c 5# or s Ÿ c 7# ;
solution intervals ˆc_ß c 7# ‘ r
ùùùùùùủqqqqqqủùùùùùùợ s
c7/2
c5/2

31. 1 c x 1 or c(" c x) € 1 Ê cx € 0 or x € 2
Ê x  0 or x € 2; solution intervals (c_ß !) r (2ß _)
32. 2 c 3x € 5 or c(2 c 3x) € 5 Ê c3x € 3 or 3x € 7
Ê x  c1 or x € 73 ;
solution intervals (c_ß c1) r ˆ 73 ò _
33.

rb"
#

ùùùùùùqqqqqqùùùùùùợ x
c1
7/3

1 or c rb# 1 ‰   1 Ê r b 1   2 or r b 1 Ÿ c2

Ê r   1 or r Ÿ c3; solution intervals (c_ß c3] r [1ß _)
34.

3r
5

c"€

Ê

or c ˆ 3r5 c "‰ €

2
5

or c 3r5 € c 53 Ê r € 37 or r  1
solution intervals (c_ß ") r 73 ò _
3r
5



2
5
7
5

ùùùùùùqqqqqqùùùùùùợ r
1
7/3

35. x#  # Ê kxk  È2 Ê cÈ2  x  È2 ;
solution interval ŠcÈ2ß È2‹

qqqqqqðïïïïïïðqqqqqqp x

È#
cÈ #

36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x   2 or x Ÿ c2;
solution interval (c_ò c2] r [2ò _)

ùùùùùùủqqqqqqủùùùùùùợ r
c2
2

37. 4  x#  9 Ê 2  kxk  3 Ê 2  x  3 or 2  cx  3
Ê 2  x  3 or c3  x  c2;
solution intervals (c3ß c2) r (2ß 3)
38.

"
9

 x# 

Ê

x

"
#

"
3

 kxk 

"
#

Ê

"
3

x

or c #"  x  c 3" ;
solution intervals ˆc "# ß c 3" ‰ r ˆ 3" ß #" ‰
Ê

"
3

"
4

"
#

or

"
3

 cx 

qqqqðïïïïðqqqqðïïïïðqqqp x
c3
c2
2
3
"
#

qqqqðïïïïðqqqqðïïïïðqqqp x
c1/2 c1/3
1/3
1/2

39. (x c 1)#  4 Ê kx c 1k  2 Ê c2  x c 1  2
Ê c1  x  3; solution interval (c"ß $)

qqqqqqðïïïïïïïïðqqqqp x
c1
3

40. (x b 3)#  # Ê kx b 3k  È2
Ê cÈ2  x b 3  È2 or c3 c È2  x  c3 b È2 ;
solution interval Šc3 c È2ß c3 b È2‹

qqqqqqðïïïïïïïïðqqqqp x
c3 c È #
c3 b È #

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3


4

Chapter 1 Preliminaries

41. x# c x  0 Ê x# c x +

1
4

<

1
4

2
Ê ˆx c 12 ‰ <

1
4

ʹx c

1
2

¹<

Ê c 12 < x c

1
2

1
2

<

1
2

Ê 0 < x < 1.

So the solution is the interval (0ß 1)
42. x# c x c 2   0 Ê x# c x +

1
4

 

9
4

Ê ¹x c

1
2

¹  

3
2

Ê xc

1
2

 

3
2

or cˆx c 12 ‰  

3
2

Ê x   2 or x Ÿ c1.

The solution interval is (c_ß c1] r [2ß _)
43. True if a   0; False if a  0.
44. kx c 1k œ 1 c x Í kc(x c 1)k œ 1 c x Í 1 c x   0 Í x Ÿ 1
45. (1) ka b bk œ (a b b) or ka b bk œ c(a b b);
both squared equal (a b b)#

(2) ab Ÿ kabk œ kak kbk
(3) kak œ a or kak œ ca, so kak# œ a# ; likewise, kbk# œ b#
(4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka b bk and
y œ kak b kbk so that ka b bk# Ÿ akak b kbkb# Ê ka b bk Ÿ kak b kbk .
46. If a   0 and b   0, then ab   0 and kabk œ ab œ kak kbk .
If a  0 and b  0, then ab € 0 and kabk œ ab œ (ca)(cb) œ kak kbk .
If a   0 and b  0, then ab Ÿ 0 and kabk œ c(ab) œ (a)(cb) œ kak kbk .
If a  0 and b   0, then ab Ÿ 0 and kabk œ c(ab) œ (ca)(b) œ kak kbk .
47. c3 Ÿ x Ÿ 3 and x € c "# Ê c

"
#

 x Ÿ 3.

48. Graph of kxk b kyk Ÿ 1 is the interior
of “diamond-shaped" region.

49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | xc1 | < $ . Then | xc1 | < $ Ê 2| xc1 | < 2$ Ê
| 2x c # | < 2$ Ê | (2x + 1) c 3 | < 2$ Ê | f(x) c f(1) | < 2$
50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x c 0 | < % /2. Then 2| x c 0 | < % and
| 2x + 3 c3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) c f(0) | < %.
51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | œ a by definition. Now, a > 0 Ê ca < 0. Let ca = b. By definition, | b | œ cb. Since b = ca,
| ca | œ c(ca) œ a and | a | œ | ca | œ a.
ii) For a < 0, | a | œ ca. Now, a < 0 Ê ca > 0. Let ca œ b. By definition, | b | œ b and thus |ca| œ ca. So again
| a | œ |ca|.
iii) By definition | 0 | œ 0 and since c0 œ 0, | c0 | œ 0. Thus, by i), ii), and iii) | a | œ | ca | for any real number.

