Page 1
Section 5.1
1. 25
2. 2187
3. 1024
4. 256
5.
64
27
6. 729
9. 2.139  105 10. 9.72 × 106
14. 2.5 × 101
21.
1
12
15. x6
18
22. b
23. x y
24.
y 6n2
x
z
2n4
512
19, 683
8.
1
125
12. 2.0 × 103
13. 1.0 × 106
9
27r 6
18. 1 19. 14 2 20. 3 3
q s
x y
11. 1.6 × 105
2m7
45c13
16.  6 17.
n
d5
n2 2n2
7.
25. x n  4 y n  4
26.
4
3
3
1
ab
9b
3b 1 2
8
2n
10
a) 4 b) 2
c) x  1 d) 5 e) 2 f) x g) 3  12 x h)
 ab i) 3a-4a 2
2
ab
a b
4 a 2
a3
2 2
1
1
j )2 a  (30 4a )13 k) 2 l) 1/2 m) 25/9 n) 1/4 o) -8 27. a )a 2 b 2 (a  b)
Section 5.2 through 5.9
1.
2 2  3 1 0  6 20
4 2 6 12 12
2 1 3 6
6 32
2.
3  8 5 0 14
24  87 261
1

1
b)a 2 b 2 (1  ab)
3.
As x  , f ( x)  
As x  , f ( x)  
 8 29  87 275
f (2)  32
4.
As x  , f ( x)  
f (3)  275
5.
As x  , f ( x)  
As x  , f ( x)  
As x  , f ( x)  
6.
As x  , f ( x)  
As x  , f ( x)  
7. 11x 2  2 x  2 8a. 4 x3 ; 16 x3 ; 36 x3 8b. 56 x3  448
8c. x  2
8d. r  4, h  2; r  8, h  2; r  12, h  2
( x  2)( x  1)( x  2)  0
( x  4)( x  3)( x  3)  0
10.
9.
x3  x 2  4 x  4  0
x 4  4 x3  9 x 2  36 x  0
( x  6  5)( x  6  5)  0
x( x  5  2 3)( x  5  2 3)  0
12.
11.
x 2  12 x  31  0
x3  10 x 2  13 x  0
28
8 x  71
14. 5 x 2  20  2
13. 2 x 2  4 x  9 
2x  3
x 4
2
107
10 x  10 x  9
2
2
16.
3
x

6
x

1
17.
x

x


15. 2 x  3
2
8
28
x4
x  x2  5
34
1
18. 6 x 2  12 x  17 
19. 2 x 2  5 x  1 20. 3x 4  3x3  5 x 2  5 x 
x 1
x2
2
21. f ( x)  ( x  2)(2 x  5)(2 x  5) 22. f ( x)  ( x  4)( x  2)( x  2 x  4)
Page 2
23. f ( x)  ( x  1) 2 ( x  4)( x 2  x  1)
24. no other zeros 25. other zeros 
2
2
and 
3
3
1
3
27. remaining factors are (2 x  3) and ( x  1)
and 
3
2
28. remaining factors are (4 x  3) and (2 x  1)
29. remaining factors are (2 x  1) and ( x  5)
30. remaining factors are (4 x  1) and (3x  2) 31. k  210 32. k  42
26. other zeros 
34. f ( x)  ( x  1)( x  2) 2
33. f ( x)  2( x  2)( x  1)( x  2)
Section 6.1
1. 633/5 2. (25)4/3 3. 1247/6 4.
 3 57 
4
5.
 13 
3
6.
 8 204 
5
7. 9 8. 64 9. 16
1
1
1
1
12. 2 13. 
14.
15. 
16. 186.01 17. 22.89
125
8
59, 049
3,125
18. 47.29 19. 132.26 20. 0.14 21. 32 22. 32.15 23. 0.09 24. 64311.53
25. 4.21 26. 2.34 27. 0.18, 5.82 28. 3.21 29.  2.06 30. 1.97, 0.63
10. 216 11.
31. 1.32 32. 1.78 33. 3.58
34. 6.19 cm
3
 3  3i 3 
37. a. 
  27
2


