Chapter 1 Visualizing the Graph

H
B
D
A
G
I
C
J
F
E

Exercise Set 1.1

1. A: (−5, 4); B: (2, −2); C: (0, −5); D: (3, 5); E: (−5, −4); F: (3, 0)
3.
5.
7. (2008, 10), (2009, 15), (2010, 9), (2011, 9), (2012, 12), (2013, 5)
9. Yes; no
11. Yes; no
13. No; yes
15. No; yes
17. x-intercept: (−3, 0); y-intercept: (0, 5);
19. x-intercept: (2, 0); y-intercept: (0, 4);
21. x-intercept: (−4, 0); y-intercept: (0, 3);
23.
25.
27.
29.
31.
33.
35.
37.
39.
41. $\sqrt{10}, 3.162$
43. 13
45. $\sqrt{45}, 6.708$
47. 16
49. $\frac{14}{3}$
51. $\sqrt{128.05}, 11.316$
53. $\sqrt{a^{2} + b^{2}}$
55. 6.5
57. Yes
59. No
61. (−4, −6)
63. $(- \frac{1}{5}, \frac{1}{4})$
65. (4.95, −4.95)
67. $(- 6, \frac{13}{2})$
69. $(- \frac{5}{12}, \frac{13}{40})$
71. $(- \frac{1}{2}, \frac{3}{2}), (\frac{7}{2}, \frac{1}{2}), (\frac{5}{2}, \frac{9}{2}), (- \frac{3}{2}, \frac{11}{2});$ no
73. $(\frac{\sqrt{7} + \sqrt{2}}{2}, - \frac{1}{2})$
75. ${(x - 2)}^{2} + {(y - 3)}^{2} = \frac{25}{9}$
77. ${(x + 1)}^{2} + {(y - 4)}^{2} = 25$
79. ${(x - 2)}^{2} + {(y - 1)}^{2} = 169$
81. ${(x + 2)}^{2} + {(y - 3)}^{2} = 4$
83. (0, 0); 2;
85. (0, 3); 4;
87. (1, 5); 6;
89. (−4, −5); 3;
91. ${(x + 2)}^{2} + {(y - 1)}^{2} = 3^{2}$
93. ${(x - 5)}^{2} + {(y + 5)}^{2} = 15^{2}$
95. Third
97. $\sqrt{h^{2} + h + 2 a - 2 \sqrt{a^{2} + a h}}, (\frac{2 a + h}{2}, \frac{\sqrt{a} + \sqrt{a + h}}{2})$
99. ${(x - 2)}^{2} + {(y + 7)}^{2} = 36$
101. (0, 4)
103.
1. (0, −3);
2. 5 ft
105. Yes
107. Yes
109. Let $P_{1} = (x_{1}, y_{1}), P_{2} = (x_{2}, y_{2})$ , and $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ . Let $d (A B)$ denote the distance from point A to point B.

$\begin{array}{rcl} d (P_{1} M) & = & \sqrt{{(\frac{x_{1} + x_{2}}{2 - x_{1}})}^{2} + {(\frac{y_{1} + y_{2}}{2 - y_{1}})}^{2}} \\ = & \frac{1}{2} \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}; \\ d (P_{2} M) & = & \sqrt{{(\frac{x_{1} + x_{2}}{2} - x_{2})}^{2} + {(\frac{y_{1} + y_{2}}{2} - y_{2})}^{2}} \\ = & \frac{1}{2} \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} \\ = & \frac{1}{2} \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = d (P_{1} M) \end{array}$

Exercise Set 1.2

1. Yes
3. Yes
5. No
7. Yes
9. Yes
11. Yes
13. No
15. Function; domain: ${2, 3, 4};$ range: ${10, 15, 20}$
17. Not a function; domain: ${- 7, - 2, 0};$ range: ${3, 1, 4, 7}$
19. Function; domain: ${- 2, 0, 2, 4, - 3};$ range: ${1}$
21.
1. 1;
2. 6;
3. 22;
4. $3 x^{2} + 2 x + 1;$
5. $3 t^{2} - 4 t + 2$
23.
1. 8;
2. −8;
3. $- x^{3};$
4. $27 y^{3};$
5. $8 + 12 h + 6 h^{2} + h^{3}$
25.
1. $\frac{1}{8};$
2. 0;
3. does not exist;
4. $\frac{81}{53}$ , or approximately 1.5283;
5. $\frac{x + h - 4}{x + h + 3}$
27. 0; does not exist; does not exist as a real number; $\frac{1}{\sqrt{3}}$ , or $\frac{\sqrt{3}}{3}$
29.
31.
33.
35. $h (1) = - 2;$ $h (3) = 2;$ $h (4) = 1$
37. $s (- 4) = 3;$ $s (- 2) = 0; s (0) = - 3$
39. $f (- 1) = 2;$ $f (0) = 0;$ $f (1) = - 2$
41. No
43. Yes
45. Yes
47. No
49. All real numbers, or $(- \infty, \infty)$
51. All real numbers, or $(- \infty, \infty)$
53. ${x | x \neq 0}$ , or $(- \infty, 0) \cup (0, \infty)$
55. ${x | x \neq 2}$ , or $(- \infty, 2) \cup (2, \infty)$
57. ${x | x \neq - 1 and x \neq 5}$ , or $(- \infty, - 1) \cup (- 1, 5) \cup (5, \infty)$
59. All real numbers, or $(- \infty, \infty)$
61. ${x | x \neq 0 and x \neq 7}$ , or $(- \infty, 0) \cup (0, 7) \cup (7, \infty)$
63. All real numbers, or $(- \infty, \infty)$
65. Domain: [0, 5]; range: [0, 3]
67. Domain: $[- 2 π, 2 π];$ range: [−1, 1]
69. Domain: $(- \infty, \infty);$ range: ${- 3}$
71. Domain: [−5, 3]; range: [−2, 2]
73. Domain: all real numbers, or $(- \infty, \infty);$ range: $[0, \infty)$
75. Domain: all real numbers, or $(- \infty, \infty);$ range: all real numbers, or $(- \infty, \infty)$
77. Domain: $(- \infty, 3) \cup (3, \infty);$ range: $(- \infty, 0) \cup (0, \infty)$
79. Domain: all real numbers, or $(- \infty, \infty);$ range: all real numbers, or $(- \infty, \infty)$
81. Domain: $(- \infty, 7];$ range: $[0, \infty)$
83. Domain: all real numbers, or $(- \infty, \infty);$ range: $(- \infty, 3]$
85.
1. 2018: $25.21; 2025: $28.23;
2. about 49 years after 1985, or in 2034
87. 645 m; 0 m
89. (−3, −2), yes; (2, −3), no
90. (0, −7), no; (8, 11), yes
91. $(\frac{4}{5}, - 2)$ , yes; $(\frac{11}{5}, \frac{1}{10})$ , yes
92.
93.
94.
95.
97. ${x | x \geq - 1 and x \neq 0}$ , or $[- 1, 0) \cup (0, \infty)$
99. ${x | 0 \leq x \leq 4}$ , or [0, 4]
101.
103.
1. $f (x) = - 7;$
2. $f (x) = 2 x - 1;$
3. $f (x) = 7$

Visualizing the Graph

E
D
A
J
C
F
H
G
B
I

Exercise Set 1.3

1.
1. Yes;
2. yes;
3. yes
3.
1. Yes;
2. no;
3. no
5. $\frac{6}{5}$
7. $- \frac{3}{5}$
9. 0
11. $\frac{1}{5}$
13. Not defined
15. 0.3
17. 0
19. $- \frac{6}{5}$
21. $- \frac{1}{3}$
23. Not defined
25. −2
27. 5
29. 0
31. 1.3
33. Not defined
35. $- \frac{1}{2}$
37. −1
39. 0
41. The average rate of change in sales of electric bicycles from 2013 to 2020 is expected to be $0.343 billion, or $343 million.
43. The average rate of change in the population of Cleveland, Ohio, over the 12-year period was about −7290 people per year.
45. The average rate of change in the number of acres used for growing almonds in California from 2003 to 2012 was about 28,889 acres per year.
47. The average rate of change in per-capita consumption of whole milk from 1970 to 2011 was about −0.5 gal per year.
49. $\frac{3}{5};$ $(0, - 7)$
51. Slope is not defined; there is no y-intercept.
53. $- \frac{1}{2};$ $(0, 5)$
55. $- \frac{3}{2};$ $(0, 5)$
57. 0; (0, −6)
59. $\frac{4}{5}; (0, \frac{8}{5})$
61. $\frac{1}{4}; (0, - \frac{1}{2})$
63.
65.
67.
69.
71. 1 atm, 2 atm, $31 \frac{10}{33}$ atm, $152 \frac{17}{33}$ atm, $213 \frac{4}{33}$ atm
73.
1. $\frac{11}{10}$ . For each mile per hour faster that the car travels, it takes $\frac{11}{10}$ ft longer to stop;
2. 6 ft, 11.5 ft, 22.5 ft, 55.5 ft, 72 ft;
3. ${r | r > 0}$ , or $(0, \infty)$ . If r is allowed to be 0, the function says that a stopped car has a reaction distance of $\frac{1}{2}$ ft.
75. $C (t) = 2250 + 3380 t; C (20) = $69, 850$
77. $C (t) = 750 + 15 x; C (32) = $1230$
79. $- \frac{5}{4}$
80. 10
81. 40
82. $a^{2} + 3 a$
83. $a^{2} + 2 a h + h^{2} - 3 a - 3 h$
85. $2 a + h$
87. False
89. $f (x) = x + b$

Mid-Chapter Mixed Review: Chapter 1

Exercise Set 1.4

1. 4, $(0, - 2); y = 4 x - 2$
3. $- 1, (0, 0); y = - x$
5. 0, $(0, - 3); y = - 3$
7. $y = \frac{2}{9} x + 4$
9. $y = - 4 x - 7$
11. $y = - 4.2 x + \frac{3}{4}$
13. $y = \frac{2}{9} x + \frac{19}{3}$
15. $y = 8$
17. $y = - \frac{3}{5} x - \frac{17}{5}$
19. $y = - 3 x + 2$
21. $y = - \frac{1}{2} x + \frac{7}{2}$
23. $y = \frac{2}{3} x - 6$
25. $y = 7.3$
27. Horizontal: $y = - 3;$ vertical: $x = 0$
29. Horizontal: $y = - 1;$ vertical: $x = \frac{2}{11}$
31. $h (x) = - 3 x + 7;$ 1
33. $f (x) = \frac{2}{5} x - 1; - 1$
35. Perpendicular
37. Neither parallel nor perpendicular
39. Parallel
41. Perpendicular
43. $y = \frac{2}{7} x + \frac{29}{7};$ $y = - \frac{7}{2} x + \frac{31}{2}$
45. $y = - 0.3 x - 2.1; y = \frac{10}{3} x + \frac{70}{3}$
47. $y = - \frac{3}{4} x + \frac{1}{4}; y = \frac{4}{3} x - 6$
49. $x = 3; y = - 3$
51. True
53. True
55. False
57. No
59. Yes
61.
1. Using (1, 1.319) and (7, 2.749) gives us $y = 0.238 x + 1.081$ , where x is the number of years after 2006 and y is in billions;
2. 2017: about 3.699 billion Internet users; 2020: about 4.413 billion Internet users
63. Using (0, 11,504) and (3, 10,819) gives us $y = - 228 x + 11, 504$ , where x is the number of years after 2010 and y is in kilowatt-hours; 2019: about 9452 kilowatt-hours
65. Using (1, 28.3) and (3, 30.8) gives us $y = 1.25 x + 27.05$ , where x is the number of years after 2009 and y is in gallons; 2017: about 37.1 gal
67. Not defined
68. −1
69. $x^{2} + {(y - 3)}^{2} = 6.25$
70. ${(x + 7)}^{2} + {(y + 1)}^{2} = \frac{81}{25}$
71. −7.75
73. 6.7% grade; $y = 0.067 x$

Exercise Set 1.5

Exercise Set 1.6

Review Exercises: Chapter 1

Test: Chapter 1

Chapter 2 Exercise Set 2.1

1.
1. (−5, 1);
2. (3, 5);
3. (1, 3)
3.
1. (−3, −1), (3, 5);
2. (1, 3);
3. (−5, −3)
5.
1. $(- \infty, - 8), (- 3, - 2)$ ;
2. (−8, −6);
3. (−6, −3), $(- 2, \infty)$
7. Domain: [−5, 5]; range: [−3, 3]
9. Domain: $[- 5, - 1] \cup [1, 5]$ ; range: [−4, 6]
11. Domain: $(- \infty, \infty)$ ; range: $(- \infty, 3]$
13. Relative maximum: 3.25 at $x = 2.5$ ; increasing: $(- \infty, 2.5)$ ; decreasing: $(2.5, \infty)$
15. Relative maximum: 2.370 at $x = - 0.667$ ; relative minimum: 0 at $x = 2$ ; increasing: $(- \infty, - 0.667)$ , $(2, \infty)$ ; decreasing: (−0.667, 2)
17. Increasing: $(0, \infty)$ ; decreasing: $(- \infty, 0)$ ; relative minimum: 0 at $x = 0$
19. Increasing: $(- \infty, 0)$ ; decreasing: $(0, \infty)$ ; relative maximum: 5 at $x = 0$
21. Increasing: $(3, \infty)$ ; decreasing: $(- \infty, 3)$ ; relative minimum: 1 at $x = 3$
23. $A (x) = x (240 - x)$ , or $240 x - x^{2}$
25. $h (d) = \sqrt{d^{2} - 3500^{2}}$
27. $A (w) = 10 w - \frac{w^{2}}{2}$
29. $d (s) = \frac{14}{s}$
31.
1. $A (x) = x (240 - 4 x)$ , or $240 x - 4 x^{2}$ ;
2. ${x | 0 < x < 60}$ ;
3. 120 ft by 30 ft
33.
1. $V (x) = x (12 - 2 x) (12 - 2 x)$ , or $4 x {(6 - x)}^{2}$ ;
2. ${x | 0 < x < 6}$ ;
3. 8 cm by 8 cm by 2 cm
35. $g (- 4) = 0; g (0) = 4$ ; $g (1) = 5$ ; $g (3) = 5$
37. $h (- 5) = 1; h (0) = 1; h (1) = 3; h (4) = 6$
39.
41.
43.
45.
47.
49.
51.
53. Domain: $(- \infty, \infty)$ ; range: $(- \infty, 0) \cup [3, \infty)$
55. Domain: $(- \infty, \infty)$ ; range: $[- 1, \infty)$
57. Domain: $(- \infty, \infty)$ ; range: ${y | y \leq - 2 or y = - 1 or y \geq 2}$
59. Domain: $(- \infty, \infty)$ ; range: ${- 5, - 2, 4}$ ;

$f (x) = {\begin{array}{l} - 2, & for x < 2, \\ - 5, & for x = 2, \\ 4, & for x > 2 \end{array}$
61. Domain: $(- \infty, \infty)$ ; range: $(- \infty, - 1] \cup [2, \infty)$ ;

$g (x) = {\begin{array}{l} x, & for x \leq - 1, \\ 2, & for - 1 < x < 2, \\ x, & for x \geq 2 \end{array}$

or

$g (x) = {\begin{array}{l} x, & for x \leq - 1, \\ 2, & for - 1 < x \leq 2, \\ x, & for x \geq 2 \end{array}$
63. Domain: [−5, 3]; range: (−3, 5);

$h (x) = {\begin{array}{l} x + 8, & for - 5 \leq x < - 3, \\ 3, & for - 3 < x \leq 1, \\ 3 x - 6, & for 1 < x \leq 3 \end{array}$
65.
1. 38;
2. 38;
3. $5 a^{2} - 7$ ;
4. $5 a^{2} - 7$
66.
1. 22;
2. −22;
3. $4 a^{3} - 5 a$ ;
4. $- 4 a^{3} + 5 a$
67. $y = - \frac{1}{8} x + \frac{7}{8}$
68. Slope: $\frac{2}{9}$ ; y-intercept: $(.0, \frac{1}{9})$
69.
2. $C (t) = 3 (〚 t 〛 + 1), t > 0$
71. ${x | - 5 \leq x < - 4$ or $5 \leq x < 6}$
73.
1. $h (r) = \frac{30 - 5 r}{3}$ ;
2. $V (r) = π r^{2} (\frac{30 - 5 r}{3})$ ;
3. $V (h) = π h {(\frac{30 - 3 h}{5})}^{2}$