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Section 1.2 Lines, Circles and Parabolas
Prove | x | > 0 Ê x > a or x < ca for any positive number, a.
For x   0, | x | œ x. | x | > a Ê x > a.
For x < 0, | x | œ cx. | x | > a Ê cx > a Ê x < ca.
ii) Prove x > a or x < ca Ê | x | > 0 for any positive number, a.
a > 0 and x > a Ê | x | œ x. So x > a Ê | x | > a.
For a > 0, ca < 0 and x < ca Ê x < 0 Ê | x | œ cx. So x < ca Ê cx > a Ê | x | > a.

52. i)

53. a)

1=1 Ê |1|=1 ʹb

b)

lal
lbl

œ ¹a

†

"
b

¹ œ ¹ a¹

† "b ¹ œ

†¹

"
b

l bl
lbl

¹ œ ¹ a¹

Ê ¹ b¹

† l bl
"

† ¹ b" ¹ œ

œ

lbl
lbl

Ê

†

¹ b ¹ ¹ "b ¹
¹ b¹

œ

¹ b¹

†

¹ b¹ ¹ b¹

Ê ¹ b" ¹ œ "

¹ b¹

lal
lbl

54. Prove Sn œ kan k œ kakn for any real number a and any positive integer n.
ka" k œ kak " œ a, so S" is true. Now, assume that Sk œ ¸ak ¸ œ kak k is true form some positive integer 5 .
Since ka" k œ kak " and ¸ak ¸ œ kak k , we have ¸akb" ¸ œ ¸ak † a" ¸ œ ¸ak ¸ka" k œ kak k kak " œ kak k+" . Thus,
Skb" œ ¸akb" ¸ œ kak k+" is also true. Thus by the Principle of Mathematical Induction, Sn œ l an l œ l a ln
is true for all n positive integers.
1.2 LINES, CIRCLES, AND PARABOLAS
1. ?x œ c1 c (c3) œ 2, ?y œ c2 c 2 œ c4; d œ È(?x)# b (?y)# œ È4 b 16 œ 2È5
2. ?x œ c$ c (c1) œ c2, ?y œ 2 c (c2) œ 4; d œ È(c2)# b 4# œ 2È5
3. ?x œ c8.1 c (c3.2) œ c4.9, ?y œ c2 c (c2) œ 0; d œ È(c4.9)# b 0# œ 4.9
#

4. ?x œ 0 c È2 œ cÈ2, ?y œ 1.5 c 4 œ c2.5; d œ ÊŠcÈ2‹ b (c2.5)# œ È8.25
5. Circle with center (!ß !) and radius 1.

6. Circle with center (!ß !) and radius È2.

7. Disk (i.e., circle together with its interior points) with center (!ß !) and radius È3.
8. The origin (a single point).
9. m œ

?y
?x

œ

c1 c 2
c2 c (c1)

œ3

perpendicular slope œ c "3

10. m œ

?y
?x

œ

c# c "
2 c (c2)

œ c 34

perpendicular slope œ

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4
3

5


6

Chapter 1 Preliminaries

11. m œ

?y
?x

œ

3c3
c1 c 2

œ0

12. m œ

14. (a) x œ È2

(b) y œ

c# c 0
c# c (c#)

; no slope

15. (a) x œ 0

16. (a) x œ c1

(b) y œ cÈ2

(b) y œ c1.3

4
3

œ

perpendicular slope œ 0

perpendicular slope does not exist

13. (a) x œ c1

?y
?x

(b) y œ 0

17. P(c1ß 1), m œ c1 Ê y c 1 œ c1ax c (c1)b Ê y œ cx
18. P(2ß c3), m œ

"
#

Ê y c (c3) œ

19. P(3ß 4), Q(c2ß 5) Ê m œ

?y
?x

20. P(c8ß 0), Q(c1ß 3) Ê m œ

œ

?y
?x

"
#

(x c 2) Ê y œ

5c4
c2 c 3

œ

"
#

xc4

œ c "5 Ê y c 4 œ c "5 (x c 3) Ê y œ c "5 x b

3c0
c1 c (c8)

œ

3
7

Ê yc0œ

3
7

ax c (c8)b Ê y œ

3
7

23
5

xb

21. m œ c 54 , b œ 6 Ê y œ c 54 x b 6

22. m œ "# , b œ c3 Ê y œ

"
#

23. m œ 0, P(c12ß c9) Ê y œ c9

24. No slope, P ˆ "3 ß %‰ Ê x œ

24
7

xc3
"
3

25. a œ c1, b œ 4 Ê (0ß 4) and (c"ß 0) are on the line Ê m œ

?y
?x

œ

0c4
c1 c 0

œ 4 Ê y œ 4x b 4

26. a œ 2, b œ c6 Ê (2ß 0) and (!ß c6) are on the line Ê m œ

?y
?x

œ

c6 c 0
0c2

œ 3 Ê y œ 3x c 6

27. P(5ß c1), L: 2x b 5y œ 15 Ê mL œ c 25 Ê parallel line is y c (c1) œ c 25 (x c 5) Ê y œ c 25 x b 1
È
È
È
28. P ŠcÈ2ß 2‹ , L: È2x b 5y œ È3 Ê mL œ c 52 Ê parallel line is y c 2 œ c 52 Šx c ŠcÈ2‹‹ Ê y œ c 52 x b

8
5

29. P(4ß 10), L: 6x c 3y œ 5 Ê mL œ 2 Ê m¼ œ c "# Ê perpendicular line is y c 10 œ c "# (x c 4) Ê y œ c "# x b 12
30. P(!ß 1), L: 8x c 13y œ 13 Ê mL œ

8
13

13
Ê m¼ œ c 13

8 Ê perpendicular line is y œ c 8 x b 1

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Section 1.2 Lines, Circles and Parabolas
31. x-intercept œ 4, y-intercept œ 3

32. x-intercept œ c4, y-intercept œ c2

33. x-intercept œ È3, y-intercept œ cÈ2

34. x-intercept œ c2, y-intercept œ 3

35. Ax b By œ C" Í y œ c AB x b

C"
B

and Bx c Ay œ C# Í y œ

B
A

xc

C#
A.