and n is even
35. 1.81 in.
36.
n
a n  a when a < 0
3
 5  5i 3 
b. 
  125
2


c. cube roots of 1
2  2i 3
1  i 3
 1  i 3 ; cube roots of 64 are 4,
; cube roots of 8 are 2,
2
2
4  4i 3
 2  2i 3
2
Section 6.2
6
70
1. 400 4 5 2. 33/2 3. 76/5
4. 51/2 5. 4 6. 24 6 7.
9.
8. 9 5 81
3
35
44
3
5
3 2
45
x x
16. x1 4
10. 4 1331  113 4 11.
12. 2
13.
14. x 3 3 15.
y2
4
3
are 1,
17. x1 3 y1 6
4
18.
8 x 4 y1 5
z3
2
23. 2 3
24. 3x z
29. 2 4 2
30.
5
Section 6.3
1. 2x2 + x  x1/2  4
5. 2x2  4x1/2 + 5
3
19.
xyz
2
z1 4 3
x1 2 y1 3
20.
y9 2
8 x3 2 z 3 2
3
25.
8
x
31. y 8 8 x
2. 6x2 + 2x1/2  5
26.
21. 9 x 4 4 x 3
50 x  25 x
5
27. 2 3 x
22. x 23 120
28. 4x 2 z xyz
32. 3ab 6 2ab3
3. 4x2 + x  5x1/2 + 1 4. 2x2  x + 7x1/2  6
3
6. 4x2 + x  3x1/2 + 1 7. 3x5 3  1 3
8. x5/2 + x1/2 9. 3x1/6
x
Page 3
 x 7 3  x1 3
1
 x5 6
2  x1 2
11. x3 2  1 2 12.
13.
; positive real numbers
3
x
3
x
12
2x2  7 x  5
 3 
14. 
; all real numbers
 ; all real numbers greater than 1 15.
9
 x 1 
1
1
2x2  x 1
;
all
real
numbers
<
0
and
<>
17.
; all real numbers
16.
12
2
2
3

x
2x


10.
18. x1 4 ; positive real numbers
3
25. Sample answer:f(x) = x , g(x) = x + 1
26. Sample answer: f(x) = x3 + 2, g(x) = x
27. Sample f(x) =
x 1
g(x) = x  1
x2
Section 6.4
1.
1
1
1
)
 1  x  1  1  x, x  1; f ( f 1 ( x))  f (  1) 
1
x 1
x
x 1
1
1
1
1
is f -1 ( x)   1.
  x, x  0 Therefore, the inverse of f(x)=
1
1
x
x-1
11
x
x
f 1 ( f ( x))  f 1 (
2.
2x  3
4
3
x
2

3
4(2 x  3)  3( x  4) 8 x  12  3 x  12 5 x
g 1 ( g ( x))  g 1 (
) x4



 x, x  4;
2x  3
x4
2( x  4)  (2 x  3)
2x  8  2x  3
5
2
x4
4x  3
2
3
4x  3
2(4 x  3)  3(2  x) 8 x  6  6  3 x 5 x
g( g 1 ( x))  g (
)  2 x



 x, x  2
4x  3
2 x
(4 x  3)  4(2  x) 4 x  3  8  4 x 5
4
2 x
2x  3
4x  3
Therefore, the inverse of g(x)=
is g -1 ( x) 
.
x4
2 x
3. a) f 1 ( x)  3  x , x  0; y  3
b) f 1 ( x)  3  x , x  0; y  3
f
f
f^-1
f^-1
O
O
Page 4
4. a) f 1 ( x)  2  x  2, x  2; y  2
b) f 1 ( x)  2  x  2, x  2; y  2
f
f^-1
f
O
O
f^-1
5. a ) f 1 ( x )   4  x 2 , 0  x  2; 2  y  0
b) f 1 ( x )  4  x 2 , 2  x  0; 0  y  2
f^-1
f
O
O
f^-1
6. a ) f 1 ( x ) 
f
5
x2  5
, x  0; y 
2
2
5
x2  5
, x  0; y 
2
2
b ) f 1 ( x ) 
f^-1
f^-1
O
f
f
O
7.
a ) f ( x)   x  2, x  2, y  0; f 1 ( x)   x  2, x  0; y  2
b) f ( x)  x  2, x  2, y  0;
8. f ( x) 
x2
, x  2, y  1;
x2
9. f ( x) 
3x  4
3
3
,x  ,y  ;
2x  3
2
2
f 1 ( x)  x  2, x  0; y  2
f 1 ( x) 
2( x  1)
, x  1, y  2
x 1
f 1 ( x) 
3x  4
3
3
,x  ,y 
2x  3
2
2
10. k = 3
11. f (0)  3, f (1)  0, f (1)  4, f 1 (0)  1, f 1 (3)  0, and f 1 (4)  1
Inverse Functions
1-6 Show that f (g(x)) = x and g( f (x)) = x
1
7. f (x)
 31 x
2 2
1
8. f (x) = 5x15
1 2 5
1
10. f ( x)  2 x  2 , x ≥ 0
1
7 x
11. f ( x)  4
1
2
9. f (x) = x + 3, x ≥ 0
x 1
1
12. f (x)  4 , x ≥ 1
Page 5
13. f
1. 3.
1
( x) 
2.
4
3x  1
1

