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f (1) = 3; f (2) = 9 75. 20 ; f (−1) = 9 3 77. The range for this viewing window is [0, 100]. 79. The range for this viewing window is [−0.001, 0.001]. y 100 80 60 40 20 –5 –20 –40 –60 –80 –100 –10 x 5 10 –0.1 y 0.001 0.0008 0.0006 0.0004 0.0002 –0.05 –0.0002 –0.0004 –0.0006 –0.0008 –0.001 0.05 0.1 x 83. The range for this viewing window is [0, 10]. y 10 8 6 4 2 x –20 20 40 60 80 100 x 50 100 10.105 8.105 6.105 4.105 2.105 –50 –2.105 –4.105 –6.105 –8.105 –10.105 –100 85. The range for this viewing window is [−0.1, 0.1]. y 87. The range for this viewing window is [−100, 100]. 0.1 0.08 0.06 0.04 0.02 –0.0005 –0.02 –0.04 –0.06 –0.08 –0.1 –0.0001 0.0005 0.0001 x –10.105 y 100 80 60 40 20 –5.105 –20 –40 –60 –80 –100 5.105 10.105 x 89. a. g(5000) = 50 b. The number of cubic yards of dirt 91. a. The required for a garden of 100 square feet is 1. height of the rocket above ground after 1 second is 200 ft. b. The height of the rocket above ground after 2 seconds is 350 ft. Section 1.2 — 3 √ 1. The domain of a function depends upon what values of the independent variable make the function undefined or imaginary. — 3. There is no restriction on x for f (x) = x because you can take the cube root of any real number. So the domain is all real numbers, (−∞, ∞). When dealing with the set of real numbers, you cannot take the square root of negative numbers. So x-values are restricted for f (x) = √ domain is [0, ∞). function over its corresponding domain. Use the
same scale for the x-axis and y-axis for each graph. Indicate included endpoints with a solid circle and excluded endpoints with an open circle. Use an arrow to indicate −∞ or ∞. Combine the graphs to find the graph of the piecewise function. x to nonnegative numbers and the 5. Graph each formula of the piecewise 9. (−∞, 3]  ∪  − 1 15.  −∞, − 1, ∞  _ _ 2 2 17. (−∞, −11)∪(−11, 2)∪(2, ∞) 19. (−∞, −3)∪(−3, 5)∪(5, ∞) 21. (−∞, 5) 25. (−∞, −9)∪(−9, 9)∪(9, ∞) 27. domain: (2, 8], range: [6, 8) 29. domain: [−4, 4], range: [0, 2] 31. domain: [−5, 3), range: [0, 2] 33. domain: (−∞, 1], range: [0, ∞) 11. (−∞, ∞) 13. (−∞, ∞) 7. (−∞, ∞) 23. [6, ∞) C-1 C-2 1, 6 , range:  −6, − 1 1 35. domain:  −6 37. domain: [−3, ∞), range is [0, ∞) 39. domain:(−∞, ∞) 41. domain: (−∞, ∞) y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x –2 –1 y 5 4 3 2 1 –1 –2 –3 –4 –5 1 2 x 43. domain: (−∞, ∞) 45. domain: (−∞, ∞) y 5 4 3 2 1 –1 –2 –3 –4 –5 –2 –1 1 2 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x 47. f (−3) =
1; f (−2) = 0; f (−1) = 0; f (0) = 0 49. f (−1) = −4; f (0) = 6; f (2) = 20; f (4) = 34 51. f (−1) = −5; f (0) = 3; f (2) = 3; f (4) = 16 53. (−∞, 1)∪(1, ∞) 55. y y 104 96 88 80 72 64 56 48 40 32 24 16 8 –8 –0.5 –0.4 –0.3 –0.2 –0.1 104 96 88 80 72 64 56 48 40 32 24 16 8 x 0.1 –0.1 –8 0.1 0.2 0.3 0.4 0.5 x The viewing window: [−0.5, −0.1] has a range: [4, 100]. The viewing window: [0.1, 0.5] has a range: [4, 100]. 59. Many answers; one function is f (x) = 1 _______ 57. [0, 8]. — x − 2 √ 61. a. The fixed cost is $500. b. The cost of making 25 items is $750. c. The domain is [0, 100] and the range is [500, 1500]. Section 1.3 1. Yes, the average rate of change of all linear functions is constant. 3. The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region in an open interval. −1 _________ 11. 13(13 + h) 4 _ 7. 3 19. Increasing on (−∞, −2.5)∪(1, ∞) and decreasing 5. 4(b + 1) 13. 3h2 + 9h + 9 15. 4x + 2h − 3 9. 4x + 2h 7. 3 on (−2.5, 1) on (1, 3)∪(4, ∞) minimum: (3, 50) 21. Increasing on (−∞, 1)∪(3, 4) and decreasing 23. Local maximum: (−3, 50) and local 33. ≈ −0.167 b. −1,250 people per year 25. absolute maximum at approximately (7, 150) and
absolute minimum at approximately (−7.5, −220) 27. a. −3,000 people per year 29. −4 31. 27 (3, −22), decreasing on (−∞, 3), increasing on (3, ∞) 37. local minimum: (−2, −2), decreasing on (−3, −2), increasing on (−2, ∞) minima: (−3.25, −47) and (2.1, −32), decreasing on (−∞, −3.25) and (−0.5, 2.1), increasing on (−3.25, −0.5) and (2.1, ∞) 43. b = 5 41. A 47. ≈ −0.6 milligrams per day 39. local maximum: (−0.5, 6), local 45. ≈ 2.7 gallons per minute 35. local minimum: Section 1.4 1. Find the numbers that make the function in the denominator g equal to zero, and check for any other domain restrictions on f and g, such as an even-indexed root or zeros in the denominator. 3. Yes, sample answer: Let f (x) = x + 1 and g(x) = x − 1. Then f (g(x)) = f (x − 1) = (x − 1)+ 1 = x and g ( f (x)) = g (x + 1) = (x + 1)− 1 = x so f ∘ g = g ∘ f. 5. (f + g)(x) = 2x + 6; domain: (−∞, ∞) (f − g)(x) = 2x 2 + 2x − 6; domain: (−∞, ∞) (fg)(x) = −x 4 − 2x 3 + 6x 2 + 12x; domain: (−∞, ∞) f  g  (x) = _ 7. (f + g)(x) = ; domain: (−∞, 0)∪(0, ∞) 6 )∪(− √ 6 )∪( √ 6, ∞) 6, √ — — — — x 2 + 2x ______ 6 − x 2 ; domain: (−∞, − √ 4x3 + 8x2 + 1 ___________ 2x 4x3 + 8
x2 − 1 ___________ 2x ; domain: (−∞, 0)∪(0, ∞) (f − g)(x) = — — (fg)(x) = x + 2; domain: (−∞, 0)∪(0, ∞) f  g  (x) = 4x 3 + 8x 2; domain: (−∞, 0)∪(0, ∞) _ 9. (f + g)(x) = 3x 2 + √ x − 5 ; domain: [5, ∞) (f − g)(x) = 3x 2 − √ (fg)(x) = 3x 2 √ f  3x2 g  (x) = _ _ x − 5 √ b. f ( g(x)) = 18x 2 − 60x + 51 d. ( g ∘ g)(x) = 9x − 20 e. ( f ∘ f )(−2) = 163 x − 5 ; domain: [5, ∞) x − 5 ; domain: [5, ∞) ; domain: (5, ∞) — — 11. a. f (g(2)) = 3 c. g( f (x)) = 6x 2 − 2 13. f ( g(x)) = √ — x2 + 3 + 2 ; g( f (x)) = ( f (x)) = 15. f ( g(x)) = x x __, x ≠ 0; g( f (x)) = 2x − 4, x ≠ 4 17. f ( g(x)) = 2 1 _ (x + 3)2 + 1 3 — 19. f ( g(h(x))) = 21. a. (g ∘ f )(x) = − 1 __  b.  −∞, 2 _ 2 − 4x √ 23. a. (0, 2)∪(2, ∞) except x = −2 b. (0, ∞) c. (0, ∞) 25. (1, ∞) 27. Many solutions; one possible answer: f (x) = x3; g(x(x) = (x + 2)2 29. Many solutions; one possible answer: f (x) = 3 x ; g
(x) = √ 31. Many solutions; one possible answer: f (x) = — 33. Many solutions; one possible answer: f (x) = 4 √ 35. Many solutions; one possible answer: f (x) = √ 37. Many solutions; one possible answer: f (x) = 39. Many solutions; one possible answer: f (x) = x 3; g(x) = — 3 √ x ; g(x _____ 2x − 3 3x − 2 ______ x + 5 x ; g(x) = 2x + 6 x ; g(x) = — ODD ANSWERS — 41. Many solutions; one possible answer: f (x) = √ 51. 2 43. 2 55. 4 63. 2 67. 11 45. 5 57. 4 69. 0 1 _ 75. f (g(0)) =, g(f (0)) = 5 5 47. 4 59. 9 71. 7 49. 0 61. 4 73. f (g(0)) = 27, g(f (0)) = −94 x ; g(x) = 53. 1 65. 3 77. f (g(x)) = 18x2 + 60x + 51 2x − 1 _ 3x + 4 87. (f ∘ g )(6) = 6; (g ∘ f )(6) = 6 85. False 81. ( f ∘ g)(x) = 2, (g ∘ f )(x) = 2 79. g ∘ g(x) = 9x + 20 83. (−∞, ∞) 89. ( f ∘ g )(11) = 11; (g ∘ f )(11) = 11 93. A(t) = π  25 √ square inches 97. a. N(T(t)) = 575t 2 + 65t − 31.25 b. ≈ 3.38 hours t + 2  2 and A(2) = π  25 √ 95. A(5) = 121π square units 91. C — 4  2 = 2,500π — Section 1.5 1. A horizontal shift results when a constant is added to or subtracted from the input. A vertical shift results when a constant 3. A horizontal is added to or subtracted from the output.
compression results when a constant greater than 1 multiplies the input. A vertical compression results when a constant between 0 5. For a function f, substitute and 1 multiplies the output. (−x) for (x) in f (x) and simplify. If the resulting function is the same as the original function, f (−x) = f (x), then the function is even. If the resulting function is the opposite of the original function, f (−x) = −f (x), then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even. 1 _ 9. g(x) = (x + 4)2 + 2 7. g(x) = ∣ x − 1 ∣ − 3 11. The graph of f (x + 43) is a horizontal shift to the left 43 units of the graph of f. 13. The graph of f (x − 4) is a horizontal shift to the right 4 units 15. The graph of f (x) + 8 is a vertical shift of the graph of f. up 8 units of the graph of f. 19. The graph of vertical shift down 7 units of the graph of f. f (x + 4) − 1 is a horizontal shift to the left 4 units and a vertical shift 21. Decreasing on (−∞, −3) and down 1 unit of the graph of f. increasing on (−3, ∞) 25. 23. Decreasing on (0, ∞) y 17. The graph of f (x) − 7 is a y 27. 10 1 0 –1 –2 –3 –4 h 21 3 4 x –5 –4 –3 –2 10 1 –1 –2 –3 –4 f x 21 3 –9 –8 –7 –6 –5 –4 –3 –2 C-3 k 21 3 4 5 6 29. –6 –5 –4 –3 –1 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12 –13 –14 — x x — x + 3 − 1 31. g(x) = f (x − 1), h(x) = f (x) + 1 33. f (x) = ∣ x − 3 ∣ − 2 35. f (x) = √ 37. f (x) = (x − 2)2 39. f
(x) = ∣ x + 3 ∣ − 2 41. f (x) = − √ 43. f (x) = −(x + 1)2 + 2 45. f (x) = √ 47. Even 51. Even of g is a vertical reflection (across the x-axis) of the graph 55. The graph of g is of f. a vertical stretch by a factor of 4 of the graph of f. −x + 1 49. Odd 53. The graph — 1 _ 57. The graph of g is a horizontal compression by a factor of 5 59. The graph of g is a horizontal stretch of the graph of f. by a factor of 3 of the graph of f. horizontal reflection across the y-axis and a vertical stretch by a 61. The graph of g is a factor of 3 of the graph of f. 65. g(x) = 1 _ 3(x + 2)2 − 3 63. g(x) = ∣ −4x ∣ 67. g(x) = 1 _ (x − 5)2 + 1 2 69. This is a parabola shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units. y 71. This is an absolute value function stretched vertically by a factor of 2, shifted 4 units to the right, reflected across the horizontal axis, and then shifted 3 units up. y 10 1 –1 –2 –3 –4 –5 –5 –4 –3 –2 g –6 –5 –4 –3 –2 21 3 4 5 x h 21 1 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 73. This is a cubic function compressed vertically by a 1 _ factor of. 2 y 75. The graph of the function is stretched horizontally by a factor of 3 and then shifted downward by 3 units. y –5 –4 –3 –1 –1 –2 –3 –4 –5 –6 –7 –8 m 21 3 4 5 x –6 –5 –4 –3 –2 p 21 1 –1 –2 –3 –4 –5 –6 –7 –8 ODD ANSWERS C-4 77. The graph of — f (x) = √ x is shifted right 4 units and then reflected across the y-axis. y –8 –7 –6 –5 –4 –3 –2 81.