Since ˆc AB ‰ ˆ AB ‰ œ c1 is the

product of the slopes, the lines are perpendicular.
36. Ax b By œ C" Í y œ c AB x b
slope

c AB ,

C"
B

and Ax b By œ C# Í y œ c AB x b

C#
B.

Since the lines have the same

they are parallel.

37. New position œ axold b ?xß yold b ?yb œ (c# b &ß 3 b (c6)) œ ($ß c3).
38. New position œ axold b ?xß yold b ?yb œ (6 b (c6)ß 0 b 0) œ (0ß 0).
39. ?x œ 5, ?y œ 6, B(3ß c3). Let A œ (xß y). Then ?x œ x# c x" Ê 5 œ 3 c x Ê x œ c2 and
?y œ y# c y" Ê 6 œ c3 c y Ê y œ c9. Therefore, A œ (c#ß c9).
40. ?x œ " c " œ !, ?y œ ! c ! œ !

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7

8

Chapter 1 Preliminaries

41. C(!ß 2), a œ 2 Ê x# b (y c 2)# œ 4

42. C(c$ß 0), a œ 3 Ê (x b 3)# b y# œ 9

43. C(c1ß 5), a œ È10 Ê (x b 1)# b (y c 5)# œ 10

44. C("ß "), a œ È2 Ê (x c 1)# b (y c 1)# œ 2
x œ 0 Ê (0 c 1)# b (y c 1)# œ 2 Ê (y c 1)# œ 1
Ê y c 1 œ „ 1 Ê y œ 0 or y œ 2.
Similarly, y œ 0 Ê x œ 0 or x œ 2

#

45. C ŠcÈ3ß c2‹ , a œ 2 Ê Šx b È3‹ b (y b 2)# œ 4,
#

x œ 0 Ê Š0 b È3‹ b (y b 2)# œ 4 Ê (y b 2)# œ 1
Ê y b 2 œ „ 1 Ê y œ c1 or y œ c3. Also, y œ 0
#

#

Ê Šx b È3‹ b (0 b 2)# œ 4 Ê Šx b È3‹ œ 0
Ê x œ cÈ 3
#
46. C ˆ3ß "# ‰, a œ 5 Ê (x c 3)# b ˆy c "# ‰ œ 25, so

#
x œ 0 Ê (0 c 3)# b ˆy c "# ‰ œ 25
#
Ê ˆy c "# ‰ œ 16 Ê y c

"
#

œ „4 Ê yœ

9
#

#
or y œ c 7# . Also, y œ 0 Ê (x c 3)# b ˆ0 c "# ‰ œ 25

Ê (x c 3)# œ
Ê xœ3„

99
4
3È11
#

Ê xc3œ „

3È11
#

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Section 1.2 Lines, Circles and Parabolas
47. x# b y# b 4x c 4y b % œ 0
Ê x# b %B b y# c 4y œ c4
Ê x# b 4x b 4 b y# c 4y b 4 œ 4
Ê (x b 2)# b (y c 2)# œ 4 Ê C œ (c2ß 2), a œ 2.

48. x# b y# c 8x b 4y b 16 œ 0
Ê x# c 8x b y# b 4y œ c16
Ê x# c 8x b 16 b y# b 4y b 4 œ 4
Ê (x c 4)# b (y b 2)# œ 4
Ê C œ (%ß c2), a œ 2.

49. x# b y# c 3y c 4 œ 0 Ê x# b y# c 3y œ 4
Ê x# b y# c 3y b 94 œ 25
4
#
Ê x# b ˆy c 3# ‰ œ

25
4

Ê C œ ˆ0ß 3# ‰ ,

a œ 5# .

50. x# b y# c 4x c
#

9
4
#

œ0

Ê x c 4x b y œ

9
4
#

Ê x# c 4x b 4 b y œ
Ê (x c 2)# b y# œ

25
4

25
4

Ê C œ (2ß 0), a œ 5# .

51. x# b y# c 4x b 4y œ 0
Ê x# c 4x b y# b 4y œ 0
Ê x# c 4x b 4 b y# b 4y b 4 œ 8
Ê (x c 2)# b (y b 2)# œ 8
Ê C(2ß c2), a œ È8.

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9


10

Chapter 1 Preliminaries

52. x# b y# b 2x œ 3
Ê x# b 2x b 1 b y# œ 4
Ê (x b 1)# b y# œ 4
Ê C œ (c1ß 0), a œ 2.

c2
53. x œ c #ba œ c 2(1)
œ1

Ê y œ (1)# c 2(1) c 3 œ c4
Ê V œ ("ß c4). If x œ 0 then y œ c3.
Also, y œ 0 Ê x# c 2x c 3 œ 0
Ê (x c 3)(x b 1) œ 0 Ê x œ 3 or
x œ c1. Axis of parabola is x œ 1.

4
54. x œ c #ba œ c 2(1)
œ c2

Ê y œ (c2)# b 4(c2) b 3 œ c1
Ê V œ (c2ß c1). If x œ 0 then y œ 3.
Also, y œ 0 Ê x# b 4x b 3 œ 0

Ê (x b 1)(x b 3) œ 0 Ê x œ c1 or
x œ c3. Axis of parabola is x œ c2.

55. x œ c #ba œ c 2(c4 1) œ 2
Ê y œ c(2)# b 4(2) œ 4
Ê V œ (2ß 4). If x œ 0 then y œ 0.
Also, y œ 0 Ê cx# b 4x œ 0
Ê cx(x c 4) œ 0 Ê x œ 4 or x œ 0.
Axis of parabola is x œ 2.

56. x œ c #ba œ c 2(c4 1) œ 2
Ê y œ c(2)# b 4(2) c 5 œ c1
Ê V œ (2ß c1). If x œ 0 then y œ c5.
Also, y œ 0 Ê cx# b 4x c 5 œ 0
Ê x# c 4x b 5 œ 0 Ê x œ

4 „È c 4
#

Ê no x intercepts. Axis of parabola is x œ 2.