Exercise Set 2.2

1. 33
3. −1
5. Does not exist
7. 0
9. 1
11. Does not exist
13. 0
15. 5
17.
1. Domain of f, g, $f + g, f - g, f g$ , and $f f : (- \infty, \infty)$ ; domain of $f / g : (- \infty, \frac{3}{5}) \cup (\frac{3}{5}, \infty)$ ; domain of $g / f : (- \infty, - \frac{3}{2}) \cup (- \frac{3}{2}, \infty)$ ;
2. $(f + g) (x) = - 3 x + 6$ ; $(f - g) (x) = 7 x$ ; $(f g) (x) = - 10 x^{2} - 9 x + 9$ ; $(f f) (x) = 4 x^{2} + 12 x + 9$ ; $(f / g) (x) = \frac{2 x + 3}{3 - 5 x}$ ; $(g / f) (x) = \frac{3 - 5 x}{2 x + 3}$
19.
1. Domain of $f : (- \infty, \infty)$ ; domain of $g : [- 4, \infty)$ ; domain of $f + g, f - g$ , and $f g : [- 4, \infty)$ ; domain of $f f : (- \infty, \infty)$ ; domain of $f / g : (- 4, \infty)$ ; domain of $g / f : [- 4, 3) \cup (3, \infty)$ ;
2. $(f + g) (x) = x - 3 + \sqrt{x + 4}$ ; $(f - g) (x) = x - 3 - \sqrt{x + 4}$ ; $(f g) (x) = (x - 3) \sqrt{x + 4}$ ; $(f f) (x) = x^{2} - 6 x + 9$ ; $(f / g) (x) = \frac{x - 3}{\sqrt{x + 4}}$ ; $(g / f) (x) = \frac{\sqrt{x + 4}}{x - 3}$
21.
1. Domain of f, g, $f + g, f - g$ , fg, and $f f : (- \infty, \infty)$ ; domain of $f / g :$ $(- \infty, 0) \cup (0, \infty)$ ; domain of $g / f : (- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$ ;
2. $(f + g) (x) = - 2 x^{2} + 2 x - 1$ ; $(f - g) (x) = 2 x^{2} + 2 x - 1$ ; $(f g) (x) = - 4 x^{3} + 2 x^{2}$ ; $(f f) (x) = 4 x^{2} - 4 x + 1$ ; $(f / g) (x) = \frac{2 x - 1}{- 2 x^{2}}$ ; $(g / f) (x) = \frac{- 2 x^{2}}{2 x - 1}$
23.
1. Domain of $f : [3, \infty)$ ; domain of $g : [- 3, \infty)$ ; domain of $f + g, f - g, f g$ , and $f f : [3, \infty)$ ; domain of $f / g :$ $[3, \infty)$ ; domain of $g / f : (3, \infty)$ ;
2. $(f + g) (x) = \sqrt{x - 3} + \sqrt{x + 3}$ ; $(f - g) (x) = \sqrt{x - 3} - \sqrt{x + 3}$ ; $(f g) (x) = \sqrt{x^{2} - 9}$ ; $(f f) (x) = | x - 3 |$ ; $(f / g) (x) = \frac{\sqrt{x - 3}}{\sqrt{x + 3}}$ ; $(g / f) (x) = \frac{\sqrt{x + 3}}{\sqrt{x - 3}}$
25.
1. Domain of f, g, $f + g, f - g$ , fg, and $f f : (- \infty, \infty)$ ; domain of $f / g :$ $(- \infty, 0) \cup (0, \infty)$ ; domain of $g / f : (- \infty, - 1) \cup (- 1, \infty)$ ;
2. $(f + g) (x) = x + 1 + | x |$ ; $(f - g) (x) = x + 1 - | x |$ ; $(f g) (x) = (x + 1) | x |$ ; $(f f) (x) = x^{2} + 2 x + 1$ ; $(f / g) (x) = \frac{x + 1}{| x |}$ ; $(g / f) (x) = \frac{| x |}{x + 1}$
27.
1. Domain of f, g, $f + g, f - g, f g$ , and $f f : (- \infty, \infty)$ ; domain of $f / g :$ $(- \infty, - 3) \cup (- 3, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$ ; domain of $g / f : (- \infty, 0) \cup (0, \infty)$ ;
2. $(f + g) (x) = x^{3} + 2 x^{2} + 5 x - 3$ ; $(f - g) (x) = x^{3} - 2 x^{2} - 5 x + 3$ ; $(f g) (x) = 2 x^{5} + 5 x^{4} - 3 x^{3}$ ; $(f f) (x) = x^{6}$ ; $(f / g) (x) = \frac{x^{3}}{2 x^{2} + 5 x - 3}$ ; $(g / f) (x) = frac 2 x^{2} + 5 x - 3 \ x^{3}$
29.
1. Domain of $f : (- \infty, - 1) \cup (- 1, \infty)$ ; domain of $g : (- \infty, 6) \cup (6, \infty)$ ; domain of $f + g, f - g$ , and $f g : (- \infty, - 1) \cup (- 1, 6) \cup (6, \infty)$ ; domain of $f f : (- \infty, - 1) \cup (- 1, \infty)$ ; domain of $f / g$ and $g / f : (- \infty, - 1) \cup (- 1, 6) \cup (6, \infty)$ ;
2. $(f + g) (x) = \frac{4}{x + 1} + \frac{1}{6 - x}$ ; $(f - g) (x) = \frac{4}{x + 1} - \frac{1}{6 - x}$ ; $(f g) (x) = \frac{4}{(x + 1) (6 - x)}$ ; $(f f) (x) = \frac{16}{{(x + 1)}^{2}}$ ; $(f / g) (x) = \frac{4 (6 - x)}{x + 1}$ ; $(g / f) (x) = \frac{x + 1}{4 (6 - x)}$
31.
1. Domain of $f : (- \infty, 0) \cup (0, \infty)$ ; domain of $g : (- \infty, \infty)$ ; domain of $f + g, f - g, f g$ , and $f f : (- \infty, 0) \cup (0, \infty)$ ; domain of $f / g : (- \infty, 0) \cup (0, 3) \cup (3, \infty)$ ; domain of $g / f : (- \infty, 0) \cup (0, \infty)$ ;
2. $(f + g) (x) = \frac{1}{x} + x - 3$ ; $(f - g) (x) = \frac{1}{x} - x + 3$ ; $(f g) (x) = 1 - \frac{3}{x}$ ; $(f f) (x) = \frac{1}{x^{2}}$ ; $(f / g) (x) = \frac{1}{x (x - 3)}$ ; $(g / f) (x) = x (x - 3)$
33.
1. Domain of $f : (- \infty, 2) \cup (2, \infty)$ ; domain of $g : [1, \infty)$ ; domain of $f + g$ , $f - g$ , and $fg : [1, 2) \cup (2, \infty)$ ; domain of $ff : (- \infty, 2) \cup (2, \infty)$ ; domain of $f / g :$ $(1, 2) \cup (2, \infty)$ ; domain of $g / f :$ $[1, 2) \cup (2, \infty)$ ;
2. $(f + g) (x) = \frac{3}{x - 2} + \sqrt{x - 1}$ ; $(f - g) (x) = \frac{3}{x - 2} - \sqrt{x - 1}$ ; $(f g) (x) = \frac{3 \sqrt{x - 1}}{x - 2}$ ; $(f f) (x) = \frac{9}{{(x - 2)}^{2}}$ ; $(f / g) (x) = \frac{3}{(x - 2) \sqrt{x - 1}}$ ; $(g / f) (x) = \frac{(x - 2) \sqrt{x - 1}}{3}$
35. Domain of F: [2, 11]; domain of G: [1, 9]; domain of $F + G : [2, 9]$
37. $[2, 3) \cup (3, 9]$
39.
41. Domain of F: [0, 9]; domain of G: [3, 10]; domain of $F + G : [3, 9]$
43. $[3, 6) \cup (6, 8) \cup (8, 9]$
45.
47.
1. $P (x) = - 0.4 x^{2} + 57 x - 13$ ;
2. $R (100) = 2000$ ; $C (100) = 313; P (100) = 1687$
49. 3
51. 6
53. $\frac{1}{3}$
55. $\frac{- 1}{3 x (x + h)}$ , or $- \frac{1}{3 x (x + h)}$
57. $\frac{1}{4 x (x + h)}$
59. $2 x + h$
61. $- 2 x - h$
63. $6 x + 3 h - 2$
65. $\frac{5 | x + h | - 5 | x |}{h}$
67. $3 x^{2} + 3 x h + h^{2}$
69. $\frac{7}{(x + h + 3) (x + 3)}$
71.
72.
73.
74.
75. $f (x) = \frac{1}{x + 7}, g (x) = \frac{1}{x - 3}$ ; answers may vary
77. $(- \infty, - 1) \cup (- 1, 1) \cup (1, \frac{7}{3}) \cup (\frac{7}{3}, 3) \cup (3, \infty)$

Exercise Set 2.3

Mid-Chapter Mixed Review: Chapter 2

1. True
2. False
3. True
4.
1. (2, 4);
2. (−5, −3), (4, 5);
3. (−3, −1)
5. Relative maximum: 6.30 at $x = - 1.29$ ; relative minimum: −2.30 at $x = 1.29$ ; increasing: $(- \infty, - 1.29), (1.29, \infty)$ ; decreasing: (−1.29, 1.29)
6. Domain: $[- 5, - 1] \cup [2, 5]$ ; range: [−3, 5]
7. $A (h) = \frac{h^{2}}{2 + 2 h}$
8. −10; −8; 1; 3
9.
10. 1
11. −4
12. 5
13. Does not exist
14.
1. Domain of f, g, $f + g$ , $f - g$ , fg, and $f f : (- \infty, \infty)$ ; domain of $f / g : (- \infty, - 4) \cup (- 4, \infty)$ ; domain of $g / f : (- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \infty)$ ;
2. $(f + g) (x) = x + 1$ ; $(f - g) (x) = 3 x + 9$ ; $(f g) (x) = - 2 x^{2} - 13 x - 20$ ; $(f f) (x) = 4 x^{2} + 20 x + 25$ ; $(f / g) (x) = \frac{2 x + 5}{- x - 4}$ ; $(g / f) (x) = \frac{- x - 4}{2 x + 5}$
15.
1. Domain of $f : (- \infty, \infty)$ ; domain of g, $f + g$ , $f - g$ , and $f g : [- 2, \infty)$ ; domain of $f f : (- \infty, \infty)$ ; domain of $f / g : (- 2, \infty)$ ; domain of $g / f : [- 2, 1) \cup (1, \infty)$ ;
2. $(f + g) (x) = x - 1 + \sqrt{x + 2}$ ; $(f - g) (x) = x - 1 - \sqrt{x + 2}$ ; $(f g) (x) = (x - 1) \sqrt{x + 2}$ ; $(f f) (x) = x^{2} - 2 x + 1$ ; $(f / g) (x) = \frac{x - 1}{\sqrt{x + 2}}$ ; $(g / f) (x) = \frac{\sqrt{x + 2}}{x - 1}$
16. 4
17. $- 2 x - h$
18. 6
19. 28
20. −24
21. 102
22. $(f \circ g) (x) = 3 x + 2$ ; $(g \circ f) (x) = 3 x + 4$ ; domain of $f \circ g$ and $g \circ f : (- \infty, \infty)$
23. $(f \circ g) (x) = 3 \sqrt{x} + 2$ ; $(g \circ f) (x) = \sqrt{3 x + 2}$ ; domain of $f \circ g : [0, \infty)$ ; domain of $g \circ f : [- \frac{2}{3}, \infty)$
24. The graph of $y = (h - g) (x)$ will be the same as the graph of $y = h (x)$ moved down b units.
25. Under the given conditions, $(f + g) (x)$ and $(f / g) (x)$ have different domains if $g (x) = 0$ for one or more real numbers x.
26. If f and g are linear functions, then any real number can be an input for each function. Thus the domain of $f \circ g =$ the domain of $g \circ f = (- \infty, \infty)$ .
27. This approach is not valid. Consider Exercise 23 in Section 2.3, for example. Since $(f \circ g) (x) = \frac{4 x}{x - 5}$ , an examination of only this composed function would lead to the incorrect conclusion that the domain of $f \circ g$ is $(- \infty, 5) \cup (5, \infty)$ . However, we must also exclude from the domain of $f \circ g$ those values of x that are not in the domain of g. Thus the domain of $f \circ g$ is $(- \infty, 0) \cup (0, 5) \cup (5, \infty)$ .

Exercise Set 2.4

1. x-axis: no; y-axis: yes; origin: no
3. x-axis: yes; y-axis: no; origin: no
5. x-axis: no; y-axis: no; origin: yes
7. x-axis: no; y-axis: yes; origin: no
9. x-axis: no; y-axis: no; origin: no
11. x-axis: no; y-axis: yes; origin: no
13. x-axis: no; y-axis: no; origin: yes
15. x-axis: no; y-axis: no; origin: yes
17. x-axis: yes; y-axis: yes; origin: yes
19. x-axis: no; y-axis: yes; origin: no
21. x-axis: yes; y-axis: yes; origin: yes
23. x-axis: no; y-axis: no; origin: no
25. x-axis: no; y-axis: no; origin: yes
27. x-axis: (−5, −6); y-axis: (5, 6); origin: (5, −6)
29. x-axis: (−10, 7); y-axis: (10, −7); origin: (10, 7)
31. x-axis: (0, 4); y-axis: (0, −4); origin: (0, 4)
33. Even
35. Odd
37. Neither
39. Odd
41. Even
43. Odd
45. Neither
47. Even
49.
50. University of California–Berkeley: 3576 volunteers; University of Wisconsin–Madison: 3112 volunteers
51. Odd
53. x-axis: yes; y-axis: no; origin: no
55. $E (- x) = \frac{f (- x) + f (- (- x))}{2} = \frac{f (- x) + f (x)}{2} = E (x)$
57.
1. $E (x) + O (x) = \frac{f (x) + f (- x)}{2} + \frac{f (x) - f (- x)}{2} = \frac{2 f (x)}{2} = f (x)$ ;
2. $f (x) = \frac{- 22 x^{2} + \sqrt{x} + \sqrt{- x} - 20}{2} + \frac{8 x^{3} + \sqrt{x} - \sqrt{- x}}{2}$
59. True

Visualizing the Graph

C
B
A
E
G
D
H
I
F

Exercise Set 2.5

Review Exercises: Chapter 2

1. True
2. False
3. True
4. True
5.
1. (−4, −2);
2. (2, 5);
3. (−2, 2)
6.
1. $(- 1, 0), (2, \infty)$ ;
2. (0, 2);
3. $(- \infty, - 1)$
7. Increasing: $(0, \infty)$ ; decreasing: $(- \infty, 0)$ ; relative minimum: −1 at $x = 0$
8. Increasing: $(- \infty, 0)$ ; decreasing: $(0, \infty)$ ; relative maximum: 2 at $x = 0$
9. $A (x) = x (48 - 2 x)$ , or $48 x - 2 x^{2}$
10. $A (x) = 2 x \sqrt{4 - x^{2}}$
11.
1. $A (x) = x^{2} + \frac{432}{x}$ ;
2. $(0, \infty)$ ;
3. $x = 6$ in., $height = 3$ in.
12.
13.
14.
15.
16.
17. $f (- 1) = 1$ ; $f (5) = 2$ ; $f (- 2) = 2$ ; $f (- 3) = - 27$
18. $f (- 2) = - 3$ ; $f (- 1) = 3$ ; $f (0) = - 1$ ; $f (4) = 3$
19. −33
20. 0
21. Does not exist
22.
1. Domain of $f : (- \infty, 0) \cup (0, \infty)$ ; domain of g: $(- \infty, \infty)$ ; domain of $f + g, f - g$ , and $f g : (- \infty, 0) \cup (0, \infty)$ ; domain of $f / g : (- \infty, 0) \cup (0, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$ ;
2. $(f + g) (x) = \frac{4}{x^{2}} + 3 - 2 x$ ; $(f - g) (x) = \frac{4}{x^{2}} - 3 + 2 x$ ; $(f g) (x) = \frac{12}{x^{2}} - \frac{8}{x}$ ; $(f / g) (x) = \frac{4}{x^{2} (3 - 2 x)}$
23.
1. Domain of $f, g, f + g, f - g$ , and $f g : (- \infty, \infty)$ ; domain of $f / g : (- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$ ;
2. $(f + g) (x) = 3 x^{2} + 6 x - 1$ ; $(f - g) (x) = 3 x^{2} + 2 x + 1$ ; $(f g) (x) = 6 x^{3} + 5 x^{2} - 4 x$ ; $(f / g) (x) = \frac{3 x^{2} + 4 x}{2 x - 1}$
24. $P (x) = - 0.5 x^{2} + 105 x - 6$
25. 2
26. $- 2 x - h$
27. $\frac{- 4}{x (x + h)}$ , or $- \frac{4}{x (x + h)}$
28. 9
29. 5
30. 128
31. 580
32. 7
33. −509
34. $4 x - 3$
35. $- 24 + 27 x^{3} - 9 x^{6} + x^{9}$
36.
1. $(f \circ g) (x) = \frac{4}{{(3 - 2 x)}^{2}}$ ; $(g \circ f) (x) = 3 - \frac{8}{x^{2}}$ ;
2. domain of $f \circ g : (- \infty, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$ ; domain of $g \circ f : (- \infty, 0) \cup (0, \infty)$
37.
1. $(f \circ g) (x) = 12 x^{2} - 4 x - 1$ ; $(g \circ f) (x) = 6 x^{2} + 8 x - 1$ ;
2. domain of $f \circ g$ and $g \circ f : (- \infty, \infty)$
38. $f (x) = \sqrt{x}, g (x) = 5 x + 2$ ; answers may vary.
39. $f (x) = 4 x^{2} + 9, g (x) = 5 x - 1$ ; answers may vary.
40. x-axis: yes; y-axis: yes; origin: yes
41. x-axis: yes; y-axis: yes; origin: yes
42. x-axis: no; y-axis: no; origin: no
43. x-axis: no; y-axis: yes; origin: no
44. x-axis: no; y-axis: no; origin: yes
45. x-axis: no; y-axis: yes; origin: no
46. Even
47. Even
48. Odd
49. Even
50. Even
51. Neither
52. Odd
53. Even
54. Even
55. Odd
56. $f (x) = {(x + 3)}^{2}$
57. $f (x) = - \sqrt{x - 3} + 4$
58. $f (x) = 2 | x - 3 |$
59.
60.
61.
62.
63. $y = 4 x$
64. $y = \frac{2}{3} x$
65. $y = \frac{2500}{x}$
66. $y = \frac{54}{x}$
67. $y = \frac{48}{x^{2}}$
68. $y = \frac{1}{10} \cdot \frac{x z^{2}}{w}$
69. 20 min
70. 75
71. 500 watts
72. A
73. C
74. B
75. Let $f (x)$ and $g (x)$ be odd functions. Then by definition, $f (- x) = - f (x)$ , or $f (x) = - f (- x)$ , and $g (- x) = - g (x)$ , or $g (x) = - g (- x)$ . Thus, $(f + g) (x) = f (x) + g (x) = - f (- x) + [- g (- x)] = - [f (- x) +$ $g (- x)] =$ $- (f + g) (- x)$ and $f + g$ is odd.
76. Reflect the graph of $y = f (x)$ across the x-axis and then across the y-axis.
77.
1. $4 x^{3} - 2 x + 9;$
2. $4 x^{3} + 24 x^{2} + 46 x + 35;$
3. $4 x^{3} - 2 x + 42$ .
1. adds 2 to each function value;
2. adds 2 to each input before finding a function value;
3. adds the output for 2 to the output for x
78. In the graph of $y = f (c x)$ , the constant c stretches or shrinks the graph of $y = f (x)$ horizontally. The constant c in $y = c f (x)$ stretches or shrinks the graph of $y = f (x)$ vertically. For $y = f (c x)$ , the x-coordinates of $y = f (x)$ are divided by c; for $y = c f (x)$ , the y-coordinates of $y = f (x)$ are multiplied by c.
79. The graph of $f (x) = 0$ is symmetric with respect to the x-axis, the y-axis, and the origin. This function is both even and odd.
80. If all the exponents are even numbers, then $f (x)$ is an even function. If $a_{0} = 0$ and all the exponents are odd numbers, then $f (x)$ is an odd function.
81. Let $y (x) = k x^{2}$ . Then $y (2 x) = k {(2 x)}^{2} = k \cdot 4 x^{2} = 4 \cdot k x^{2} = 4 \cdot y (x)$ . Thus doubling x causes y to be quadrupled.
82. Let $y = k_{1} x$ and $x = \frac{k_{2}}{z}$ . Then $y = k_{1} \cdot \frac{k_{2}}{z}$ , or $y = \frac{k_{1} k_{2}}{z}$ , so y varies inversely as z.