2
14.
1
4
f ( x)  x 5 
5
5
1
3. Sample answer:
1
15. f (x) = x  1, x ≥ 1
4. Sample answer:
1
f 1 = x  2; domain x ≥ 0
; f = x  1 + 3;
domain x ≥ 1
Section 6.5
1.
domain: x ≥ 4, range: y ≥ 2
3.
2.
domain: x ≥ 2, range: y ≥ 3
4.
domain: x ≥ 2, range: y ≤ 2
domain: x ≥ 0, range: y ≤ 3
5.
6.
domain and range: all real numbers
domain and range: all real numbers
8.
domain and range: all real numbers
domain and range: all real numbers
7.
Page 6
9.
;
; domain and range: all real numbers
10. (0, 0) and (1, 1)
11. On the interval 0 to 1, the larger the root, the steeper the graph. On the interval 1 to ,
the larger the root, the less steep the graph.
13. (1, 1), (0, 0) and (1, 1)
12.
14. On the interval 1 to 1, the larger the root, the steeper the graph. On the intervals  
to 1 and 1 to , the larger the root, the less steep the graph.
15.
16.
O
b)
a)
17.
O
a)
O
O
c)
O
b)
O
c)
Page 7
18.
a)
O
O
O
b)
c)
O
d)
Section 6.6
15
1.
2. 21 3. 4 4. 5 5.  39
2
397
11.
12. 2 3 3 13. 1,000 14. 6
81
7
18.  23 19. 3 20.
21. −5
3
3
25
26. 
27. no solution
28.
4
12
3
3. 1
4. 0, ,1 5. a) k =2 b) k<2
5
Section 7.1
1.
2.
6.  3 7. 6
8. 27
9. 120
10. 169
15. 9 16. 3  6 3 6 17. no solution
22. 3
23. no solution
29. 12 in.
c) k>2
30. 24 in
6.
81
25. 0
16
3 2
1 5 5
1. 
2.
2
2
24.
9  4 66
 1.566
15
3.
D :  ,   ; R :  3,  
D :  ,   ; R : 1,  
D :  ,   ; R :  ,3
4.
5.
6.
Page 8
D :  ,   ; R :  5,  
D :  ,   ; R :  , 2 
D :  ,   ; R :  1.5,  
7. sample answer: y  3x 1  3 8. y-intercept of 1. As a get larger the graph gets
steeper. 9. $8,396.71 10.$8,719.45 11.$8,121.19 12. 2.13% 13. y = 12,941,197
(1.0213)t
1
14. 19,726,093 1. y  64(8) x 2. y  3(4) x 3. y  (5) x 4. y  3(3) x
2
Section 7.2
1. decay
2. growth 3. growth 4. decay 5. decay 6. growth
7.
8.
9.
D :  ,   ; R :  2,  
D :  ,   ; R :  3,  
D :  ,   ; R : 1,  
10.
11.
12.
D :  ,   ; R :  3,  
D :  ,   ; R :  5,  
D :  ,   ; R :  , 2.5 
t
13. V = 175,000(0.82)
16. after 5 yr
18. a. V  6000t  30,000
c.
14. $24,053.41
15.
17. V= 1600(0.8)t
b. V  30,000(0.775)t
During the first 2 years, the exponential model represents a faster depreciation.
d. Book value after 1 year: Linear model: $24,000Exponential model: $23,250Book
value after 3 years: Linear model: $12,000Exponential model: $13,965e. According to
the linear model, the car will have no value after 5 years. According to the exponential
model, there will never be a time when the car will have no value because the x-axis is a
horizontal asymptote of the graph of the model. f. Linear model: An advantage for the
buyer is the older the car, the less the buyer would have to pay compared to the
exponential model. A disadvantage is if the buyer would resell the car, they would lose
more money compared to the exponential model. An advantage for the seller is during
the first two years, the seller would be able to sell the car for more money compared to
the exponential model. A disadvantage is after 5 years the car is not worth anything and
would then be difficult to sell.
Page 9
Exponential model: An advantage for the buyer is during the first two years, the buyer
would have to pay less for the car compared to the linear model. A disadvantage is if the
buyer buys the car after two years, they would pay more compared to the linear model.
An advantage for the buyer is that after 5 years the car is still valuable compared to the
linear model. A disadvantage is that if the car were sold during the first two years, the
seller would get less money for it compared to the linear model.
Section 7.3
1. 4e10
2.
9
e
11.
3. 27e12
4
4. 32e6 x
5.
e9 x 3
8
3
6. 2e3 x 2e x
12.
13.
D :  ,   ; R :  3,  
D :  ,   ; R : 1,  
D :  ,   ; R :  2,  
14.
15.
16.
D :  ,   ; R : 1,  
D :  ,   ; R :  4,  
D :  ,   ; R :  3,  
17. exponential growth
18.
19. 3
2
20. 23
x
x
 e x  e x 
e 2 x  2  e 2 x
2  e e
; [ g ( x)]  
22. [ f ( x)]  
 