a –8 –7 –6 –5 –4 –3 –2 2 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.2 0 0.5 1 1.5 2 2.5 3 3.5 4 x Section 1.6 79. 41. 43. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 g 21 3 4 5 6 7 8 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x –6 –5 –4 –3 –2 g 21 1 –1 –2 –3 –4 –5 –6 –7 –8 45. 471 –1 –2 –3 –4 –5 –8 –7 –6 –5 –4 –3 –2 21 3 4 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –1 –1 –2 –3 –4 –5 21 3 4 5 6 x 21 3 4 5 x 1. Isolate the absolute value term so that the equation is of the form ∣ A ∣ = B. Form one equation by setting the expression inside the absolute value symbol, A, equal to the expression on the other side of the equation, B. Form a second equation by setting A equal to the opposite of the expression on the other side of the equation, −B. Solve each equation for the variable. 3. The graph of the absolute value function does not cross the x-axis, so the graph is either completely above or completely below the 5. First determine the boundary points by finding x-axis. the solution(s) of the equation. Use the boundary points to form possible solution intervals. Choose a test value in each interval to 1 7. ∣ x + 4 ∣ = _ 2 13.  − 9 13  _ _ 4, 4 5 7 19.   _ _, 2 2 determine which values satisfy the inequality. 9. ∣ f (x) − 8 ∣ < 0.03 29 10 20 15 21. No solution 23. {−57, 27} 25. (0, −8); (−
6, 0) and (4, 0) 29. ( −∞, − 8)∪(12, ∞) 27. (0, −7); no x-intercepts.  ∪[6, ∞) 35.  −∞, − 8, 4  33.  −∞, − 8 31.  − 4  ∪[16, ∞) _ _ _ 3 3 3 11 17.  _ 5 11. {1, 11},, 37. y 39. y 49. 51. 321 4 5 x –7 –6 –5 –4 –3 –2 y –1 –1 –2 –3 –4 –5 –6 –7 21 1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 53. range: [0, 20] y 55. y f –100 –75 –50 2 1.5 1 0.5 –25 –0.5 –1 –1.5 –2 25 50 75 100 x 20 18 16 14 12 10 8 6 4 2 21 3 4 5 x –4 –3 –2 –1 –1 –2 59. There is no value for a that will keep the 57. (−∞, ∞) function from having a y-intercept. The absolute value function always crosses the y-intercept when x = 0. 61. ∣ p − 0.08 ∣ ≤ 0.015 63. ∣ x − 5.0 ∣ ≤ 0.01 (–1, 2) (3, 2) (0, 1) (2, 1) (1, 0) x (–2, 3) (2, 3) SeCtion 1.7 (–1, 2) (1, 2) (0, 1) x 1. Each output of a function must have exactly one input for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y-values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y-values repeat and the function is one-to-one. ODD ANSWERS 1 _ x is its own inverse. 3. Yes. For example, f (x) = 7. f
−1(x) = x − 3 9. f −1(x) = 2 − x 13. Domain of f (x): [−7,∞); f −1 (x) = √ 5. y = f −1(x) 11. f −1(x) = − 2x _ x − 1 x − 7 — 15. Domain of f (x): [0, ∞); f −1 (x) = √ 17. f ( g(x)) = x and g( f (x)) = x 21. One-to-one 23. Not one-to-one — x + 5 19. One-to-one 25. 3 27. 2 33. 6 37. 0 39. 1 31. [2, 10] 35. −4 41. x f −1(x) 1 3 4 6 7 9 12 13 16 14 5 _ 45. f −1 (x) = (x − 32) 9 d _ ; t(180) = 50 47. t(d) = 180 _ 50. The time for the car to travel 180 miles is 3.6 hours. 29. y 10 8 6 4 2 –2 –2 –4 f –1 f 42 6 8 10 x –10 –8 –6 –4 1 _ 43. f −1 (x) = (1 –1 –2 –3 –4 –5 –5 –4 –3 –2 f f –1 21 3 4 5 x Chapter 1 Review exercises 1. Function 3. Not a function 5. f (−3) = −27; f (2) = −2; f (−a) = −2a2 − 3a; −f (a) = 2a2 −3a; f (a + h) = −2a2 − 4ah − 2h2 + 3a + 3h 7. One-to-one 11. Function 9. Function 13. –5 –4 –3 –2 23. –5 –4 –3 –2 y 3 2 1 –1 –1 –2 –3 y 3 2 1 –1 –1 –2 –3 21 3 4 5 21 3 4 5 x x 15. 2 19. 17. −1.8 or 1.8 −64 + 80a − 16a2 __ −1 + a = −16a + 64; a ≠ 1 25. 31 27. Increasing on (2, �
�), decreasing on (−∞, 2) 29. Increasing on (−3, 1), constant on (−∞, −3) and (1, ∞) 31. Local minimum: (−2, −3); local maximum: (1, 3) 33. Absolute maximum: 10 35. ( f ∘ g )(x) = 17 − 18x, ( g ∘ f )(x) = −7 −18x 37. ( f ∘ g )(x) = √ ______ )(x) = 1 _ — x + 2 √ 39. (f ∘ g )(x) = = 1 + x _____ 1 + 4x  ∪  − 1 ; Domain:  −∞, − 1, 0  ∪(0, ∞) _ _ 4 4 1 ____ 1 __ x + 1 _ 1 ____ 1 __ x + 4 C-5 x x 2x −1 _ 3x + 4 and 41. ( f ∘ g )(x) = 1 _ ; Domain: (0, ∞) — x √ 43. Many solutions; one possible answer: g(x) = f (x) = √ — x. 45 476 –5 –4 –3 –2 –1 –1 –2 21 3 4 5 6 x –6 –5 –4 –3 –2 21 3 4 5 6 –1 –1 –2 21 3 4 5 x 49. –5 –4 –3 –2 53. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 21 3 4 5 6 x 67. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –4 –3 –2 21 3 4 5 76 x 321 4 5 6 7 8 51. y –4 –3 –2 2 –1 –2 –4 –6 –8 –10 –12 –14 –16 –18 –20 –22 –24 55. f (x) = ∣ x − 3 ∣ 57. Even 59. Odd 61. Even 1 ∣ x + 2 ∣ + 1 _ 63. f (x) = 2 65. f (x) = −3 ∣ x − 3 ∣ + 3 69. {−22, 14}
71.  − 5, 3  _ 3 73. f −1(x) = 75. f −1(x) = √ x − 9 _ 10 — x − 1 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 1 2 3 4 5 x Chapter 1 practice test 1. Relation is a function parabola and the graph fails the horizontal line test. — 9. −2(a + b) + 1; b ≠ a 7. 2a2 − a 11. √ 2 3. −16 5. The graph is a 21. (−∞, −2)∪(−2, 6)∪(6, ∞) 77. The function is one-to-one. y 79. 5 ODD ANSWERS C-6 13. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 21 3 4 5 6 x ChapteR 2 Section 2.1 15. Even 19. {−7, 10} 17. Odd 21. f −1(x) = x + 5 _ 3 23. (−∞, −1.1) and (1.1, ∞) 25. (1.1, −0.9) 29. f (x) =  \|x\| if x ≤ 2 if x > 2 3 31. x = 2 33. Yes x − 11 _ 2 35. f −1(x) = − or 27. f (2) = 2 11 − x _ 2 9. No 11. No 1. Terry starts at an elevation of 3,000 feet and descends 70 feet per second 3. 3 miles per hour 7. Yes 17. Decreasing 5. d(t) = 100 − 10t 13. No 15. Increasing 21. Increasing 29. 4 1 _ _ 27. − 5 3 1 __ x + 35. y = − 33. y = 2x + 3 3 19. Decreasing 23. Decreasing 22 __ 3 25. 3 1 7 __ __ 31 39. − 5 4 __ __ 37. y = 5 4 45. y = 3 49. Linear, f(x) = 5x − 5 53. Linear, f (x) = 10x − 24 2 __ x + 1 41. y = 3 43. y
= −2x + 3 47. Linear, g(x) = −3x + 5 51. Linear, g(x) = − 25 __ 2 55. f (x) = −58x + 17.3 x + 6 642 8 10 x 57. y 30,000 25,000 20,000 15,000 10,000 5,000 −10 −4−6−8 −2 −5,000 −10,000 −15,000 −20,000 −25,000 −30,000 61. y 30 20 10 –0.1 –0.05 0.05 0.1 x –10 –20 –30 59. a. a = 11,900, b = 1001.1 b. q(p) = 1000p − 100 x − 65. x = a 63. x = − 16 __ 3 d ad ____ ____ 67. y = c − a c − a 69. $45 per training session 71. The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24. 73. The slope is −400. this means for every year between 1960 and 1989, the population dropped by 400 per year in the city. 75. C Section 2.2 1. The slopes are equal; y-intercepts are not equal. 3. The point of intersection is (a, a). This is because for the horizontal line, all of the y-coordinates are a and for the vertical line, all of the x coordinates are a. The point of intersection is on both lines and therefore will have these two characteristics. 7. Neither 11. Parallel 5. First, find the slope of the linear function. Then take the negative reciprocal of the slope; this is the slope of the perpendicular line. Substitute the slope of the perpendicular line and the coordinate of the given point into the equation y = mx + b and solve for b. Then write the equation of the line in the form y = mx + b by substituting in m and b. 9. Perpendicular 1 15.  __, 0 , (0, 1) 5 19. Line 1: m = 8, Line 2: m = −6, neither 21. Line 1: m = − 1 _, Line 2: m = 2
, perpendicular 2 23. Line 1: m = −2, Line 2: m = −2, parallel 25. g(x) = 3x − 3 5 31.  − 17  _ _, 5 3 27. p(t) = − 1 __ t + 2 29. (−2, 1) 3 35. C 13. (−2, 0), (0, 4) 17. (8, 0), (0, 28) 33. F 37. A 39. –6 –5 –4 –3 –2 43. –6 –5 –4 –3 –2 47. –6 –5 –4 –3 –2 51. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 321 4 5 6 x 321 4 5 6 x 41. 321 4 5 6 x –6 –5 –4 –3 –2 45. 321 4 5 6 x –6 –5 –4 –3 –2 49. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 321 4 5 6 x –6 –5 –4 –3 –2 321 4 5 6 x –1 –1 –2 –3 –4 –5 –6 53. 321 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 321 4 5 6 x ODD ANSWERS C-7 57. 321 4 5 6 x –6 –5 –4 –3 –2 55. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 321 4 5 6 x 9. No 150 125 100 75 50 0 0 1.6 3.2 4.8 6
.4 8 59. a. g(x) = 0.75x − 5.5 b. 0.75 c. (0, −5.5) 11. No 61. y = 3 63. x = −3 65. no point of intersection 67. (2, 7) 69. (−10, −5) 71. y = 100x − 98 73. x < 1999 _ 201, x > 1999 _ 201 75. Greater than 3,000 texts 250 200 150 100 50 0 Section 2.3 19. W(t) = 0.5t + 7.5 7. 20.012 square units 25. C(t) = 12,025 − 205t 5. 6 square 11. 64,170 23. At age 5.8 27. (58.7, 0) In 58.7 1. Determine the independent variable. This is the variable upon 3. To determine the initial value, which the output depends. find the output when the input is equal to zero. 9. 2,300 units. 13. P(t) = 2500t + 75,000 15. (−30, 0) 30 years before the start of this model, the town has no citizens. (0, 75,000) Initially, 17. Ten years after the the town had a population of 75,000. 21. (−15, 0) The model began. x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. (0, 7.5) the baby weighed 7.5 pounds at birth. months years, the number of people afflicted with the common cold would be zero (0, 12,025) Initially, 12,025 people were afflicted 29. 2063 with the common cold 33. In 2070, the company’s profits will be zero 35. y = 30t − 300 37. (10, 0) In the year 1990, the company’s profits were zero 39. Hawaii 43. $105,620 45. a. 696 people b. 4 years c. 174 people per year d. 305 people e. P(t) = 305 + 174t f. 2,219 people 47. a. C(x) = 0.15x + 10 b. The flat monthly fee is $10 and there is a $0.15 fee for each additional minute used c. $113.05 49. a. P(
t) = 190t + 4,360 b. 6,640 moose cubic feet c. During the year 2017 133 minutes 57. More than $66,666.67 in sales 55. More than $42,857.14 worth of jewelry 51. a. R(t)= −2.1t + 16 b. 5.5 billion 41. During the year 1933 31. y = −2t + 180 53. More than Section 2.4 1. When our model no longer applies, after some value in the 3. We predict a value domain, the model itself doesn’t hold. 5. The closer the outside the domain and range of the data. number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data. 7. 61.966 years 0 2.5 5 7.5 10 12.5 13. Interpolation, about 60° F 15. C 17. B 68 64 60 56 52 48 44 0 10 19. 8 6 4 2 0 21. 10 8 6 4 2 0 0 15 20 25 30 35 0 2 4 6 8 10 0 2 4 6 8 10 29. y = −1.981x + 60.197; r = −0.998 23. Yes, trend appears linear; during 2016 25. y = 1.640x + 13.800, r = 0.987 27. y = −0.962x + 26.86, r = −0.965 31. y = 0.121x − 38.841, r = 0.998 (5, −20), (6, −22), (9, −28) sells 18,980 units, its profits will be zero dollars 37. y = 0.00587x + 1985.41 41. y = −10.75x + 742.50 35. (189.8, 0) If the company 39. y = 20.25x − 671.5 33. (−2, −6), (1, −12), ODD ANSWERS C-8 Chapter 2 Review exercises ChapteR 3 0 2 4 6 8 10 x 49. 3i 51. 0 53. 5 − 5i 3. Increasing 5. y = −3x + 26 1. Yes 9. y = 2x − 2 11. Not linear 13. Parallel 15. (−9, 0); (0, −7) 17.
Line 1: m −2, Line 2: m = −2, parallel 19. y = −0.2x + 21 21. 7. 3 y 23. More than 250 25. 118,000 27. y = −300x + 11,500 29. a. 800 b. 100 students per year c. P(t) = 100t + 1700 31. 18,500 33. y = $91, 625 –6 –5 –4 –3 –2 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 321 4 5 6 x 35. Extrapolation y 37. y 120 100 80 60 40 20,900 6,800 6,700 6,600 6,500 6,400 6,300 6,200 6,100 6,000 5,900 5,800 5,700 5,600 0 1985 1990 1995 2000 x 2005 2010 Year 39. Midway through 2023 41. y = −1.294x + 49.412; r = −0.974 43. Early in 2027 45. 7, 660 Chapter 2 practice test 1. Yes 3. Increasing 5. y = −1.5x − 6 7. y = −2x − 1 9. No 15. y = −0.25x + 12 13. (−7, 0); (0, −2) 11. perpendicular 21. 165,000 19. 150 23. y = 875x + 10,625 25. a. 375 b. dropped an average of 46.875, or about 47 people per year c. y = −46.875t + 1250 17. Slope = −1 and y-intercept = 3 –2 –1–1 –2 –3 –4 321 4 5 6 7 8 9 x 29. Early in 2018 31. y = 0.00455x + 1979.5 33. r = 0.999 27. y 35 30 25 20 15 10 5 0 0 2 4 6 8 10 12 x Section 3.1 1. Add the real parts together and the imaginary parts together. 3. i times i equals −1, which is not imaginary. (answers vary) 23 __ 5. −8 + 2i 7. 14 + 7i 9. − 29 i 11. 2 real and 0 nonreal 15 __ 29 + 13. −2−3−4−5 i 5 4 3 2 1 −1−1 −2 −3
−4 −5 15. 321 4 5 r −2−3−4−5 i 5 4 3 2 1 −1−1 −2 −3 −4 −5 321 4 5 r 17. 8 − i 19. −11 + 4i 21. 2 − 5i 25. −16 + 32i 27. −4 − 7i 29. 25 23. 6 + 15i 2 __ 31. 2 − i 3 — 2 __ + 33. 4 − 6i 35. 5 11 __ i 5 41. 1 43. −1 45. 128i — 3 39. 1 + i √ 37. 15i 6 471 + 2 2 9 9 __ __ 55. −2i − 57. i 2 2 Section 3.2 37  _ 12 − 37 _ 12 17. Minimum is − 1. When written in that form, the vertex can be easily identified. 3. If a = 0 then the function becomes a linear function. 5. If possible, we can use factoring. Otherwise, we can use the 7. g(x) = (x + 1)2 − 4; vertex: (−1, −4) 33  _ 4 quadratic formula. 2 5 − 33 5 ; vertex:  −  9. f (x)=  11. k(x) = 3(x − 1)2 − 12; vertex: (1, −12) 2 13. f (x) = 3  x − 5 5 ; vertex:   _ _, − 6 6 17 5 5 _ _ _ 15. Minimum is − ; axis of symmetry: x = and occurs at 2 2 2 17 ; axis of symmetry: x = − 1 1 _ _ _ and occurs at − 8 8 16 7 _ 19. Minimum is − and occurs at −3; axis of symmetry: x = −3 2 21. Domain: (−∞, ∞); range: [2, ∞) range: [−5, ∞) 27.  2i √ 31. {2 + i, 2 − i} 2  33. {2 + 3i, 2 − 3i} 35. {5 + i, 5 − i} 3 3 39.  − 1 i 37, − 3 41  43 47. f (x) = x 2 + 1 45. f (x)