Test: Chapter 2

Chapter 3 Exercise Set 3.2

Visualizing the Graph

C
B
A
J
F
D
I
G
H
E

Exercise Set 3.3

1.
1. $(- \frac{1}{2}, - \frac{9}{4});$
2. $x = - \frac{1}{2};$
3. minimum: $- \frac{9}{4}$
3.
1. (4, −4);
2. $x = 4;$
3. minimum: −4;
5.
1. $(\frac{7}{2}, - \frac{1}{4});$
2. $x = \frac{7}{2};$
3. minimum: $- \frac{1}{4};$
7.
1. (−2, 1);
2. $x = - 2;$
3. minimum: 1;
9.
1. (−4, −2);
2. $x = - 4;$
3. minimum: −2;
11.
1. $(- \frac{3}{2}, \frac{7}{2});$
2. $x = - \frac{3}{2};$
3. minimum: $\frac{7}{2};$
13.
1. (−3, 12);
2. $x = - 3;$
3. maximum: 12;
15.
1. $(\frac{1}{2}, \frac{3}{2});$
2. $x = \frac{1}{2};$
3. maximum: $\frac{3}{2};$
17. (f )
19. (b)
21. (h)
23. (c)
25. True
27. False
29. True
31.
1. (3, −4);
2. minimum: −4;
3. $[- 4, \infty);$
4. increasing: $(3, \infty);$ decreasing: $(- \infty, 3)$
33.
1. (−1, −18);
2. minimum: −18;
3. $[- 18, \infty);$
4. increasing: $(- 1, \infty);$ decreasing: $(- \infty, - 1)$
35.
1. $(5, \frac{9}{2});$
2. maximum: $\frac{9}{2};$
3. $(- \infty, \frac{9}{2}];$
4. increasing: $(- \infty, 5);$ decreasing: $(5, \infty)$
37.
1. (−1, 2);
2. minimum: 2;
3. $[2, \infty);$
4. increasing: $(- 1, \infty);$ decreasing: $(- \infty, - 1)$
39.
1. $(- \frac{3}{2}, 18);$
2. maximum: 18;
3. $(- \infty, 18];$
4. increasing: $(- \infty, - \frac{3}{2});$ decreasing: $(- \frac{3}{2}, \infty)$
41. 0.625 sec; 12.25 ft
43. 3.75 sec; 305 ft
45. 4.5 in.
47. Base: 10 cm; height: 10 cm
49. 350 doghouses
51. $797; 40 units
53. 4800 $y d^{2}$
55. About 60.5 ft
57. 3
58. $4 x + 2 h - 1$
59.
60.
61. −236.25
63.
65. Pieces should be $\frac{24 π}{4 + π}$ in. and $\frac{96}{4 + π}$ in.

Mid-Chapter Mixed Review: Chapter 3

1. True
2. False
3. True
4. False
5. 6i
6. $\sqrt{5} i$
7. −4i
8. $4 \sqrt{2} i$
9. $- 1 + i$
10. $- 7 + 5 i$
11. $23 + 2 i$
12. $- \frac{1}{29} - \frac{17}{29} i$
13. i
14. 1
15. −i
16. −64
17. −4, 1
18. $- 2, - \frac{3}{2}$
19. $\pm \sqrt{6}$
20. $\pm 10 i$
21. $4 x^{2} - 8 x - 3 = 0; 4 x^{2} - 8 x = 3;$

$x^{2} - 2 x = \frac{3}{4};$ $x^{2} - 2 x + 1 = \frac{3}{4} + 1; {(x - 1)}^{2} = \frac{7}{4};$

$x - 1 = \pm \frac{\sqrt{7}}{2};$ $x = 1 \pm \frac{\sqrt{7}}{2} = \frac{2 \pm \sqrt{7}}{2}$
22.
1. 29; two real;
2. $\frac{3 \pm \sqrt{29}}{2}; - 1.193, 4.193$
23.
1. 0; one real;
2. $\frac{3}{2}$
24.
1. −8; two nonreal;
2. $- \frac{1}{3} \pm \frac{\sqrt{2}}{3} i$
25. $\pm 1, \pm \sqrt{6} i$
26. $\frac{1}{4}, 4$
27. 5 and 7; −7 and −5
28.
1. (3, −2);
2. $x = 3;$
3. minimum: −2;
4. $[- 2, \infty);$
5. increasing: $(3, \infty);$ decreasing: $(- \infty, 3);$
29.
1. (−1, −3);
2. $x = - 1;$
3. maximum: −3;
4. $(- \infty, - 3];$
5. increasing: $(- \infty, - 1);$ decreasing: $(- 1, \infty);$
30. Base: 8 in., height: 8 in.
31. The sum of two imaginary numbers is not always an imaginary number. For example, $(2 + i) + (3 - i) = 5$ , a real number.
32. Use the discriminant. If $b^{2} - 4 a c < 0$ , there are no x-intercepts. If $b^{2} - 4 a c = 0$ , there is one x-intercept. If $b^{2} - 4 a c > 0$ , there are two x-intercepts.
33. Completing the square was used in Section 3.2 to solve quadratic equations. It was used again in Section 3.3 to write quadratic functions in the form $f (x) = a {(x - h)}^{2} + k$ .
34. The x-intercepts of $g (x)$ are also $(x_{1}, 0)$ and $(x_{2}, 0)$ . This is true because $f (x)$ and $g (x)$ have the same zeros. Consider $g (x) = 0$ , or $- a x^{2} - b x - c = 0$ . Multiplying by −1 on both sides, we get an equivalent equation $a x^{2} + b x + c = 0$ , or $f (x) = 0$ .

Exercise Set 3.5

Review Exercises: Chapter 3

Chapter 4 Exercise Set 4.1

Visualizing the Graph

H
D
J
B
A
C
I
E
G
F

Exercise Set 4.2

Exercise Set 4.3

1.
1. No;
2. yes;
3. no
3.
1. Yes;
2. no;
3. yes
5. $P (x) = (x + 2) (x^{2} - 2 x + 4) - 16$
7. $P (x) = (x + 9) (x^{2} - 3 x + 2) + 0$
9. $P (x) = (x + 2) (x^{3} - 2 x^{2} + 2 x - 4) + 11$
11. $Q (x) = 2 x^{3} + x^{2} - 3 x + 10, R (x) = - 42$
13. $Q (x) = x^{2} - 4 x + 8, R (x) = - 24$
15. $Q (x) = 3 x^{2} - 4 x + 8, R (x) = - 18$
17. $Q (x) = x^{4} + 3 x^{3} + 10 x^{2} + 30 x + 89, R (x) = 267$
19. $Q (x) = x^{3} + x^{2} + x + 1, R (x) = 0$
21. $Q (x) = 2 x^{3} + x^{2} + \frac{7}{2} x + \frac{7}{4}, R (x) = - \frac{1}{8}$
23. 0; −60; 0
25. 10; 80; 998
27. 5,935,988; −772
29. 0; 0; 65; $1 - 12 \sqrt{2}$
31. Yes; no
33. Yes; yes
35. No; yes
37. No; no
39. $f (x) = (x - 1) (x + 2) (x + 3); 1, - 2, - 3$
41. $f (x) = (x - 2) (x - 5) (x + 1); 2, 5, - 1$
43. $f (x) = (x - 2) (x - 3) (x + 4); 2, 3, - 4$
45. $f (x) = {(x - 3)}^{3} (x + 2); 3, - 2$
47. $f (x) = (x - 1) (x - 2) (x - 3) (x + 5); 1, 2, 3, - 5$
49.
51.
53.
55. $\frac{5}{4} \pm \frac{\sqrt{71}}{4} i$
56. $- 1, \frac{3}{7}$
57. −5, 0
58. 10
59. $- 3, - 2$
60. $f (x) = 0.172 x + 2.69$ ; 1995: $5.27; 2018: $9.23
61. $b = 15 in ., h = 15 in$ .
63.
1. $x + 4, x + 3, x - 2, x - 5$ ;
2. $P (x) = (x + 4) (x + 3) (x - 2) (x - 5)$ ;
3. yes; two examples are $f (x) = c \cdot P (x)$ for any nonzero constant c; and $g (x) = (x - a) P (x)$ ;
4. no
65. $\frac{14}{3}$
67. 0, −6
69. Answers may vary. One possibility is $P (x) = x^{15} - x^{14}$ .
71. $x^{2} + 2 i x + (2 - 4 i), R - 6 - 2 i$
73. $x - 3 + i, R 6 - 3 i$

Mid-Chapter Mixed Review: Chapter 4

Exercise Set 4.4

1. $f (x) = x^{3} - 6 x^{2} - x + 30$
3. $f (x) = x^{3} + 3 x^{2} + 4 x + 12$
5. $f (x) = x^{3} - 3 x^{2} - 2 x + 6$
7. $f (x) = x^{3} - 6 x - 4$
9. $f (x) = x^{3} + 2 x^{2} + 29 x + 148$
11. $f (x) = x^{3} - \frac{5}{3} x^{2} - \frac{2}{3} x$
13. $f (x) = x^{5} + 2 x^{4} - 2 x^{2} - x$
15. $f (x) = x^{4} + 3 x^{3} + 3 x^{2} + x$
17. $- \sqrt{3}$
19. $i, 2 + \sqrt{5}$
21. −3i
23. $- 4 + 3 i, 2 + \sqrt{3}$
25. $- \sqrt{5}, 4 i$
27. $2 + i$
29. $- 3 - 4 i, 4 + \sqrt{5}$
31. $4 + i$
33. $f (x) = x^{3} - 4 x^{2} + 6 x - 4$
35. $f (x) = x^{2} + 16$
37. $f (x) = x^{3} - 5 x^{2} + 16 x - 80$
39. $f (x) = x^{4} - 2 x^{3} - 3 x^{2} + 10 x - 10$
41. $f (x) = x^{4} + 4 x^{2} - 45$
43. $- \sqrt{2}, \sqrt{2}$
45. i, 2, 3
47. $1 + 2 i, 1 - 2 i$
49. $\pm 1$
51. $\pm 1, \pm \frac{1}{2}, \pm 2, \pm 4, \pm 8$
53. $\pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{2}{3}, \pm \frac{2}{5}, \pm \frac{1}{15}, \pm \frac{2}{15}$
55.
1. Rational: −3; other: $\pm \sqrt{2}$ ;
2. $f (x) = (x + 3) (x + \sqrt{2}) (x - \sqrt{2})$
57.
1. Rational: $\frac{1}{3}$ ; other: $\pm \sqrt{5}$ ;
2. $f (x) = 3 (x - \frac{1}{3}) (x + \sqrt{5}) (x - \sqrt{5})$
59.
1. Rational: $- 2, 1$ ; other: none;
2. $f (x) = (x + 2) {(x - 1)}^{2}$
61.
1. Rational: $- \frac{3}{2}$ ; other: $\pm 3 i$ ;
2. $f (x) = 2 (x + \frac{3}{2}) (x + 3 i) (x - 3 i)$
63.
1. Rational: $- \frac{1}{5}, 1$ ; other: $\pm 2 i$ ;
2. $f (x) = 5 (x + \frac{1}{5}) (x - 1) (x + 2 i) (x - 2 i)$ , or $(5 x + 1) (x - 1) (x + 2 i) (x - 2 i)$
65.
1. Rational: $- 2, - 1$ ; other: $3 \pm \sqrt{13}$ ;
2. $f (x) = (x + 2) (x + 1) (x - 3 - \sqrt{13}) (x - 3 + \sqrt{13})$
67.
1. Rational: 2; other: $1 \pm \sqrt{3}$ ;
2. $f (x) = (x - 2) (x - 1 - \sqrt{3}) (x - 1 + \sqrt{3})$
69.
1. Rational: −2; other: $1 \pm \sqrt{3} i$ ;
2. $f (x) = (x + 2) (x - 1 - \sqrt{3} i) (x - 1 + \sqrt{3} i)$
71.
1. Rational: $\frac{1}{2}$ ; other: $\frac{1 \pm \sqrt{5}}{2}$ ;
2. $f (x) = \frac{1}{3} (x - \frac{1}{2}) (x - \frac{1 + \sqrt{5}}{2}) (x - \frac{1 - \sqrt{5}}{2})$
73. 1, −3
75. No rational zeros
77. No rational zeros
79. −2, 1, 2
81. 3 or 1; 0
83. 0; 3 or 1
85. 2 or 0; 2 or 0
87. 1; 1
89. 1; 0
91. 2 or 0; 2 or 0
93. 3 or 1;1
95. 1; 1
97.
99.
101.
1. (4, −6);
2. $x = 4$ ;
3. minimum: −6 at $x = 4$
102.
1. (1, −4);
2. $x = 1$ ;
3. minimum: −4 at $x = 1$
103. 10
104. −3, 11
105. Cubic; $- x^{3}; - 1; 3$ ; as $x \to \infty, g (x) - \infty$ , and as $x \to - \infty, g (x) \to \infty$
106. Quadratic; $- x^{2}; - 1; 2$ ; as $x \to \infty, f (x) - \infty$ , and as $x \to - \infty, f (x) - \infty$
107. Constant; $- \frac{4}{9}; - \frac{4}{9}$ ; zero degree; for all x, $f (x) = - \frac{4}{9}$
108. Linear; x; 1; 1; as $x \to \infty, h (x) \to \infty$ , and as $x \to - \infty, h (x) \to - \infty$
109. Quartic; $x^{4}$ ; 1; 4; as $x \to \infty, g (x) \to \infty$ , and as $x \to - \infty, g (x) \to \infty$
110. Cubic; $x^{3}$ ; 1; 3; as $x \to \infty, h (x) \to \infty$ , and as $x \to - \infty, h (x) \to - \infty$
111.
1. $- 1, \frac{1}{2}, 3$ ;
2. $0, \frac{3}{2}, 4$ ;
3. $- 3, - \frac{3}{2}, 1$ ;
4. $- \frac{1}{2}, \frac{1}{4}, \frac{3}{2}$
113. $- 8, - \frac{3}{2}$ , 4, 7, 15

Visualizing the Graph

A
C
D
H
G
F
B
I
J
E

Exercise Set 4.5

Exercise Set 4.6

Review Exercises: Chapter 4

Test: Chapter 4

[4.1] $- x^{4}, - 1, 4;$ quartic
[4.1] $- 4.7 x, - 4.7, 1;$ linear
[4.1] $0, \frac{5}{3}$ , each has multiplicity 1; 3, multiplicity 2; −1, multiplicity 3
[4.1] 2008: 329,277 hybrid automobiles; 2011: 275,779 hybrid automobiles
[4.2]
[4.2]
[4.2] $f (0) = 3$ and $f (2) = - 17$ . Since $f (0)$ and $f (2)$ have opposite signs, $f (x)$ has a zero between 0 and 2.
[4.2] $g (- 2) = 5$ and $g (- 1) = 1$ . Both $g (- 2)$ and $g (- 1)$ are positive. We cannot use the intermediate value theorem to determine if there is a zero between −2 and −1.
[4.3] $Q (x) = x^{3} + 4 x^{2} + 4 x + 6, R (x) = 1;$ $P (x) = (x - 1) (x^{3} + 4 x^{2} + 4 x + 6) + 1$
[4.3] $3 x^{2} + 15 x + 63, R 322$
[4.3] −115
[4.3] Yes
[4.4] $f (x) = x^{4} - 27 x^{2} - 54 x$
[4.4] $- \sqrt{3}, 2 + i$
[4.4] $f (x) = x^{3} + 10 x^{2} + 9 x + 90$
[4.4] $f (x) = x^{5} - 2 x^{4} - x^{3} + 6 x^{2} - 6 x$
[4.4] $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}$
[4.4] $\pm \frac{1}{10}, \pm \frac{1}{5}, \pm \frac{1}{2}, \pm 1, \pm \frac{5}{2}, \pm 5$
[4.4]
1. Rational: −1; other: $\pm \sqrt{5};$
2. $f (x) = (x + 1) (x - \sqrt{5}) (x + \sqrt{5})$
[4.4]
1. Rational: $- \frac{1}{2}$ , 1, 2, 3; other: none;
2. $f (x) = 2 (x + \frac{1}{2}) (x - 1) (x - 2) (x - 3)$
[4.4]
1. Rational: −4; other: $\pm 2 i;$
2. $f (x) = (x - 2 i) (x + 2 i) (x + 4)$
[4.4]
1. Rational: $\frac{2}{3}, 1;$ other: none;
2. $f (x) = 3 (x - \frac{2}{3}) {(x - 1)}^{3}$
[4.4] 2 or 0; 2 or 0
[4.5] Domain: $(- \infty, 3) \cup (3, \infty);$ x-intercepts: none, y-intercept: $(0, \frac{2}{9});$
[4.5] Domain: $(- \infty, - 1) \cup (- 1, 4) \cup (4, \infty);$ x-intercept: (−3, 0), y-intercept: $(0, - \frac{3}{4});$
[4.5] Answers may vary; $f (x) = \frac{x + 4}{x^{2} - x - 2}$
[4.6] $(- \infty, - \frac{1}{2}) \cup (3, \infty)$
[4.6] $(- \infty, 4) \cup [\frac{13}{2}, \infty)$
1. [4.1] 6 sec;
2. [4.1], [4.6] (1, 3)
[4.2] D
[4.1], [4.6] $(- \infty, - 4] \cup [3, \infty)$

Chapter 5 Exercise Set 5.1

1. ${(8, 7), (8, - 2), (- 4, 3), (- 8, 8)}$
3. ${(- 1, - 1), (4, - 3)}$
5. $x = 4 y - 5$
7. $y^{3} x = - 5$
9. $y = x^{2} - 2 x$
11.
13.
15.
17. Assume $f (a) = f (b)$ for any numbers a and b in the domain of f. Since $f (a) = \frac{1}{3} a - 6$ and $f (b) = \frac{1}{3} b - 6$ , we have

$\begin{array}{rcl} \frac{1}{3} a - 6 & = & \frac{1}{3} b - 6 \\ \frac{1}{3} a & = & \frac{1}{3} b & Adding 6 \\ a & = & b . & Multiplying by 3 \end{array}$

Thus, if $f (a) = f (b)$ , then $a = b$ and f is one-to-one.
19. Assume $f (a) = f (b)$ for any numbers a and b in the domain of f. Since $f (a) = a^{3} + \frac{1}{2}$ and $f (b) = b^{3} + \frac{1}{2}$ , we have

$\begin{array}{rcll} a^{3} + \frac{1}{2} & = & b^{3} + \frac{1}{2} \\ a^{3} & = & b^{3} & Subtracting \frac{1}{2} \\ a & = & b . & Taking the cube root \end{array}$