2
4
2



e2 x  2  e2 x e 2 x  2  e 2 x 4
[ f ( x)]2  [ g ( x)]2 

 1
4
4
4
Section 7.4
2
1/2
1. 4
9.
3
2
2
4
2. 3  81
10. a)
b) 3
1
3.  
4
c) 2
3
d)
4. 3
 64
125
x4
e)
et
k
5.
3
2
f )1000 2
21. 21
2

e 2 x  2  e 2 x
 
4

6.
5
2
7.
2
3
13. f 1 ( x)  7 x
8. 
1
4
Page 10
3x
1
15. f 1 ( x)   
14. f
4
2
1
1
18. f 1 ( x)  x 
3
3
11.
12.
1
x2
16. f 1 ( x)  e x 2
( x) 
D :  2,   ; R :  ,  
1. log5 125 = 3
2.
a) x  3 b) x  3
13.
D :  2,   ; R :  ,  
1
 3
64
log 4
c) x  1
17. f 1 ( x)  e x 3  1
D : 1,   ; R :  ,  
3. ln 7.3890 = 2
4. ln 0.6065 = 0.5
d ) x  0 and x  1 or(0,1)  (1,)
Section 7.5
1. 3.332 2.0.560 3. 2.772
4. 3.738
3
7. log 4 x  8. log x
2
10. 3 log2 x  log2 y  2 log2 z
5. 1.659
6. 0.980
 log y  log z
9. log7 y  log7 z  log7 3
3
1
11. ln x  ln y 12. log 2  log x  2 log y 13. ln 7  ln y  2 ln x
2
2
14. 5[log9 2  2 log9 x  log9 y  log9 z] 15. 2 log6(x  3)  log6 3  3 log6 y
16. log3
5
3
21. ln
3x 4
( x  1) 2