= x 2 − 4x + 4 297 6 ___ __ 49 49 25. Domain: (−∞, ∞); range: [−12, ∞) 51. f (x) = −x 2 + 1 29.  3i √ 49. f (x) = 2, −2i √ 3, −3i √ 60 __ 49  23. Domain: (−∞, ∞); ODD ANSWERS C-9 51. y-intercept: (0, 0); x-intercepts: (0, 0) and (2, 0); as x → −∞, f (x ) → ∞, as x → ∞, f (x) → ∞ 53. y-intercept: (0, 0); x-intercepts: (0, 0), (5, 0), (7, 0); as x → −∞, f (x ) → −∞, as x → ∞, f (x) → ∞ 53. Vertex: (1, −1), axis of symmetry: x = 1, intercepts: (0, 0), (2, 0) 49 5 55. Vertex:  , axis of _ _, − 4 2 5 _ symmetry: x =, intercepts: 2 (6, 0), (−1, 0) y = f (x1−1 −2 −3 −4 −5 −6 −2−3−4−5−6 321 4 5 6 x −10 −5 y 15 12 9 6 3 0 −3 −6 −9 −12 −15 y = f (x) 5 10 x −5−6 −4 y 5 4 3 2 1 −3 −2 −1−1 −2 −3 −4 −5 y 200 160 120 80 40 −1−2 −40 −80 −120 −160 −200 21 3 4 5 6 x 21 3 4 5 6 7 8 9 10 x 93. $10.70 59. y-intercept: (0, 0); x-intercepts: (−3, 0), (0, 0), (5, 0); as x → −∞, f (x ) → −∞, as x → ∞, f (x) → ∞ 39 5 57. Vertex:  , axis of _ _, − 4
8 5 _ symmetry: x =, intercept: 4 (0, −8) −10 −5 y 12 8 4 −4 −8 −12 −16 −20 −24 5 10 x y = f (x) 59. f (x) = x2 − 4x + 1 61. f (x) = −2x 2 + 8x − 1 1 7 __ __ 63. f (x) = x 2 − 3x + 2 2 65. f (x) = x 2 + 1 67. f (x) = 2 − x 2 69. f (x) = 2x 2 71. The graph is shifted up or down (a vertical shift). 73. 50 feet 75. Domain: (−∞, ∞); range: [−2, ∞) 77. Domain: (−∞, ∞); range: (−∞, 11] 81. f (x) = 3x 2 − 9 83. f (x) = 5x 2 − 77 85. 50 feet by 50 feet 89. 6 and −6; product is −36 91. 2909.56 meters 87. 125 feet by 62.5 feet 79. f (x) = 2x 2 − 1 Section 3.3 13. Degree: 2, coefficient: −2 15. Degree: 4, 1. The coefficient of the power function is the real number that is multiplied by the variable raised to a power. The degree is the highest power appearing in the function. 3. As x decreases without bound, so does f (x). As x increases without bound, so does f (x). 5. The polynomial function is of even degree and leading coefficient is negative. 7. Power function 9. Neither 11. Neither coefficient: −2 17. As x → ∞, f (x) → ∞, as x → −∞, f (x) → ∞ 19. As x → −∞, f (x) → −∞, as x → ∞, f (x) → −∞ 21. As x → −∞, f (x) → −∞, as x → ∞, f (x) → −∞ 23. As x → ∞, f (x) → ∞, as x → −∞, f (x) → −∞ 25. y-intercept is (0, 12), t-intercepts are (1,
0), (−2, 0), and (3, 0) 27. y-intercept is (0, −16), x-intercepts are (2, 0), and (−2, 0) 29. y-intercept is (0, 0), x-intercepts are (0, 0), (4, 0), and (−2, 0) 31. 3 33. 5 least possible degree: 3 degree: 2 45. Yes, 0 turning points, least possible degree: 1 47. As x → −∞, f (x ) → ∞, as x → ∞, f (x) → ∞ 37. 5 39. Yes, 2 turning points, 41. Yes, 1 turning point, least possible 49. As x → −∞, f (x ) → ∞, as x → ∞, f (x) → −∞ 43. Yes, 0 turning points, least possible degree: 1 35. 3 x 10 100 −10 −100 f (x) 9,500 99,950,000 9,500 99,950,000 x 10 100 −10 −100 f (x) −504 −941,094 1,716 1,061,106 55. y-intercept: (0, 0); x-intercepts: (−4, 0), (0, 0), (4, 0); as x → −∞, f (x ) → −∞, as x → ∞, f (x) → ∞ 57. y-intercept: (0, −81); x-intercepts: (−3, 0), and (3, 0); as x → −∞, f (x ) → ∞, as x → ∞, f (x) → ∞ y 500 400 300 200 100 −10 −2−4−6−8 −100 −200 −300 −400 −500 42 6 8 10 x y 100 80 60 40 20 21 3 4 5 6 x −1−2−3−4−5−6 −20 −40 −60 −80 −100 y 50 40 30 20 10 −1−2−3−4−5 −10 −20 −30 −40 −50 21 3 4 5 6 x 61. f (x ) = x 2 − 4 63. f (x ) = x 3 − 4x 2 + 4x 65. f (x ) = x 4
+ 1 67. V(m ) = 8m 3 + 36m 2 + 54m + 27 69. V(x ) = 4x 3 − 32x 2 + 64x Section 3.4 9. (3, 0), (−1, 0), (0, 0) 7. (−2, 0), (3, 0), (−5, 0) 3. If we evaluate the function at a and at b and 1. The x-intercept is where the graph of the function crosses the x-axis, and the zero of the function is the input value for which f (x) = 0. the sign of the function value changes, then we know a zero exists 5. There will be a factor raised to an even between a and b. power. 11. (0, 0), (−5, 0), (2, 0) 15. (2, 0), (−2, 0), (−1, 0) 19. (1, 0), (−1, 0) 23. (0, 0), (1, 0), (−1, 0), (2, 0), (−2, 0) 25. f (2) = −10, f (4) = 28; sign change confirms 27. f (1) = 3, f (3) = −77; sign change confirms 29. f (0.01) = 1.000001, f (0.1) = −7.999; sign change confirms 31. 0 with multiplicity 2, − 3 _ multiplicity 5, 4 multiplicity 2 2 33. 0 with multiplicity 2, −2 with multiplicity 2 35. − 2 _ with multiplicity 5, 5 with multiplicity 2 3 13. (0, 0), (−5, 0), (4, 0) 17. (−2, 0), (2, 0),  1 _ 3, 0  3, 0 ,  − √ 21. (0, 0),  √ 2, 0  — — ODD ANSWERS C-10 37. 0 with multiplicity 4, 2 with multiplicity 1, −1 with multiplicity 1 3 _ 39. with multiplicity 2, 0 with multiplicity 3 2 41. 0 with 2 _ with multiplicity 2 multiplicity 6, 3 43. x-intercept: (1, 0) with multiplicity 2, (−4, 0) with multiplicity 1; y
-intercept: (0, 4); as x → −∞, g (x) → −∞, as x → ∞, g (x) → ∞ 45. x-intercept: (3, 0) with multiplicity 3, (2, 0) with multiplicity 2; y-intercept: (0, −108); as x → −∞, k (x) → −∞, as x → ∞, k (x) → ∞ k(x) g(x) 20 16 12 8 4 −1−2−3−4−5 −4 − 24 12 −1−2−3−4−5−6 −12 −24 −36 −48 −60 −72 −84 −96 −108 −120 47. x-intercepts: (0, 0), (−2, 0), (4, 0) with multiplicity 1; y-intercept: (0, 0); as x → −∞, n (x) → ∞, as x → ∞, n (x) → −∞ n(x) 75 60 45 30 15 49. f (x) = − 2 _ (x − 3)(x + 1)(x + 3) 9 1 _ 51. f (x) = (x + 2)2(x − 3) 4 53. −4, −2, 1, 3 with multiplicity 1 55. −2, 3 each with multiplicity 2 57. f (x) = − 2 _ (x + 2)(x − 1)(x − 3) 3 1 _ 59. f (x) = (x − 3)2(x − 1)2(x + 3) 3 61. f (x) = −15(x − 1)2(x − 3)3 1 2 3 4 5 x 63. f (x) = −2(x + 3)(x + 2)(x − 1) −1−2−3−4−5 −15 −30 −45 −60 −75 41. Yes, 35. x 3 − 9x 2 + 27x − 27 39. Yes, (x − 2)(3x 3 − 5) 43. No 33. x 3 − 6x 2 + 12x − 8 37. 2x 3 − 2x + 2 (x − 2)(4x 3 + 8x 2 + x + 2) 45. (x
− 1)(x 2 + 2x + 4) 49. Quotient: 4x 2 + 8x + 16, remainder: −1 51. Quotient 53. Quotient is x 3 − 2x 2 + is 3x 2 + 3x + 5, remainder: 0 4x − 8, remainder: −6 55 57 47. (x − 5)(x 2 + x + 1) 61. 1 + 59 63. x 2 + ix − 1 + 1 − i _ x − i 65. 2x 2 + 3 67. 2x + 3 69. x + 2 71. x − 3 73. 3x 2 − 2 Section 3.6 1. The theorem can be used to evaluate a polynomial. 3. Rational zeros can be expressed as fractions whereas real zeros include irrational numbers. can have repeated zeros, so the fact that number is a zero doesn’t preclude it being a zero again. 5. Polynomial functions 7. −106 9. 0 11. 255 13. −1 21. − 5 _, √ 2 1 _ 15. −2, 1, 2 — 6, − √ — 6 17. −2 3 _ 23. 2, −4, − 2 — 5 19. −3 25. 4, −4, −5 — 5 3 _ 31. 2 33. 2, 3, −1, −2 1 _ 29., 2 1 _ 27. 5, −3, −3, − 35. 2 2 5 39. − 3 1 _ _ 37. −1, −1, √ 41. 2, 3 + 2i, 3 − 2i, − 4 2 43. − 2, 1 + 2i, 1 − 2i 45. − 1 _ _, 1 + 4i, 1 − 4i 2 3 49. 1 positive, 0 negative 47. 1 positive, 1 negative 5, − √ — — f (x) 5 4 3 2 1 f (x) 5 4 3 2 1 −3 −2 −1−1 −2 −3 −4 −5 21 3 4 5 6 x 21 3 4 5 6 x −1−2−3−4−5−6 −1 −2 −3 −4 −5 65. f (x) = − 3 _ (2x − 1)2(x − 6)(x + 2) 2 67. Local max: (−0.58, −
0.62); local min: (0.58, −1.38) 69. Global min: (−0.63, −0.47) 71. Global min: (0.75, −1.11) 73. f (x) = (x − 500)2(x + 200) 75. f (x) = 4x 3 − 36x 2 + 80x −5−6 −4 77. f (x) = 4x 3 − 36x 2 + 60x + 100 1 _ π (9x 3 + 45x 2 + 72x + 36) 79. f (x) = Section 3.5 1. The binomial is a factor of the polynomial. 3. x + 6 +, quotient: x + 6, remainder: 5 5 _ x − 1 5. 3x + 2, quotient: 3x + 2, remainder: 0 9. 2x − 7 + x − 5, remainder: 0 6 _ 3x + 1 remainder 16 11. x − 2 + 7. x − 5, quotient: 16 _, quotient: 2x − 7, x + 2, quotient: x − 2, remainder: 6 13. 2x 2 − 3x + 5, quotient: 2x 2 − 3x + 5, remainder: 0 15. 2x 2 + 2x + 1 + 10 _ x − 4 17. 2x 2 − 7x + 1 − 2 _ 2x + 1 51. 0 positive, 3 negative 53. 2 positive, 2 negative f (x) −14 −12 −10 −8 −6 −4 100 80 60 40 20 −2−20 −40 −60 −80 −100 42 6 8 10 x f (x) 40 32 24 16 8 21 3 4 5 6 x −1−2−3−4−5−6 −8 −16 −24 −32 −40 19. 3x 2 − 11x + 34 − 23. 4x 2 − 21x + 84 − 106 _ x + 3 323 _ x + 4 21. x 2 + 5x + 1 25. x 2 − 14x + 49 27. 3x 2 + x + 29. x 3 − 3x + 1 31 3x − 1 ODD ANSWERS 55. 2 positive, 2 negative f (x) 15 12 9 6 3 21 3 4 5 6 x −1−2−3−4−5−6
−3 −6 −9 −12 −15, ±1, ±5, ± 5 57. ± 1 _ _ 2 2 59. ±1 61. 1, 3 2, − 3 1 _ _ 63. 2, 4 2 4 _ 67. f (x) = (x 3 + x 2 − x − 1) 9 69. f (x) = − 1 __ (4x 3 − x) 5 71. 8 by 4 by 6 inches 5 _ 65. 4 73. 5.5 by 4.5 by 3.5 inches 77. Radius: 6 meters; height: 2 meters meters, height: 4.5 meters 75. 8 by 5 by 3 inches 79. Radius: 2.5 Section 3.7 3. The numerator and denominator 5. Yes. The numerator of the 1. The rational function will be represented by a quotient of polynomial functions. must have a common factor. formula of the functions would have only complex roots and/or factors common to both the numerator and denominator. 7. All reals except x = −1, 1 9. All reals except x = −1, 1, −2, 2 11. Vertical asymptote: x = − 2 _ ; horizontal asymptote: y = 0; 5 domain: all reals except x = − 2 _ 5 x = 4, −9; horizontal asymptote: y = 0; domain: all reals except x = 4, −9 15. Vertical asymptotes: x = 0, 4, −4; horizontal asymptote: y = 0; domain: all reals except x = 0, 4, −4 17. Vertical asymptotes: x = −5; horizontal asymptote: y = 0; domain: all reals except x = 5, −5 13. Vertical asymptotes: 1 2 _ _ 19. Vertical asymptote: x = ; horizontal asymptote: y = − ; 3 3 1 _ domain: all reals except x = 21. None 3 1 23. x-intercepts: none, y-intercept:  0,  _ 4 − + 25. Local behavior: x → − 1, f (x) → −∞, x → − 1 _ _, f (x) → ∞ 2 2 1 _ End behavior: x → ±∞,
f (x) → 2 27. Local behavior: x → 6+, f (x) → −∞, x → 6−, f (x) → ∞ End behavior: x → ±∞, f (x) → −2 − + 1 1 _ _ 29. Local behavior: x → −, f (x) → −∞,, f (x) → ∞, x) → ∞, x → − 5 _ _, f (x) → −∞ 2 2 1 _ End behavior: x → ±∞, f (x) → 3 33. y = 2x 31. y = 2x + 4 35. Vertical asymptote at x = 0, horizontal asymptote at y = 2 y 10 8 6 4 2 y = 2 −10 −4−6−8 −2−2 −4 −6 −8 −10 42 6 8 10 x x = 0 37. Vertical asymptote at x = 2, horizontal asymptote at y = 0 y 10 8 6 4 2 −10 −4−6−8 y = 0 −2−2 −4 −6 −8 −10 x 8 10 42 6 x = 2 39. Vertical asymptote at x = −4; horizontal asymptote 3 3 , 0 ,  0, − at y = 2;  _ _ 4 2 p(x) C-11 41. Vertical asymptote at x = 2; horizontal asymptote at y = 0; (0, 1) s(x) 10 8 6 4 2 −2−2 −4 −6 −8 −10 −10 −4−6−8 x = −4 y = 2 x 42 6 8 10 12 10 10 −4−6−8 −2−1 −2 x = 2 y = 0 42 6 8 10 x 43. Vertical asymptote 4 _ at x = −4, ; horizontal 3 asymptote at y = 1; (5, 0), 1, 0 ,  0,  − _ 3 5  _ 16 f(x) 12 10 8 6 4 2 −2−2 −4 −6 −8 −10 −12 −12 −4−6−8−10 x = −4 x = 4 3 y = 1 42 6 8 10 12 x 47
. Vertical asymptote at x = 4; slant asymptote at 1 y = 2x + 9; (−1, 0), , 0 , _ 2 1   0, _ 4 h(x) 50 40 30 20 10 −50 −40 −30 −20 −10 −10 −20 −30 −40 −50 y = 2x + 9 x = 4 10 20 30 40 50 45. Vertical asymptote at x = −1; horizontal asymptote at y = 1; (−3, 0), (0, 3) a(x) x =−1 15 12 9 6 3 −15 −6−9−12 −3−3 −6 −9 −12 −15 y = 1 15 x 63 9 12 49. Vertical asymptote at x = −2, 4; horizontal asymptote at y = 1; 15 (1, 0), (5, 0), (−3, 0),  0, −  _ 16 w(x) x = 4 x = −2 12 10 8 6 4 2 −12 −4−6−8−10 x −2−2 −4 −6 −8 −10 y = 1 42 6 8 10 12 x 51. f (x) = 50 53. f (x) = 7 x2 − x − 2 ________ x2 − 25 x2 + 2x − 24 __ x2 + 9x + 20 1 _ 55. f (x) = ⋅ 2 x2 − 4x + 4 __ x + 1 57. f (x) = 4 x − 3 __ x2 − x − 12 59. f (x) = −9 x − 2 __ x 2 − 9 1 _ 61. f (x 63. f (x) = −6 (x − 1)2 ____________ (x + 3)(x − 2)2 65. Vertical asymptote at x = 2; horizontal asymptote at y = 0 x y x y 2.01 100 10 2.001 1,000 100 0.125 0.0102 2.0001 10,000 1,000 0.001 1.99 1.999 −100 −1,000 10,000 100,000 0.0001 0.00001 ODD ANSWERS –10 –8 –6 –4 10 8 6 4 2 –2–2
–4 –6 –8 –10 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2–2 –4 –6 –8 –10 42 6 8 10 x 21 3 4 5 x −1−2−3−4−5 −1 −2 −3 45. [−4, 2) ∪ [5, ∞) y 5 4 3 2 1 16 14 12 8 4 84 12 14 16 −16 −14 −4−8−12 −4 −8 −12 −14 −16 47. (−2, 0), (4, 2), (22, 3) y C-12 67. Vertical asymptote at x = −4; horizontal asymptote at y = 2 x y x y −4.1 82 10 −4.01 802 100 1.4286 1.9331 −4.001 −3.99 −3.999 −7998 −798 8,002 1,000 1.992 10,000 100,000 1.9992 1.999992 69. Vertical asymptote at x = −1; horizontal asymptote at y = 1 −0.999 998,001 −1.01 10,201 −0.99 9,801 −0.9 81 −1.1 121 x y x y 10 0.82645 100 0.9803 1,000 0.998 10,000 0.9998 100,000 3, ∞  71.  _ 2 f(x) 73. (−∞, 1) ∪ (4, ∞) f(x) 75. (2, 4) 77. (2, 5) 79. (−1, 1) 81. C(t) = 8 + 2t _ 300 + 20t 83. After about 6.12 hours 87. radius 2.52 meters 85. 2 by 2 by 5 feet Section 3.8 1. It can be too difficult or impossible to solve for x in terms of y. 3. We will need a restriction on the domain of the answer. 7. f −1(x) = √ x + 3 −1 x + 4 — — 9. f −1(x) = − √ 5. f −1 (x) = √ ______ 13. f −1(x) = 11. f −1(x) = √ — 9 − x 15
. f −1(x) = √ 3 — 4 − x 17. f −1(x) =, [0, ∞) 19. f −1(x) = (x − 9)2 + 4 __________ 4, [9, ∞) √ x 2 − 1 ______ 2 x − 9 _____  2 7x − 3 ______ 27. f −1(x) = 1 − x 3 23. f −1(x) = 2 − 8x ______ x 5x − 4 ______ 4x + 3 31. f −1(x) = √ — x + 6 + 3 21. f −1(x) =  25. f −1(x) = 29. f −1(x) = √ 33. f −1(x — 10 8 6 4 2 −10 −2−4−6−8 −2 −4 −6 −8 −10 42 6 8 10 x 35. f −1(x) = √ y — x + 4 12 10 8 6 4 2 42 6 8 10 12 x −12 −2−4−6−8−10 −2 −4 −6 −8 37. f −1(x) = 3 √ y — 1 − x 39. f −1 (x2−1 −1 −2 −3 −4 −5 −3−4−5 1 2 3 4 5 x 15 12 9 6 3 −15 −3−6−9−12 −3 −6 −9 −12 −15 63 9 12 15 x __ 41. f −1 (x) = √ 1 _ x 43. [−2, 1) ∪ [3, ∞) y x x x f (x) 10 8 6 4 2 −10 −6−8 −4−2 −2 −4 −6 −8 −10 42 6 8 10 x −3−4−5 40 32 24 16 8 −2 –1 −8 −16 −24 −32 −40 1 2 3 4 5 49. (−4, 0), (0, 1), (10, 2) y 51. (−3, −1), (1, 0), (7, 1) y 40 32 24 16 8 −2 –1 −8 −16 −24 −32 −40 −3−4−5 1 2 3 4 5 x −3−4−5 40 32 24 16 8 −