Thus, if $f (a) = f (b)$ , then $a = b$ and f is one-to-one.
21. Find two numbers a and b for which $a \neq b$ and $g (a) = g (b)$ . Two such numbers are −2 and 2, because $g (- 2) = g (2) = - 3$ . Thus, g is not one-to-one.
23. Find two numbers a and b for which $a \neq b$ and $g (a) = g (b)$ . Two such numbers are −1 and 1, because $g (- 1) = g (1) = 0$ . Thus, g is not one-to-one.
25. Yes
27. No
29. No
31. Yes
33. Yes
35. No
37. No
39. Yes
41. No
43. No
45.
1. One-to-one;
2. $f^{- 1} (x) = x - 4$
47.
1. One-to-one;
2. $f^{- 1} (x) = \frac{x + 1}{2}$
49.
1. One-to-one;
2. $f^{- 1} (x) = \frac{4}{x} - 7$
51.
1. One-to-one;
2. $f^{- 1} (x) = \frac{3 x + 4}{x - 1}$
53.
1. One-to-one;
2. $f^{- 1} (x) = \sqrt{x + 1}$
55.
1. Not one-to-one;
2. does not have an inverse that is a function
57.
1. One-to-one;
2. $f^{- 1} (x) = \sqrt{\frac{x + 2}{5}}$
59.
1. One-to-one;
2. $f^{- 1} (x) = x^{2} - 1, x \geq 0$
61. $\frac{1}{3} x$
63. $- x$
65. $x^{3} + 5$
67.
69.
71.
73. $f^{- 1} (f (x)) = f^{- 1} (\frac{7}{8} x) = \frac{8}{7} \cdot \frac{7}{8} x = x;$ $f (f^{- 1} (x)) = f (\frac{8}{7} x) = \frac{7}{8} \cdot \frac{8}{7} x = x$
75. $f^{- 1} (^{} f (x)) = f^{- 1} (\frac{1 - x}{x}) = \frac{1}{\frac{1 - x}{x} + 1} = \frac{1}{\frac{1 - x + x}{x}} = \frac{1}{\frac{1}{x}} = 1 \cdot \frac{x}{1} = x;$ $f (f^{- 1} (x)) = f (\frac{1}{x + 1}) = \frac{1 - \frac{1}{x + 1}}{\frac{1}{x + 1}} = \frac{\frac{x + 1 - 1}{x + 1}}{\frac{1}{x + 1}} = \frac{x}{x + 1} \cdot \frac{x + 1}{1} = x$
77. $f^{- 1} (f (x)) = f^{- 1} (\frac{2}{5} x + 1) = \frac{5 (\frac{2}{5} x + 1) - 5}{2} = \frac{2 x + 5 - 5}{2} = \frac{2 x}{2} = x;$ $f (f^{- 1} (x)) = f (\frac{5 x - 5}{2}) = \frac{2}{5} (\frac{5 x - 5}{2}) + 1 = x - 1 + 1 = x$
79. $f^{- 1} (x) = \frac{1}{5} x + \frac{3}{5};$ domain of f and $f^{- 1} : (- \infty, \infty);$ range of f and $f^{- 1} : (- \infty, \infty);$
81. $f^{- 1} (x) = \frac{2}{x};$ domain of $f and f^{- 1} : (- \infty, 0) \cup (0, \infty);$ range of f and $f^{- 1} : (- \infty, 0) \cup (0, \infty);$
83. $f^{- 1} (x) = \sqrt{3 x + 6};$ domain of $f and f^{- 1} : (- \infty, \infty);$ range of $f and f^{- 1} : (- \infty, \infty);$
85. $f^{- 1} (x) = \frac{3 x + 1}{x - 1};$ domain of $f : (- \infty, 3) \cup (3, \infty);$ range of $f : (- \infty, 1) \cup (1, \infty);$ domain of $f^{- 1} : (- \infty, 1) \cup (1, \infty);$ range of $f^{- 1} : (- \infty, 3) \cup (3, \infty);$
87. 5; a
89.
1. $38, $16.40, $11;
2. $C^{- 1} (x) = \frac{72}{x - 2};$ $C^{- 1} (x)$ represents the number of players in the group lesson, where x is the cost per player, in dollars;
3. 1 player, 4 players, 8 players
91.
1. 2010: $40.86 billion; 2013: $60.6 billion;
2. $H^{- 1} (x) = \frac{x - 27.7}{6.58};$ $H^{- 1} (x)$ represents the number of years after 2008, where x is the e-commerce holiday season sales, in billions of dollars.
93. (b), (d), (f), (h)
94. (a), (c), (e), (g)
95. (a)
96. (d)
97. (f)
98. (a), (b), (c), (d)
99. $f (x) = x^{2} - 3$ , for inputs $x \geq 0;$ $f^{- 1} (x) = \sqrt{x + 3}$ , for inputs $x \geq - 3$
101. Answers may vary; $f (x) = 3 / x, f (x) = 1 - x, f (x) = x$

Exercise Set 5.2

Visualizing the Graph

J
F
H
B
E
A
C
I
D
G

Exercise Set 5.3

1.
3.
5.
7.
9. 4
11. 3
13. −3
15. −2
17. 0
19. 1
21. 4
23. $\frac{1}{4}$
25. −7
27. $\frac{1}{2}$
29. $\frac{3}{4}$
31. 0
33. $\frac{1}{2}$
35. $\log_{10} 1000 = 3$ , or $\log 1000 = 3$
37. $\log_{8} 2 = \frac{1}{3}$
39. $\log_{e} t = 3$ , or $ln t = 3$
41. $\log_{e} 7.3891 = 2$ , or $\ln 7.3891 = 2$
43. $\log_{p} 3 = k$
45. $5^{1} = 5$
47. $10^{- 2} = 0.01$
49. $e^{3.4012} = 30$
51. $a^{- x} = M$
53. $a^{x} = T^{3}$
55. 0.4771
57. 2.7259
59. −0.2441
61. Does not exist
63. 0.6931
65. 6.6962
67. Does not exist
69. 3.3219
71. −0.2614
73. 0.7384
75. 2.2619
77. 0.5880
79.
81.
83. Shift the graph of $y = \log_{2} x$ left 3 units. Domain: $(- 3, \infty);$ vertical asymptote: $x = - 3;$
85. Shift the graph of $y = \log_{3} x$ down 1 unit. Domain: $(0, \infty);$ vertical asymptote: $x = 0;$
87. Stretch the graph of $y = \ln x$ vertically. Domain: $(0, \infty)$ ; vertical asymptote: $x = 0$ ;
89. Reflect the graph of $y = ln x$ across the x-axis and shift it up 2 units. Domain: $(0, \infty);$ vertical asymptote: $x = 0;$
91. Shift the graph of $\log x$ right 1 unit, shrink it vertically, and shift it down 2 units.
93.
95.
1. 2.5 ft/sec;
2. 2.8 ft/sec;
3. 2.0 ft/sec;
4. 2.4 ft/sec;
5. 2.2 ft/sec;
6. 2.5 ft/sec;
7. 2.3 ft/sec;
8. 3.1 ft/sec
97.
1. 7.7;
2. 9.5;
3. 6.6;
4. 7.6;
5. 8.0;
6. 7.9;
7. 5.1;
8. 9.3
99.
1. $10^{- 7};$
2. $4.0 \times 10^{- 6};$
3. $6.3 \times 10^{- 4};$
4. $1.6 \times 10^{- 5}$
101.
1. 140 decibels;
2. 115 decibels;
3. 40 decibels;
4. 65 decibels;
5. 120 decibels;
6. 194 decibels
102. $m = \frac{3}{10};$ y-intercept: $(0, - \frac{7}{5})$
103. $m = 0;$ y-intercept: (0, 6)
104. Slope is not defined; no y-intercept
105. −280
106. −4
107. $f (x) = x^{3} - 7 x$
108. $f (x) = x^{3} - x^{2} + 16 x - 16$
109. 3
111. $(0, \infty)$
113. $(- \infty, 0) \cup (0, \infty)$
115. $(- \frac{5}{2}, - 2)$
117. (d)
119. (b)

Mid-Chapter Mixed Review: Chapter 5

Exercise Set 5.4

Exercise Set 5.5

1. 4
3. $\frac{3}{2}$
5. 5.044
7. $\frac{5}{2}$
9. $- 3, \frac{1}{2}$
11. 0.959
13. 0
15. 0
17. 6.908
19. 84.191
21. −1.710
23. 2.844
25. −1.567, 1.567
27. 1.869
29. −1.518, 1.518
31. 625
33. 0.0001
35. e
37. $- \frac{1}{3}$
39. $\frac{22}{3}$
41. 10
43. 4
45. $\frac{1}{63}$
47. 2
49. $\frac{2}{5}$
51. 5
53. $\frac{21}{8}$
55. $\frac{8}{7}$
57. 6
59. 6.192
61. 0
63.
1. (0, −6);
2. $x = 0;$
3. minimum: −6 when $x = 0$
64.
1. (3, 1);
2. $x = 3;$
3. maximum: 1 when $x = 3$
65.
1. (−1, −5);
2. $x = - 1;$
3. maximum: −5 when $x = - 1$
66.
1. (2, 4);
2. $x = 2;$
3. minimum: 4 when $x = 2$
67. $\frac{\ln 2}{2}$ , or $0.347$
69. $1, e^{4}$ or 1, 54.598
71. $\frac{1}{3}, 27$
73. $1, e^{2}$ or 1, 7.389
75. $0, \frac{\ln 2}{\ln 5}$ or 0, 0.431
77. $e^{- 2}$ , $e^{2}$ or 0.135, 7.389
79. $\frac{7}{4}$
81. $a = \frac{2}{3} b$

Exercise Set 5.6

1.
1. $P (t) = 6.18 e^{0.0214 t}$ , where t is the number of years after 2012 and P is in millions;
2. 7.0 million;
3. about 12.1 years after 2012;
4. about 32.4 years
3.
1. 0.90%;
2. 1.63%;
3. 20.9 years;
4. 62.4 years;
5. 0.18%;
6. 29.9 years;
7. 54.2 years;
8. 0.46%;
9. 2.64%;
10. 177.7 years
5. About 819 years after 2013
7.
1. $P (t) = 10, 000 e^{0.054 t};$
2. $10,554.85; $11,140.48; $13,099.64; $17,160.07;
3. about 12.8 years
9. About 12,320 years
11.
1. 22.4% per minute;
2. 3.1% per year;
3. 60.3 days;
4. 10.7 years;
5. 2.4% per year;
6. 1.0% per year;
7. 0.0029% per year
13.
1. $k \approx 0.0069, M (t) = 72.2 e^{- 0.0069 t};$
2. 2015: 49.4%; 2018: 48.4%;
3. in 2046
15.
1. $k \approx 0.0536, C (t) = 1.85 e^{0.0536 t};$
2. 7.07 million barrels of oil per day;
3. 12.9 years;
4. 36.4 years after 1980
17.
1. 167;
2. 500; 1758; 3007; 3449; 3495;
3. as $t \to \infty, N (t) \to 3500;$ the number approaches 3500 but never actually reaches it.
19. 46.7°F
21. 59.6°F
23. Multiplication principle for inequalities
24. Product rule
25. Principle of zero products
26. Principle of square roots
27. Power rule
28. Multiplication principle for equations
29. $166.16
31. $19,609.67
33. $t = - \frac{L}{R} [ln (1 - \frac{i R}{V})]$
35. Linear

Review Exercises: Chapter 5

1. True
2. False
3. False
4. True
5. False
6. True
7. ${(- 2.7, 1.3), (- 3, 8), (3, - 5), (- 3, 6), (- 5, 7)}$
8.
1. $x = - 2 y + 3;$
2. $x = 3 y^{2} + 2 y - 1;$
3. $0.8 y^{3} - 5.4 x^{2} = 3 y$
9. No
10. No
11. Yes
12. Yes
13.
1. Yes;
2. $f^{- 1} (x) = \frac{- x + 2}{3}$
14.
1. Yes;
2. $f^{- 1} (x) = \frac{x + 2}{x - 1}$
15.
1. Yes;
2. $f^{- 1} (x) = x^{2} + 6$ , $x \geq 0$
16.
1. Yes;
2. $f^{- 1} (x) = \sqrt[3]{x + 8}$
17.
1. No
18.
1. Yes;
2. $f^{- 1} (x) = \ln x$
19. $f^{- 1} (f (x)) = f^{- 1} (6 x - 5) = \frac{6 x - 5 + 5}{6} = \frac{6 x}{6} = x;$ $f (f^{- 1} (x)) = f (\frac{x + 5}{6}) = 6 (\frac{x + 5}{6}) - 5 = x + 5 - 5 = x$
20. $f^{- 1} (f (x)) = f^{- 1} (\frac{x + 1}{x}) = \frac{1}{\frac{x + 1}{x} - 1} = \frac{1}{\frac{x + 1 - x}{x}} = \frac{1}{\frac{1}{x}} = x;$ $f (f^{- 1} (x)) = f (\frac{1}{x - 1}) = \frac{\frac{1}{x - 1} + 1}{\frac{1}{x - 1}} = \frac{\frac{1 + x - 1}{x - 1}}{\frac{1}{x - 1}} = \frac{x}{x - 1} \cdot \frac{x - 1}{1} = x$
21. $f^{- 1} (x) = \frac{2 - x}{5};$ domain of f and $f^{- 1} : (- \infty, \infty);$ range of f and $f^{- 1} : (- \infty, \infty);$
22. $f^{- 1} (x) = \frac{- 2 x - 3}{x - 1};$ domain of $f : (- \infty, - 2) \cup (- 2, \infty);$ range of $f : (- \infty, 1) \cup (1, \infty);$ domain of $f^{- 1} : (- \infty, 1) \cup (1, \infty);$ range of $f^{- 1} : (- \infty, - 2) \cup (- 2, \infty);$
23. 657
24. a
25.
26.
27.
28.
29.
30.
31. (c)
32. (a)
33. (b)
34. (f)
35. (e)
36. (d)
37. 3
38. 5
39. 1
40. 0
41. $\frac{1}{4}$
42. $\frac{1}{2}$
43. 0
44. 1
45. $\frac{1}{3}$
46. −2
47. $4^{2} = x$
48. $a^{k} = Q$
49. $\log_{4} \frac{1}{64} = - 3$
50. $\ln 80 = x$ , or $\log_{e} 80 = x$
51. 1.0414
52. −0.6308
53. 1.0986
54. −3.6119
55. Does not exist
56. Does not exist
57. 1.9746
58. 0.5283
59. $\log_{b} \frac{x^{3} \sqrt{z}}{y^{4}}$
60. $ln (x^{2} - 4)$
61. $\frac{1}{4} ln w + \frac{1}{2} ln r$
62. $\frac{2}{3} \log M - \frac{1}{3} \log N$
63. 0.477
64. 1.699
65. −0.699
66. 0.233
67. 5k
68. −6t
69. 16
70. $\frac{1}{5}$
71. 4.382
72. 2
73. $\frac{1}{2}$
74. 5
75. 4
76. 9
77. 1
78. 3.912
79.
1. $A (t) = 30, 000 {(1.0105)}^{4 t};$
2. $30,000; $38,547.20; $49,529.56, $63,640.87
80. 2005: 59.8 GW; 2010: 189.9 GW; 2016: 760.1 Gw
81. 15.4 years
82. 2.7%
83. About 2623 years
84. 5.6
85. 6.3
86. 30 decibels
87.
1. 2.2 ft/sec;
2. 8,553,143
88.
1. $k \approx 0.1392;$
2. $S (t) = 0.035 e^{0.1392 t}$ , where t is the number of years after 1940 and S is in billions of dollars;
3. 1970: $2.279 billion; 2000: $148.353 billion; 2015: $1197.023 billion, or about $1.197 trillion;
4. in 2019
89.
1. $P (t) = 15.2 e^{0.0167 t}$ , where t is the number of years after 2013 and P is in millions;
2. 2017: 16.3 million; 2020: 17.1 million;
3. about 10 years after 2013;
4. 41.5 years
90. D
91. A
92. D
93. B
94. $\frac{1}{64}, 64$
95. 1
96. 16
97. Measure the atmospheric pressure P at the top of the building. Substitute that value in the equation $P = 14.7 e^{- 0.00005 a}$ , and solve for the height, or altitude, a of the top of the building. Also measure the atmospheric pressure at the base of the building and solve for the altitude of the base. Then subtract to find the height of the building.
98. Reflect the graph of $f (x) = \ln x$ across the line $y = x$ to obtain the graph of $h (x) = e^{x}$ . Then shift this graph right 2 units to obtain the graph of $g (x) = e^{x - 2}$ .
99. The inverse of a function $f (x)$ is written $f^{- 1} (x)$ , whereas ${[f (x)]}^{- 1}$ means $\frac{1}{f (x)}$ .
100. $\log_{a} a b^{3} \neq (\log_{a} a) (\log_{a} b^{3})$ . If the first step had been correct, then the second step would be as well. The correct procedure follows: $\log_{a} a b^{3} = \log_{a} a + \log_{a} b^{3} = 1 + 3 \log_{a} b$ .