17. log
25 y
18. log 2
x2
22. 1.869
26. log 4 x5 ( x  2)( x  2)3

9
x3 y
19. ln
23. 1.032
27. log
( x  3) 2
6( x  2)3
24. 2.665
1036 ( x 2  4)6 ( x  1)12
x2
2
20. log
25. ln 3
( x  4)3
6 x3
( x  1) 2 ( x  8)
28. ln 4
x( x 2  1)
x3 y 3 z 3
125w9v 6
1. –0.8 2. –0.5 3. 2 4. 16 5. 4 6. –2.5 7. ln (x  1)  4 ln x
1
1
1
8. [3log 6 x  log 6 ( x 2  9)] 9. log 2  4 log x  log 6  log y  log 3  5log w  log z
4
2
2
1
10. ln 4  2 ln y  ln(2 x 2  y 2 ) 15. ln 2 0.6931, ln 3  1.0986, ln 4  1.3862, ln 5 
2
1.6094, ln 6  1.7917, ln 8  2.0793, ln 9  2.1972, ln 10  2.3025, ln 12  2.4848, ln 15
 2.7080, ln 16  2.7724, ln 18  2.8903, ln 20  2.9956
Page 11
Section 7.6
1) 2 2) 4 3) 1  1  e 4
4) 4,
7
7
10) e, e 2 11)
2
3
2 ln 5  2 ln 3
17) 1
16)
ln 5  3ln 3
9) 1,
1
4
5)
15
2
6)
11
2
13) ln(2  5)
12) 0
18) 16
19) 81
7) 1, e 2 8) 1, 9, 5  34
14) 6  log 38 15) 2 ln 75
20) a) 0.567
b) 2.787
Section 7.7 – Practice B
1. y 
1 x
(3)
2
31
6. y   
43
11. y = 3x3
1
2. y  (2) x
5
3. y  3(4) x
1
4. y  8  
2
x
2
5. y  5  
3
x
x
7. y = 37.86(6.85)x
12. y 
1 2
x
2
8. y = 0.0019(1.33)x
13. y = 2x1.5
1 2.5
x
17. y = 12.94x3.3
2
20. 0.81x0.79;$6.14
16. y 
14. y = 4x2
18. y = 0.014x1.05
9. y = 62.12(0.03)x
15. y 
1 0.5
x
4
19. y = 93.34x2
Section 7.7 – Practice A
7. yes 8. yes 9. no 19. yes 20. no
Section 7.7 – Practice C
4.
; y = 3(1.8)x
5.
; y = 0.9(4x)
Exponential and Logarithmic Application Problems
1. $789.24 2. $156.83 3. $5913.70 4. 7.123%(semiannually); 7.251%(cont.)
5. 6.882%( semiannually); 6.7661%(cont.) 6. 7.177% 7. 6.960
1
( t ln 2)
3
b) 4.755 hr c) 6 hr 9. 33062.445 years 10a). 18.616 min later
8. a) N (t )  N 0 e
10b) The temperature of the object will approach the room’s temperature of 30°C.
11. b) N (t )  0.033860(1.94737) x c) N (t )  0.033860e0.66648 x e) 1.847 f) 6.193 hr
later
Exponent and Logarithm Practice
1. $6,414.27 2. $4,429.41 3. 7.727 years 4. $4,931.94 5. $4,204.80 6. 19.56%
7. 0.0488 ml 8. 10,400 years 9. 209 days 10. 75.044 pascals 11. no, it is 3,561
years
Page 12
12. 1971 earthquake was 100.7 times more intense than 1987 earthquake
(Note: 100.7  5.012)
ln 0.8
13. 1,396.7 volts 14. pH = 7 15. 8.71 minutes 16a. k 
16b. 1:42 p.m.
60
17. 232 P 18. 6,031,455,620 years
Section 8.1
1. direct
2. neither
3. inverse 4. inverse
5. y 
8.64
; 34.556
x
6. y 
15
; 60
x
5
; 2.5 8. direct 9. neither 10. inverse 11. z  18 xy; 540
8x
15
25
125
43725
xy;
14. 10 yr 15. 43,725 16. d 
12. z  xy; 225 13. z 
2
84
14
p
17. 7950 units
18. 0.22; H = 0.22mT 19. 4.541 kilocalories
7. y 
Variation Word Problems
1890
1. a. Inversely; f 
b. inch-pounds
l
31.5 pounds
1e.
0
d. 6.3 inches, 37.8 inches
2b.
c. 630 pounds, 126 pounds, 63 pounds,
2. a. v 
16560
c. 60 pounds
p
d. no, 16560/p can never equal
e. Yes. The equation can be rewritten in the form vp  16560 so any product of volume
and pressure pairs will be equal to the same number and therefore to each other.
Page 13
3. a. s 
4000
d2
b. 40 units
c. 400,000 units
proportional to the square of the distance; I 
320
d2
4. a. The intensity is inversely
b. 1.25 mr/hr; 3.2 mr/hr; 32,000
1 2
7
s ; r
s
b. b(60) = 24 m; r(60) = 14
150
30
m; b(120) = 96 m; r(120) = 28 m; b(180) = 216 m; r(180) = 42 m
1 2 7
s  s ; t(60) = 38 m; t(120) = 124 m; t(180) = 258 m
c. t 
150
30
mr/hr c. 25.2982 meters
5 a. b 
4d.
5d.
940 1 football field
m
 2.856 fields 6 a. s  6.492 7 p
b. 33.123 knots
3
109.7 m
c. No, because the power is inside the 7th root d. It would require too much power
7. a. d  1.297 h3 2 b. 458.56 cm; more than 10 times as big
c. 83.24 m
8a. because the time divided by the number of slices is not a constant
b. t  1.225n0.5146 c. 3.572 minutes; 1.225 minutes
125bd 2