2 –1 −8 −16 −24 −32 −40 59. r(V) =, 5.53 seconds b2 + 4x √ _ 55. f −1(x) =, ≈ 3.63 feet 61. n(C) = — 3V _ 4π ___ V _ 6π 53. f −1(x) = − b _ + 2 ________ 57. t(h) = √ 200 − h _ 4.9 √ 63. r(V) = √ Section 3.9 1. The graph will have the appearance of a power function. 5. y = 5x 2 3. No. Multiple variables may jointly vary. 18 _ x 2, ≈ 3.99 m 65. r(V) = √ 100C − 25 _ 0.6 − C ___ V _ 4π 9. y = 6x 4 13. y = 11. y = 17. y = 10xzw 19. y = 10x √ — z 81 _ x 4, ≈ 1.99 inches, 250 mL 23. y = 40 25. y = 256 xz _ — w t2 √ 31. y = 27 27. y = 6 29. y = 6 33. y = 3 15. y = 7. y = 10x 3 20 _ — 3 x √ xz _ w 21. y = 4 ODD ANSWERS 35. y = 18 37. y = 90 3 _ 41. y = x 2 4 y 39. y = 81 _ 2 1 __ 43. y = √ 3 — x y 75 60 45 30 15 −10 −2−4−6−8 −15 −30 −45 −60 −75 4 __ 45. y = x2 y 10 8 6 4 2 −10 −2−4−6−8 −2 −4 −6 −8 −10 42 6 8 10 x 10 8 6 4 2 −25 −20 −15 −5−10 −2 −4 −6 −8 −10 5 10 15 20 25 x 47. ≈ 1.89 years 49. ≈ 0.61 years 51. 3 seconds 53. 48 inches 55. ≈ 49.75 pounds 57. ≈ 33.33 amperes 59. ≈ 2.88 42 6 8 10 x Chapter 3 Review exercises 1. 2 − 2i 3. 24 + 3i 7. f (x) = (x −
2)2 −9; vertex: (2, −9); intercepts: (−1, 0), (5, 0), (0, −5) f(x) −2−3−4−5 10 8 6 4 2 −1−2 −4 −6 −8 −10 321 4 5 6 7 x 5. {2 + i, 2 − i} 3 _ 25 9. f (x) = (x + 2)2 + 3 11. 300 meters by 150 meters, the longer side parallel to the river 13. Yes; degree: 5, leading coefficient: 4 15. Yes; degree: 4; leading coefficient: 1 17. As x → −∞, f (x) → −∞, as x → ∞, f (x) → ∞ 1 _ 19. −3 with multiplicity 2, − with multiplicity 1, −1 with 2 21. 4 with multiplicity 1 1 _ 23. with 2 multiplicity 3 multiplicity 1, 3 with multiplicity 3 25. x 2 + 4 with remainder 12 27. x 2 − 5x + 20 − 61 _____ x + 3 29. 2x 2 − 2x − 3, so factored form is (x + 4)(2x 2 − 2x − 3) 1 31.  −2, 4, − 1 33.  1, 3, 4,   _ _ 2 2 35. 2 or 0 positive, 1 negative 37. Intercepts: (−2, 0),  0, − 2 , _ 5 asymptotes: x = 5 and y = 1 y 39. Intercepts: (3, 0), (−3, 0), 27  0,  ; asymptotes: x = 1, −2 _ 2 and y = 3 25 20 15 10 5 y = 1 −25 −20 −15 −5−10 −5 −10 −15 −20 −25 5 10 15 20 25 30 x x = 5 y 40 32 24 16 8 −15 −12 −3−6−9 −8 −16 −24 −32 −40 x = −2 y = 3 x 3 6 9 12 15 x = 1 41. y = x − 2 43. f −1(x) = √ — x + 2 45. f −1(x) = √ — x + 11 − 3 C-13
47. f −1(x) =, x ≥ −3 49. y = 64 51. y = 72 (x + 3)2 − 5 __ 4 53. ≈ 148.5 pounds Chapter 3 practice test 1. 20 − 10i 3. {2 + 3i, 2 − 3i} 5. As x → −∞, f (x) → −∞, as x → ∞, f (x) → ∞ 7. f (x) = (x + 1)2 − 9, vertex: (−1, −9), intercepts: (2, 0), (−4, 0)(0, −8) y 9. 60,000 square feet 11. 0 with multiplicity 4, 3 with multiplicity 2 13. 2x 2 − 4x + 11 − 26 _ x + 2 15. 2x 2 − x − 4, so factored form is (x + 3)(2x 2 − x − 4) 100 80 60 40 20 –4 0 –2 –20 –40 –60 –80 –100 –10 –8 –6 42 6 8 10 x g(x) = 2x3 – 50x — 15 −1 ± i √ __ 2 17. − 1 _ (has multi plicity 2), 2 21. f(x) = 2(2x − 1)3(x + 3) 19. −2 (multiplicity 3), ±i 23. Intercepts: (−4, 0),  0, − 4  ; _ 3 asymptotes: x = 3, −1 and y = 0 25. y = x + 4 — 27. f −1(x) = x + 4 _ 3 3 √ 31. 4 seconds y 10 8 6 4 2 −2 −4 −6 −8 −10 −10 −2−4−6−8 x = −1 29. y = 18 y = 0 2 4 6 8 10 x x = 3 ChapteR 4 Section 4.1 5. Exponential; the population 7. Not exponential; 1. Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original. 3. When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal. decreases by
a proportional rate. the charge decreases by a constant amount each visit, so the statement represents a linear function. 11. After 20 years forest A will have 43 more trees than forest B. 13. Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors. 9. Forest B ODD ANSWERS C-14 − 3 – 5 23. Linear x – 5 ≈ 2.93(0.699)x 15. Exponential growth; the growth factor, 1.06, is greater than 1. 17. Exponential decay; the decay factor, 0.97, is between 0 and 1. 19. f (x) = 2000(0.1)x 1 1  21. f (x) =  __ __   6 6 27. Linear 25. Neither −nt r 33. P = A(t) ⋅  1 + _ n  39. Continuous growth; the growth rate is greater than 0. 41. Continuous decay; the growth rate is less than 0. 43. $669.42 49. f (3) ≈ 483.8146 55. y ≈ 0.2 ⋅ 1.95x 45. f (−1) = −4 47. f (−1) ≈ −0.2707 31. $13,268.58 37. 4% 29. $10,250 35. $4,569.10 53. y ≈ 18 ⋅ 1.025x 51. y = 3 ⋅ 5x 57. APY = A(t) − a _ a 365(1) a  1 + r ___  365 ___ a − a = − 1; n 365 365 − 1 − 1  r _  365 ___  365 ___ = a r I(n) =  1 + _ n  59. Let f be the exponential decay function f (x such that b > 1. Then for some number n > 0, f (x 61
. 47,622 foxes 67. $82,247.78; $449.75 63. 1.39%; $155,368.09 x = a(e−n)x = a(e)−nx. = a(b−1)x = a ((en)−1 ) 65. $35,838.76 Section 4.2 1. An asymptote is a line that the graph of a function approaches, as x either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely 3. g(x) = 4(3)−x; y-intercept: (0, 4); large or extremely small. domain: all real numbers; range: all real numbers greater than 0. 5. g(x) = −10x + 7; y-intercept: (0, 6); domain: all real numbers; range: all real numbers less than 7. 1 7. g(x) = 2   _ 4 range: all real numbers greater than 0. ; y-intercept: (0, 2); domain: all real numbers; x 9. y-intercept: (0, −2) y 11. 5 4 3 2 1 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 g(x) = −2(0.25)x x 21 3 4 5 13. B 23. g(−x) = −2(0.25)−x 17. E 15. A 19. D 25. y 5 5 4 3 2 1 f(x) = 1 2(4)x x −5 −4 5 3 21 4 −f(x) = −1 2 (4)x −5 −4 −3 −2 −1−1 −2 −3 −4 −5 y g(x) = 3(2)x h(x) = 3(4)x 4( )x f (x) = 3 1 5 4 3 2 1 21 3 4 5 x −5 −4 −3 −2 −1−1 −2 −3 −4 −5 213 −2 −1−1 −2 −3 −4 −5 −f (x) = 4(2)x − 2 21 3 4 5 x f (x) = −
4(2)x + 2 27. Horizontal asymptote: h(x) = 3; domain: all real numbers; range: all real numbers strictly greater than 3. h(x1−2−3−4−5 21 3 4 5 x 29. As x → ∞, f (x) → −∞; as x → −∞, f (x) → −1 31. As x → ∞, f (x) → 2; as x → −∞, f (x) → ∞ 33. f (x) = 4x − 3 35. f (x) = 4x − 5 37. f (x) = 4−x 39. y = −2x + 3 41. y = −2(3)x + 7 43. g(6) ≈ 800.3 49. x ≈ −0.222 47. x ≈ −2.953 45. h(−7) = −58 x 51. The graph of g(x) =  1  _ is the reflection about the y-axis b of the graph of f (x) = bx; for any real number b > 0 and function x 1  f (x) = bx, the graph of  _ is the reflection about the y-axis, f (−x). b 53. The graphs of g(x) and h(x) are the same and are a horizontal shift to the right of the graph of f (x). For any real number n, real number b > 0, and function f (x) = bx, the graph of  1 bn  bx is the _ horizontal shift f (x − n). Section 4.3 (x), the 9. x y = 64 1. A logarithm is an exponent. Specifically, it is the exponent to which a base b is raised to produce a given value. In the expressions given, the base b has the same value. The exponent, y, in the expression by can also be written as the logarithm, logbx, and the value of x is the result of raising b to the power of y. 3. Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation b y = x, and then properties of exponents can
be applied to solve for x. 5. The natural logarithm is a special case of the logarithm with base b in that the natural log always has base e. Rather than notating the natural logarithm as loge notation used is ln(x). 7. ac = b 15. e n = w 21. logn(103) = 4 1 _ 27. x = 8 35. x = e2 45. 4 55. ≈ 2.708 defined value for x = 0. To verify, suppose x = 0 is in the domain of the function f (x) = log(x). Then there is some number n such that n = log(0). Rewriting as an exponential equation gives: 10n = 0, which is impossible since no such real number n exists. Therefore, x = 0 is not the domain of the function f (x) = log(x). 61. Yes. Suppose there exists a real number, x such that ln(x) = 2. Rewriting as an exponential equation gives x = e2, which is a real number (x = e2 ≈ 7.389056099). To verify, let x = e 2. Then, by 19. log19 (y) = x  = x 57. ≈ 0.151 59. No, the function has no 25. ln(h) = k 1 _ 216 17. logc(k) = d 23. logy  47. −3 49. −12 51. 0 53. 10 39 _ 100 31. x = 3 1 _ 43. 2 13. 13a = 142 41. 14.125 11. 15b = a 29. x = 27 39. 1.06 33. x = 37. 32 definition, ln(x) = ln(e2) = 2. is undefined. 65. 2 63. No; ln(1) = 0, so ln(e1.725) _ ln(1) ODD ANSWERS Section 4.4 1. Since the functions are inverses, their graphs are mirror images about the line y = x. So for every point (a, b) on the graph of a logarithmic function, there is a corresponding point (b, a) on 3. Shifting the the graph of its inverse exponential function. function right or left and reflecting the function about the
y-axis 5. No. A horizontal asymptote would will affect its domain. suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers. 1 7. Domain:  −∞,  ; range: (−∞, ∞) _ 2 9. Domain:  − 17, ∞  ; range: (−∞, ∞ ) _ 4 11. Domain: (5, ∞); vertical asymptote: x = 5, ∞  ; vertical asymptote: x = − 1 13. Domain:  − 1 _ _ 3 3 15. Domain: (−3, ∞); vertical asymptote: x = −3 17. Domain:  3 3, ∞  ; vertical asymptote: x = _ _ ; end behavior: 7 7 3 as x →   _ 7 19. Domain: (−3, ∞); vertical asymptote: x = −3; end behavior: as x → −3+, f (x) → −∞ and as x → ∞, f (x) → ∞ 21. Domain: (1, ∞); range: (−∞, ∞); vertical asymptote: x = 1; 5 x-intercept: , 0  ; y-intercept: DNE _ 4 23. Domain: (−∞, 0); range: (−∞, ∞); vertical asymptote: x = 0; x-intercept: (−e2, 0); y-intercept: DNE 25. Domain: (0, ∞); range: (−∞, ∞) vertical asymptote: x = 0; x-intercept: (e3, 0); y-intercept: DNE 27. B 35. 5 4 3 2 1, f (x) → −∞ and as x → ∞, f (x) → ∞ 33. C 37. f(x) = log(x) 29. C 31. B x f (x) = e 391−1 −2 −3 −4 −5 41. −5 −4 −3 −2 −1−1 −2 −3 −4 −5 21 3 4 5 876 9
10 x −5 −4 g(x) = log (x) l 2 43. f(x) 5 4 3 2 1 21 3 4 5 x −3−4−5−6−7−8 −3 −2 −1−1 −2 −3 −4 −5 x 4 5 21 3 g(x) = ln(x) y 5 4 3 2 1 −2 −1−1 −2 −3 −4 −5 x 21 47. f (x) = log2(−(x − 1)) 49. f (x) = 3log4(x + 2) 51. x = 2 53. x ≈ 2.303 55. x ≈ −0.472 45. g(x) 5 4 3 2 1 21 3 4 x −3−4−5−6 −2 −1−1 −2 −3 −4 −5 C-15 57. The graphs of f (x) = log (x) and g(x) = −log 2(x) appear to be 1 _ 2 the same; conjecture: for any positive base b ≠ 1, logb(x) = −log 1 _ b (x). 59. Recall that the argument of a logarithmic function must be positive, so we determine where > 0. From the graph of x + 2 _ x − 4 the function f (x) =, note that the graph lies above the x + 2 _ x − 4 x-axis on the interval (−∞, −2) and again to the right of the vertical asymptote, that is (4, ∞). Therefore, the domain is (−∞, −2)∪(4, ∞). Section 4.5 1. Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, logb(x 1 3. logb (2) + logb (7) + logb(x) + logb(y) 5. logb(13) − logb(17) 7. −kln(4) 11. logb(4) 13. logb(7) 15. 15log(x) + 13log(y) − 19log(z) 8 3 _ _ log(x) + log(x) − 2log(y) 19. log(y)
21. ln(2x7) 17. 3 2 1 _ n logb(x). n ) = 9. ln(7xy) 14 _ 3 _ 23. log  xz3  _ — y √ 1 _ 27. log11(5) = b 25. log7(15) = 6 29. log11  _ 11 35. ≈ 0.93913 ln(15) _ ln(7)  = or 37. ≈ −2.23266 33. ≈ 2.81359 39. x = 4, By the quotient rule: log6(x + 2) − log6(x − 3) = log6  Rewriting as an exponential equation and solving for x: 61 = x + 2 _____ x − 3  = 1 31. 3 − 6 x + 2 _____ x − 3 x + 2 _____ x − 3 6(x − 3) x + 2 _____ _______ − (x − 3) x − 3 x + 2 − 6x + 18 _____________ _____ x − 3 x = 4 Checking, we find that log6(4 + 2) − log6 (4 − 3) = log6(6) − log6(1) is defined, so x = 4. 41. Let b and n be positive integers greater than 1. Then, by the change-of-base formula, logb(n) = logn(n) _ logn(b) = 1 _. logn(b) Section 4.6 1. Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use 3. The one-to-one property properties of logarithms to solve. can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base. 1 _ 5. x = − 3 11. x = 10
13. No solution 17. k = − ln(38) _ 3 19. x = 6 _ 9. b = 7. n = −1 5 17 15. p = log   − 7 _ 8 38 ln   − 8 _ 3 __ 9 21. x = ln(12) ODD ANSWERS C-16 23. x = 25. No solution 27. x = ln(3) y 67. About 5 years 3 ln   − 3 _ 5 __ 8 1 _ 100 29. 10−2 = 31. n = 49 33. k = 1 _ 36 35. x = 9 − e _ 8 37. n = 1 43. x = ± 10 _ 3 51. x = 9 y 39. No solution 41. No solution 45. x = 10 47. x = 0 3 _ 49. x = 4 53. x = ≈ 2.5 e2 _ 3 321 4 5 6 7 8 9 10 11 12 x 1 −1 −2 −3 −4 −5 −6 −7 y 5 4 3 2 1 −2 −1−1 −2 −3 −4 −5 21 3 4 5 x e + 10 _ 4 ≈ 3.2 57. x = y 321 4 5 6 7 8 9 10 x 1 −1−1 −2 −3 −4 −5 −6 x –1 61. x = 11 _ 5 ≈ 2.2 y 55. x = −5 y 5 4 3 2 1 −2−3−4−5−6−7−8 −1 −1 −2 −3 59. No solution y 3 2 1 71. ≈ 2.078 69. ≈ 0.567 73. ≈ 2.2401 75. ≈ −44655.7143 77. About 5.83 1 _ k ) 79. t = ln (  y ______ A  81. t = ln (  1 _ − T − Ts k  ______ T0 − Ts ) (5, 20,000) 25,000 20,000 15,000 10,000 5,000 0 1 2 3 4 5 6 x Section 4.7 1. Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half