Test: Chapter 5

[5.1] ${(5, - 2), (3, 4), (- 1, 0), (- 3, - 6)}$
[5.1] No
[5.1] Yes
[5.1]
1. Yes;
2. $f^{- 1} (x) = \sqrt[3]{x - 1}$
[5.1]
1. Yes;
2. $f^{- 1} (x) = 1 - x$
[5.1]
1. Yes;
2. $f^{- 1} (x) = \frac{2 x}{1 + x}$
[5.1]
1. No
[5.1] $f^{- 1} (f (x)) = f^{- 1} (- 4 x + 3) = \frac{3 - (- 4 x + 3)}{4} = \frac{4 x}{4} = x;$ $f (f^{- 1} (x)) = f (\frac{3 - x}{4}) = - 4 (\frac{3 - x}{4}) + 3 = - 3 + x + 3 = x$
[5.1] $f^{- 1} (x) = \frac{4 x + 1}{x};$ domain of $f : (- \infty, 4) \cup (4, \infty);$ range of $f : (- \infty, 0) \cup (0, \infty);$ domain of $f^{- 1} : (- \infty, 0) \cup (0, \infty);$ range of $f^{- 1} : (- \infty, 4) \cup (4, \infty);$
[5.2]
[5.3]
[5.2]
[5.3]
[5.3] −5
[5.3] 1
[5.3] 0
[5.3] $\frac{1}{5}$
[5.3] $x = e^{4}$
[5.3] $x = \log_{3} 5.4$
[5.3] 2.7726
[5.3] −0.5331
[5.3] 1.2851
[5.4] $\log_{a} \frac{x^{2} \sqrt{z}}{y}$
[5.4] $\frac{2}{5} ln x + \frac{1}{5} ln y$
[5.4] 0.656
[5.4] −4t
[5.5] $\frac{1}{2}$
[5.5] 1
[5.5] 1
[5.5] 4.174
[5.3] 6.6
[5.6] 0.0154
[5.6]
1. 4.5%;
2. $P (t) = 1000 e^{0.045 t};$
3. $1433.33;
4. 15.4 years
[5.2] C
[5.5] $\frac{27}{8}$

Chapter 6 Exercise Set 6.1

Exercise Set 6.3

Mid-Chapter Mixed Review: Chapter 6

Exercise Set 6.4

Exercise Set 6.5

1.
1. $(- \frac{3}{4}, - \frac{\sqrt{7}}{4});$
2. $(\frac{3}{4}, \frac{\sqrt{7}}{4});$
3. $(\frac{3}{4}, - \frac{\sqrt{7}}{4})$
3.
1. $(\frac{2}{5}, \frac{\sqrt{21}}{5});$
2. $(- \frac{2}{5}, - \frac{\sqrt{21}}{5});$
3. $(- \frac{2}{5}, \frac{\sqrt{21}}{5})$
5. $(\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2})$
7. 0
9. $\sqrt{3}$
11. 0
13. $- \frac{\sqrt{3}}{2}$
15. Not defined
17. $\frac{\sqrt{3}}{2}$
19. $- \frac{\sqrt{2}}{2}$
21. 0
23. 0
25. 0.4816
27. 1.3065
29. $- 2.1599$
31. 1
33. $- 1.1747$
35. $- 1$
37. $- 0.7071$
39. 0
41. 0.8391
43.
3. same as (b);
4. the same
45.
1. See Exercise 43(a);
3. same as (b);
4. the same
47.
3. same as (b);
4. the same
49.
3. same as (b);
4. the same
51. Even: cosine, secant; odd: sine, tangent, cosecant, cotangent
53. Positive: I, III; negative: II, IV
55. Positive: I, IV; negative: II, III
57.

Stretch the graph of f vertically, then shift it down 3 units.
58.

Shift the graph of f to the right 2 units.
59.

Shift the graph of f to the right 4 units, shrink it vertically, then shift it up 1 unit.
60.

Reflect the graph of f across the x-axis.
61. $y = - {(x - 2)}^{3} - 1$
62. $y = \frac{1}{4 x} + 3$
63. cos x
65. sin x
67. sin x
69. $- \cos x$
71. $- \sin x$
73.
1. $\frac{π}{2} + 2 k π, k an integer;$
2. $π + 2 k π, k an integer;$
3. $k π, k$ an integer
75. $[- \frac{π}{2} + 2 k π, \frac{π}{2} + 2 k π], k$ an integer
77. ${x | x \neq \frac{π}{2} + k π, k an integer}$
79.
81.
83.
1. $△ O P A \sim △ O D B;$
  
  Thus, $\frac{A P}{O A} = \frac{B D}{O B}$
  
  $\frac{\sin θ}{\cos θ} = \frac{B D}{1}$
  
  $\tan θ = B D$
2. $△ O P A \sim △ O D B;$
  
  $\frac{O D}{O P} = \frac{O B}{O A}$
  
  $\frac{O D}{1} = \frac{1}{\cos θ}$
  
  $O D = \sec θ$
3. $△ O A P \sim △ E C O;$
  
  $\frac{O E}{P O} = \frac{C O}{A P}$
  
  $\frac{O E}{1} = \frac{1}{\sin θ}$
  
  $O E = \csc θ$
4. $△ O A P \sim △ E C O$
  
  $\frac{C E}{A O} = \frac{C O}{A P}$
  
  $\frac{C E}{\cos θ} = \frac{1}{\sin θ}$
  
  $C E = \frac{\cos θ}{\sin θ}$
  
  $C E = \cot θ$

Visualizing the Graph

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Exercise Set 6.6

ReviewExercises: Chapter 6

72.

Function	Domain	Range
Sine	$(- \infty, \infty)$	$[- 1, 1]$
Cosine	$(- \infty, \infty)$	$[- 1, 1]$
Tangent	All real numbers except $(π / 2) + k π$ , where `k` is an integer	$(- \infty, \infty)$

73.

Function	I	II	III	IV
Sine	$+$	$+$	$-$	$-$
Cosine	$+$	$-$	$-$	$+$
Tangent	$+$	$-$	$+$	$-$

Test: Chapter 6

Chapter 7 Exercise Set 7.1

Exercise Set 7.2

Exercise Set 7.3

1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31. Sine sum and difference identities:

$\begin{array}{l} sin (x + y) = sin x cos y + cos x sin y, \\ sin (x - y) = sin x cos y - cos x sin y . \end{array}$

Add the sum and difference identities:

$\begin{array}{l} sin (x + y) + sin (x - y) = 2 sin x cos y \\ \frac{1}{2} [sin (x + y) + sin (x - y)] = sin x cos y . (3) \end{array}$

Subtract the difference identity from the sum identity:
$\begin{array}{l} \sin (x + y) - \sin (x - y) = 2 \cos x \sin y \\ \frac{1}{2} [\sin (x + y) - \sin (x - y)] = \cos x \sin y . (4) \end{array}$
33. $\sin 3 θ - \sin 5 θ = 2 \cos \frac{8 θ}{2} \sin \frac{- 2 θ}{2} = - 2 \cos 4 θ \sin θ$
35. $\sin 8 θ + \sin 5 θ = 2 \sin \frac{13 θ}{2} \cos \frac{3 θ}{2}$
37. $\sin 7 u \sin 5 u = \frac{1}{2} (\cos 2 u - \cos 12 u)$
39. $\begin{matrix} 7 \cos θ \sin 7 θ & = \frac{7}{2} [\sin 8 θ - \sin (- 6 θ)] \\ = \frac{7}{2} (\sin 8 θ + \sin 6 θ) \end{matrix}$
41. $\cos 55 ° \sin 25 ° = \frac{1}{2} (\sin 80 ° - \sin 30 °) = \frac{1}{2} \sin 80 ° - \frac{1}{4}$
43.
45.
47.
49.
51.
1. (a), (d)
2. (b) yes;
3. (c) $f^{- 1} (x) = \frac{x + 2}{3}$
52.
1. (a), (d)
2. (b) yes;
3. (c) $f^{- 1} (x) = \sqrt[3]{x - 1}$
53.
1. (a), (d)
2. (b) yes;
3. (c) $f^{- 1} (x) = \sqrt{x + 4}$
54.
1. (a), (d)
55. $0, \frac{5}{2}$
56. $- 4, \frac{7}{3}$
57. $\pm 2, \pm 3 i$
58. $5 \pm 2 \sqrt{6}$
59. 27
60. 9
61.
63. $\begin{matrix} \log (\cos x - \sin x) & + \log (\cos x + \sin x) \\ = \log [(\cos x - \sin x) (\cos x + \sin x)] \\ = \log (\cos^{2} x - \sin^{2} x) = \log \cos 2 x \end{matrix}$
65. $\begin{matrix} \frac{1}{ω C (\tan θ + \tan ϕ)} & = \frac{1}{ω C (\frac{\sin θ}{\cos θ} + \frac{\sin ϕ}{\cos ϕ})} \\ = \frac{1}{ω C (\frac{\sin θ \cos ϕ + \sin ϕ \cos θ}{\cos θ \cos ϕ})} \\ = \frac{\cos θ \cos ϕ}{ω C \sin (θ + ϕ)} \end{matrix}$

Mid-Chapter Mixed Review: Chapter 7

Visualizing the Graph

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Exercise Set 7.5

Review Exercises: Chapter 7

Chapter 8 Exercise Set 8.1

1. $A = 121 °$ , $a \approx 33$ , $c \approx 14$
3. $B \approx 57.4 °$ , $C \approx 86.1 °$ , $c \approx 40$ , or $B \approx 122.6 °$ , $C \approx 20.9 °$ , $c \approx 14$
5. $B \approx 44 ° 24'$ , $A \approx 74 ° 26'$ , $a \approx 33.3$
7. $A = 110.36 °$ , $a \approx 5 mi$ , $b \approx 3.4 mi$
9. $B \approx 83.78 °$ , $A \approx 12.44 °$ , $a \approx 12.30 yd$
11. $B \approx 14.7 °$ , $C \approx 135.0 °$ , $c \approx 28.04 cm$
13. No solution
15. $B = 125.27 °$ , $b \approx 302 m$ , $c \approx 138 m$
17. $8.2 {ft}^{2}$
19. $12 {yd}^{2}$
21. $596.98 {ft}^{2}$
23. About 3894 ft
25. $787 {ft}^{2}$
27. About 51 ft
29. From A: about 35 mi; from B: about 66 mi
31. About 102 mi
33. 1.348, $77.2 °$
34. No angle
35. $18.24 °$
36. $125.06 °$
37. $\frac{\sqrt{2}}{2}$
38. $\frac{\sqrt{3}}{2}$
39. $- \frac{1}{2}$
40. $- \frac{\sqrt{3}}{2}$
41. Use the formula for the area of a triangle and the law of sines.

$K = \frac{1}{2} b c sin A and b = \frac{c sin B}{sin C},$

so $K = \frac{c^{2} sin A sin B}{2 sin C} .$

$K = \frac{1}{2} a b sin C and b = \frac{a sin B}{sin A},$

so $K = \frac{a^{2} sin B sin C}{2 sin A} .$

$K = \frac{1}{2} b c sin A and c = \frac{b sin C}{sin B},$

so $K = \frac{b^{2} sin A sin C}{2 sin B} .$
43.
For the quadrilateral ABCD, we have

$\begin{matrix} Area & = & \frac{1}{2} b d sin θ + \frac{1}{2} a c sin θ + \frac{1}{2} a d (sin 180 ° - θ) \\ + \frac{1}{2} b c sin (180 ° - θ) Note: \sin θ = \sin (180 ° - θ) . \\ = & \frac{1}{2} (b d + a c + a d + b c) sin θ \\ = & \frac{1}{2} (a + b) (c + d) sin θ \\ = & \frac{1}{2} d_{1} d_{2} sin θ, \end{matrix}$

where $d_{1} = a + b$ and $d_{2} = c + d$ .
45. $44. 1^{″}$ from wall 1 and $104. 3^{″}$ from wall 4

Exercise Set 8.2

Exercise Set 8.3

Mid-Chapter Mixed Review: Chapter 8

1. True
2. True
3. False
4. True
5. $B = 63 °$ , $b \approx 9.4 in .$ , $c \approx 9.5 in$ .
6. No solution
7. $A \approx 40.5 °$ , $B \approx 28.5 °$ , $C \approx 111.0 °$
8. $B \approx 70.2 °$ , $C \approx 67.1 °$ , $c \approx 39.9 cm or B \approx 109.8 °$ , $C \approx 27.5 °$ , $c \approx 20.0 cm$
9. $a \approx 370 yd$ , $B \approx 16.6 °$ , $C \approx 15.4 °$
10. $C \approx 45 °$ , $A \approx 107 °$ , $a \approx 37 ft$ , or $C \approx 135 °$ , $A \approx 17 °$ , $a \approx 11 ft$
11. About $446 {in}^{2}$
12.
; $\sqrt{34}$
13.
; 1
14.
; 4
15.
; $\sqrt{26}$
16. $\sqrt{2} (cos \frac{π}{3} + i sin \frac{π}{3}), or \sqrt{2} (cos 60 ° + i sin 60 °)$
17. $2 (cos \frac{5 π}{3} + i sin \frac{5 π}{3}), or 2 (cos 300 ° + i sin 300 °)$
18. $5 (cos \frac{π}{2} + i sin \frac{π}{2}), or 5 (cos 90 ° + i sin 90 °)$
19. $2 \sqrt{2} (cos \frac{5 π}{4} + i sin \frac{5 π}{4}), or 2 \sqrt{2} (cos 225 ° + i sin 225 °)$
20. $\sqrt{2} - \sqrt{2} i$
21. $6 \sqrt{3} + 6 i$
22. $\sqrt{5}$
23. $4 i$
24. $16 (cos 45 ° + i sin 45 °)$
25. $9 [cos (\frac{π}{12}) + i sin (\frac{π}{12})]$
26. $2 \sqrt{2} (cos 285 ° + i sin 285 °)$
27. $\sqrt{2} (cos 255 ° + i sin 255 °)$
28. $8 \sqrt{2} (cos 45 ° + i sin 45 °)$
29. $8 + 8 \sqrt{3} i$
30. $2 (\cos 120 ° + i \sin 120 °)$ and $2 (\cos 300 ° + i \sin 300 °)$ , or $- 1 + \sqrt{3} i$ and $1 - \sqrt{3} i$
31. $1 (cos 60 ° + i sin 60 °)$ , $1 (\cos 180 ° + i \sin 180 °),$ and $1 (\cos 300 ° + i \sin 300 °),$ or $\frac{1}{2} + \frac{\sqrt{3}}{2} i, - 1,$ and $\frac{1}{2} - \frac{\sqrt{3}}{2} i$
32. Using the law of cosines, it is necessary to solve the quadratic equation:

$\begin{matrix} {(11.1)}^{2} & = & a^{2} + {(28.5)}^{2} - 2 a (28.5) cos 19 °, or \\ 0 & = & a^{2} - [2 (28.5) cos 19 °] a + [{(28.5)}^{2} - {(11.1)}^{2}] . \end{matrix}$

The law of sines requires less complicated computations.
33. A nonzero complex number has n different complex nth roots. Thus, 1 has three different complex cube roots, one of which is the real number 1. The other two roots are complex conjugates. Since the set of real numbers is a subset of the set of complex numbers, the real cube root of 1 is also a complex root of 1.
34. The law of sines involves two angles of a triangle and the sides opposite them. Three of these four values must be known in order to find the fourth. Given SAS, only two of these four values are known.
35. The law of sines involves two angles of a triangle and the sides opposite them. Three of these four values must be known in order to find the fourth. Thus we must know the measure of one angle in order to use the law of sines.
36. Trigonometric notation is not unique because there are infinitely many angles coterminal with a given angle. Standard notation is unique because any point has a unique ordered pair $(a, b)$ associated with it.
37.
$\begin{matrix} x^{6} - 2 x^{3} + 1 & = & 0 \\ {(x^{3} - 1)}^{2} & = & 0 \\ x^{3} - 1 & = & 0 \\ x^{3} & = & 1 \\ x & = & 1^{1 / 3} \end{matrix}$

This equation has three distinct solutions because there are three distinct cube roots of 1.

$\begin{matrix} x^{6} - 2 x^{3} & = & 0 \\ x^{3} (x^{3} - 2) & = & 0 \\ x^{3} = 0 & or & x^{3} - 2 = 0 \\ x^{3} = 0 & or & x^{3} = 2 \\ x = 0 & or & x = 2^{1 / 3} \end{matrix}$

This equation has four distinct solutions because 0 is one solution and the three distinct cube roots of 2 provide an additional three solutions.

$\begin{matrix} x^{6} - 2 x & = & 0 \\ x (x^{5} - 2) & = & 0 \\ x = 0 & o r & x^{5} - 2 & = 0 \\ x = 0 & o r & x^{5} & = 2 \\ x = 0 & o r & x & = 2^{1 / 5} \end{matrix}$

This equation has six distinct solutions because 0 is one solution and the five fifth roots of 2 provide an additional five solutions.

Visualizing the Graph

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Exercise Set 8.4

13.
$A : (4, 30 °), (4, 390 °), (- 4, 210 °)$ ; $B : (5, 300 °), (5, - 60 °), (- 5, 120 °)$ ; $C : (2, 150 °), (2, 510 °), (- 2, 330 °)$ ; $D : (3, 225 °), (3, - 135 °), (- 3, 45 °)$ ; answers may vary
15. $(3, 270 °), (3, \frac{3 π}{2})$
17. $(6, 300 °), (6, \frac{5 π}{3})$
19. $(8, 330 °), (8, \frac{11 π}{6})$
21. $(2, 225 °), (2, \frac{5 π}{4})$
23. $(2, 60 °), (2, \frac{π}{3})$
25. $(5, 315 °), (5, \frac{7 π}{4})$
27. $(\frac{5}{2}, \frac{5 \sqrt{3}}{2})$
29. $(- \frac{3 \sqrt{2}}{2}, - \frac{3 \sqrt{2}}{2})$
31. $(- \frac{3}{2}, - \frac{3 \sqrt{3}}{2})$
33. $(- 1, \sqrt{3})$
35. $(- \sqrt{3}, - 1)$
37. $(3 \sqrt{3}, - 3)$
39. $r (3 cos θ + 4 sin θ) = 5$
41. $r cos θ = 5$
43. $r = 6$
45. $r^{2} {cos}^{2} θ = 25 r sin θ$
47. $r^{2} {sin}^{2} θ - 5 r cos θ - 25 = 0$
49. $r^{2} = 2 r cos θ$
51. $x^{2} + y^{2} = 25$
53. $y = 2$
55. $y^{2} = - 6 x + 9$
57. $x^{2} - 9 x + y^{2} - 7 y = 0$
59. $x = 5$
61. $y = - \sqrt{3} x$
63.
65.
67.
69.
71. 12
72. $\frac{1}{5}$
73.
74.
75.
76.
77. (d)
79. (g)
81. (j)
83. (b)
85. (e)
87. (k)
89. $y^{2} = - 4 x + 4$

Exercise Set 8.6

Review Exercises: Chapter 8

Chapter 9 Visualizing the Graph

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Exercise Set 9.1

Exercise Set 9.2

1. (3, −2, 1)
3. (−3, 2, 1)
5. $(2, \frac{1}{2}, - 2)$
7. No solution
9. Infinitely many solutions; $(\frac{11 y + 19}{5}, y, \frac{9 y + 11}{5})$
11. $(\frac{1}{2}, \frac{2}{3}, - \frac{5}{6})$
13. (−1, 4, 3)
15. (1, −2, 4, −1)
17. Russian Federation: 80 medals; Ukraine: 25 medals; United States: 18 medals
19. In a restaurant: 78 meals; in a car: 34 meals; at home: 58 meals
21. Bacon: $142.4 million; Warhol: $105.4 million; Koons: $58.4 million
23. Dogs: 69.9 million; cats: 74.1 million; birds: 8.3 million
25. United States: $31.5 billion; Great Britain: $17.9 billion; Germany: $14.1 billion
27. $1 \frac{1}{4}$ servings of beef, 1 baked potato, $\frac{3}{4}$ serving of strawberries
29. 2%: $300; 3%: $800; 4%: $2400
31. Orange juice: $2.00; bagel: $2.50; coffee: $1.80
33.
1. $f (x) = \frac{3}{40} x^{2} - \frac{7}{4} x + 43;$
2. 2007: 33.625% 2014: 40.8%
35.
1. $f (x) = - 7 x^{2} + 55 x + 291;$
2. 333,000 deportations
37. Perpendicular
38. The leading-term test
39. Vertical line
40. One-to-one function
41. Rational function
42. Inverse variation
43. Vertical asymptote
44. Horizontal asymptote
45. $(- 1, \frac{1}{5}, - \frac{1}{2})$
47. $180 °$
49. $3 x + 4 y + 2 z = 12$
51. $y = - 4 x^{3} + 5 x^{2} - 3 x + 1$
53. Adults: 5; students: 1; children: 94