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Section 1.2 Lines, Circles and Parabolas
57. x œ c #ba œ c 2(cc61) œ c3
Ê y œ c(c3)# c 6(c3) c 5 œ 4
Ê V œ (c3ß %). If x œ 0 then y œ c5.
Also, y œ 0 Ê cx# c 6x c 5 œ 0
Ê (x b 5)(x b 1) œ 0 Ê x œ c5 or

x œ c1. Axis of parabola is x œ c3.

c1
58. x œ c #ba œ c 2(2)
œ

"
4

#

Ê y œ 2 ˆ "4 ‰ c 4" b 3 œ 23
8
‰
Ê V œ ˆ "4 ß 23
.
If
x
œ
0
then
y œ 3.
8

Also, y œ 0 Ê 2x# c x b 3 œ 0
Ê xœ

1„Èc23
4

Ê no x intercepts.

Axis of parabola is x œ "4 .

1
59. x œ c #ba œ c 2(1/2)
œ c1
"
#

(c1)# b (c1) b 4 œ 72
Ê V œ ˆc"ß 72 ‰ . If x œ 0 then y œ 4.
Ê yœ

Also, y œ 0 Ê
Ê xœ

c1 „ È c 7
1

"
#

x# b x b 4 œ 0
Ê no x intercepts.

Axis of parabola is x œ c1.
60. x œ c #ba œ c 2(c21/4) œ 4
Ê y œ c "4 (4)# b 2(4) b 4 œ 8
Ê V œ (4ß 8) . If x œ 0 then y œ 4.

Also, y œ 0 Ê c "4 x# b 2x b 4 œ 0
Ê xœ

c2 „ È 8
c1/2

œ 4 „ 4È2.

Axis of parabola is x œ 4.

61. The points that lie outside the circle with center (!ß 0) and radius È7.
62. The points that lie inside the circle with center (!ß 0) and radius È5.
63. The points that lie on or inside the circle with center ("ß 0) and radius 2.
64. The points lying on or outside the circle with center (!ß 2) and radius 2.
65. The points lying outside the circle with center (!ß 0) and radius 1, but inside the circle with center (!ß 0),
and radius 2 (i.e., a washer).

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11


12

Chapter 1 Preliminaries

66. The points on or inside the circle centered
at (!ß !) with radius 2 and on or inside the
circle centered at (c2ß 0) with radius 2.

67. x# b y# b 6y  0 Ê x# b (y b 3)#  9.
The interior points of the circle centered at
(!ß c3) with radius 3, but above the line
y œ c3.

68. x# b y# c 4x b 2y € 4 Ê (x c 2)# b (y b 1)# € 9.
The points exterior to the circle centered at
(2ß c1) with radius 3 and to the right of the
line x œ 2.

69. (x b 2)# b (y c 1)#  6

70. (x b 4)# b (y c 2)# € 16

71. x# b y# Ÿ 2, x   1

72. x# b y# € 4, (x c 1)# b (y c 3)#  10

73. x# b y# œ 1 and y œ 2x Ê 1 œ x# b 4x# œ 5x#
Ê Šx œ

"
È5

and y œ

2
È5 ‹

or Šx œ c È"5 and y œ c È25 ‹ .

Thus, A Š È"5 ß È25 ‹ , B Šc È"5 ß c È25 ‹ are the
points of intersection.

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Section 1.2 Lines, Circles, and Parabolas
74. x b y œ 1 and (x c 1)# b y# œ 1
Ê 1 œ (cy)# b y# œ 2y#
Ê Šy œ

"
È2

and x œ " c

Šy œ c È"2 and x œ 1 b
A Š" c

"
È2

"
È2 ‹

"
È2 ‹ .

ß È"2 ‹ and B Š1 b

or

Thus,

"
È2

ß c È"2 ‹

are intersection points.

75. y c x œ 1 and y œ x# Ê x# c x œ 1
1 „È 5
.
#
1 bÈ 5
3 bÈ 5
If x œ # , then y œ x b 1 œ # .
È
È
If x œ 1c# 5 , then y œ x b 1 œ 3c# 5 .
È
È
È
È
Thus, A Š 1b# 5 ß 3b# 5 ‹ and B Š 1c# 5 ß 3c# 5 ‹

Ê x# c x c 1 œ 0 Ê x œ

are the intersection points.

76. y œ cx and C œ c(x c 1)# Ê (x c 1)# œ x
3 „È 5
.
#
È 5 c3
3 cÈ 5
x œ # , then y œ cx œ # . If
È
È
x œ 3b# 5 , then y œ cx œ c 3b# 5 .
È
È
È
Thus, A Š 3c# 5 ß 5#c3 ‹ and B Š 3b# 5

Ê x# c 3x b " œ 0 Ê x œ

If

È

ß c 3b# 5 ‹

are the intersection points.

77. y œ 2x# c 1 œ cx# Ê 3x# œ 1
Ê x œ È"3 and y œ c 3" or x œ c È"3 and y œ c 3" .
Thus, A Š È"3 ß c 3" ‹ and B Šc È"3 ß c 3" ‹ are the

intersection points.

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13


14

Chapter 1 Preliminaries

78. y œ

x#
4

œ (x c 1)# Ê 0 œ
#

3x#
4

c 2x b 1

Ê 0 œ 3x c 8x b 4 œ (3x c 2)(x c 2)
Ê x œ 2 and y œ
yœ

#

x
4

x#
4

œ 1, or x œ

œ 9" . Thus, A(2ß 1) and

2
3 and
2
B ˆ 3 ß 9" ‰

are the intersection points.

79. x# b y# œ 1 œ (x c 1)# b y#
Ê x# œ (x c 1)# œ x# c 2x b 1
Ê 0 œ c2x b 1 Ê x œ "# . Hence
y# œ " c x # œ
A Š "# ß

È3
# ‹

and

3
4

or y œ „

È3
#

È
B Š "# ß c #3 ‹

. Thus,

are the

intersection points.