of the initial amount of that 3. Doubling time is a measure substance or quantity to decay. of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity 5. An order of magnitude is the nearest to double in size. power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about 102 times, or 2 orders of magnitude greater, than the mass of Earth. 7. f (0) ≈ 16.7; the amount initially present is about 16.7 units. 9. 150 13. Logarithmic 11. Exponential; f (x) = 1.2x 15. Logarithmic 5 4 3 2 1 −3 −2 −1−1 −2 −3 −4 −5 f(x) 21 3 4 5 x 10 10 x y 13 12 11 10 9 0 4 5 6 7 8 9 10 11 12 13 x 21 3 4 5 6 7 8 9 10 x −5 −4 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 63. x = y 101 _ 11 ≈ 9.2 6 5 4 3 2 1 −2 −1 −2 42 6 8 10 12 14 x 65. About $27,710.24 y 35,000 30,000 25,000 20,000 15,000 10,000 5,000 −1 0 −1 (20, 27710.24) f(x) = 6500e0.0725x 2 4 6 8 10 12 14 16 18 20 22 24 x 17. P(t) 1,000 900 800 700 600 500 400 300 200 100 −20 −18 −16 −14 −12 −4−6−8−10 −2 −100 42 6 8 10 12 14 16 18 20 t ODD ANSWERS 19. About 1.4 years 23. Four half-lives; 8.18 minutes 21. About 7.3 years __ __ 3 _ 2 10 3M 2 _ 25. M = 3 27. Let y = b x for some non-negative real
number b such that b ≠ 1. Then, ln (y) = ln (b x) ln (y) = x ln (b) e ln(y) = e xln(b) y = e xln(b) log  S _  S0 M = log  S _  S0 2 =  S _  S0 S010 3M 2 = S 29. A = 125e (−0.3567t); A ≈ 43mg 33. f (t) = 250e −0.00914t; half-life: about 76 minutes 35. r ≈ − 0.0667; hourly decay rate: about 6.67% 37. f (t) = 1350 e 0.034657359t; after 3 hours; P (180) ≈ 691,200 39. f (t) = 256 e (0.068110t); doubling time: about 10 minutes 41. About 88minutes 43. T(t) = 90 e (−0.008377t) + 75, where t is in minutes 45. About 113 minutes 47. log10x = 1.5; x ≈ 31.623 49. MMS Magnitude: ≈ 5.82 31. About 60 days 51. N(3) ≈ 71 53. C Section 4.8 1. Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic 3. Regression analysis is model best describes populations. the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu. 5. The y-intercept on the graph of a logistic equation corresponds to the initial population for the population model. 7. C 13. p ≈ 2.67 19. About 6.8 months. 21. 11. P (0) = 22; 175 15. y-intercept: (0, 15) 17. 4 koi 9
. B y 23. About 38 wolves 25. About 8.7 years 27. f (x) = 776.682 (1.426)x 29. y C-17 43. f (10) ≈ 2.3 45. When f (x) = 8, x ≈ 0.82 47. f (x) = 25.081 __ 1 + 3.182e−0.545x 49. About 25 41. y 10 51. y 140 130 120 110 100 90 80 70 60 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 53. 0 y 140 130 120 110 100 90 80 70 60 50 40 30 20 10 10 11 12 13 14 15 16 17 18 x x 55. When f (x) = 68, x ≈ 4.9 g (x) = 4.035510; the regression curves are symmetrical about y = x, so it appears that they are inverse functions. 57. f (x) = 1.034341(1.281204) x; 59. f −1(x) = c ln (a) − ln  _ − 1  x __ b 600 550 500 450 400 350 300 250 200 150 100 50 5 10 15 20 x 0 y 31. 600 550 500 450 400 350 300 250 200 150 100,000 6,000 5,000 4,000 3,000 2,000 1,000 0 1 2 3 4 5 6 7 x 33. f (x) = 731.92e −0.3038x 35. When f (x) = 250, x ≈ 3.6 37. y = 5.063 + 1.934 log(x) 39. y 10 Chapter 4 Review exercises 5. $42,888.18 1. Exponential decay; the growth factor, 0.825, is between 0 and 1. 3. y = 0.25(3) x 7. Continuous decay; the growth rate is negative y 9. Domain: all real numbers; range: all real numbers strictly greater than zero; y-intercept: (0, 3.5) 10 1−2−3−4−5−6−7−8 −1 1 2 3 4 x ODD ANSWERS 19. x = ln(1000) _______ + 5 ln(16) __ 3 ≈ 2.497 21
. a = ln (4) + 8 ________ 10 23. No solution 25. x = ln(9) — 27. x = ± 3 √ 3 ____ 2 29. f (t) = 112e−0.019792t; half-life: about 35 days 31. T(t) = 36 e−0.025131t + 35; T(60) ≈ 43° F 33. Logarithmic C-18 11. g(x) = 7(6.5)−x; y-intercept: (0, 7); domain: all real numbers; range: all real numbers greater than 0. 2 _ 15. loga b = − 5 21. ln(e−0.8648) = −0.8648 23. 19. log(0.000001) = −6 13. 17x = 4,913 17. x = 4 g(x) 1 2 3 4 5 x 1 −1−2−3−4−5 −1 −2 −3 −4 −5 −6 −7 −8 25. Domain: x > −5; vertical asymptote: x = −5; end behavior: as x → −5+, f (x) → −∞ and as x → ∞, f (x) → ∞ 27. log 8(65xy) z 29. ln  _ xy  31. logy(12) 33. ln(2) + ln(b) + ln(b + 1) − ln(b − 1) __ 2 35. log 7  v 3w 6 _ — 3 u √  5 _ 37. x = 39. x = −3 41. No solution 43. No solution 3 45. x = ln(11) 51. About 5.45 years 47. a = e4 − 3 53. f −1(x) = 9 __ 49. x = ± 5 24x − 1 — 3 √ 55. f (t) = 300(0.83)t ; f (24) ≈ 3.43 g 57. About 45 minutes y 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 59. About 8.5 days 61. Exponential y 100 90 80 70 60 50 40 30 20 10 10 11 x
y 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 63. y = 4(0.2)x; y = 4e–1.609438x 65. About 7.2 days 67. Logarithmic y = 16.68718 − 9.71860ln(x 10 11 x 35. Exponential; y = 15.10062(1.24621)x y 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 10 x 37. Logistic; y = 18.41659 __ 1 + 7.54644 e−0.68375x y 20 19 18 17 16 15 14 13 12 11 10.5 1 1.5 2 2.5 3 3.5 4 x Chapter 4 practice test 1. About 13 dolphins 5. y-intercept: (0, 5) 3. $1,947 y f(−x) = 5(0.5)−x f(x) = 5(0.5)− 49 7. 8.5a = 614.125 9. x = 11. ln(0.716) ≈ − 0.334 13. Domain: x < 3; vertical asymptote: x = 3; end behavior: as x → 3−, f (x) → −∞ and as x → −∞, f (x) → ∞ −1−2−3−4−5 −1 1 2 3 4 5 x 15. log t(12) 17. 3ln(y) + 2ln(z) + ln(x − 4 ChapteR 5 Section 5.1 1. Terminal side Vertex Initial side 3. Whether the angle is positive or negative determines the direction. A positive angle is drawn in the counterclockwise direction, and a negative angle is drawn in the clockwise direction. ODD ANSWERS 5. Linear speed is a measurement found by calculating distance of an arc compared to time. Angular speed is a measurement found by calculating the angle of an arc compared to time. 7. 9. Section 5.2 1. The unit circle is a circle of radius 1 centered at the origin. 3. Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, t, formed by the terminal side of the angle t and the horizontal axis. C-19 11
. 13. 15. 17. 240° 19. 4π ___ 3 21. 2π ___ 3 35. −3π radians ≈ 12.72 cm2 27. 20° 23. 7π ___ ≈ 11.00 in2 2 81π ____ 25. 20 π __ 29. 60° 31. −75° 33. radians 2 radians 39. 37. π radians 25π ___ 9 43. 47. 104.7198 cm2 5π ___ 6 5.02π _____ 3 41. 21π ___ 10 49. 0.7697 in2 51. 250° 53. 320° ≈ 6.60 meters 45. ≈ 5.26 miles ≈ 8.73 centimeters 55. 4π ___ 3 57. 8π ___ 9 59. 1320 rad/min 210.085 RPM 61. 7 in/s, 4.77 RPM, 28.65 deg/s 63. 1,809,557.37 mm/min = 30.16 m/s 67. 120° 73. 11.5 inches 69. 794 miles per hour 71. 2,234 miles per hour 65. 5.76 miles — √ 3 _ 2 1 __ 13. 2 5. The sine values are equal. 7. I 9. IV 11. 15 __ 33. 8 17. 0 19. −1 21. 23. 60°, cos (135° ) = − π __ 29. 3 25. 80° 27. 45° 39. 60°, Quadrant II, sin(120°) = 37. 45°, Quadrant II, sin(135°) = 35. 60°, Quadrant IV, sin(300°) = − 1 __, cos (300 __ 31 __ _, cos (120 __ ____ 41. 30°, Quadrant II, sin(150°) =, cos (150 7π 7π 1, cos , Quadrant III, sin  ___ __ ___ __  = −  = − _ 43 3π 2, cos  ___ ____  = − _ 4 2 2 — 1 2π √ 3, cos  __ ___ ____  = − 3 2 2 — √ 7π 2, cos  ___ ____  = 4 2 2π π, Quadrant II, sin  ___ __ �
� = 47. 3 3 π 7π, Quadrant IV, sin  ___ __  = − 49. 4 4 — √ 15 ____ 3  4 1 __, cos t = − 57. (−2.778, 15.757) 59. [−1, 1] 61. sin t = 2 — √ 3 ____ 2 3π π, Quadrant II, sin  ___ __  = 45. 4 4 55.  −10, 10 √ — √ 77 ____ 9 65. sin t = 63. sin t = − — √ 2 ____ 2, cos t = − 53. − 51. — 67. sin t = − — √ 2 ____ 2 — √ 2 ____ 2 — √ 2 ____ 2 — √ 2 ____ 2 71. sin t = −0.596, cos t = 0.803 75. sin t = − 1 _, cos t = 2, cos t = 79. sin t = 1, cos t = 0 85. −0.7071 — √ 6 _ 4 87. −0.1392 — √ 2 _ 4 95. 93. − — √ 3 _ 2 81. −0.1736 — √ 2 _ 4 69. sin t = 0, cos t = −1 — √ 3 ____ 2 1 __ 73. sin t =, cos t = 2 77. sin t = 0.761, cos t = −0.649 83. 0.9511 89. −0.7660 91. — √ 2 _ 4 97. 99. 0 101. (0, −1) — √ 3 ____ 2 1 __, cos t = − 2 103. 37.5 seconds, 97.5 seconds, 157.5 seconds, 217.5 seconds, 277.5 seconds, 337.5 seconds Section 5.3 π _ 1. Yes, when the reference angle is and the terminal side of the 4 π _ angle is in quadrants I or III. Thus, at x =, 4 5π _ 4, the sine and cosine values are equal. in for y in the Pythagorean Theorem x 2+ y take the negative solution. cotangent will repeat every π units. 3. Substitute the sine of the angle 2 = 1. Solve for x
and 5. The outputs of tangent and — — 3 7. 9. √ — 2 √ 3 _____ 3 19. − 2 √ √ 3 3 ____ ____ 3 3 — 27. −2 29. − √ 3 ____ 3 17. — 11. √ — 2 13. 1 15. 2 21. √ — 3 23. − √ — 2 25. −1 31. 2 33. 35. −2 37. −1 — √ 3 ____ 3 ODD ANSWERS C-20 — —, sec t = − 3, csc t = − 3 √ 39. sin t = − 2 √ 2 2 ____ ____, 4 3 2, cot t = tan t = 2 √ 41. sec t = 2, csc t = — — 2 √ 3 ____, 3 tan t = √ 3, cot t = — 43. − 45. 3.1 — √ 2 ____ 4 — √ 3 ____ 3 — √ 2 ____ 2 — √ 2 ____ 2 — √ 2 ____ 2 47. 1.4 49. sin t =, cos t =, tan t = 1, cot t = 1, sec t = √ 51. sin t = − — t = √ 3, cot t =, sec t = − 2, csc t = − — 2 — 2, csc t = √ — √ 3 ____ 3 57. 1.414 55. −2.414 53. −0.228 61. 1.556 63. sin(t) ≈ 0.79 65. csc(t) ≈ 1.16 69. Even 71. 75. 7.73 inches sin t _ cos t = tan t 73. 13.77 hours, period: 1000π — √ 1 3 __ ____, cos t = −, tan 2 2 — 2 √ 3 ____ 3 59. 1.540 67. Even Section 5.4 1. Opposite side Hypotenuse Adjacent side 5. For example, the sine of 3. The tangent of an angle is the ratio of the opposite side to the adjacent side. an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement. π _ 7. 6 π _ 9. 4
11. b = — 20 √ 3 ______ 3, c = — 40 √ 3 ______ 3 13. a = 10,000, c = 10,000.5 15. b = 17. — 5 √ 29 _____ 29 5 __ 19. 2 21. — √ 29 ____ 2 — 5 √ 3 ____ 3, c = 23. — 10 √ 3 _____ 3 — 5 √ 41 _____ 41 29. c = 14, b = 7 √ — 3 33. b = 9.9970, c = 12.2041 27. 5 __ 25. 4 — √ 41 ____ 4 31. a = 15, b = 15 35. a = 2.0838, b = 11.8177 39. a = 46.6790, b = 17.9184 43. 188.3159 49. 1,060.09 ft. 55. 368.7633 ft. 45. 200.6737 51. 27.372 ft. 37. a = 55.9808, c = 57.9555 41. a = 16.4662, c = 16.8341 47. 498.3471 ft. 53. 22.6506 ft. Chapter 5 Review exercises 1. 45° 3. − 7π ___ 5. 10.385 meters 7. 60° 9. 6 11. 13. 2π ___ 11 15. 1,036.73 miles per hour 17. 19. −1 23. − 31. √ — √ 2 ____ 2 — 2 25. [−1, 1] 33. 0.6 35. 37. Sine, cosecant, tangent, cotangent π __ 21. 4 29. √ — 2 — √ 3 ____ 2 27. 1 — √ 2 ____ 2 or − — √ 2 ____ 2 — √ 3 ____ 3 39. 47. a = 4, b = 4 41. 0 43. b = 8, c = 10 45. 49. 14.0954 ft. — 11 √ 157 _______ 157 Chapter 5 practice test 1. 150° 7. 3. 6.283 centimeters 5. 15° 2π _ 75 13. [–1, 1] 11. − 9. 3.351 feet per second, radians per second — √ 3 ____ 2 — 3 — √ 3 ____ 2 — √ 3 _
3 π __ 21 _____ __ 23. a = 2 2 17. 15. √ 19. ChapteR 6 Section 6.1 1. The sine and cosine functions have the property that f (x + P) = f (x) for a certain P. This means that the function 3. The absolute values repeat for every P units on the x-axis. value of the constant A (amplitude) increases the total range and the constant D (vertical shift) shifts the graph vertically. 5. At the point where the terminal side of t intersects the unit circle, you can determine that the sin t equals the y-coordinate of the point. 2 _ 7. Amplitude: ; period: 2π; 3 midline: y = 0; maximum: 2 _ occurs at x = 0; y = 3 minimum: y = − 2 _ occurs 3 at x = π; for one period, the graph starts at 0 and ends at 2π. f(x) 9. Amplitude: 4; period: 2π; midline: y = 0; maximum: π _ y = 4 occurs at x = ; 2 minimum: y = −4 occurs at x = ; for one period, the graph starts at 0 and ends at 2π. f(x) 3π _ 2 1 2 3 1 3 −π 0 π− 3 1− 2 2− 3 −1 π 2 π 3π 2 2π x −2π 3π− 2 −π 4 3 2 1 0 π− 2 −1 −2 −3 −4 π 2 π 3π 2 2π x −2π − 3π 2 11. Amplitude: 1; period: π; midline: y = 0; maximum: y = 1 occurs at x = π; minimum: y = −1 occurs at π _ x = ; for one period, the 2 graph starts at 0 and ends at π. f(t) 1 13. Amplitude: 4; period: 2; midline: y = 0; maximum: y = 4 occurs at x = 0; minimum: y = −4 occurs at x = 1; for one period, the graph starts at 0 and ends at π. f(t) 4 3 2 1 −π − 3π 4 π− 0 π−2 4 −1 π 4 π 2 3π 4 π
t 0.5 1 1.5 2 t 0 −2−1.5−1−0.5 −1 −2 −3 −4 ODD ANSWERS π _ 15. Amplitude: 3; period: ; 4 midline: y = 5; maximum: y = 8 occurs at x = 0.12; minimum: y = 2 occurs at x = 0.516; horizontal shift: −4; vertical translation: 5; for one period, the graph starts at 0 π _. and ends at 16 π 8 3π 16 π 4 x π− 4 3π− 16 π− 8 0 π− 16 −1 −2 −3 2π _ ; 5 17. Amplitude: 5; period: midline: y = −2; maximum: y = 3 occurs at x = 0.08; minimum: y = −7 occurs at x = 0.71; phase shift: −4; vertical translation: −2; for one period, the graph starts at 2π _. 0 and ends at 5 y 2π− 3π−5 10 π 10 π 5 3π 10 2π 5 x 5 4 3 2 1 π− 5 0 π− 10 −1 −2 −3 −4 −5 −6 −7 19. Amplitude: 1; period: 2π ; midline: y = 1; maximum: y = 2 occurs at t = 2.09; minimum: y = 0 occurs at t = 5.24; phase shift: π _ − ; vertical translation: 1; for one period, the graph starts at 0 and 3 ends at 2π. f(t) 43. A linear function is added to a periodic sine function. The graph does not have an amplitude because as the linear function increases without bound the combined function h(x) = x + sin x will increase without bound as well. The graph is bounded between the graphs of y = x + 1 and y = x − 1 because sine oscillates between −1 and 1. 45. There is no amplitude because the function is not bounded. f (x) 5 4 3 2 1 C-21 h(t 3π 2 2π t −2π 3π− 0 π−2 −π 2 −1 −2 −3 −4 −5 −6 47. The graph is symmetric with respect to the y-axis and there is no amplitude because the function�
�s bounds decrease as \|x\| grows. There appears to be a horizontal asymptote at y = 0. π 2 π 3π 2π 2 x 3π−− 2π 2 −π −2 π −1 −2 −3 −4 −5 y 5 4 3 2 1 π 2π 3π 4π 5π x −4π 0 −5π −3π−2π−π −1 −2 −3 −4 −5 5π−−2π 4π−3 2π−3 −π 3 5 4 3 2 1 0 π− 3 −1 −2 −3 −4 −5 −3π 8π− 7π−3 3 π 3 2π 3 π 4π 3 5π 3 2π 7π 3 8π 3 3π t Section 6.2 21. Amplitude: 1; period: 4π; midline: y = 0; maximum: y = 1 occurs at t = 11.52; minimum: y = −1 occurs at t = 5.24; phase shift: − 10π _ ; vertical shift: 0; for 3 one period, the graph starts at 0 and ends at 4π. f(t) 6 5 4 3 2 1 −4π −2π 0 −1 −2 −3 −4 −5 −6 1. Since y = csc x is the reciprocal function of y = sin x, you can plot the reciprocal of the coordinates on the graph of y = sin x to obtain the y-coordinates of y = csc x. The x-intercepts of the graph y = sin x are the vertical asymptotes for the graph of y = csc x. 3. Answers will vary. Using the unit circle, one can show that tan (x + π) = tan x. 5. The period is the same: 2π 7. IV 9. III 11. Period: 8; horizontal shift: 1 unit to the left 17. −cot x cos x − sin x 13. 1.5 15. 5 2π 4π 6π 8π t 19. Stretching factor: 2; π _ period: ; asymptotes: 4 π 1  _ _ + πk  + 8, where x = 4 2 k is an integer f (x) 8 6 4 2 25. Amplitude: 2, midline: y =
3; 2π x  + 3 _ 5 23. Amplitude: 2, midline: y = −3; period: 4; equation: π f (x) = 2sin  _ x  −3 2 period: 5; equation: f (x) = −2cos  27. Amplitude: 4, midline: y = 0; period: 2; equation: π f (x) = −4cos  π  x − _   2 period: 2; equation: f (x) = 2cos(πx) + 1 π π 33. sin  _ _  = 1 35. 2 2 with respect to the origin. 31. 0, π 37. f (x) = sin x is symmetric 29. Amplitude: 2, midline: y = 1; π _ 39., 3 5π _ 3 41. Maximum: 1 at x = 0; minimum: −1 at x = π x π 16 π 8 3π 16 π 4 3π− π− 4 16 π− π− 16 8 −2 −4 −6 −8 21. Stretching factor: 6; period: 6; asymptotes: x = 3k, where k is an integer 23. Stretching factor: 1; period: π; asymptotes: x = πk, where k is an integer m(x) 16 12 8 4 0 −4 −8 −12 −16 −6 −3 3 6 x p(x) 8 6 4 2 π 4 π 2 3π 4 π x −p 3π− 4 0 π− π− 4 2 −4 −6 −8 ODD ANSWERS C-22 25. Stretching factor: 1; period: π; asymptotes: π _ x = + πk, where k is an 4 integer f(x) 16 12 8 4 π 4 π 2 3π 4 π 5π 4 x −π 3π− 4 0 2 π− π− 4 −8 −12 −16 27. Stretching factor: 2; period: 2π; asymptotes: x = πk, where k is an integer π 37. y = �
� tan 39. f (x) = csc(2x) 41. f (x) = csc(4x) 43. f (x) = 2csc x 1 _ 45. f (x) = tan(100πx) 2 f (x) 47. 8 6 4 2 5π− 12 π 2 π 3π 2 2π x π− 4 π− π− 6 3 0 π− 12 −2 −4 −6 −8 π 12 π 6 π 4 π 3 5π 12 x 8 6 4 2 f(x) 16 12 8 4 −2π − 3π 2 −π 0 π− 2 −8 −12 −16 π 4 π 2 3π 4 π x 49. f(x) −π 8 6 4 2 3π− 4 2 0 π− π− 4 −2 −4 −6 −8 period: ; asymptotes: x = π 2π _ _ 5 10 where k is an integer k, π 4 π 2 3π 4 π x −π 3π− 4 2 0 π− π− 4 −2 −4 −6 −8 29. Stretching factor: 4; 31. Stretching factor: 7; period: ; asymptotes: x = π 2π _ _ k, 3 6 where k is an integer f (x) 16 12 8 4 π 6 π 3 π 2 2π 3 x 2π 4 −8 −12 −16 f(x) 51. 14 7 0 π− 10 −7 π 10 π 5 3π 10 2π 5 x 2π− 3π−5 π−10 5 −14 33. Stretching factor: 2; period: 2π ; asymptotes: x = − π _ + πk, 4 where k is an integer −2000π −1500π −1000π f (x500π 0 −1 −2 −3 −4 −5 −6 −7 −8 500π 1000π 1500π 2000π x 3π 2 − 5π 4 − −π − 3π 4 f(x1 −2 −3 −4 −5 −6 −7 −8 −9 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 x 9π 4 − −2π
− 7π 4 53. y 4 3 2 1 π 2 π 3π 2 5π 2π 2 x 5π− 2 −2π 3π− 2 −π 0 π− 2 −1 −2 −3 −4 ; period: 2π ; asymptotes: x = π 7 _ _ + πk, 35. Stretching factor: 5 4 where k is an integer f (x) π 55. a. f(x) 7π −4 5π −4 − 7 6 5 4 3 2 1 3π −4 0 π 4 −1 −2 −3 −4 −5 −6 −7 π 4 3π 4 5π 4 7π 4 9π 4 x 16 14 12 10 − 2 π− 3 π− −2 6 −4 −6 −8 c. x = − π π _ _ and x = ; the distance 2 2 grows without bound as ∣ x ∣ π _ approaches –i.e., at right angles 2 to the line representing due north, the boat would be so far away, the fisherman could not see it d. 3; when x = − π _, the boat is 3 km away 3 π _ e. 1.73; when x =, the boat is about 6 1.73 km away f. 1.5 km; when x = 0 ODD ANSWERS 3. Amplitude: 3; period: is 2π; midline: y = 0; no asymptotes C-23 f (x) 4 3 2 1 13π− −4π 3 11π− 3 10π− 3 −3π 8π− 3 7π− 3 −2π 5π− 3 4π− 3 −π 2π− 3 π− −1 3 −2 −3 −4 π 3 2π 3 π 4π 3 5π 3 2π 7π 3 8π 3 3π 10π 3 11π 3 x 5. Amplitude: 3; period: is 2π; midline: y = −4; no asymptotes f(x) π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 −2π 9π 4 x 7π− 4 3π− 2 5π− 4 −π 3π− 4 1 π− π− 4 −1 2 −2 −3 −
4 −5 −6 −7 7. Amplitude: 6; period: is 2π _ ; 3 midline: y = −1; no asymptotes f(x1 −2 −3 −4 −5 −6 −7 2π π − − 9 3 π 9 2π 9 π 3 4π 9 5π 9 2π 3 x − 2π 3 5π −9 4π 9 − 9. Stretching factor: none; period: π; midline: y = −4; π _ asymptotes: x = + πk, 2 where k is an integer f(x) 10 8 6 4 2 x π 2 π −π π− 2 −2 −4 −6 −8 −10 11. Stretching factor: 3; π _ ; midline: y = −2; period: 4 π π _ _ + asymptotes: x = k, 4 8 where k is an integer f(x) 13. Amplitude: none; period: 2π; no phase shift; asymptotes: π _ x = k, where k is an odd integer 2 f(x) 5 4 3 2 1 0 π− 2 −1 −2 −3 −4 −5 −2π 3π− 2 −π π 2 π 3π 2 2π x π 16 π 8 3π 16 π 4 x 10 16 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 3π π − −4 − 16 π 8 57. a. h(x) = 2 tan  b. f (x) π _ 120 x  100 90 80 70 60 50 40 30 20 10 0 −5 −10 −20 5 10 15 20 25 30 35 40 45 50 55 60 x d. As x approaches c. h(0) = 0: after 0 seconds, the rocket is 0 mi above the ground; h(30) = 2: after 30 seconds, the rockets is 2 mi high; 60 seconds, the values of h(x) grow increasingly large. As x → 60 the model breaks down, since it assumes that the angle of elevation continues to increase with x. In fact, the angle is bounded at 90 degrees. Section 6.3 1. The function y = sin x is one-to-one on
 − π π  ; thus, this _ _, 2 2 interval is the range of the inverse function of y = sin x, f (x) = sin−1 x. The function y = cos x is one-to-one on [0, π]; thus, this interval is the range of the inverse function of π _ y = cos x, f (x) = cos−1 x. 3. is the radian measure of an 6 π angle between − π _ _ 5. In order for whose sine is 0.5. and 2 2 any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval  − π π  so that it is one-to-one and possesses an _ _, 2 2 7. True. The angle, θ1 that equals arccos (−x), inverse. x > 0, will be a second quadrant angle with reference angle, θ2, where θ2 equals arccos x, x > 0. Since θ2 is the reference angle for θ1, θ2 = π − θ1 and arccos(−x) = π − arccos x 3π _ 17. 1.98 11. 4 21. 1.41 29. −0.71 radians 9. − π _ 6 19. 0.93 27. 0.71 radians 4 _ 35. 33. 5 13. − π π _ _ 15. 3 3 23. 0.56 radians 25. 0 31. − π _ radians 4 — x 2 − 1 √ _ x 39. 41. 5 _ 13 37. x + 0.5 __ − __ − x 2 + 2x √ 2x + 1 √ _ x + 1 — 45. 43. 49. Domain: [−1, 1]; range: [0, π] y — 2x + 1 √ _ x 47. t 51. x = 0 53. 0.395 radians 55. 1.11 radians 57. 1.25 radians 59. 0.405 radians 61. No. The angle the ladder makes
with the horizontal is 60 degrees. π −1 1 x −π Chapter 6 Review exercises 1. Amplitude: 3; period: is 2π; midline: y = 3; no asymptotes f (x) 6 5 4 3 2 1 π− 2 −2 −3 −4 −5 −6 −2π 3π− 2 −π x π 2 π 3π 2 2π ODD ANSWERS C-24 15. Amplitude: none; period: no phase shift; asymptotes: π _ x = k, where k is an integer 5 f(x) 2π _ ; 5 17. Amplitude: none; period: 4π ; no phase shift; asymptotes: x = 2π k, where k is an integer 5. Amplitude: 1; period: 2π ; midline: y = 1 7π− 3 −2π 5π− 3 4π− 3 f(x) 2 1.5 1 0.5 −π π− 2π− 3 3 −0.5 −1 −1.5 −2 π 3 2π 3 π 4π 3 5π 3 x f(x) 5 4 3 2 1 π 10 π 5 3π 10 x π 2π 3π 4π x −4π 0 −3π −π −2π −1 −2 −3 −4 −5 7. Amplitude: 3; period: 6π ; midline: y = 0 f (x) 3π 10 − π 5 − − 10 8 6 4 2 π 10 0 −2 −4 −6 −8 −10 21. Amplitude: 8,000; 19. Largest: 20,000; smallest: 4,000 period: 10; phase shift: 0 23. In 2007, the predicted population is 4,413. In 2010, the population will be 11,924. π _ 33. 3 π _ 31. 4 27. 10 seconds 25. 5 in. 35. No solution π _ 29. 6 12 _ 37. 5 39. The graphs are not symmetrical with respect to the line y = x. They are symmetrical with respect to the y-axis 3π 2 2π x −2π 3π− 2 −π 0 π− 2 −1 −2 −3 −4 −5 5π− 6 2π− 3 π
− 2 π− 3 41. The graphs appear to be identical. y 1 0.8 0.6 0.4 0.2 −0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1 −0.6 0.2 0.6 1 x −1 −0.6 y 1 0.8 0.6 0.4 0.2 0 −0.2 −0.2 −0.4 −0.6 −0.8 −1 0.2 0.6 1 x 9π− 4 −2π 7π− 4 3π− 2 5π− 4 −π 3π− 4 4 3 2 1 7π− −3π 2 5π− 2 −2π 3π− 2 −π π − 2 −1 −2 −3 −4 π 2 π 3π 2 2π 5π 2 3π 7π 2 4π 9π 2 5π 2 11π 6π 2 13π 7π 2 15π 8π 2 17π x 9. Amplitude: none; period: π; midline: y = 0; asymptotes: + πk, where k is some x = 2π _ 3 integer 11. Amplitude: none; period: midline: y = 0; asymptotes: π _ x = k, where k is some integer 3 2π _ ; 3 f (x) f(x) 8 6 4 2 π− 6 −2 −4 −6 −8 π 6 π 3 π 2 2π 3 5π 6 π 7π 6 x 2π− 3 π− 2 π− 3 20 15 10 5 π− 6 −5 −10 −15 −20 π 6 π 3 π 2 2π 3 x 13. Amplitude: none; period: 2π; midline: y = −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 x f (x) 6 5 4 3 2 1 π− π− 4 −1 2 −2 −3 −4 −5 −6 −7 −8 −9 −10 15. Amplitude; 2; period: 2; midline: y = 0; f (x) = 2sin(π(x − 1)) 17. Amplitude; 1; period: 12; phase shift:
−6; midline: y = −3 19. D(t) = 68 − 12sin  π _ 21. Period: ; horizontal 6 shift: −7 23. f (x) = sec(πx); period: 2; phase shift: 0 25. 4 π _ 12 x  Chapter 6 practice test 1. Amplitude: 0.5; period: 2π ; midline: y = 0 3. Amplitude: 5; period: 2π ; midline: y = 0 y y 0.5 0.25 0 π− −π 2 −0.25 −0.5 −2π 3π− 2 π 2 π 3π 2 2π x −2π 3π− 2 −π 6 5 4 3 2 1 0 π− 2 −1 −2 −3 −4 −5 −6 27. The views are different because the period of the wave is Over a bigger domain, there will be more cycles of the graph. n(x) n(x) 1 _. 25 π 2 π 3π 2 2π x 0.02 0.01 −0.2 −0.01 −0.02 0.2 0.4 0.6 0.8 1 1.2 1.4 0.02 0.01 x −0.2 −0.01 −0.02 0.2 0.4 0.6 0.8 1 x ODD ANSWERS 3 _ 29. 5 2π π π π 5π ,  , , π ,  _ _ _ _ _ 31 33. f (x) = 2 cos  12  x) 5 4 3 2 1 π 24 π 12 π 8 π 6 5π 12 π 4 x π− 4 5π− 24 π− 6 π− π− 8 12 π− 24 −1 −2 −3 −4 −5 3π _, 2 4π ,  _ 3 5π ,  _ 3 7π _, 6 35. This graph is periodic with a period of 2π. 11π _ 6, 2π  y 4 3.5 3 2.5 2 1.5 1 0.5 π 2 π 3π 2 2
π x −2π 3π− 2 −π π− 2 −0.5 −1 −1.5 −2 π _ 37. 3 π _ 39. 2 x + 1 _ x 41. √ — 1 − (1 − 2x)2 1 — 43. _ 1 + x 4 √ 49. 0.07 radians 45. csc t = 47. False ChapteR 7 Section 7.1 1. All three functions, F, G, and H, are even. This is because F(−x) = sin(−x)sin(−x) = (−sin x)(−sin x) = sin 2 x = F(x), G(−x) = cos(−x)cos(−x) = cos x cos x = cos 2 x = G(x) and H(−x) = tan(−x)tan(−x) = (−tan x)(−tan x) = tan 2 x = H(x). 1 _ 3. When cos t = 0, then sec t =, which is undefined. 0 9. csc t 13. sec 2 x 7. sec x 5. sin x 15. sin 2 x + 1 17. 21. tan x 11. −1 1 _ cot x 19. 1 _ sin x + 1 27. ± — 1 − sin 2 x √ ___________ sin x 23. −4sec x tan x 25. ± √ _________ 1 _____ cot 2 x 29. Answers will vary. Sample proof: cos x − cos 3x = cos x(1 − cos 2 x) = cos x sin 2 x 31. Answers will vary. Sample proof: 1 + sin 2 x _ = cos 2 x = sec 2 x + tan 2 x sin 2 x _ cos 2 x 1 _ cos 2 x + = tan 2 x + 1 + tan 2 x = 1 + 2tan 2 x 33. Answers will vary. Sample proof: cos 2 x − tan 2 x = 1 − sin 2 x −(sec 2 x − 1) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x 37. False 35. False 39. Proved with negative and Pythagorean identities 41. True 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3(sin
2 θ + cos 2 θ)+ cos 2 θ = 3 + cos 2 θ Section 7.2 1. The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those π _ − x. Then angles measures x, the second angle measures 2 π − x . The same holds for the other cofunction sin x = cos  _ 2 identities. The key is that the angles are complementary. 3. sin(−x) = −sin x, so sin x is odd. cos(−x) = cos(0 − x ) = cos x, so cos x is even. C-25 9. −2 − √ — 3 7. — — 6 − √ √ 2 _________ 4 — 3 √ _ 2 sin x 1 _ 13. − cos x − 2 x  19. tan  ___ 10 2 − √ 6 √ _________ 4 23. — — y 25. sin x 1 0.8 0.6 0.4 0.2 −π π− 2 −0.4 −0.6 −0.8 −1 −2π 3π− 2 π 2 π 3π 2 2π x 5. — — 2 + √ √ 6 _________ 4 — 2 √ _ 2 11. − sin x − cos x — 2 √ _ 2 15. csc θ 17. cot x 21. sin(a − b = 15 — —  2 √ 2 _ 3 cos(a + b 15 = — —  2 √ 2 _ 3 π − x  27. cot  _ 6 −2π 5π− 3 4π− 3 −π 2π− 3 π + x  29. cot  _ 4 y 10 8 6 4 2 π− 3 −2 −4 −6 −8 −10 π 3 3π 3 π 4π 3 5π 3 2π x y 10 8 6 4 2 π− 2 π− 4 −4 −6 −8 −10 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x −2π 7π− 4 3π− 2 5π− 4 −π 3π− 4 31. (sin x +
cos x) — √ 2 _ 2 y 1 0.8 0.6 0.4 0.2 π− π− 4 2 −0.4 −0.6 −0.8 −1 −2π 7π− 4 3π− 2 5π− 4 −π 3π− 4 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 33. They are the same. g(x) = sin(9x) − cos(3x) sin(6x) 35. They are the different, try 37. They are the same. 39. They are the different, try g(θ) = 41. They are different, try g(x) = 2tan θ _ 1 − tan 2 θ tan x − tan(2x) __ 1 + tan x tan(2x), or 0.9659 43. − — 3 − 1 √ _______ or − 0.2588 2 2 √ — π  = 47. tan  45. — 2 √ 2 π  tan x + tan  _ 4 _________________ π  1 − tan x tan  _ 4 tan x + 1 tan x + 1 _ __ = 1 − tan x 1 − tan x(1) = ODD ANSWERS C-26 49. cos(a + b) _ = cos a cos b cos a cos b _ − cos a cos b sin a sin b _ cos a cos b = 1 − tan a tan b 51. cos (x + h) − cos x __ = h cos x(cos h − 1) − sin x sin h ___ h cos x cos h − sin x sin h − cos x ___ h cos h − 1 _ = cos x h − sin x = sin h _ h 53. True expand the right hand side. 55. True. Note that sin(α + β) = sin(π − γ) and Section 7.3 1. Use the Pythagorean identities and isolate the squared term. 3. 1 − cos x sin x ________ ________, 1 + cos x sin x — 1 − cos x and √ √ — 1 + cos x, respectively., multiplying the top and bottom by 5. a. — 3 √ 7 _ 32 c. — 3 √ 7 _ 31 7.