Exercise Set 9.4

1. $x = - 3, y = 5$
3. $x = - 1, y = 1$
5. $[\begin{array}{r} - 2 & 7 \\ 6 & 2 \end{array}]$
7. $[\begin{array}{r} 1 & 3 \\ 2 & 6 \end{array}]$
9. $[\begin{array}{r} 9 & 9 \\ - 3 & - 3 \end{array}]$
11. $[\begin{array}{r} 11 & 13 \\ 5 & 3 \end{array}]$
13. $[\begin{array}{r} - 4 & 3 \\ - 2 & - 4 \end{array}]$
15. $[\begin{array}{r} 17 & 9 \\ - 2 & 1 \end{array}]$
17. $[\begin{array}{r} 0 & 0 \\ 0 & 0 \end{array}]$
19. $[\begin{array}{r} 1 & 2 \\ 4 & 3 \end{array}]$
21. $[\begin{array}{r} 1 \\ 40 \end{array}]$
23. $[\begin{array}{r} - 10 & 28 \\ 14 & - 26 \\ 0 & - 6 \end{array}]$
25. Not defined
27. $[\begin{array}{r} 3 & 16 & 3 \\ 0 & - 32 & 0 \\ - 6 & 4 & 5 \end{array}]$
29.
1. $[\begin{matrix} 40 & 20 & 30 \end{matrix}];$
2. $[\begin{matrix} 44 & 22 & 33 \end{matrix}];$
3. $[\begin{matrix} 84 & 42 & 63 \end{matrix}];$ the total amount of each type of produce ordered for both weeks
31.
1. $C = [\begin{matrix} 140 & 27 & 3 & 13 & 64 \end{matrix}]$ , $P = [\begin{matrix} 180 & 4 & 11 & 24 & 662 \end{matrix}]$ , $B = [\begin{matrix} 50 & 5 & 1 & 82 & 20 \end{matrix}];$
2. $[\begin{matrix} 650 & 50 & 28 & 307 & 1448 \end{matrix}];$ the total nutritional value of a meal of 1 serving of chicken, 1 cup of potato salad, and 3 broccoli spears
33.
1. $[\begin{matrix} 1.50 & 0.30 & 0.36 & 0.45 & 0.64 \\ 1.55 & 0.28 & 0.48 & 0.57 & 0.75 \\ 1.62 & 0.52 & 0.65 & 0.38 & 0.53 \\ 1.70 & 0.43 & 0.40 & 0.42 & 0.68 \end{matrix}];$
2. $[\begin{matrix} 65 & 48 & 93 & 57 \end{matrix}];$
3. $[\begin{matrix} 419.46 & 105.81 & 129.69 & 115.89 & 165.65 \end{matrix}];$
4. the total cost, in dollars, for each item for the day’s meals
35.
1. $[\begin{matrix} 8 & 15 \\ 6 & 10 \\ 4 & 3 \end{matrix}];$
2. $[\begin{matrix} 4 & 2.5 & 3 \end{matrix}];$
3. $[\begin{matrix} 59 & 94 \end{matrix}];$
4. the total cost, in dollars, of ingredients for each coffee shop
37.
1. $[\begin{matrix} 7.50 & 4.80 & 6.25 \end{matrix}];$
2. $PS = [\begin{matrix} 113.80 & 179.25 \end{matrix}]$
39. $[\begin{array}{r} 2 & - 3 \\ 1 & 5 \end{array}] [\begin{array}{r} x \\ y \end{array}] = [\begin{array}{r} 7 \\ - 6 \end{array}]$
41. $[\begin{array}{r} 1 & 1 & - 2 \\ 3 & - 1 & 1 \\ 2 & 5 & - 3 \end{array}] [\begin{array}{r} x \\ y \\ z \end{array}] = [\begin{array}{r} 6 \\ 7 \\ 8 \end{array}]$
43. $[\begin{array}{r} 3 & - 2 & 4 \\ 2 & 1 & - 5 \end{array}] [\begin{array}{r} x \\ y \\ z \end{array}] = [\begin{array}{r} 17 \\ 13 \end{array}]$
45. $[\begin{array}{r} - 4 & 1 & - 1 & 2 \\ 1 & 2 & - 1 & - 1 \\ - 1 & 1 & 4 & - 3 \\ 2 & 3 & 5 & - 7 \end{array}] [\begin{array}{r} w \\ x \\ y \\ z \end{array}] = [\begin{array}{r} 12 \\ 0 \\ 1 \\ 9 \end{array}]$
47.
1. $(\frac{1}{2}, - \frac{25}{4});$
2. $x = \frac{1}{2};$
3. minimum: $- \frac{25}{4};$
48.
1. $(\frac{5}{4}, - \frac{49}{8});$
2. $x = \frac{5}{4};$
3. minimum: $- \frac{49}{8};$
49.
1. $(- \frac{3}{2}, \frac{17}{4});$
2. $x = - \frac{3}{2};$
3. maximum: $\frac{17}{4};$
50.
1. $(\frac{2}{3}, \frac{16}{3});$
2. $x = \frac{2}{3};$
3. maximum: $\frac{16}{3};$
51. $(A + B) (A - B) = [\begin{array}{r} - 2 & 1 \\ 2 & - 1 \end{array}];$ $A^{2} - B^{2} = [\begin{array}{r} 0 & 3 \\ 0 & - 3 \end{array}]$
53. $\begin{array}{rcl} (A + B) (A - B) & = & [\begin{array}{r} - 2 & 1 \\ 2 & - 1 \end{array}] \\ = & A^{2} + BA - AB - B^{2} \end{array}$ $= A^{2} + B A - A B - B^{2}$

Mid-Chapter Mixed Review: Chapter 9

Exercise Set 9.7

Review Exercises: Chapter 9

Test: Chapter 9

[9.1] (−3, 5); consistent, independent
[9.1] Infinitely many solutions; $(x, 2 x - 3)$ or $(\frac{y + 3}{2}, y);$ consistent, dependent
[9.1] No solution; inconsistent, independent
[9.1] (1, −2); consistent, independent
[9.2] (−1, 3, 2)
[9.1] Student: 462 tickets; nonstudent: 158 tickets
[9.2] Hui: 120 orders; Ashlyn: 104 orders; Sheriann: 128 orders
[9.4] $[\begin{array}{r} - 2 & - 3 \\ - 3 & 4 \end{array}]$
[9.4] Not defined
[9.4] $[\begin{array}{r} - 7 & - 13 \\ 5 & - 1 \end{array}]$
[9.4] Not defined
[9.4] $[\begin{array}{r} 2 & - 2 & 6 \\ - 4 & 10 & 4 \end{array}]$
[9.5] $[\begin{array}{r} 0 & - 1 \\ - \frac{1}{4} & - \frac{3}{4} \end{array}]$
[9.4]
1. $[\begin{array}{r} 1.55 & 1.00 & 0.99 \\ 1.70 & 0.95 & 1.01 \\ 1.65 & 0.99 & 0.96 \end{array}]$ ;
2. $[\begin{array}{r} 26 & 18 & 23 \end{array}]$ ;
3. $[\begin{matrix} 108.85 & 65.87 & 66.00 \end{matrix}]$ ;
4. the total cost, in dollars, for each type of menu item served on the given day
[9.4] $[\begin{array}{r} 3 & - 4 & 2 \\ 2 & 3 & 1 \\ 1 & - 5 & - 3 \end{array}] [\begin{array}{r} x \\ y \\ z \end{array}] = [\begin{array}{r} - 8 \\ 7 \\ 3 \end{array}]$
[9.5] (−2, 1, 1)
[9.6] 61
[9.6] −33
[9.6] $(- \frac{1}{2}, \frac{3}{4})$
[9.7]
[9.7] Maximum: 15 when $x = 3$ and $y = 3$ ; minimum: 2 when $x = 1$ and $y = 0$
[9.7] Maximum profit of $750 is achieved when 25 pound cakes and 75 carrot cakes are prepared.
[9.8] $- \frac{2}{x - 1} + \frac{5}{x + 3}$
[9.7] D
[9.2] $A = 1, B = - 3, C = 2$

Chapter 10 Exercise Set 10.2

Exercise Set 10.3

Visualizing the Graph

B
J
F
I
H
G
E
D
C
A

Exercise Set 10.4

Mid-Chapter Mixed Review: Chapter 10

Exercise Set 10.6

1. (b)
3. (a)
5. (d)
7.
1. Parabola;
2. vertical, 1 unit to the right of the pole;
3. $(\frac{1}{2}, 0);$
9.
1. Hyperbola;
2. horizontal, $\frac{3}{2}$ units below the pole;
3. $(- 3, \frac{π}{2}), (1, \frac{3 π}{2});$
11.
1. Ellipse;
2. vertical, $\frac{8}{3}$ units to the left of the pole;
3. $(\frac{8}{3}, 0), (\frac{8}{9}, π);$
13.
1. Hyperbola;
2. horizontal, $\frac{4}{3}$ units above the pole;
3. $(\frac{4}{5}, \frac{π}{2}), (- 4, \frac{3 π}{2});$
15.
1. Ellipse;
2. vertical, 3 units to the right of the pole;
3. $(1, 0), (3, π);$
17.
1. Parabola;
2. horizontal, $\frac{3}{2}$ units below the pole;
3. $(\frac{3}{4}, \frac{3 π}{2});$
19.
1. Ellipse;
2. vertical, 4 units to the left of the pole;
3. $(4, 0), (\frac{4}{3}, π);$
21.
1. Hyperbola;
2. horizontal, $\frac{7}{10}$ units above the pole;
3. $(\frac{7}{12}, \frac{π}{2}), (- \frac{7}{8}, \frac{3 π}{2});$
23. $y^{2} + 2 x - 1 = 0$
25. $x^{2} - 3 y^{2} - 12 y - 9 = 0$
27. $27 x^{2} + 36 y^{2} - 48 x - 64 = 0$
29. $4 x^{2} - 5 y^{2} + 24 y - 16 = 0$
31. $3 x^{2} + 4 y^{2} + 6 x - 9 = 0$
33. $4 x^{2} - 12 y - 9 = 0$
35. $3 x^{2} + 4 y^{2} - 8 x - 16 = 0$
37. $4 x^{2} - 96 y^{2} + 140 y - 49 = 0$
39. $r = \frac{6}{1 + 2 \sin θ}$
41. $r = \frac{4}{1 + \cos θ}$
43. $r = \frac{2}{2 - \cos θ}$
45. $r = \frac{15}{4 + 3 \sin θ}$
47. $r = \frac{8}{1 - 4 \sin θ}$
49. $f (t) = {(t - 3)}^{2} + 4, or t^{2} - 6 t + 13$
50. $f (2 t) = {(2 t - 3)}^{2} + 4, or 4 t^{2} - 12 t + 13$
51. $f (t - 1) = {(t - 4)}^{2} + 4, or t^{2} - 8 t + 20$
52. $f (t + 2) = {(t - 1)}^{2} + 4, or t^{2} - 2 t + 5$
53. $r = \frac{1.5 \times 10^{8}}{1 + \sin θ}$

Review Exercises: Chapter 10

Test: Chapter 10

Chapter 11 Exercise Set 11.1

Visualizing the Graph

J
A
C
G
F
H
E
D
B
I

Exercise Set 11.3

1. 2
3. −1
5. −2
7. 0.1
9. $\frac{a}{2}$
11. 128
13. 162
15. $7 {(5)}^{40}$
17. $3^{n - 1}$
19. ${(- 1)}^{n - 1}$
21. $\frac{1}{x^{n}}$
23. 762
25. $\frac{4921}{18}$
27. True
29. True
31. True
33. 8
35. 125
37. Does not exist
39. $\frac{2}{3}$
41. $S_{11} \approx 29.65317$
43. 2
45. Does not exist
47. $4545. $\bar{45}$
49. $\frac{160}{9}$
51. $\frac{13}{99}$
53. 9
55. $\frac{34, 091}{9990}$
57. $2,684,354.55
59.
1. About 297 ft;
2. 300 ft
61. $39,505.71
63. 10,485.76 in.
65. $19,694.01
67. $86,666,666,667
69. $(f \circ g) (x) = 16 x^{2} + 40 x + 25$ ; $(g \circ f) (x) = 4 x^{2} + 5$
70. $(f \circ g) (x) = x^{2} + x + 2$ ; $(g \circ f) (x) = x^{2} - x + 3$
71. 2.209
72. $\frac{1}{16}$
73. $(4 - \sqrt{6}) / (\sqrt{3} - \sqrt{2}) = 2 \sqrt{3} + \sqrt{2}$ ,

$(6 \sqrt{3} - 2 \sqrt{2}) / (4 - \sqrt{6}) = 2 \sqrt{3} + \sqrt{2}$ ; there exists a common ratio, $2 \sqrt{3} + \sqrt{2}$ ; thus the sequence is geometric.
75.
1. $\frac{13}{3}; \frac{22}{3}, \frac{34}{3}, \frac{46}{3}, \frac{58}{3}$ ;
2. $- \frac{11}{3}; - \frac{2}{3}, \frac{10}{3}, - \frac{50}{3}, \frac{250}{3}$ or 5; 8, 12, 18, 27
77. $S_{n} = \frac{x^{2} (1 - {(- x)}^{n})}{x + 1}$

Exercise Set 11.4

1. $1^{2} < 1^{3}$ , false; $2^{2} < 2^{3}$ , true; $3^{2} < 3^{3}$ , true; $4^{2} < 4^{3}$ , true; $5^{2} < 5^{3}$ , true
3. A polygon of 3 sides has $\frac{3 (3 - 3)}{2}$ diagonals. True; A polygon of 4 sides has $\frac{4 (4 - 3)}{2}$ diagonals.

True; A polygon of 5 sides has $\frac{5 (5 - 3)}{2}$ diagonals. True; A polygon of 6 sides has $\frac{6 (6 - 3)}{2}$ diagonals. True; A polygon of 7 sides has $\frac{7 (7 - 3)}{2}$ diagonals. True.
5.
$\begin{array}{l} S_{n} : & 2 + 4 + 6 + \dots + 2 n = n (n + 1) \\ S_{1} : & 2 = 1 (1 + 1) \\ S_{k} : & 2 + 4 + 6 + \dots + 2 k = k (k + 1) \\ S_{k + 1} : & 2 + 4 + 6 + \dots + 2 k + 2 (k + 1) \\ = (k + 1) (k + 2) \end{array}$
1. Basis step: $S_{1}$ true by substitution.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Starting with the left side of $S_{k + 1}$ , we have
  
  $\begin{array}{l} 2 + 4 + 6 + \dots + 2 k + 2 (k + 1) \\ = k (k + 1) + 2 (k + 1) By S_{k} \\ = (k + 1) (k + 2) . Distributive law \end{array}$
7.
$\begin{array}{l} S_{n} : & 1 + 5 + 9 + \dots + (4 n - 3) = n (2 n - 1) \\ S_{1} : & 1 = 1 (2 \cdot 1 - 1) \\ S_{k} : & 1 + 5 + 9 + \dots + (4 k - 3) = k (2 k - 1) \\ S_{k + 1} : & 1 + 5 + 9 + \dots + (4 k - 3) + [4 (k + 1) - 3] \\ = (k + 1) [2 (k + 1) - 1] \\ = (k + 1) (2 k + 1) \end{array}$
1. Basis step: $S_{1}$ true by substitution.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Starting with the left side of $S_{k + 1}$ , we have
$\begin{array}{l} 1 + 5 + 9 + \dots + (4 k - 3) + [4 (k + 1) - 3] \\ = k (2 k - 1) + [4 (k + 1) - 3] By S_{k} \\ = 2 k^{2} - k + 4 k + 4 - 3 \\ = 2 k^{2} + 3 k + 1 \\ = (k + 1) (2 k + 1) \end{array} .$
9.
$\begin{array}{l} S_{n} : & 2 + 4 + 8 + \dots + 2^{n} = 2 (2^{n} - 1) \\ S_{1} : & 2 = 2 (2 - 1) \\ S_{k} : & 2 + 4 + 8 + \dots + 2^{k} = 2 (2^{k} - 1) \\ S_{k + 1} : & 2 + 4 + 8 + \dots + 2^{k} + 2^{k + 1} = 2 (2^{k + 1} - 1) \end{array}$
1. Basis step: $S_{1}$ true by substitution.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Starting with the left side of $S_{k + 1}$ , we have
$\begin{array}{l} \underset{}{\underset{︸}{2 + 4 + 8 + \dots + 2^{k}}} + 2^{k + 1} \\ = 2 (2^{k} - 1) + 2^{k + 1} By S_{k} \\ = 2^{k + 1} - 2 + 2^{k + 1} \\ = 2 \cdot 2^{k + 1} - 2 \\ = 2 (2^{k + 1} - 1) \end{array}$ .
11.
$\begin{array}{l} S_{n} : & n < n + 1 \\ S_{1} : & 1 < 1 + 1 \\ S_{k} : & k < k + 1 \\ S_{k + 1} : & k + 1 < (k + 1) + 1 \end{array}$
1. Basis step: Since $1 < 1 + 1, S_{1}$ is true.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Now
$\begin{array}{rcll} k & < & k + 1 & By S_{k} \\ k + 1 & < & k + 1 + 1 & Adding 1 \\ k + 1 & < & k + 2. & Simplifying \end{array}$
13.
$\begin{array}{l} S_{n} : & 2 n \leq 2^{n} \\ S_{1} : & 2 \cdot 1 \leq 2^{1} \\ S_{k} : & 2 k \leq 2^{k} \\ S_{k + 1} : & 2 (k + 1) \leq 2^{k + 1} \end{array}$
1. Basis step: Since $2 = 2, S_{1}$ is true.
2. Induction step: Let k be any natural number. Assume $S_{k}$ . Deduce $S_{k + 1}$ .
$\begin{array}{rcl} 2 k & \leq & 2^{k} & By S_{k} \\ 2 \cdot 2 k & \leq & 2 \cdot 2^{k} & Multiplying by 2 \\ 4 k & \leq & 2^{k + 1} \end{array}$

Since $1 \leq k, k + 1 \leq k + k$ , or $k + 1 \leq 2 k$ .