80. x# b y# œ 1 œ x# b y Ê y# œ y
Ê y(y c 1) œ 0 Ê y œ 0 or y œ 1.
If y œ 1, then x# œ " c y# œ 0 or x œ 0.
If y œ 0, then x# œ 1 c y# œ 1 or x œ „ 1.
Thus, A(0ß 1), B("ß 0), and C(c1ß 0) are the
intersection points.

81. (a) A á (69ò 0 in), B á (68ò .4 in) ấ m
(b) A á (68ò .4 in), B á (10ò 4 in) ấ m
(c) A á (10ò 4 in), B á (5ò 4.6 in) Ê m œ
82. The time rate of heat transfer across a material,
to the temperature gradient across the material,
of the material.

?U

?>

œ

X
-kA ?
?B

Ê

?U ỴA
k = c ??> X .
?B

68° c 69°
.4 c 0 ¸ c2.5°/in.
10° c 68°
4 c .4 ¸ c16.1°/in.
5° c 10°
4.6 c 4 ¸ c8.3°/in.
?U
?> , is directly
?X
?B (the slopes

Note that

?U
?>

proportional to the cross-sectional area, A, of the material,
from the previous problem), and to a constant characteristic

and

?X
?B

are of opposite sign because heat flow is toward lower

temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good
insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are
X
not changing), we may define another constant, K, characteristics of the material: K œ c ?"X Þ Using the values of ?
?B from
?B

the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the
poorest insulator, with K œ 0.4.
83. p œ kd b 1 and p œ 10.94 at d œ 100 Ê k œ

10.94c"
100

œ 0.0994. Then p œ 0.0994d b 1 is the diver's

pressure equation so that d œ 50 Ê p œ (0.0994)(50) b 1 œ 5.97 atmospheres.
84. The line of incidence passes through (!ß 1) and ("ß 0) Ê The line of reflection passes through ("ß 0) and (#ß ")
c0
Ê m œ 1#c

1 œ 1 Ê y c 0 œ 1(x c 1) Ê y œ x c 1 is the line of reflection.

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Section 1.2 Lines, Circles, and Parabolas
85. C œ

5
9

(F c 32) and C œ F Ê F œ

86. m œ

37.1
100

œ

14
?x

Ê ?x œ

14
.371 .

5
9

Fc

160
9

Ê

4
9

15

F œ c 160
9 or F œ c40° gives the same numerical reading.

#

14 ‰
Therefore, distance between first and last rows is É(14)# b ˆ .371
¸ 40.25 ft.

87. length AB œ È(5 c 1)# b (5 c 2)# œ È16 b 9 œ 5
length AC œ È(4 c 1)# b (c# c #)# œ È9 b 16 œ 5
length BC œ È(4 c 5)# b (c# c 5)# œ È1 b 49 œ È50 œ 5È2 Á 5
#

88. length AB œ Ê(1 c 0)# b ŠÈ3 c 0‹ œ È1 b 3 œ 2
length AC œ È(2 c 0)# b (0 c 0)# œ È4 b 0 œ 2
#

length BC œ Ê(2 c 1)# b Š0 c È3‹ œ È1 b 3 œ 2
89. Length AB œ È(?x)# b (?y)# œ È1# b 4# œ È17 and length BC œ È(?x)# b (?y)# œ È4# b 1# œ È17.
Also, slope AB œ c41 and slope BC œ "4 , so AB ¼ BC. Thus, the points are vertices of a square. The coordinate
increments from the fourth vertex D(xß y) to A must equal the increments from C to B Ê 2 c x œ ?x œ 4 and
c1 c y œ ?y œ " Ê x œ c2 and y œ c2. Thus D(c#ß c2) is the fourth vertex.

90. Let A œ (xß 2) and C œ (9ß y) Ê B œ (xß y). Then 9 c x œ kADk and 2 c y œ kDCk Ê 2(9 c x) b 2(2 c y) œ 56
and 9 c x œ 3(2 c y) Ê 2(3(2 c y)) b 2(2 c y) œ 56 Ê y œ c5 Ê 9 c x œ 3(2 c (c5)) Ê x œ c12.
Therefore, A œ (c12ß 2), C œ (9ß c5), and B œ (c12ß c5).
91. Let A(c"ß "), B(#ß $), and C(2ß !) denote the points.
Since BC is vertical and has length kBCk œ 3, let
D" (c"ß 4) be located vertically upward from A and
D# (c"ß c2) be located vertically downward from A so
that kBCk œ kAD" k œ kAD# k œ 3. Denote the point
D$ (xß y). Since the slope of AB equals the slope of
3
"
CD$ we have yxc
c2 œ c 3 Ê 3y c 9 œ cx b 2 or

x b 3y œ 11. Likewise, the slope of AC equals the slope
0
2
of BD$ so that yx c
c 2 œ 3 Ê 3y œ 2x c 4 or 2x c 3y œ 4.

Solving the system of equations

x b 3y œ ""

we find x œ 5 and y œ 2 yielding the vertex D$ (5ß #).
2x c 3y œ 4 I

92. Let ax, yb, x Á ! and/or y Á ! be a point on the coordinate plane. The slope, m, of the segment a!ß !b to ax, yb is yx . A 90‰

rotation gives a segment with slope mw œ c m" œ c xy . If this segment has length equal to the original segment, its endpoint
will be acy, xb or ay, cxb, the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise
rotation.
(a) (c"ß 4);
(b) (3ß c2);
(c) (5ß 2);
(d) (0ß x);

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16

Chapter 1 Preliminaries
(e) (cyß 0);

(f) (cyß x);

(g) (3ß c10)

93. 2x b ky œ 3 has slope c 2k and 4x b y œ 1 has slope c4. The lines are perpendicular when c 2k (c4) œ c1 or
k œ c8 and parallel when c 2k œ c4 or k œ "# .