a. — √ 1 3 __ ____ c. − √ b. − 2 2 — 3 b. 31 _ 32 — 5 2 √ _ 5 9. cos θ = −, sin θ = sec θ = −, cot θ = − − √ __ 3 2 15. — 5 √ _ 5 1 _, tan θ = −, csc θ = √ 2 — 5, π  11. 2sin  __ 2 — — 2 − √ √ __ 2 13. 2 17. 2 + √ — 3 19. −1 − √ — 2 21. a. 23. a. b. − 3 __ c. − 2 — 2 √ 13 _ 13 — 3 √ 13 _ 13 — 6 √ √ 10 _ _ 4 4 — 13 13 2 √ 3 √ 2 _ _ _ 29. cos(74°) 31. cos(18x) 33. 3sin(10x),, 3 13 13 — √ 15 _ 3 119 _ 169 120 _ 169 120 _ 119 25. b. c., −, − — — 27. 35. −2sin(−x)cos(−x) = −2(−sin(x)cos(x)) = sin(2x) 37. tan 2(θ ) = sin(2θ) __ 1 + cos(2θ) 2sin(θ )cos(θ ) __ 2cos 2(θ ) 1 + cos(12x) __ 2 2sin(θ )cos(θ ) __ 1 + cos 2(θ ) − sin 2(θ ) tan 2(θ ) = sin(θ ) _ cos(θ ) 3 + cos(12x) − 4cos(6x) ___ 8 39. 41. tan 2(θ ) = tan 2(θ ) = cot(θ )tan 2(θ ) = tan(θ ) 43. 2 + cos(2x) − 2cos(4x) − cos(6x) ____ 32 45. 3 + cos(4x) − 4cos(2x) ___ 3 + cos(4x) + 4cos(2x) 47. 1 − cos(4x) __ 8 49. 3 + cos(4x) − 4cos(2x
) ___ 4(cos(2x) + 1) 53. 4sin x cos x (cos 2 x − sin 2 x) 51. (1 + cos(4x)) sin x __ 2 55. 2tan x _ = 1 + tan 2 x 2sin x _ cos x _ sin 2 x _ 1 + cos 2 x = 2sin x _ cos x __ cos 2 x + sin 2 x __ cos 2 x = cos 2 x _ 1 2sin x _ cos x. sin(2x) _ cos(2x) = tan(2x) 57. 2sin x cos x __ = 2cos 2 x − 1 = 2sin x cos x = sin(2x) 59. sin(x + 2x) = sin x cos(2x) + sin(2x)cos x = sin x(cos 2 x − sin 2 x) + 2sin x cos x cos x = sin x cos 2 x − sin 3 x + 2sin x cos 2 x = 3sin x cos 2 x − sin 3 x 61. 1 + cos(2t) __ sin(2t) − cos t = 1 + 2cos 2 t − 1 __ 2sin t cos t − cos t 2cos 2 t __ cos t(2sin t − 1) = = 2cos t _ 2sin t − 1 63. (cos 2 (4x) − sin 2(4x) − sin(8x))(cos 2(4x) − sin 2(4x) + sin(8x)) = (cos(8x) − sin(8x))(cos(8x) + sin(8x)) = cos 2(8x) − sin 2(8x) = cos(16x) Section 7.4 1. Substitute α into cosine and β into sine and evaluate. 3. Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a sin(3x) + sin x __ cos x product are easier to solve. For example: = 1. When converting the numerator to a product the equation — — — 3 − 1) 5. 8(cos(5x) − cos(27x)) 2sin(2x)cos x __ = 1. cos x becomes: 1 _ 7. sin(2x) + sin(8x) (cos(6x) − cos(4x)) 9. 2
13. 2cos(7x) 11. 2cos(5t)cos t 15. 2cos(6x)cos(3x) 1 1 1 3 − 2) ( √ 3 ) (1 + √ ( √ _ _ _ 17. 19. 21. 4 4 4 1 _ 23. cos(80°) − cos(120°) (sin(221°) + sin(205°)) 25. 2 — 27. √ 2 cos(31°) 31. 2sin(−1.5°)cos(0.5°) 33. 2sin(7x) − 2sin x = 2sin(4x + 3x) − 2sin(4x − 3x) = = 2(sin(4x)cos(3x) + sin(3x)cos(4x)) − 2(sin(4x)cos(3x) − sin(3x)cos(4x)) = 2sin(4x)cos(3x) + 2sin(3x)cos(4x)) − 2sin(4x)cos(3x) + 2sin(3x)cos(4x)) = 4sin(3x)cos(4x) 35. sin x + sin(3x) = 2sin  −2x ____ 2 = 2(2sin x cos x)cos x = 4sin x cos 2 x  = 2sin(2x)cos x 29. 2cos(66.5°)sin(34.5°)  cos  4x ___ 2 37. 2tan x cos(3x) = = 2sin x cos(3x) __ cos x 2(0.5 (sin (4x) − sin (2x))) ___ cos x 1 _ cos x (sin(4x) − sin(2x)) = = sec x (sin(4x) − sin(2x)) 45. It is an identity. 39. 2cos(35°)cos(23°), 1.5081 41. −2sin(33°)sin(11°), − 0.2078 1 __ (cos(99°) − cos(71°)), −0.2410 43. 2 47. It is not an identity, but 2cos 3 x is. 49. tan(3t) 55. Start with cos x
+ cos y. Make a substitution and let x = α + β and let y = α − β, so cos x + cos y becomes cos(α + β) + cos(α − β) = 53. −sin(14x) 51. 2cos(2x) = cos α cos β − sin α sin β + cos α cosβ + sin α sin β = 2cos α cos β Since x = α + β and y = α − β, we can solve for α and β in terms of x and y and substitute in for 2cos α cos β and get x − y 2cos  . _ 2 x + y _ 2  cos  ODD ANSWERS 57. cos(3x) + cos x __ cos(3x) − cos x = 2cos(2x) cos x __ −2sin(2x) sin x = −cot(2x) cot x 59. cos(2y) − cos(4y) __ = sin(2y) + sin(4y) −2sin(3y) sin(−y) __ 2sin(3y) cos y = 2sin(3y) sin(y) __ 2sin(3y) cos y = tan y 61. cos x − cos(3x) = −2sin(2x)sin(−x) = 2(2sin x cos x)sin x π − t  = 63. tan  _ 4 = 4sin2 x cos x π  − tan t tan  _ 4 __ = π  tan(t) 1 + tan  _ 4 1 − tan t ________ 1 + tan t Section 7.5 1. There will not always be solutions to trigonometric function equations. For a basic example, cos(x) = −5. 3. If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the
equation has no solution. 7π _ 4 29π _ 18 19. 7. 37 ___ 6 17. 15. 29. 0, π π _ 9., 4 5π _ 4 π _, 18 3π _, 4 7π _, 6 13π ____, 12, 5π ____, 4 25π _, 18 29 ___, 6 3π π _ _ 11., 4 4 17π _, 18 25 ___, 6 π 5π 2π _ _ _ 5., 4 3 3 π 7π 11π 13π 5π _ _ _ _ _ 13.,,, 4 4 18 18 6 3π 19π 11π 5π 17 21π 13 5 1 ___ ____ ____ ___ ___ ____ ___ __ __ 21.,,,,,,,, 12 12 12 12 6 12 6 6 6 π π 5π 5π 5π 3π π __ __ ___ ___ ___ ___ __ 23. 0, 25. 27., π,, π 11π 7π 1 1 ___ ____ __ __ 31. π − sin−1  − , , 2π + sin− 2π π 1 1 1  ,  sin−1   ,  sin−1   ,  sin− − + 33. 3 10 3 3 10 3 10 3 9 9 9 5π 4π 1 1 1  ,  ,  sin−1   sin−1   ,  sin− 10 3 10 3 10 2 2 2 , π + sin−1  , π − sin−1  37. θ = sin−1  , _ _ _ 35. 0 3 3 3 4π π 5π π 3π 2 2π − sin−1  ___ __ ___ __ ___  _ 41. 0, 39., π,,, 3 3 3 6 6 2 43. There are no solutions. 1 45. cos−1  __  1 − √ 3 1 47. tan−1 �
� _  √ 2 1 π + tan−1  _  √ 2 49. There are no solutions. 2π 3π 5π 7π 4π π _ _ _ _ _ _ 53. 0, 55 3π ,, π − sin−1  , 57. sin−  , 2π − cos−1  − 59. cos− 5π , , 2π − cos−1  −, cos−1  − _ _ _ _ 61. 4 4 3 3 3 3 2 2  , 2π − cos−1  , cos−1  − 63. cos−1  , 2π − cos−  , 2π − cos−1  __  1 − √ 3 1 29 − 5  , π + tan−1  _  − √ 2 1 29 − 5  , 2π + tan−1  _  − √ 2 29 − 5   51. There are no solutions. 7   29 − -27 5π _ 3 π _ 65. 0,, π, 2 π 1 1 , , 2π − cos−1  −, cos−1  − _ _ _ 67. 4 4 3 71. π + tan−1(−2), 3π _ 2 69. There are no solutions. 3 3 __ π + tan−1  − , 2π + tan−1(−2), 2π + tan−1  −  _ 2 2 73. 2πk + 0.2734, 2πk + 2.8682 75. πk − 0.3277 77. 0.6694, 1.8287, 3.8110, 4.9703 79. 1.0472, 3.1416, 5.2360 3π 1 1 81. 0.5326, 1.7648, 3.6742, 4.9064 83. sin−1
 , , π − sin−1  _ _ _ 4 4 2 π 3π π _ _ _ 85. 87. There are no solutions. 89. 0,, π,, 2 2 2 3π _ 2 91. There are no solutions. 93. 7.2° 95. 5.7° 97. 82.4° 99. 31.0° 103. 59.0° 101. 88.7° 105. 36.9° Section 7.6 1. Physical behavior should be periodic, or cyclical. cumulative rainfall is always increasing, a sinusoidal function would not be ideal here. 7. 5sin(2x) + 2 π x  − 1 5. y = −3cos  _ 6 11. y = tan  9. y = 4−6cos  xπ ___  2 3. Since xπ ___  8 13. Answers will vary. Sample answer: This function could model the average monthly temperatures for a city in the northern hemisphere. 15. Answers will vary. Sample answer: This function could model the population of an invasive fish species in thousands over the next 80 years. 8 16 24 32 40 48 56 64 72 80 x 60 40 20 y 0 y 100 80 60 40 20 0 2 4 6 8 10 12 x 21. 2:49 23. 4:30 27. Floods: April 16 to 17. 75 °F 19. 8 a.m. 25. From June 15 to November 16 July 15. Drought: October 16 to January 15. 1 _ 29. Amplitude: 8, period:, frequency: 3 Hz 3 1 _ 31. Amplitude: 4, period: 4, frequency: Hz 4 t  + 800 + 160 π 33. P(t) = −19cos  _ _ t 12 6 π 35. P(t) = −33cos  _ t  + 900(1.07)t 6 37. D(t) = 10(0.85)t cos(36πt) 39. D(t) = 17(0.9145)tcos(28πt) 41. 6 years 45. Spring 2 comes to rest first after 8.0 seconds. π 49. y = 6(4)x + 5sin  _ 
2 x π 53. y = 3 (2)xcos  _ x  + 1 2 π 51. y = 4(−2)x + 8sin  _ x  2 47. 234.3 miles, at 72.2° 43. 15.4 seconds Chapter 7 Review exercises 1. sin−1  7π ___, 6 3. — — — 3 3 3 √ √ √ , 2π − sin−1  , π + sin−1  , π − sin−1  _ _ _ 3 3 3 11π 1 1  7. 1 , π − sin−1  5. sin−1  ____ _ _ 4 4 6 — √ 2 ____ 13. 2 9. Yes — 3 11. −2 − √ — 3 √  _ 3 ODD ANSWERS C-28 15. cos(4x) − cos(3x)cosx = cos(2x + 2x) − cos(x + 2x)cos x = cos(2x)cos(2x) − sin(2x)sin(2x) − cos x cos(2x)cos x + sin x sin(2x) cos x = (cos 2 x − sin 2 x) 2 − 4cos 2 x sin 2 x − cos 2 x(cos 2 x − sin 2 x) + sin x (2)sin x cos x cos x = (cos 2 x − sin 2 x)2 − 4cos 2 x sin 2 x − cos 2 x(cos 2 x − sin 2 x) + 2 sin 2 x cos 2 x = cos 4 x − 2cos 2 x sin 2 x + sin 4 x − 4cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x (sin 2 x + cos 2 x) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x 5 x  17. tan  _ 8 24 _ 25 — 10 27. cot x cos(2x) = cot x (1 − 2sin
2 x, 10 — √ 3 _ 3 21. − 7 _, 25 25. 19. 23. √ 24 _ 7 —, − 2  — (2)sin 2 x = cot x − cos x _ sin x = −2sin x cos x + cot x = −sin (2x) + cot x 29. 10sin x − 5sin(3x) + sin(5x) ___ 8(cos(2x) + 1) 1 __ (sin(6x) + sin(12x)) 35. 2, π 39. 43. 7π ___ 4 3π ___, 4 5π π ___ __ 41. 0,, 6 6 47. 0.2527, 2.8889, 4.7124 51. 3sin  1 _ 57. Amplitude: 3, period: 2, frequency: Hz 2 59. C(t) = 20sin(2πt) + 100(1.4427)t xπ  − 2 _ 2 53. 71.6 33. − 31. 2 9 13 x  x  cos  37. 2sin  _ _ 2 2 3π ___ 2 45. No solution 49. 1.3694, 1.9106, 4.3726, 4.9137 π t  55. P(t) = 950 − 450sin  _ 6 Chapter 7 practice test 1. 1 3 11. 2cos(3x)cos(5x) — — 3 2 − √ 5. − √ 1 13. x = cos−1  __  5 7. 0, π π _ 9., 2,− 3,− 4 3 __ __ __ 15. 4 5 5 3π _ 2 17. tan3 x−tan x sec2 x = tan x (tan2 x − sec2 x) = tan x (tan2 x − (1 + tan2 x)) = tan x (tan2 x − 1 − tan2 x) = −tan x = tan(−x) = tan(−x) 19. sin(2x) _ − sin x cos(2x) _ cos x = 2sin xcos x _ sin x − 2 cos2 x−1 _ cos x 1 _ cos x = 2cos x−2cos x + = 1 _
cos x = sec x = sec x 1 __ 60, frequency: 60 Hz 1 __ 21. Amplitude:, period 4 23. Amplitude: 8, fast period: 1 ___ 500, slow frequency: 10 Hz period: 1 __ 10 cos (4πt), 31 second, fast frequency: 500 Hz, slow 25. D(t) = 20 (0.9086)t ChapteR 8 Section 8.1 11. b ≈ 3.78 7. β = 72°, a ≈ 12.0, 5. A triangle with two 9. γ = 20°, b ≈ 4.5, c ≈ 1.6 1. The altitude extends from any vertex to the opposite side or to 3. When the line containing the opposite side at a 90° angle. the known values are the side opposite the missing angle and another side and its opposite angle. given sides and a non-included angle. b ≈ 19.9 13. c ≈ 13.70 15. One triangle, α ≈ 50.3°, β ≈ 16.7°, a ≈ 26.7 17. Two triangles, γ ≈ 54.3°, β ≈ 90.7°, b ≈ 20.9 or γ′ ≈ 125.7°, 19. Two triangles, β ≈ 75.7°, γ ≈ 61.3°, β′ ≈ 19.3°, b′ ≈ 6.9 b ≈ 9.9 or β′ ≈ 18.3°, γ′ ≈ 118.7°, b′ ≈ 3.2 21. Two triangles, α ≈ 143.2°, β ≈ 26.8°, a ≈ 17.3 or α′ ≈ 16.8°, β′ ≈ 153.2°, a′ ≈ 8.3 23. No triangle possible 27. 8.6 37. 29.7° 43. A ≈ 39.4, C ≈ 47.6, BC ≈ 20.7 49. 430.2 51. 10.1 57. L ≈ 49.7, N ≈ 56.1, LN ≈ 5.8 61. The distance from the satellite to station A is approximately 1,716 miles. The satellite is approximately 1,706 miles above the ground. 63. 2.6 ft 65. 5
.6 km 67. 371 ft 69. 5,936 ft 71. 24.1 ft 73. 19,056 ft 2 75. 445,624 square miles 77. 8.65 ft 2 25. A ≈ 47.8° or A′ ≈ 132.2° 33. 12.2 47. 42.0 55. AB ≈ 2.8 39. x = 76.9°or x = 103.1° 53. AD ≈ 13.8 59. 51.4 feet 41. 110.6° 29. 370.9 45. 57.1 31. 12.3 35. 16.0 Section 8.2 9. 34.7 7. 11.3 11. 26.7 17. 95.5° 19. 26.9° 13. 257.4 21. B ≈ 45.9°, 1. Two sides and the angle opposite the missing side. 3. s is the semi-perimeter, which is half the perimeter of the triangle. 5. The Law of Cosines must be used for any oblique (non-right) triangle. 15. Not possible C ≈ 99.1°, a ≈ 6.4 25. A ≈ 37.8°, B ≈ 43.8°, C ≈ 98.4° 27. 177.56 in 2 29. 0.04 m 2 39. 70.7° 31. 0.91 yd 2 35. 29.1 49. 1.41 41. 77.4° 51. 0.14 59. 7.62 65. 99.9 ft 67. 37.3 miles 69. 2,371 miles 71. 43. 25.0 53. 18.3 61. 85.1 23. A ≈ 20.6°, B ≈ 38.4°, c ≈ 51.1 37. 0.5 47. 43.52 63. 24.0 km 55. 48.98 45. 9.3 33. 3.0 57. 528 6 5 BO 9.18 PH 150.28 x DC 20.78 73. 599.8 miles 75. 65.4 cm 2 77. 468 ft 2 Section 8.3 1. For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in
the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations. 3. Determine θ for the point, then move r units from the pole to plot the point. If r is negative, move r units from the pole in the opposite direction but along the same angle. The point is a distance of r away from the origin at an angle of θ from the π  has a positive angle but a 5. The point  −3, __ 2 and π __ negative radius and is plotted by moving to an angle of 2 polar axis. ODD ANSWERS then moving 3 units in the negative direction. This places the π  has a point 3 units down the negative y-axis. The point  3, − _ 2 negative angle and a positive radius and is plotted by first moving π __ to an angle of − and then moving 3 units down, which is the 2 positive direction for a negative angle. The point is also 3 units down the negative y-axis. 7. (−5, 0) 9.  − 13.  √ — 34, 5.253  — 3 3 √ _____ 2 3  __, − 2 15.  8 √ π —  __ 2, 4 11. (2 √ — 5, 0.464) 17. r = 4csc θ 19. r = 21. r = 3cos θ 23. r = 3sin θ _ cos (2θ) — sin θ _ 2cos 4 θ 3 √ 9sin θ _ cos 2 θ 25. r = 29. x 2 + y 2 = 4x or 27. r = √ __________ 1 __________ 9cos θ sin θ y 2 ___ 4 = 1; circle (x − 2)2 _______ 4 + 61. r = 3cosθ 63. x 2 + y 2 = 16 y C-29 1 2 3 4 5 652 −2 −4 −6 −2 −2 −4 −6 −8 −4−6−8 2 4 6 8 x 67. x 2 + (y + 5)2 = 25 y 10 8 6 4 2 2 4 6 8 10 x −10 −4−6−8 −2 −2 −4 −6 −8 −10 31.
3y + x = 6; line 33. y = 3; line 37. x 2 + y 2 = 4; circle 3π  41.  3, ___ 4 45. 43. (5, π) 47. 39. x − 5y = 3; line 35. xy = 4; hyperbola −4−6−8 69. (1.618, −1.176) 71. (10.630, 131.186°) 73. (2, 3.14) or (2, π) 75. A vertical line with a units left of the y-axis. 77. A horizontal line with a units below the x-axis. 79. 81. p u = 4 49. 51. 83. 53. 55. r = 6 __ 5cos θ − sin θ 2 4 6 8 10 Section 8.4 57. r = 2sin θ 59. r = 2 _ cos. Symmetry with respect to the polar axis is similar to symmetry about the x-axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the is similar to symmetry about the y-axis. π __ line θ = 2 3. Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, π __ limaçon, lemniscate, etc., then plot points at θ = 0,, π and 2 5. The shape of the polar graph is and sketch the graph. determined by whether or not it includes a sine, a cosine, and 7. Symmetric with respect to constants in the equation. 9. Symmetric with respect to the polar axis, the polar axis π __ symmetric with respect to the line θ = 2, symmetric with respect 3π ___, 2 to the pole 15. Symmetric with respect to the pole 11. No symmetry 13. No symmetry ODD ANSWERS C-30 17. Circle (θ from 0 to 2π) 19. Cardioid (θ from 0 to 2π) 41. Archimedes’ spiral 43. Archimedes’ spiral (θ from 0 to 3π) (θ from 0 to 3π 10 10 20 30 21. Cardioid 23. One-loop/dimpled
limaçon 45. (θ from 0 to 8) 47. (θ from −π to π) (θ from 0 to 2π) (θ from 0 to 2π 11 0.5 1 1.5 2 0.5 1 1.5 2 25. One-loop/dimpled limaçon (θ from 0 to 2π) 27. Inner loop/two-loop limaçon 49. (θ from 0 to 2π) 51. (θ from 0 to 3π) 1 2 3 2 4 6 8 10 1 3 5 7 9 1 2 3 4 53. (θ from 0 to 2π) 29. Inner loop/two-loop limaçon 31. Inner loop/two-loop limaçon (θ from 0 to 2π) (θ from 0 to 2π 12 1 3 5 7 9 11 33. Lemniscate (θ from −π to π) 35. Lemniscate (θ from −π to π) 1 2 3 4 1 2 3 37. Rose curve 39. Rose curve (θ from 0 to 2π) 1 2 3 4 1 2 3 4 55. They are both spirals, but not quite the same. 57. Both graphs are curves with 2 loops. The equation with a coefficient of θ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to 4π to get a better picture. 59. When the width of the domain is increased, more petals of 61. The graphs are three-petal, rose the flower are visible. curves. The larger the coefficient, the greater the curve’s distance from the pole. coefficient, the tighter the spiral. 5π 3 π 3  ,  67.  ___ __ __ __,, 3 2 3 2 — — 4 4 ,  71.  √ √ π 8 8 ____ __ ____,, 4 2 2 63. The graphs are spirals. The smaller the 5π π  ,  4, 65.  4, ___ __ 3 3 3π π , (0, 2π) , (0, π),  0,
69.  0, ___ __ 2 2  and at θ = 7π 5π ___ ___ 4 4 since r is squared 3π ___, 4 Section 8.5 — −1 1. a is the real part, b is the imaginary part, and i = √ 3. Polar form converts the real and imaginary part of the complex number in polar form using x = r cos θ and y = r sin θ. 5. zn = rn(cos (nθ) + i sin (nθ)) It is used to simplify polar form when a number has been raised to a power. — 9. √ 14.45 — 5 cis(333.4°) 13. 4 √ 11. √ 7. 5 √ — 38 2 — — 17. π  15. 2cis  __ 6 — 3 √ 3 ____ 21. −1.5 − i 2 7 √ 3 _____ 2 7 __ + i 2 — 3 cis(198°) 23. 4 √ 19. −2 √ — 3 − 2i 3 __ 25. cis(180°) 4 ODD ANSWERS 31. 5cis(80°) — 29. 7cis(70°) 37. 9cis(240°) 17π  ____ 24 35. 125cis(135°) 3 cis  27. 5 √ π 33. 5cis   __ 3 3π 39. cis   ___ 41. 3cis(80°), 3cis(200°), 3cis(320°) 4 8π 2π , 2 4 cis  , 2 4 cis  3 3 ___ ___ 43. 2 √ √ 9 9 15π 7π  2 cis  , 2 √ 2 cis  ____ ___ 8 8 49. 14π  ____ 9 4 cis  45. 2 √ Imaginary Imaginary 47 Real −2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 −2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 1 2 3 4 5 6 Real 51. Imaginary 53.