d. about 38 slices 9. a. s 
b. 250 pounds
l
c. 12,000 pounds d. i. 2 ii. 4 iii. ½ 10. a. f  2500 pr 4
b. 163.84 cubic millimeters per second c. 244.14 units
k
k
80, 000
b. g  2 c. g 
11. a. w  k1s 2 ; g  2
s
s2
w
d. 5.5556 km/l; 3.125 km/l; 2 km/l
e. No because the km/l would be too large
e.
Section 8.2
1. x = 1; y = 5
2
6. x   ; y  5
3
7.
2. x = 2; y = 3
8.
1
3
3. x  ; y 
2
2
1
4. x  ; y  2
4
9.
5. x = 2; y = 1
Page 14
D : x  4; R : y  3
D : x  0; R : y  2
2
D: x  ; R: y  4
5
10.
11.
12.
D : x  1; R : y  1.5
D : x  0.25; R : y  1.25
13. y 
2 x  12
x4
14. y 
6 x  14
x3
2
4
D: x  ; R: y 
7
7
0.75 x  12.5
b) 0  x  950
15. a) C ( x) 
x  50
d) As the tank is filled, the rate at which the concentration
of brine is increasing slows. The concentration of brine
appears to approach 75%.
c)
Section 8.3
1.
3
3
5
a ) D : x  4, x  4; x  int : 2, ; y  int : ; H . A : y  ; V . A : x  4;
5
16
2
f(x) does not have a removable discontinuity point;
74
The graph of f(x) crosses its horizontal asymptote at x = .
7
3
b) D : x  3, x  7; x  int : 3; y  int : ; H . A : y  1; V . A : x  7;
7
3
f(x) has a removable discontinuity point (3, );
5
The graph of f(x) never crosses its horizontal asymptote.
Page 15
3
c) D : x  2; x  int : 3; y  int : ; H . A : y  0; V . A : None;
4
5
f(x) has a removable discontinuity point (2, );
12
The graph of f(x) crosses its horizontal asymptote at (-3, 0).
d ) D : x  0; x  int : None; y  int : None; H . A : None; V . A : x  0; Oblique Asy: y 
1
x 1
4
3
1
2
3
1
e) D : x  ; x  int :  , 2; y  int : ; H . A : None; V . A : x  ; Oblique Asy: y  2 x 
2
4
3
2
2
f ) D : x  0, x  2, x  1; x  int : 2; y  int : None; H . A : None; V . A : x  0, x  1; Oblique Asy: y  x  1;
g ) D : x  1; x  int : 1.109(calculator needed ); y  int : 2; H . A : None; V . A : x  1; Oblique Asy: None
O
a)
O
b)
O
c)
O
d)
O
O
e)
f)
Page 16
O
g)
20
c)3.169 cm d )9.466 cents
r
( x  3)(2 x  54)
3. a) A( x) 
b) (0, ) c) 9in. by 6in. d ) 96 in 2
x
5
4. a. The graph of f(x) crosses its horizontal asymptote at ( , 0) .
2
41
b. The graph of g(x) crosses its horizontal asymptote at ( , 2) .
6
c. The graph of h(x) never crosses its horizontal asymptote.
Section 8.4
x7
x 2  3x  9
15 y
55 x 7 y 4
1.
5.
2. not possible 3.
4.
2x
x3
4x
32 z 3
2. a)C (r )  0.1 r 2 
( x  5)( x  4)
7.
3 x( x  7)
13.
2.
8.
x( x  6)
(2 x  1)( x  2)
5(t  5)
4t 3
3.
2(2 x  3)
9. 1
x 2 ( x  2)
( x  2)( x  6)
14.
2x
x4
( x n  1) 2
4.
5. horizontal asymptote: y = 1;
vertical asymptote: x = 1
6. horizontal asymptote: y = 0;
vertical asymptote: x = 3
(2 x  5)( x  4)
10.
3( x  2)
15.
2x
x 1
( x n  8)( x n  5)
x n ( x n  1)
1.
6. 2x
x
x2  1
11.
12.
x 1
4 x( x  2)
(4 x  1)(5 x  6)( x  2)
(4 x  5)( x  1)( x  6)
Page 17
7. horizontal asymptote: y = 0;
vertical asymptotes:x   5
8. horizontal asymptote: y = 0;
vertical asymptotes: x  1
9.
10.
1
xh
13. a) 
8
11.
1
xh  x
x
b) about 6,594 gal
Section 8.5
1. 2x(x + 3)(x  3)
1
x  h 1  x 1
12.
1
xh2  x2
c) 30 ft by 15 ft by 5 ft
2. 2(x  3)(x+ 4)
3. 3(x  1)(x  2)(x  3)
x4
x 2  x  13
3x 2  2 x  3
6. 3x  3x  2  7.
4. (x + 1)(x  3)(x + 5)(x + 6) 5. x  5x  2 
x2
2
6
x 11x  33
6 x
2
2
8.
9.
10.
11. 7 x  22 x  9 12. 7 x  13 x  1
2
x 1
3xx  2
x  1x  1
32 x  1x  3
2x  1x 2  x  1
13. x  6
14. 5 x  1
15. x  3x  5 16.
2x 2  4 x  3
6
2
x  1
x 2 10 x  26
xx 3  x 2  2 x  2
3
x
4. 2x  2x  4
3
5. a.domain: all real numbers except x = 1, range: all real numbers except y = 0
1.
2x 1
2x
2. 
1
t2 t2 1
3.
6
x
b. domain: all real numbers except x = 0 and x = 1 ; f ( f ( x)) 
x 1
x
c. f ( f ( f ( x)))  x ; the graph is not a line because the graph has holes at x= 0 and x = 1.
6. A = 4, B = 2, C = 2
4
2
2
4( x  1)( x  1)  2 x( x  1)  2 x( x  1)