94. At the point of intersection, 2x b 4y œ 6 and 2x c 3y œ c1. Subtracting these equations we find 7y œ 7 or
y œ 1. Substitution into either equation gives x œ 1 Ê (1ß 1) is the intersection point. The line through (1ß 1)

and ("ß #) is vertical with equation x œ 1.
95. Let M(aß b) be the midpoint. Since the two triangles
shown in the figure are congruent, the value a must
lie midway between x" and x# , so a œ x" b#x# .
Similarly, b œ

y " by #
# .

96. (a) L has slope 1 so M is the line through P(2ß 1) with slope c1; or the line y œ cx b 3. At the intersection
point, Q, we have equal y-values, y œ x b 2 œ cx b 3. Thus, 2x œ 1 or x œ "# . Hence Q has coordinates
ˆ "# ß 5# ‰ . The distance from P to L œ the distance from P to Q œ Ɉ #3 ‰# b ˆc 3# ‰# œ É 18
4 œ
(b) L has slope c 43 so M has slope

3
4

3È 2
# .

and M has the equation 4y c 3x œ 12. We can rewrite the equations of

84
the lines as L: x b y œ 3 and M: cB b 43 y œ 4. Adding these we get 25
12 y œ 7 so y œ 25 . Substitution
12
‰
ˆ 12 84 ‰
into either equation gives x œ 43 ˆ 84

25 c 4 œ 25 so that Q 25 ß 25 is the point of intersection. The distance
3
4

from P to L œ Ɉ4 c

12 ‰#
25

b ˆ6 c

84 ‰#
25

œ

22
5 .

(c) M is a horizontal line with equation y œ b. The intersection point of L and M is Q(c"ß b). Thus, the
distance from P to L is È(a b 1)# b 0# œ ka b 1k .
(d) If B œ 0 and A Á 0, then the distance from P to L is ¸ AC c x! ¸ as in (c). Similarly, if A œ 0 and B Á 0, the
distance is ¸ CB c y! ¸ . If both A and B are Á 0 then L has slope c AB so M has slope AB . Thus,
L: Ax b By œ C and M: cBx b Ay œ c Bx! b Ay! . Solving these equations simultaneously we find the
point of intersection Q(xß y) with x œ

ACcB aAy! cBx! b
A# bB#

P to Q equals È(?x)# b (?y)# , where (?x)# œ

œ

A# aAx! bBy! bCb#
aA# bB# b#

#

#

BCbA aAy! cBx! b
.
A# bB#
#
#
#
#
bABy! cB x!
Š x! aA bB bcAAC
‹
# bB#

and y œ

#

#

cA y! bABx!
, and (?y)# œ Š y! aA bB bcABC
‹ œ

# bB#
#

! bCb
Thus, È(?x)# b (?y)# œ É aAx!Ab#By
œ
bB#

kAx! bBy! bCk
ÈA# bB#

The distance from

B# aAx! bBy! bCb#
.
aA# bB# b#

.

1.3 FUNCTIONS AND THEIR GRAPHS
1. domain œ (c_ß _); range œ [1ß _)
3. domain œ (!ß _); y in range Ê y œ

2. domain œ [0ß _); range œ (c_ß 1]
"
Èt

, t € 0 Ê y# œ

"

t

and y € ! Ê y can be any positive real number

Ê range œ (!ß _).

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Section 1.3 Functions and Their Graphs
4. domain œ [0ß _); y in range Ê y œ

"
1 bÈ t

17

, t € 0. If t œ 0, then y œ 1 and as t increases, y becomes a smaller

and smaller positive real number Ê range œ (0ß 1].
5. 4 c z# œ (2 c z)(2 b z)   0 Í z − [c2ß 2] œ domain. Largest value is g(0) œ È4 œ 2 and smallest value is
g(c2) œ g(2) œ È0 œ 0 Ê range œ [0ß 2].
6. domain œ (c2ß 2) from Exercise 5; smallest value is g(0) œ "# and as 0  z increases to 2, g(z) gets larger and
larger (also true as z  0 decreases to c2) Ê range œ < "# ß _‰ .
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.
9. y œ Ɉ "x ‰ c " Ê
(a) No (x € !Ñ;

(c) No; if x   ",

"
x

"
x

c "   ! Ê x Ÿ 1 and x € !. So,

"Ê

"
x

(b) No; division by ! undefined;
(d) Ð!ß "Ĩ

c "  !;

10. y œ É# c Èx Ê # c Èx   ! Ê Èx   ! and Èx Ÿ #. Èx   ! Ê x   ! and Èx Ÿ # Ê x Ÿ %Þ So, ! Ÿ x Ÿ %.
(a) No; (b) No; (c) Ị!ß %Ĩ
#

11. base œ x; (height)# b ˆ #x ‰ œ x# Ê height œ

È3
#

x; area is a(x) œ

"
#

(base)(height) œ

"
#

(x) Š

È3
# x‹

œ

È3
4

x# ;

perimeter is p(x) œ x b x b x œ 3x.
12. s œ side length Ê s# b s# œ d# Ê s œ

d
È2

; and area is a œ s# Ê a œ

"

#

d#

13. Let D œ diagonal of a face of the cube and j œ the length of an edge. Then j# b D# œ d# and (by Exercise 10)
D# œ 2j# Ê 3j# œ d# Ê j œ

d
È3

. The surface area is 6j# œ

6d#
3

14. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ
ˆx, Èx‰ œ ˆ m"# ,

"‰
m .

15. The domain is ac_ß _b.

#

œ 2d# and the volume is j$ œ Š d3 ‹

16. The domain is ac_ò _b.

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ẩx
x



"
ẩx

$ẻ#



d$
3ẩ 3

(x 0). Thus,

.