Then $2 (k + 1) \leq 4 k$ . Multiplying by 2

Thus, $2 (k + 1) \leq 4 k \leq 2^{k + 1}$ , so $2 (k + 1) \leq 2^{k + 1}$ .
15.
$\begin{array}{l} S_{n} : & \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots \\ + \frac{1}{n (n + 1) (n + 2)} = \frac{n (n + 3)}{4 (n + 1) (n + 2)} \\ S_{1} : & \frac{1}{1 \cdot 2 \cdot 3} = \frac{1 (1 + 3)}{4 (1 + 1) (1 + 2)} \\ S_{k} : & \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{k (k + 1) (k + 2)} \\ = \frac{k (k + 3)}{4 (k + 1) (k + 2)} \\ S_{k + 1} : & \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{k (k + 1) (k + 2)} \\ + \frac{1}{(k + 1) (k + 2) (k + 3)} \\ = \frac{(k + 1) (k + 1 + 3)}{4 (k + 1 + 1) (k + 1 + 2)} = \frac{(k + 1) (k + 4)}{4 (k + 2) (k + 3)} \end{array}$
1. Basis step: Since $\frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6}$ and $\frac{1 (1 + 3)}{4 (1 + 1) (1 + 2)} = \frac{1 \cdot 4}{4 \cdot 2 \cdot 3} = \frac{1}{6}, S_{1}$ is true.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Add $\frac{1}{(k + 1) (k + 2) (k + 3)}$ on both sides of $S_{k}$ and simplify the right side. Only the right side is shown here.
$\begin{array}{l} \frac{k (k + 3)}{4 (k + 1) (k + 2)} + \frac{1}{(k + 1) (k + 2) (k + 3)} \\ \begin{matrix} \begin{array}{l} = \frac{k (k + 3) (k + 3) + 4}{4 (k + 1) (k + 2) (k + 3)} \\ = \frac{k^{3} + 6 k^{2} + 9 k + 4}{4 (k + 1) (k + 2) (k + 3)} \\ = \frac{{(k + 1)}^{2} (k + 4)}{4 (k + 1) (k + 2) (k + 3)} \\ = \frac{(k + 1) (k + 4)}{4 (k + 2) (k + 3)} \end{array} \end{matrix} \end{array}$
17.
$\begin{array}{l} S_{n} : & 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2} \\ S_{1} : & 1 = \frac{1 (1 + 1)}{2} \\ S_{k} : & 1 + 2 + 3 + \dots + k = \frac{k (k + 1)}{2} \\ S_{k + 1} : & 1 + 2 + 3 + \dots + k + (k + 1) = \frac{(k + 1) (k + 2)}{2} \end{array}$
1. Basis step: $S_{1}$ true by substitution.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Starting with the left side of $S_{k + 1}$ , we have
$\begin{array}{l} \underset{}{\underset{︸}{1 + 2 + 3 + \dots + k}} + (k + 1) \\ \begin{array}{l} = \frac{k (k + 1)}{2} + (k + 1) & By S_{k} \\ = \frac{k (k + 1) + 2 (k + 1)}{2} & Adding \\ = \frac{(k + 1) (k + 2)}{2} . & Distributive law \end{array} \end{array}$
19.
$\begin{array}{l} S_{n} : & 1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = \frac{n^{2} {(n + 1)}^{2}}{4} \\ S_{1} : & 1^{3} = \frac{1^{2} {(1 + 1)}^{2}}{4} \\ S_{k} : & 1^{3} + 2^{3} + 3^{3} + \dots + k^{3} = \frac{{(k + 1)}^{2}}{4} \\ S_{k + 1} : & 1^{3} + 2^{3} + 3^{3} + \dots + k^{3} + {(k + 1)}^{3} \\ = \frac{{(k + 1)}^{2} {[(k + 1) + 1]}^{2}}{4} \end{array}$
1. Basis step: $S_{1} : 1^{3} = \frac{1^{2} {(1 + 1)}^{2}}{4 = 1}$ . True.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ .
$\begin{array}{l} 1^{3} + 2^{3} + \dots + k^{3} = \frac{k^{2} {(k + 1)}^{2}}{4} & S_{k} \end{array}$

$\begin{array}{rcl} 1^{3} + 2^{3} + \dots + k^{3} + {(k + 1)}^{3} & = & \frac{k^{2} {(k + 1)}^{2}}{4} + {(k + 1)}^{3} \\ Adding {(k + 1)}^{3} \\ = & \frac{k^{2} {(k + 1)}^{2} + 4 {(k + 1)}^{3}}{4} \\ = & \frac{{(k + 1)}^{2}}{4} [k^{2} + 4 (k + 1)] \\ = & \frac{{(k + 1)}^{2}}{4} (k^{2} + 4 k + 4) \\ = & \frac{{(k + 1)}^{2} {(k + 2)}^{2}}{4} \end{array}$
21.
$\begin{array}{l} S_{n} : & 2 + 6 + 12 + \dots + n (n + 1) = \frac{n (n + 1) (n + 2)}{3} \\ S_{1} : & 1 (1 + 1) = \frac{1 (1 + 1) (1 + 2)}{3} \\ S_{k} : & 2 + 6 + 12 + \dots + k (k + 1) = \frac{k (k + 1) (k + 2)}{3} \\ S_{k + 1} : & 2 + 6 + 12 + \dots + k (k + 1) + (k + 1) [(k + 1) + 1] = \frac{(k + 1) [(k + 1) + 1] [(k + 1) + 2]}{3} \end{array}$
1. Basis step: $S_{1} : 1 (1 + 1) = \frac{1 (1 + 1) (1 + 2)}{3}$ . True.
2. Induction step: Assume $S_{k} :$
  
  $2 + 6 + 12 + \dots + k (k + 1) = \frac{k (k + 1) (k + 2)}{3} .$
Then $2 + 6 + 12 + \dots + k (k + 1) + (k + 1) (k + 1 + 1)$

$\begin{array}{l} = \frac{k (k + 1) (k + 2)}{3} + (k + 1) (k + 2) \\ = \frac{k (k + 1) (k + 2) + 3 (k + 1) (k + 2)}{3} \\ = \frac{(k + 1) (k + 2) (k + 3)}{3} \\ = \frac{(k + 1) (k + 1 + 1) (k + 1 + 2)}{3} \end{array} .$
23.
$\begin{array}{l} S_{n} : & a_{1} + (a_{1} + d) + (a_{1} + 2 d) + \dots + [a_{1} + (n - 1) d] = \frac{n}{2} [2 a_{1} + (n - 1) d] \\ S_{1} : & a_{1} = \frac{1}{2} [2 a_{1} + (1 - 1) d] \\ S_{k} : & a_{1} + (a_{1} + d) + (a_{1} + 2 d) + \dots + [a_{1} + (k - 1) d] = \frac{k}{2} [2 a_{1} + (k - 1) d] \\ S_{k + 1} : & a_{1} + (a_{1} + d) + (a_{1} + 2 d) + \dots + [a_{1} + (k - 1) d] + [a_{1} + ((k + 1) - 1) d] = \frac{k + 1}{2} [2 a_{1} + ((k + 1) - 1) d] \end{array}$
1. Basis step: Since $\frac{1}{2} [2 a_{1} + (1 - 1) d] = \frac{1}{2} \cdot 2 a_{1} = a_{1}, S_{1}$ is true.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Starting with the left side of $S_{k + 1}$ , we have
$\begin{array}{l} \underset{}{\underset{︸}{a_{1} + (a_{1} + d) + \dots + [a_{1} + (k - 1) d]}} + [a_{1} + k d] \\ = \frac{k}{2} [2 a_{1} + (k - 1) d] + [a_{1} + k d] By S_{k} \\ = \frac{[2 a_{1} + (k - 1) d]}{2} + \frac{2 [a_{1} + k d]}{2} \\ = \frac{2 k a_{1} + k (k - 1) d + 2 a_{1} + 2 k d}{2} \\ = \frac{2 a_{1} (k + 1) + k (k - 1) d + 2 k d}{2} \\ = \frac{2 a_{1} (k + 1) + (k - 1 + 2) k d}{2} \\ = \frac{2 a_{1} (k + 1) + (k + 1) k d}{2} \\ = \frac{k + 1}{2} [2 a_{1} + k d] \end{array}$ .
24. (5, 3)
25. 1.5%: $800; 2%: $1600; 3%: $2000
27. $\begin{array}{l} S_{n} : & x + y is a factor of x^{2 n} - y^{2 n} . \\ S_{1} : & x + y is a factor of x^{2} - y^{2} . \\ S_{k} : & x + y is a factor of x^{2 k} - y^{2 k} . \\ S_{k + 1} : & x + y is a factor of x^{2 (k + 1)} - y^{2 (k + 1)} . \end{array}$
1. Basis step:
  
  $S_{1} : x + y$ is a factor of $x^{2} - y^{2}$ . True.
  
  $S_{2} : x + y$ is a factor of $x^{4} - y^{4}$ . True.
2. Induction step: Assume $S_{k - 1} : x + y$ is a factor of $x^{2 (k - 1)} - y^{2 (k - 1)}$ . Then $x^{2 (k - 1)} - y^{2 (k - 1)} = (x + y) Q (x)$ for some polynomial $Q (x)$ .
  
  Assume $S_{k} : x + y$ is a factor of $x^{2 k} - y^{2 k}$ . Then $x^{2 k} - y^{2 k} = (x + y) P (x)$ for some polynomial $P (x)$ .
$\begin{array}{l} x^{2 (k + 1)} - y^{2 (k + 1)} \\ \begin{array}{l} \begin{array}{l} = (x^{2 k} - y^{2 k}) (x^{2} + y^{2}) - (x^{2 (k - 1)} - y^{2 (k - 1)}) (x^{2} y^{2}) \\ = (x + y) P (x) (x^{2} + y^{2}) - (x + y) Q (x) (x^{2} y^{2}) \\ = (x + y) [P (x) (x^{2} + y^{2}) - Q (x) (x^{2} y^{2})] \end{array} \end{array} \end{array}$

so $x + y$ is a factor of $x^{2 (k + 1)} - y^{2 (k + 1)}$ .
29.
$\begin{array}{l} S_{2} : & \log_{a} (b_{1} b_{2}) = \log_{a} b_{1} + \log_{a} b_{2} \\ S_{k} : & \log_{a} (b_{1} b_{2} \dots b_{k}) = \log_{a} b_{1} + \log_{a} b_{2} + \dots + \log_{a} b_{k} \\ S_{k + 1} : & \log_{a} (b_{1} b_{2} \dots b_{k + 1}) = \log_{a} b_{1} + \log_{a} b_{2} + \dots + \log_{a} b_{k + 1} \end{array}$
1. Basis step: $S_{2}$ is true by the properties of logarithms.
2. Induction step: Let k be a natural number $k \geq 2$ . Assume $S_{k}$ . Deduce $S_{k + 1}$ .
$\begin{array}{l} \begin{array}{l} \log_{a} (b_{1} b_{2} \dots b_{k + 1}) & Left side of S_{k + 1} \end{array} \\ \begin{array}{l} \begin{array}{l} = \log_{a} (b_{1} b_{2} \dots b_{k}) + \log_{a} b_{k + 1} By S_{2} \\ = \log_{a} b_{1} + \log_{a} b_{2} + \dots + \log_{a} b_{k} + \log_{a} b_{k + 1} \end{array} \end{array} \end{array}$
31.
$\begin{array}{l} \begin{array}{l} S_{2} : \bar{z_{1} + z_{2}} = {\bar{z}}_{_{1}} + {\bar{z}}_{2} \\ \bar{(a + b i) + (c + d i)} & = & \bar{(a + c) + (b + d) i} \\ = & (a + c) - (b + d) i \\ \bar{(a + b i)} + \bar{(c + d i)} & = & a - b i + c - d i \\ = & (a + c) - (b + d) i . \end{array} \\ \begin{array}{l} S_{k} : \bar{z_{1} + z_{2} + \dots + z_{k}} = {\bar{z}}_{_{1}} + {\bar{z}}_{_{2}} + \dots + {\bar{z}}_{k} . \\ \bar{(z_{1} + z_{2} + \dots + z_{k}) + z_{k + 1}} \\ = \bar{(z_{1} + z_{2} + \dots + z_{k})} + \bar{z_{k + 1}} By S_{2} \\ = {\bar{z}}_{1} + {\bar{z}}_{2} + \dots + {\bar{z}}_{k} + z_{k + 1} By S_{K} \end{array} \end{array}$

Mid-Chapter Mixed Review: Chapter 11

1. False
2. True
3. False
4. False
5. 8, 11, 14, 17; 32; 47
6. 0, −1, 2, −3; 8; −13
7. $a_{n} = 3 n$
11. $a_{n} = {(- 1)}^{n} n^{2}$
9. $1 \frac{7}{8}$ , or $\frac{15}{8}$
10. $2 + 6 + 12 + 20 + 30 = 70$
11. $\sum_{k = 1}^{\infty} {(- 1)}^{k} 4 k$
12. 2, 6, 22, 86
13. −5
14. 22
15. 21
16. 696
17. $- \frac{1}{2}$
18.
1. 8;
2. $\frac{1023}{16}$ , or $63.9375$
19. $- \frac{16}{3}$
20. Does not exist
21. 126 plants
22. $6369.70
23.
$\begin{array}{l} S_{n} : & 1 + 4 + 7 + \dots + (3 n - 2) = \frac{1}{2} n (3 n - 1) \\ S_{1} : & 3 \cdot 1 - 2 = \frac{1}{2} \cdot 1 (3 \cdot 1 - 1) \\ S_{k} : & 1 + 4 + 7 + \dots + (3 k - 2) = \frac{1}{2} k (3 k - 1) \\ S_{k + 1} : & 1 + 4 + 7 + \dots + (3 k - 2) + [3 (k + 1) - 2] \\ = \frac{1}{2} (k + 1) [3 (k + 1) - 1] \\ = \frac{1}{2} (k + 1) (3 k + 2) \end{array}$
1. Basis step: $S_{1} : 3 \cdot 1 - 2 = \frac{1}{2} \cdot 1 (3 \cdot 1 - 1)$ . True
2. Induction step: Assume $S_{k} :$
$1 + 4 + 7 + \dots + (3 k - 2) = \frac{1}{2} k (3 k - 1) .$

Then $1 + 4 + 7 + \dots + (3 k - 2) + [3 (k + 1) - 2]$

$\begin{array}{l} = \frac{1}{2} k (3 k - 1) + [3 (k + 1) - 2] \\ = \frac{3}{2} k^{2} - \frac{1}{2} k + 3 k + 1 \\ = \frac{3}{2} k^{2} + \frac{5}{2} k + 1 \\ = \frac{1}{2} (3 k^{2} + 5 k + 2) \\ = \frac{1}{2} (k + 1) (3 k + 2) \end{array} .$
24. The first formula can be derived from the second by substituting $a_{1} + (n - 1) d$ for $a_{n}$ . When the first and last terms of the sum are known, the second formula is the better one to use. If the last term is not known, the first formula allows us to compute the sum in one step without first finding $a_{n}$ .
25. $1 + 2 + 3 + \dots + 100$

$\begin{array}{l} = (1 + 100) + (2 + 99) + (3 + 98) + \dots + (50 + 51) \\ = \underset{50 addends of 101}{\underset{︸}{101 + 101 + 101 + \dots + 101}} \\ = 50 \cdot 101 \\ = 5050 \end{array}$

A formula for the first n natural numbers is $\frac{n}{2} (1 + n)$ .
26. Answers may vary. One possibility is given. Casey invests $900 at 8% interest, compounded annually. How much will be in the account at the end of 40 years?
27. We can prove an infinite sequence of statements $S_{n}$ by showing that a basis statement $S_{1}$ is true and then that for all natural numbers k, if $S_{k}$ is true, then $S_{k + 1}$ is true.