x x 1 x 1
x( x  1)( x  1)
Page 18
4( x 2  1)  2 x 2  2 x  2 x 2  2 x
2
x( x  1)

4 x 2  4  4 x 2
3
x x

4
3
x x
Section 8.6
1. no solution
2. 23
3.
9. no solution 10. 9, 3
1. 1 2. 2, 5 3. 6
4( x  1)( x  2)
9.
( x  3)( x  4)
1
2
4.
11. 3, 1
4. 8
5. 6
9
5
5. 8
6. 1, 7
7. no solution 8. 8
1 5
,
13. 3 14. 6, 12
2 2
6. −2
8. −2; 0 and 3 are extraneous.
12.
Applications of Rational Functions
550  92 15
550  92n
1a.
 128.67 1b. C (n) 
n
15
1c.
1d. The asymptotes of the rational function are n  0 and C  92
1e. The yearly expense of electricity continues no matter how many years the
refrigerator works. The cost will never go below the $92, but the cost approaches $92.
550  92n
1200  92n
1f. Graphing the two functions C (n) 
and C2 (n) 
together or
n
n
reviewing a table of values will show the more expensive refrigerator remains more
expensive annually although both approach $92 as n approaches infinity.
2a.
2b. 13.76 micrograms which occurs at 18.2 minutes
2c. The graph shows that the concentration is 0 at time  0 . Within the first 15 minutes,
the concentration rises sharply to the maximum at 18.2 minutes. After reaching the
maximum, the kidneys begin cleansing the blood and rapidly remove the drug.
2d. There is only one asymptote in this rational function which is at C  0 . Oddly, this
value actually exists in the range of the function at C  0 ; however, the graph tends
toward C  0 as t  values increase to infinity. This characteristic gives the asymptotic
behavior.
Page 19
2e.
3a. The domains for both are x  0 .
3b. asymptotes for both are x  0 and y  5
3c. Both Tiffany and Adam are graphing the same function
3d. Answers will vary
4c.
4a. 30 mph  3.33 hours; 55 mph  1.82 hours; 65 mph  1.53 hours. 4b. y 
100
x
1
is the
x
1
reciprocal. Therefore the sum of a number and its reciprocal is represented by x  .
x
1
Graph the function y  x  for positive x-values. The minimum occurs at the point
x
(1,2).
5. If x is the number, then
4d. Faster you travel, fewer hours on the road
Section 12.1
1 2 3 4 5
1. a) 0, , , , ,
2 3 4 5 6
2. a ) an 
en
n
b) an 
3. a) 1, 2, 6, 24,120, 720
2 1 2 1
b) 2, 1, ,  , , 
3 2 5 3
1
3n 1
c) an  2n  1
b) 1,1, 2,3,5,8
1 1
c) 1, 2, , 4, , 6
3 5
d ) an  (1) n  n 2
4. a )
n
1
1
1
e) an  (1) n 1
1
1
n
1
 k  1  2  3  ...  n -1  n
k 1
b)
n
 k !  1  2! 3! ...  (n -1)! n!
k 1
5. a)
9
k
k 1
2
b)
n
1
2
k 1
n -1
6. a) 45 b) 39 c)  32
Section 12.2
1. arithmetic 2. sn 1  sn  3(n  1)  5  (3n  5)  3, s1  5, d  3 .
3. a13  50 4. a) a1  96, d  3 b) an  96  (n  1)(3)  3n  99
n(a1  an )
3
195
c) Sn 
  n2 
n
5. 1305 6. 2360 7. x  1
2
2
2
Page 20
Section 12.3
3 32 33
315 1
1 3(1  315 )
1
1
1.    ... 
 (3  32  33  ...  315 )  
  (1  315 ) or (315  1)
9 9 9
9 9
9 1 3
6
6
n 1
n
2
1 5 5
5
1
1 (1  5 )
1
1 n
2.    ... 
 (1  5  52  ...  5n 1 )  
  (1  5n ) or
(5  1)
9 9 9
9
9
9 1 5
36
36
4
(1  316 )
4(1  315 )
2
2
 2(1  315 ) or 2(315  1) 4. 3
  (1  316 ) or (316  1)
3.
1 3
3
3
1 3
3 9 27
1
1
4
5
20
 ... 