18

Chapter 1 Preliminaries

17. The domain is ac_ß _b.

18. The domain is Ðc_ß !Ĩ.

19. The domain is ac_ß !b r a!ß _b.

20. The domain is ac_ß !b r a!ß _b.

21. Neither graph passes the vertical line test
(a)

(b)

22. Neither graph passes the vertical line test
(a)

(b)

Ú xbyœ" Þ
Ú yœ1cx Þ
or
or
kx b yk œ 1 Í Û
Í Û
ß
ß
Ü x b y œ c" à
Ü y œ c" c x à

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Section 1.3 Functions and Their Graphs
23.

x
y

0
0

25. y œ œ

1
1

2
0

24.

x
y

0
1

1
0

2
0

"
, x0

26. y œ œ x
x, 0 Ÿ x

3 c x, x Ÿ 1
2x, 1  x

27. (a) Line through a!ß !b and a"ß "b: y œ x
Line through a"ß "b and a#ß !b: y œ cx b 2
x, 0 Ÿ x Ÿ 1
f(x) œ œ
cx b 2, 1  x Ÿ 2
Ú
Ý
Ý 2, ! Ÿ x  "
!ß " Ÿ x  #
(b) f(x) œ Û
Ý
Ý 2ß # Ÿ x  $
Ü !ß $ Ÿ x Ÿ %
28. (a) Line through a!ß 2b and a#ß !b: y œ cx b 2
"
Line through a2ß "b and a&ß !b: m œ !& c
c# œ
cx b #, 0  x Ÿ #
f(x) œ œ "
c $ x b &$ , #  x Ÿ &

c"
$

c$ c !
! c Ðc"Ñ œ
c" c $
c%
#c! œ #

(b) Line through ac"ß !b and a!ß c$b: m œ
Line through a!ß $b and a#ß c"b: m œ
f(x) œ œ

œ c "$ , so y œ c "$ ax c 2b b " œ c "$ x b

&
$

c$, so y œ c$x c $
œ c#, so y œ c#x b $

c$x c $, c"  x Ÿ !
c#x b $, !  x Ÿ #

29. (a) Line through ac"ß "b and a!ß !b: y œ cx
Line through a!ß "b and a"ß "b: y œ "
Line through a"ß "b and a$ß !b: m œ !c"
$c" œ
Ú
cx
c" Ÿ x  !
"
!xŸ"

f(x) œ Û
Ü c "# x b $#
"x$

c"
#

œ c "# , so y œ c "# ax c "b b " œ c "# x b

(b) Line through ac#ß c"b and a!ß !b: y œ "# x

Line through a!ß #b and a"ß !b: y œ c#x b #
Line through a"ß c"b and a$ß c"b: y œ c"

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$
#

19


20

Chapter 1 Preliminaries
Ú

"
#x

f(x) œ Û c#x b #
Ü c"

c# Ÿ x Ÿ !
!xŸ"
"xŸ$

30. (a) Line through ˆ T# ß !‰ and aTß "b: m œ

Ú
A,
Ý
Ý
Ý
cAß
f(x) œ Û
Aß
Ý
Ý
Ý
Ü cAß

! Ÿ x  T#
T
# Ÿ x T
T Ÿ x  $#T
$T
# Ÿ x Ÿ #T
x
#

31. (a) From the graph,
(b)

x
#

€1b

x € 0:

x
#

x  0:

x
2

œ T# , so y œ T# ˆx c T# ‰ b 0 œ T# x c "

!, 0 Ÿ x Ÿ T#
#
T
T x c ", #  x Ÿ T

f(x) J

(b)

"c!
TcaTẻ#b

4
x

c1c

x
#

ấ
4
x



1b

4
x

ấ x (c2ò 0) r (%ò _)

c 1 c 4x € 0
#
2xc8
0 Ê x c2x

€0 Ê

(xc4)(xb2)
#x

€0

(xc4)(xb2)
#x

0

Ê x € 4 since x is positive;
c1c

4
x

€0 Ê

x# c2xc8
2x

0 Ê

Ê x  c2 since x is negative;
sign of (x c 4)(x b 2)
b
b
c
ùùùùùùùùùùùùùùợ

c2
%
Solution interval: (c#ò 0) r (%ò _)

3
2
x c1  x b 1
3
2
x c1  x b 1

32. (a) From the graph,
(b) Case x  c1:

Ê x − (c_ß c5) r (c1ß 1)
Ê

3(xb1)
x c1

€2

Ê 3x b 3  2x c 2 Ê x  c5.
Thus, x − (c_ß c5) solves the inequality.
Case c1  x  1:

3
x c1



2
x b1

Ê

3(xb1)
x c1

2

Ê 3x b 3 € 2x c 2 Ê x € c5 which is true
if x € c1. Thus, x − (c1ß 1) solves the
inequality.
Case 1  x: xc3 1  xb2 1 Ê 3x b 3  2x c 2 Ê x  c5
which is never true if 1  x, so no solution
here.
In conclusion, x − (c_ß c5) r (c1ß 1).
33. (a) ÚxÛ œ 0 for x − [0ß 1)

(b) ÜxÝ œ 0 for x − (c1ß 0]

34. ÚxÛ œ ÜxÝ only when x is an integer.
35. For any real number x, n Ÿ x Ÿ n b ", where n is an integer. Now: n Ÿ x Ÿ n b " Ê cÐn b "Ñ Ÿ cx Ÿ cn. By
definition: ÜcxÝ œ cn and ÚxÛ œ n Ê cÚxÛ œ cn. So ÜcxÝ œ cÚxÛ for all x − d .

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Section 1.3 Functions and Their Graphs

21

36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.