Exercise Set 11.5

1. 720
3. 604,800
5. 120
7. 1
9. 3024
11. 120
13. 120
15. 1
17. 6,497,400
19. $n (n - 1) (n - 2)$
21. n
23. $6! = 720$
25. $9! = 362, 880$
27. $_{9} P_{4} = 3024$
29. $_{5} P_{5} = 120; 5^{5} = 3125$
31. $_{5} P_{5} \cdot_{4} P_{4} = 2880$
33. $8 \cdot 10^{6} = 8, 000, 000; 8 million$
35. $\frac{9!}{2! 3! 4!} = 1260$
37.
1. $_{6} P_{5} = 720$ ;
2. $6^{5} = 7776$ ;
3. $1 \cdot_{5} P_{4} = 120$ ;
4. $1 \cdot 1 \cdot_{4} P_{3} = 24$
39.
1. $10^{5}$ , or 100,000;
2. 100,000
41.
1. $10^{9} = 1, 000, 000, 000$ ;
2. yes
42. $\frac{9}{4}$ , or 2.25
43. −3, 2
44. $\frac{3 \pm \sqrt{17}}{4}$
45. −2, 1, 5
47. 8
49. 11
51. $n - 1$

Exercise Set 11.6

1. 78
3. 78
5. 7
7. 10
9. 1
11. 15
13. 128
15. 270,725
17. 13,037,895
19. n
21. 1
23. $_{36} C_{4} = 58, 905$
25. $_{13} C_{10} = 286$
27. $(\begin{array}{c} 10 \\ 7 \end{array}) \cdot (\begin{array}{c} 5 \\ 3 \end{array}) = 1200$
29. $(\begin{array}{c} 52 \\ 5 \end{array}) = 2, 598, 960$
31.
1. $_{31} P_{2} = 930$ ;
2. $31^{2} = 961$ ;
3. $_{31} C_{2} = 465$
33. $- \frac{17}{2}$
34. $- 1, \frac{3}{2}$
35. $\frac{- 5 \pm \sqrt{21}}{2}$
36. −4, −2, 3
37. $(\begin{array}{c} 13 \\ 5 \end{array}) = 1287$
39. $(\begin{array}{c} n \\ 2 \end{array}); 2 (\begin{array}{c} n \\ 2 \end{array})$
41. 4
43. 7
45. Line segments:

$_{n} C_{2} = \frac{n!}{2! (n - 2)!} = \frac{n (n - 1) (n - 2)!}{2 \cdot 1 \cdot (n - 2)!} = \frac{n (n - 1)}{2}$

Diagonals: The n line segments that form the sides of the n-gon are not diagonals. Thus the number of diagonals is

$\begin{array}{rcl} _{n} C_{2} - n & = & \frac{n (n - 1)}{2} - n \\ = & \frac{n^{2} - n - 2 n}{2} = \frac{n^{2} - 3 n}{2} \\ = & \frac{n (n - 3)}{2}, n \geq 4. \end{array}$

Let $D_{n}$ be the number of diagonals of an n-gon. Prove the result above for diagonals using mathematical induction.
$\begin{array}{l} S_{n} : & D_{n} = \frac{n (n - 3)}{2}, for n = 4, 5, 6, \dots \\ S_{4} : & D_{4} = \frac{4 \cdot 1}{2} \\ S_{k} : & D_{k} = \frac{k (k - 3)}{2} \\ S_{k + 1} : & D_{k + 1} = \frac{(k + 1) (k - 2)}{2} \end{array}$
1. Basis step: $S_{4}$ is true (a quadrilateral has 2 diagonals).
2. Induction step: Assume $S_{k}$ . Note that when an additional vertex $V_{k + 1}$ is added to the k-gon, we gain k segments, 2 of which are sides of the $(k + 1)$ -gon, and a former side $\bar{V_{1} V_{k}}$ becomes a diagonal. Thus the additional number of diagonals is $k - 2 + 1$ , or $k - 1$ . Then the new total of diagonals is $D_{k} + (k - 1)$ , or
$\begin{array}{rcl} D_{k + 1} & = & D_{k} + (k - 1) \\ = & \frac{k (k - 3)}{2} + (k - 1) & By S_{k} \\ = & \frac{(k + 1) (k - 2)}{2} \end{array}$

Exercise Set 11.7

1. $x^{4} + 20 x^{3} + 150 x^{2} + 500 x + 625$
3. $x^{5} - 15 x^{4} + 90 x^{3} - 270 x^{2} + 405 x - 243$
5. $x^{5} - 5 x^{4} y + 10 x^{3} y^{2} - 10 x^{2} y^{3} + 5 x y^{4} - y^{5}$
7. $15, 625 x^{6} + 75, 000 x^{5} y + 150, 000 x^{4} y^{2} +$ $160, 000 x^{3} y^{3} + 96, 000 x^{2} y^{4} + 30, 720 x y^{5} + 4096 y^{6}$
9. $128 t^{7} + 448 t^{5} + 672 t^{3} + 560 t + 280 t^{- 1} + 84 t^{- 3} + 14 t^{- 5} + t^{- 7}$
11. $x^{10} - 5 x^{8} + 10 x^{6} - 10 x^{4} + 5 x^{2} - 1$
13. $125 + 150 \sqrt{5} t + 375 t^{2} + 100 \sqrt{5} t^{3} + 75 t^{4} + 6 \sqrt{5} t^{5} + t^{6}$
15. $a^{9} - 18 a^{7} + 144 a^{5} - 672 a^{3} + 2016 a - 4032 a^{- 1} + 5376 a^{- 3} - 4608 a^{- 5} + 2304 a^{- 7} - 512 a^{- 9}$
17. $140 \sqrt{2}$
19. $x^{- 8} + 4 x^{- 4} + 6 + 4 x^{4} + x^{8}$
21. $21 a^{5} b^{2}$
23. $- 252 x^{5} y^{5}$
25. $- 745, 472 a^{3}$
27. $1120 x^{12} y^{2}$
29. $- 1, 959, 552 u^{5} v^{10}$
31. $2^{7}$ , or 128
33. $2^{24}$ , or 16,777,216
35. 20
37. $- 12 + 316 i$
39. $- 7 - 4 \sqrt{2} i$
41. $\sum_{k = 0}^{n} (\begin{array}{l} n \\ k \end{array}) {(- 1)}^{k} a^{n - k} b^{k}$
43. $\sum_{k = 1}^{n} (\begin{array}{l} n \\ k \end{array}) x^{n - k} h^{k - 1}$
44. $x^{2} + 2 x - 2$
45. $2 x^{3} - 3 x^{2} + 2 x - 3$
46. $4 x^{2} - 12 x + 10$
47. $2 x^{2} - 1$
49. $3, 9, 6 \pm 3 i$
51. $- \frac{35}{x^{1 / 6}}$
53. $2^{100}$
55. ${[\log_{a} (x t)]}^{23}$
57.
1. Basis step: Since $a + b = {(a + b)}^{1}, S_{1}$ is true.
2. Induction step: Let $S_{k}$ be the statement of the binomial theorem with n replaced by k. Multiply both sides of $S_{k}$ by $(a + b)$ to obtain
$\begin{array}{rcl} {(a + b)}^{k + 1} & = & [a^{k} + \dots + (\begin{array}{c} k \\ r - 1 \end{array}) a^{k - (r - 1)} b^{r - 1} + (\begin{array}{c} k \\ r \end{array}) a^{k - r} b^{r} + \dots + b^{k}] (a + b) \\ = & a^{k + 1} + \dots + [(\begin{array}{c} k \\ r - 1 \end{array}) + (\begin{array}{c} k \\ r \end{array})] a^{(k + 1) - r_{b} r} + \dots + b^{k + 1} \\ = & a^{k + 1} + \dots + (\begin{array}{c} k + 1 \\ r \end{array}) a^{(k + 1) - r_{b} r} + \dots + b^{k + 1} \end{array} .$

This proves $S_{k + 1}$ , assuming $S_{k}$ . Hence $S_{n}$ is true for $n = 1, 2, 3, \dots$ .

Exercise Set 11.8

1.
1. 0.18, 0.24, 0.23, 0.23, 0.12;
2. Opinions may vary, but it seems that people tend not to pick the first or last numbers.
3. 5187 e-mails
5.
1. $\frac{2}{7}$ ;
2. $\frac{5}{7}$ ;
3. 0;
4. 1
7. $\frac{1}{2}$
9.
1. $\frac{1}{13}$ ;
2. $\frac{2}{13}$ ;
3. $\frac{1}{4}$ ;
4. $\frac{1}{26}$
11. $\frac{1}{5525}$
13. $\frac{135}{323}$
15. $\frac{1}{108, 290}$
17. $\frac{33}{66, 640}$
19.
1. HHH, HHT, HTH, HTT, THH, THT, TTH, TTT;
2. $\frac{3}{8}$ ;
3. $\frac{7}{8}$ ;
4. $\frac{7}{8}$ ;
5. $\frac{3}{8}$
21. $\frac{9}{19}$
23. $\frac{1}{38}$
25. $\frac{18}{19}$
27. $\frac{9}{19}$
29. Zero
30. One-to-one
31. Function; domain; range; domain; range
32. Zero
33. Combination
34. Inverse variation
35. Factor
36. Geometric sequence
37.
1. $(\begin{array}{c} 13 \\ 2 \end{array}) \cdot (\begin{array}{c} 4 \\ 2 \end{array}) \cdot (\begin{array}{c} 4 \\ 2 \end{array}) \cdot (\begin{array}{c} 44 \\ 1 \end{array}) = 123, 552$
2. 0.0475
39.
1. $13 \cdot (\begin{array}{c} 4 \\ 3 \end{array}) \cdot (\begin{array}{c} 48 \\ 2 \end{array}) - 3744$ , or 54,912;
2. $\frac{54, 912}{(\begin{array}{c} 52 \\ 5 \end{array})} \approx 0.0211$

Review Exercises: Chapter 11

1. True
2. False
3. True
4. False
5. $- \frac{1}{2}, \frac{4}{17}$ , $- \frac{9}{82}, \frac{16}{257}; - \frac{121}{14, 642}; - \frac{529}{279, 842}$
6. ${(- 1)}^{n + 1} (n^{2} + 1)$
7. $\frac{3}{2} - \frac{9}{8} + \frac{27}{26} - \frac{81}{80} = \frac{417}{1040}$
11. $\sum_{k = 1}^{7} (k^{2} - 1)$
9. $\frac{15}{4}$
10. $a + 4 b$
11. 531
12. 20,100
13. 11
14. −4
15. $n = 6, S_{n} = - 126$
16. $a_{1} = 8, a_{5} = \frac{1}{2}$
17. Does not exist
18. $\frac{3}{11}$
19. $\frac{3}{8}$
20. $\frac{241}{99}$
21. $5 \frac{4}{5}, 6 \frac{3}{5}, 7 \frac{2}{5}, 8 \frac{1}{5}$
22. 167.3 ft
23. $45,993.04
24.
1. $7.38;
2. $1365.10
25. $88,888,888,889
26.
$\begin{array}{l} S_{n} : & 1 + 4 + 7 + \dots + (3 n - 2) = \frac{n (3 n - 1)}{2} \\ S_{1} : & 1 = \frac{1 (3 - 1)}{2} \\ S_{k} : & 1 + 4 + 7 + \dots + (3 k - 2) = \frac{k (3 k - 1)}{2} \\ S_{k + 1} : & 1 + 4 + 7 + \dots + (3 k - 2) + [3 (k + 1) - 2] \\ = 1 + 4 + 7 + \dots + (3 k - 2) + (3 k + 1) \\ = \frac{(k + 1) (3 k + 2)}{2} \end{array}$
1. Basis step: $\frac{1 (3 - 1)}{2} = \frac{2}{2} = 1$ is true.
2. Induction step: Assume $S_{k}$ . Add $(3 k + 1)$ on both sides.
  
  $1 + 4 + 7 + \dots + (3 k - 2) + (3 k + 1)$
$\begin{array}{l} = \frac{k (3 k - 1)}{2} + (3 k + 1) = \frac{k (3 k - 1)}{2} + \frac{2 (3 k + 1)}{2} \\ = \frac{3 k^{2} - k + 6 k + 2}{2} = \frac{3 k^{2} + 5 k + 2}{2} \\ = \frac{(k + 1) (3 k + 2)}{2} \end{array}$
27.
$\begin{array}{l} S_{n} : & 1 + 3 + 3^{2} + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2} \\ S_{1} : & 1 = \frac{3^{1} - 1}{2} \\ S_{k} : & 1 + 3 + 3^{2} + \dots + 3^{k - 1} = \frac{3^{k} - 1}{2} \\ S_{k + 1} : & 1 + 3 + 3^{2} + \dots + 3^{(k + 1) - 1} = \frac{3^{k + 1} - 1}{2} \end{array}$
1. Basis step: $\frac{3^{1} - 1}{2} = \frac{2}{2} = 1$ is true.
2. Induction step: Assume $S_{k}$ . Add $3^{k}$ on both sides.
  
  $1 + 3 + \dots + 3^{k - 1} + 3^{k}$
  
  $\begin{array}{l} = \frac{3^{k} - 1}{2} + 3^{k} = \frac{3^{k} - 1}{2} + 3^{k} \cdot \frac{2}{2} \\ = \frac{3 \cdot 3^{k} - 1}{2} = \frac{3^{k + 1} - 1}{2} \end{array}$
28.
$\begin{array}{l} S_{n} : & (1 - \frac{1}{2}) (1 - \frac{1}{3}) \dots (1 - \frac{1}{n}) = \frac{1}{n} \\ S_{2} : & (1 - \frac{1}{2}) = \frac{1}{2} \\ S_{k} : & (1 - \frac{1}{2}) (1 - \frac{1}{3}) \dots (1 - \frac{1}{k}) = \frac{1}{k} \\ S_{k + 1} : & (1 - \frac{1}{2}) (1 - \frac{1}{3}) \dots (1 - \frac{1}{k}) (1 - \frac{1}{k + 1}) = \frac{1}{k + 1} . \end{array}$
1. Basis step: $S_{2}$ is true by substitution.
2. Induction step: Assume $S_{k}$ . Deduce $S_{k + 1}$ . Starting with the left side of $S_{k + 1}$ , we have
$\begin{array}{l} \underset{}{\underset{︸}{(1 - \frac{1}{2}) (1 - \frac{1}{3}) \dots (1 - \frac{1}{k})}} (1 - \frac{1}{k + 1}) \\ \begin{array}{l} = \frac{1}{k} \cdot (1 - \frac{1}{k + 1}) & By S_{k} \\ = \frac{1}{k} \cdot (\frac{k + 1 - 1}{k + 1}) \\ = \frac{1}{k} \cdot \frac{k}{k + 1} \\ = \frac{1}{k + 1} . & Simplifying \end{array} \end{array}$
29. $6! = 720$
30. $9 \cdot 8 \cdot 7 \cdot 6 = 3024$
31. $(\begin{array}{c} 15 \\ 8 \end{array}) = 6435$
32. $24 \cdot 23 \cdot 22 = 12, 144$
33. $\frac{9!}{1! 4! 2! 2!} = 3780$
34. $3 \cdot 4 \cdot 3 = 36$
35.
1. $_{6} P_{5} = 720$ ;
2. $6^{5} = 7776$ ;
3. $_{5} P_{4} = 120$ ;
4. $_{3} P_{2} = 6$
36. $2^{8}$ , or 256
37. $m^{7} + 7 m^{6} n + 21 m^{5} n^{2} + 35 m^{4} n^{3} + 35 m^{3} n^{4} + 21 m^{2} n^{5} + 7 m n^{6} + n^{7}$
38. $x^{5} - 5 \sqrt{2} x^{4} + 20 x^{3} - 20 \sqrt{2} x^{2} + 20 x - 4 \sqrt{2}$
39. $x^{8} - 12 x^{6} y + 54 x^{4} y^{2} - 108 x^{2} y^{3} + 81 y^{4}$
40. $a^{8} + 8 a^{6} + 28 a^{4} + 56 a^{2} + 70 + 56 a^{- 2} + 28 a^{- 4} + 8 a^{- 6} + a^{- 8}$
41. $- 6624 + 16, 280 i$
42. $220 a^{9} x^{3}$
43. $- (\begin{array}{l} 18 \\ 11 \end{array}) 128 a^{7} b^{11}$
44. $\frac{1}{12}; 0$
45. $\frac{1}{4}$
46. $\frac{6}{5525}$
47. $\frac{86}{206} \approx 0.42, \frac{97}{206} \approx 0.47, \frac{23}{206} \approx 0.11$
48. B
49. A
50. D
51.
1. No (unless $a_{n}$ is all positive or all negative);
2. yes;
3. yes;
4. no (unless $a_{n}$ is constant);
5. no (unless $a_{n}$ is constant);
6. no (unless $a_{n}$ is constant)
52. $\frac{a_{k + 1}}{a_{k}} = r_{1}$ , $\frac{b_{k + 1}}{b_{k}} = r_{2}, so \frac{a_{k + 1} b_{k + 1}}{a_{k} b_{k}} = r_{1} r_{2}$ , a constant.
53. $\frac{1}{2}, - \frac{1}{6}, \frac{1}{18}$
54. −2, 0, 2, 4
55. ${(\log \frac{x}{y})}^{10}$
56. 18
57. 36
58. −9
59. For each circular arrangement of the numbers on a clock face, there are 12 distinguishable ordered arrangements on a line. The number of arrangements of 12 objects on a line is $_{12} P_{12}$ , or 12!. Thus the number of circular permutations is $\frac{_{12} P_{12}}{12} = \frac{12!}{12} = 11! = 39, 916, 800$ . In general, for each circular arrangement of n objects, there are n distinguishable ordered arrangements on a line. The total number of arrangements of n objects on a line is $_{n} P_{n}$ , or n!. Thus the number of circular permutations is $\frac{n!}{n} = \frac{n (n - 1)!}{n} = (n - 1)!$ .
60. Put the following in the form of a paragraph. First find the number of seconds in a year (365 days): $365 days \cdot \frac{24 hr}{1 day} \cdot \frac{60 min}{1 hr} \cdot \frac{60 sec}{1 min} = 31, 536, 000 sec$ . The number of arrangements possible is 15!. The time is $\frac{15!}{31, 536, 000} \approx 41, 466 years$ .
61. Order is considered in a combination lock.
62. Choosing k objects from a set of n objects is equivalent to not choosing the other $n - k$ objects.

Test: Chapter 11

[11.1] −43
[11.1] $\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}$
[11.1] $2 + 5 + 10 + 17 = 34$
[11.1] $\sum_{k = 1}^{6} 4 k$
[11.1] $\sum_{k = 1}^{\infty} 2^{k}$
[11.1] $3, 2 \frac{1}{3}, 2 \frac{3}{7}, 2 \frac{7}{17}$
[11.2] 44
[11.2] 38
[11.2] −420
[11.2] 675
[11.3] $\frac{5}{512}$
[11.3] 1000
[11.3] 510
[11.3] 27
[11.3] $\frac{56}{99}$
[11.1] $10,000, $8000, $6400, $5120, $4096, $3276.80
[11.2] $17.05
[11.3] $74,399.77
[11.4]
$\begin{array}{l} S_{n} : & 2 + 5 + 8 + \dots + (3 n - 1) = \frac{n (3 n + 1)}{2} \\ S_{1} : & 2 = \frac{1 (3 \cdot 1 + 1)}{2} \\ S_{k} : & 2 + 5 + 8 + \dots + (3 k - 1) = \frac{k (3 k + 1)}{2} \\ S_{k + 1} : & 2 + 5 + 8 + \dots + (3 k - 1) + [3 (k + 1) - 1] \\ = \frac{(k + 1) [3 (k + 1) + 1]}{2} \end{array}$
1. Basis step: $\frac{1 (3 \cdot 1 + 1)}{2} = \frac{1 \cdot 4}{2} = 2$ , so $S_{1}$ is true.
2. Induction step:
  
  $2 + 5 + 8 + \dots + (3 k - 1) + [3 (k + 1) - 1]$
  
  $\begin{array}{l} = \frac{k (3 k + 1)}{2} + [3 k + 3 - 1] By S_{k} \\ = \frac{3 k^{2}}{2} + \frac{k}{2} + 3 k + 2 \\ = \frac{3 k^{2}}{2} + \frac{7 k}{2} + 2 = \frac{3 k^{2} + 7 k + 4}{2} \\ = \frac{(k + 1) (3 k + 4)}{2} = \frac{(k + 1) [3 (k + 1) + 1]}{2} \end{array}$
[11.5] 3,603,600
[11.6] 352,716
[11.6] $\frac{n (n - 1) (n - 2) (n - 3)}{24}$
[11.5] $_{6} P_{4} = 360$
[11.5]
1. $6^{4} = 1296$ ;
2. $_{5} P_{3} = 60$
[11.6] $_{28} C_{4} = 20, 475$
[11.6] $_{12} C_{8} \cdot_{8} C_{4} = 34, 650$
[11.7] $x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 5 x + 1$
[11.7] $35 x^{3} y^{4}$
[11.7] $2^{9} = 512$
[11.8] $\frac{4}{7}$
[11.8] $\frac{48}{1001}$
[11.1] B
[11.5] 15

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