5. 1   
6.
7. 1, -4
3
3 7
1 3
4 16 64
1  ( ) 1 
1
4
4
4
4
n
5 (1  5 )
1
1

8. a) r = 5, an  54  5n 1  5n 5 b) S n 
(1  5n ) or
(5n  1)
1 5
2500
2500
Section 12.4
75
55
3
1. 2 2.
4. 10 5. doesn’t exist 6.
3. 
7. 3 8. doesn’t exist
56
2
2
9. –72 10. 162 11. doesn’t exist 12. 2  x  4 and x  3 13.
3  x  1 and x  2
1 5
15.
2
1000(1.03)n 1
14.
c)
2
16. a) an  1000(1.03) n1
b) pn  50(1.12) n1
; Since the price of the shares are increasing faster than the investment
50(1.12) n 1
amount, the number of shares you are able to buy will decrease from the initial 20 shares
each year.
Section 12.5
1. 9, 37, 149, 597, 2389, 9557 2. –4, 0, 9, 25, 50, 86
6.
4. a1  5, an  an1  7 5. a1  7, an  3an1
3. 3, 3, 3, 2, 0, –4
a1  2, a2  3, an   an 2  an 1 
7. a1  5, an  nan1 8. x0  2, x1  f ( x0 )  1, x2  f ( x1 )  8, x3  f ( x2 )  19,
x4  f ( x3 )  62 ; the four iterates are 1,8, 19, 62 9. x0  5, x1  f ( x0 )  2,
x2  f ( x1 )  12, x3  f ( x2 )  82, x4  f ( x3 )  6312 ; the four iterates are
2,12,82, 6312
5
27
47
, x4  f ( x3 ) 
; the four
10. x0  2, x1  f ( x0 )  3, x2  f ( x1 )  , x3  f ( x2 ) 
2
10
18
5 27 47
iterates are 3, , ,
11.
2 10 18
1
11
x0  3, x1  f ( x0 )  2, x2  f ( x1 )  , x3  f ( x2 )   ,
2
2
Page 21
109
1 11 109
; the four iterates are 2, ,  , 
22
2 2
22
2000
4
12a. w1  210; wn  wn1  0.0035wn1 
 0.9965wn1 
3500
7
4
4
4
4
4
12b. w20  (210)(.9965)19  (.9965)18  (.9965)17  (.9965)16  (.9965)15  ...  
7
7
7
7
7
18

4
210(.9965)19    (.9965n )   206.988
7  n0

a
45
9
13. a1  a43  23  a4  144 14. a1  a65  56 

3125 625
5
15.
1
1 3
a1  1    1.5
2
1 2
1 3 2 17
 1.416
a2    
2 2 3 12
1 17 12 577
 1.414215
a3    
2 12 17 408
1 577 408 665857


 1.414214
a4  
2 408 577 470832
x4  f ( x3 )  
1 665857 470832 8.867311011



 1.414214
2 470832 665857 6.27014 1011
As n approaches , an approaches 2.
a5 