37. v œ f(x) œ xÐ"% c 2xÑÐ22 c 2xÑ œ %x$ c 72x# b $!)x; !  x  7Þ
38. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB # b AB # œ 2# Ê AB œ È2Þ So,
#

h# b "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ c" Ê The equation of AB is
y œ f(x) œ cB b "; x − Ị!ß "Ĩ.
(b) xĐ œ 2x y œ 2xÐcx b "Đ œ c2x# b #x; x − Ị!ß "Ĩ.
39. (a) Because the circumference of the original circle was )1 and a piece of length x was removed.
x
x
(b) r œ )1#c
1 œ % c #1
(c) h œ È"' c r# œ É"' c ˆ% c
#

x‰
(d) V œ "$ 1 r# h œ "$ 1ˆ )1#c
†
1

x ‰#
#1

œ É"' c ˆ16 c

È"'1x c x#
#1

œ

4x
1

b

x# ‰
%1#

œ É 4x
1 c

x#
%1#

œ É "'%11#x c

x#
%1#

œ

È"'1xcx#
#1

a)1 c xb# È"'1x c x#
#%1#

40. (a) Note that 2 mi = 10,560 ft, so there are È)!!# b x# feet of river cable at $180 per foot and a"!ß &'! c xb feet of land
cable at $100 per foot. The cost is Caxb œ ")!È)!!# b x# b "!!a"!ß &'! c xb.
(b) Ca!b $"ò #!!ò !!!
Ca&!!b á $"ò "(&ò )"#
Ca"!!!b á $"ò ")'ò &"#
Ca"&!!b á $"ò #"#ò !!!
Ca#!!!b á $"ò #%$ò ($#
Ca#&!!b ¸ $"ß #()ß %(*
Ca$!!!b ¸ $"ß $"%ß )(!
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the
point P.
41. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, cyb lie on the same
vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !,
for any x.
42. Pick 11, for example: "" b & œ "' Ä # † "' œ $# Ä $# c ' œ #' Ä
faxb œ

#axb&bc'
#

c # œ x, the number you started with.

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#'
#

œ "$ Ä "$ c # œ "", the original number.


22

Chapter 1 Preliminaries

1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS
1. (a) linear, polynomial of degree 1, algebraic.
(c) rational, algebraic.

(b) power, algebraic.
(d) exponential.

2. (a) polynomial of degree 4, algebraic.
(c) algebraic.

(b) exponential.
(d) power, algebraic.

3. (a) rational, algebraic.
(c) trigonometric.

(b) algebraic.
(d) logarithmic.

4. (a) logarithmic.
(c) exponential.

(b) algebraic.
(d) trigonometric.

5. (a) Graph h because it is an even function and rises less rapidly than does Graph g.
(b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
6. (a) Graph f because it is linear.
(b) Graph g because it contains a!ß "b.
(c) Graph h because it is a nonlinear odd function.
7. Symmetric about the origin
Dec: c_  x  _
Inc: nowhere

8. Symmetric about the y-axis
Dec: c_  x  !
Inc: !  x  _

9. Symmetric about the origin
Dec: nowhere
Inc: c_  x  !
!x_

10. Symmetric about the y-axis
Dec: !  x  _
Inc: c_  x  !

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Section 1.4 Identifying Functions; Mathematical Models

11. Symmetric about the y-axis
Dec: c_  x Ÿ !
Inc: !  x  _

12. No symmetry
Dec: c_  x Ÿ !
Inc: nowhere

13. Symmetric about the origin
Dec: nowhere
Inc: c_  x  _

14. No symmetry
Dec: ! Ÿ x  _
Inc: nowhere

15. No symmetry
Dec: ! Ÿ x  _
Inc: nowhere

16. No symmetry
Dec: c_  x Ÿ !
Inc: nowhere

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23


24

Chapter 1 Preliminaries

17. Symmetric about the y-axis
Dec: c_  x Ÿ !
Inc: !  x  _

18. Symmetric about the y-axis
Dec: ! Ÿ x  _
Inc: c_  x  !

19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the
function is even.
20. faxb œ xc& œ

"
x&

and facxb œ acxbc& œ

"
ac x b&

œ cˆ x"& ‰ œ cfaxb. Thus the function is odd.

21. Since faxb œ x# b " œ acxb# b " œ cfaxb. The function is even.
22. Since Ịfaxb œ x# b xĨ Á Ịfacxb œ acxb# c xĨ and Ịfaxb œ x# b xĨ Á Ịcfaxb œ caxb# c xĨ the function is neither even nor
odd.
23. Since gaxb œ x$ b x, gacxb œ cx$ c x œ cax$ b xb œ cgaxb. So the function is odd.
24. gaxb œ x% b $x# b " œ acxb% b $acxb# c " œ gacxbß thus the function is even.

25. gaxb œ

"
x# c "

26. gaxb œ

x
x# c " ;

27. hatb œ

"
t c ";

œ

"
acxb# c"

œ gacxb. Thus the function is even.

gacxb œ c x#xc" œ gacxb. So the function is odd.

h a ct b œ

"
ct c " ;

ch at b œ

"
" c t.

Since hatb Á chatb and hatb Á hactb, the function is neither even nor odd.

28. Since l t$ | œ l actb$ |, hatb œ hactb and the function is even.
29. hatb œ 2t b ", hactb œ c2t b ". So hatb Á hactb. chatb œ c2t c ", so hatb Á chatb. The function is neither even nor
odd.
30. hatb œ 2l t l b " and hactb œ 2l ct l b " œ 2l t l b ". So hatb œ hactb and the function is even.
31. (a)

The graph supports the assumption that y is proportional to x. The
constant of proportionality is estimated from the slope of the
regression line, which is 0.166.

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Section 1.4 Identifying Functions; Mathematical Models
(b)

25

The graph supports the assumption that y is proportional to x"Ỵ# .
The constant of proportionality is estimated from the slope of the
regression line, which is 2.03.

32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the
regression line.

The graphs support the assumption that y is proportional to $x . The constant of proportionality is estimated from the
slope of the regression line, which is 5.00.
(b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from
the slope of the regression line, which is 2.99.

33. (a) The scatterplot of y œ reaction distance versus x œ speed is

Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is
approximately 1.1.

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