Imaginary 2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 1 2 3 4 5 6 Real −2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 1 2 3 4 5 6 Real 55. Plot of 1 − 4i in the complex plane (1 along the real axis, −4 along the imaginary axis). 59. −2 + 3.46i 61. −4.33 − 2.50i 57. 3.61e−0.59i Section 8.6 1 − y _____ 5 7. y = −2 + 2x 3. Choose one equation to 1. A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, x = f (t) and y = f (t). solve for t, substitute into the other equation and simplify. 5. Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions. ______ x − 1 x 9. y = 3 √  or y = 1 − 5ln  _____ _ 11. x = 2e 2 2 y − 3 y 13. x = 4log   y 3  15. x =  __ _ _ 17  +  19.   __ __ __ x 23. y = x 2 + 2x + 1 = 1 21  25. y =  _____ 2 31.  x(t) = t y(t) = 2sin t + 1 35.  39.  33.  37.  x(t) = 4 cos t y(t) = 6 sin t x(t) = 4 + 2t y(t) = 1 − 3t x(t) = √ y(t) = √ 41.  27. y = −3x + 14 x(t) = √ y(t) = t 10 cos t — 10 sin t 29. y = x + 3 3 − 2 ; Ellipse ; Circle t + 2t — 2 — x(t) = − 1 + 4t y(t) = − 2t 43. Yes,
at t = 2 45. 1 t x −3 y 1 2 0 7 3 5 17 C-31 47. Answers may vary: x(t) = t − 1 y(t) = t 2 x(t) = t + 1 y(t) = (t + 2)2 49. Answers may vary:,  and   x(t) = t y(t) = t 2 − 4t + 4 and  x(t) = t + 2 y(t) = t 2 Section 8.7 3. The arrows show the orientation, the direction 1. Plotting points with the orientation arrow and a graphing calculator of motion according to increasing values of t. parametric equations show the different vertical and horizontal motions over time. 5. The 7. y 9. 10 2−3−4−5−6 −1 −1 −2 11. y 6 5 4 3 2 1 −2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 −2−3−4−5−6 − 13. −2.5 −1.5 y 6 5 4 3 2 1 −1 −2 −3 −4 −5 −6 y 6 5 4 3 2 1 −0.5 −1 −2 −3 −4 −5 −t from −1 to 5) 0.5 1.5 2.5 x 1 2 3 4 5 6 x 15. y 17. 6 5 4 3 2 1 −3−4 −1−2 −1 −2 −3 −4 −5 −t from −5 to 5) −3−4−5−6 −1−2 −1 −2 −3 −4 −5 −6 19. y 0 −0.5 −1.0 −1.5 −2.0 −2.5 −3.0 0.5 1.0 1.5 (t from 0 to 360) 2.0 x 2.5 3.0 21. y 30 25 20 15 10 5 (t from 0 to 360) −2−3−4−5−6 −1 1 2 3 4 5 6 x ODD ANSWERS C-32 23. 253 −2 −2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 27. y 29. 100 80 60 40 20 −1000 −20 −
40 −60 −80 −100 −120 −140 1000 2000 3000 4000 5000 x −15 −10 (t from −1 to 5) y 3.0 2.5 2.0 1.5 1.0 0.5 −1 −0.5 −1.0 −1.5 −2.0 −2.5 −3.0 y 30 25 20 15 10 5 −5 −5 −10 −15 −20 −25 −30 31. y 33. y (t from 0 to 1000) −1000 −800 −600 −400 35 30 25 20 15 10 5 −200 −5 −10 −15 −20 −25 x 200 35. −3 −2 −1 y 3 2 1 −1 −2 −3 1 x 2 (t from −π to 0) 3 51. y 1.5 1 0.5 (t from 0 to 2π) 1 2 3 x −1.5 −1 −0.5 0.5 11.5 x −0.5 −1 −15. 57. (cont.) y 3 2.5 2 1.5 1 0.5 −2−3−4−5 −1 −0.5 −1 −1.5 −2 −2.5 −3 59. (t from 0 to 2π 20 15 10 5 (t from −4π to 6π) (t from −5 to 5) 5 10 15 x 6 5 4 3 2 1 −2−3−4−5−6−7 −1 −1 −2 −3 −4 −5 −6 π (t from − to ) 2 x π 2 1 2 3 4 5 37. y 6 4 2 −2 −1.5 −1 −0.5 0.5 1 x −5 −10 (t from−π to 0) 59. (cont.) −6 −4 −2 2 4 6 x −2 −4 −6 y 1 0.5 (t from −4π to 6π) −10 −5 5 10 15 20 x 53. a = 4, b = 3, c = 6, d = 1 55. a = 4, b = 2, c = 3, d = 3 57. y (t from 0 to 2π) x 0.5 1 6 5 4 3 2 1 −0.5 −1 −2 −3 −4 −5 −6 (t from 0 to 2π2 −1.5 −
1 y 3 2.5 2 1.5 1 0.5 −2−3−4−5 −1 −0.5 −1 −1.5 −2 −2.5 −3 y 1 0.5 (t from −4π to 6π) −10 −5 5 10 15 20 x −0.5 −1 −1.5 −2 2 61. The y-intercept changes. x x 63. y(x) = −16  + 20  _ _   15 15 65.  x(t) = 64cos(52°) y(t) = −16t2 + 64tsin(52°) 67. Approximately 3.2 seconds 69. 1.6 seconds 39. There will be 100 back-and-forth motions. opposite of the x(t) equation. 43. The parabola opens up. 41. Take the 45.  x(t) = 5 cos t y(t) = 5 sin t 47. y 1.5 1 0.5 49. y 1.5 1 0.5 (t from 0 to 2π) 71. y 73. (t from 0 to 2π) y 6 4 2 (t from 0 to 2π) −1.5 −1 −0.5 0.5 11.5 x −1.5 −1 −0.5 0.5 11.5 x −10 −5 −0.5 −1 −1.5 −0.5 −1 −1.5 5 10 x −6 −4 −2 2 4 6 x −2 −4 −6 −0.5 −1 −1.5 −2 12 10 8 6 4 2 −2 −4 −6 −8 −10 −12 ODD ANSWERS Section 8.8 Chapter 8 Review exercises C-33 3. They are unit 1. Lowercase, bold letter, usually u, v, w vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1. 5. The first number always represents the coefficient of the i, and the second represents the j. 11. Equal 19. u + v = 〈−5, 5〉, u − v = 〈−1, 3〉, 2u − 3v = 〈0, 5〉 — 23.
− 2 √ 29 _ 29 229 15 √ _ j 229 — 29 i + 5 √ _ j 29 — — 10 10 7. 〈7, − 5〉 15. −7i − 3j 9. Not equal 17. −6i − 2j 229 2 √ _ 229 21. −10i − 4j 13. Equal 27. − 25. − i + — — 3. C = 120°, a = 23.1, c = 34.1 1. Not possible 5. Distance of the plane from point A: 2.2 km, elevation of the plane: 1.6 km 7. B = 71.0°, C = 55.0°, a = 12.8 9. 40.6 km 11. 13. (0, 2) 15. (9.8489, 203.96°) 17. r = 8 19. x 2 + y 2 = 7x 29. \|v\| = 7.810, θ = 39.806° 33. −6 35. −12 37. y 3v 31. \|v\| = 7.211, θ = 236.310° 21 23. Symmetric with respect to π __ the line θ = 2 v v1 2−3−4−5−6−7 −1 −1 −2 −3 −4 −5 −6 −7 x x x 25. (θ from 0 to 2π) 27. (θ from 0 to 2π) 39. 41. u + v u − v 2u u − v u + v 2u 43. 45. — — 2 j 49. v = −7i + 3j 2 i + 3 √ — 3 j 47. 〈4, 1〉 51. 3 √ 53. i − √ 55. a. 58.7; b. 12.5 57. x = 7.13 pounds, y = 3.63 pounds 59. x = 2.87 pounds, y = 4.10 pounds 61. 4.635 miles, 17.764° N of E 63. 17 miles, 10.071 miles 65. Distance: 2.868, Direction: 86.474° North of West, or 3.526° West of North 67. 4.924°, 659 km/hr 73. 21.801°, relative to the car’s forward direction 75
. Parallel: 16.28, perpendicular: 47.28 pounds 77. 19.35 pounds, 51.65° from the horizontal 79. 5.1583 pounds, 75.8° from the horizontal 69. 4.424° 71. (0.081, 8.602 __ 33. 2.3 + 1.9i 31. cis  − 29. 5  3 3π 4π 37. 3cis   39. 25cis   ___ ___ 2 3 43. π  35. 60cis  __ 2 3π , 5cis  41. 5cis  7π ___ ___  4 4 1 __ y = 1 45. x 2 + 2 Imaginary 6 5 4 3 2 1 47.  x(t) = −2 + 6t y(t) = 3 + 4t 1 2 3 4 5 6 Real −2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 49. y = −2x 5 (t from −1 to 1) 0.5 1.0 1.5 2.0 x y 20 −20 −40 −60 −80 −100 51. a. x(t) = (80 cos (40°))t y(t) = − 16t 2 + (80 sin (40°))t + 4  b. The ball is 14 feet high and 184 feet from where it was launched. c. 3.3 seconds 53. Not equal 10 3 √ _ 10 59. Magnitude: 3 √ 61. 16 55. 4i 10 √ _ j 10 — 2, Direction: 225° 57. − i, − — — ODD ANSWERS C-34 63. u − v u + v 3v Chapter 8 practice test 1. α = 67.1°, γ = 44.9°, a = 20.9 3. 1,712 miles 5.  1, √ 7. y = −3 (θ from 0 to 2π) 9. y 6 5 4 3 2 1 −2−3−4−5−6 −1 −1 −2 −3 −4 −5 − 11. √ — 106 + 5 __ 13. − 2 — 2 cis(198°) —
5 √ 3 _____ i 2 — 2 cis(18°), 2 √ 15. 4cis (21°) 19. y = 2(x − 1)2 17. 2 √ 21. y 6 5 4 3 2 1 23. −4i − 15j 25. — 2 √ 13 ______ 13 i + — 3 √ 13 ______ j 13 (θ from 0 to 2π) 1 2 3 4 5 6 x −2−3−4−5−6 −1 −1 −2 −3 −4 −5 −6 ChapteR 9 Section 9.1 1. No, you can either have zero, one, or infinitely many. Examine 3. This means there is no realistic break-even point. graphs. By the time the company produces one unit they are already making profit. y ), graphically, or by addition. 13. (−3, 1) 2) 15.  − 3, 0  _ 5 21. (6, −6) 5. You can solve by substitution (isolating x or 11. (−1, 72 132  19.  _ _ 5 5 25. No solutions exist. 7. Yes 9. Yes, 31. (−4, 4) 17. No solutions exist 1 1 ___ __  23.  −, 10 2 x + 3   29.  x, 2 1 __ __ 27  35.   33.  _ _ _ 37. ( x, 2(7x − 6)), 8 2 6 4 5 __ __ 39.  −  41. Consistent with one solution, 3 6 43. Consistent with one solution 45. Dependent with infinitely many solutions 47. (−3.08, 4.91) 49. (−1.52, 2.29) 51  55 EC − BF _ AE − BD DC − AF _, BD − AE 59. (1,250, 100,000) 63. 24,000 67. 56 men, 74 women 53.  − 57. They never turn a profit. 61. The numbers are 7.5 and 20.5. 65. 790 sophomores, 805 freshman 69. 10 gallons of 10% solution, 15 gallons of 60% solution 71. Swan Peak: $750,000, Riverside:
$350,000 the first account, $10,500 in the second account tops: 45, Low-tops: 15 more information. 73. $12,500 in 75. High- 77. Infinitely many solutions. We need — 3  Section 9.2 1. No, there can be only one, zero, or infinitely many solutions. 3. Not necessarily. There could be zero, one, or infinitely many solutions. For example, (0, 0, 0) is not a solution to the system below, but that does not mean that it has no solution. 2x + 3y − 6z = 1 −4x − 6y + 12z = −2 x + 2y + 5z = 10 5. Every system of equations can be solved graphically, by substitution, and by addition. However, systems of three equations become very complex to solve graphically so other methods are usually preferable. 7. No 9. Yes 11. (−1, 4, 2) 13.  − 85 ____, 107 312 ____, 107 191 ____  107 1, 0  15.  1, _ 2 17. (4, −6, 1) 19.  x, 65 − 16x _______, 27 23. No solutions exist 25. (0, 0, 0) 29. (7, 20, 16) 31. (−6, 2, 1) 17 ___ 13 28 + x 45 ______ ___ , −2  21.  −, 13 27 3 1 4 __ __ __  27. , −, − 7 7 7 33. (5, 12, 15)  39.  4 1 1 __ __ __,, 5 5 2 45. (1, 1, 1) 51. 24, 36, 48 49. (6, −1, 0) 37. (10, 10, 10) 35. (−5, −5, −5)  41.  4 2 1 __ __ __ 43. (2, 0, 0),, 5 5 2  47.  28 23 128 ____ ____ ____,, 557 557 557 53. 70 grandparents, 140 parents, 190 children 55. Your share was $19.95, Sarah’s share was $40, and your other roommate�
�s share was $22.05. more information. 61. The BMW was $49,636, the Jeep was $42,636, and adults the Toyota was $47,727. 63. $400,000 in the account that pays 3% interest, $500,000 in the account that pays 4% interest, and 65. The United $100,000 in the account that pays 2% interest. States consumed 26.3%, Japan 7.1%, and China 6.4% of the world’s 67. Saudi Arabia imported 16.8%, Canada imported oil. 15.1%, and Mexico 15.0% 18.6%, and mammals were 17.1% of endangered species 57. There are infinitely many solutions; we need 59. 500 students, 225 children, and 450 69. Birds were 19.3%, fish were Section 9.3 1. A nonlinear system could be representative of two circles that overlap and intersect in two locations, hence two solutions. A nonlinear system could be representative of a parabola and a circle, where the vertex of the parabola meets the circle and the branches 3. No. There also intersect the circle, hence three solutions. does not need to be a feasible region. Consider a system that is bounded by two parallel lines. One inequality represents the region above the upper line; the other represents the region below the lower line. In this case, no points in the plane are located in both regions; hence there is no feasible region. ODD ANSWERS 5. Choose any number between each solution and plug into C(x) and R(x). If C(x) < R(x), then there is profit. — — 3 √ 3 √ 2 2 _____ _____, 2 2 — √ 62 _____ 8 9.  − 13.  1 __ 4 — —  ,  √ √ 199 199 398 398 ____ ______ ____ ______,, __, 4 — 3 √ 2 _____ 2, − —  3 √ 2 _____ 2 —  √ 62 _____ 8 17. (0, 2), (1, 3) 1 __ 5 − 1), (1 − √ 2 — __________ 5 ) ,  √ 1 — __ ( √ 2 1 __ 5 − 1), (1 − √
2 — 5 7. (0, −3), (3, 0) 11. (−3, 0), (3, 0) 15.  − __________ 19.  − √ 1 — __ ( √ 2 )  21. (5, 0) 29. No solutions exist 23. (0, 0) 25. (3, 0) 27. No solutions exist 31.  − 33. (2, 0) — √ 2 ____, − 2 — ,  − √ 2 ____ 2 35. (− √ — √ 2 ____, 2 — ,  √ 2 ____ 2 7, −3), (− √ — — — √ 2 ____, − 2 7, 3), ( √ —  √ 2 ____ 2 — √ 2 ____, 2 — ,  √ 2 ____ 2 7, −3), ( √ — — 7, 3) ___________ 1 — _ ( √ 2 1 73 −5), _ (7 − √ 2 37.  − √ ___________ 1 — __ ( √ 2 1 __ 73 − 5), (7 − √ 2 — 73 )   √ — 73 )  39. y 10 8 6 4 2 −2−3−4−5 −1 −2 −4 −6 −8 −10 41. 45. 1 2 3 4 5 x y 10 2−3−4−5 −1 −1 −2 −3 −4 − 10 x 1 2 3 4 5 x −10 −4−6−8 −2 −2 −4 −6 −8 −10 y 5 4 3 2 1 −2−3−4−5 −1 −1 −2 −3 −4 −5 2 4 6 8 x −4−6−8 −2 −2 −4 −6 −8 −10 70 ____ 383 ____ 70 ____ 383, −2 √ ___ 49.  −2 √ ____ , 35 ___ 29 ___ ,  −2 √ 35, 2 √ ___ 29 ___ ____ ,  2 √ 35 70, −2 √ ___ ____ 29 383, 2 √
___  2 √ ____ 35  ___ 29 70 ____ 383 51. No solution exists 53. x = 0, y > 0 and 0 < x < 1, √ 57. 2–20 computers — 1 _ x x < y < 55. 12,288 43. 47. Section 9.4 C-35 1. No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, cannot be 1 _ x 2 + 1 7. − + + + − + + 15 19. 11. 9. 23. − 17. 13. 21. 25. − 1 _____ x + 5 5 _____ x − 8 4 ______ 4x − 1 3 _____ x − 2 5 _____ x − 2 9 _____ x + 2 5 ________ + 2(x + 3) 9 ________ + 5(x + 2) 1 _____ x − 2 5 _______ 2(x − 3) 11 ________ 5(x − 3) 2 _______ (x − 2) 2 1 _____ x − 7 decomposed because the denominator cannot be factored. 3. Graph both sides and ensure they are equal. 5. If we choose x = −1, then the B-term disappears, letting us immediately know that A = 3. We could alternatively plug in 5 _ x = − giving us a B-value of −2. 3 8 _____ x + 3 3 ______ 5x − 2 3 _____ x + 2 8 _____ x − 3 6 ______ 4x + 5 3 4 ________ __ + − 27. 2(x + 1) x 2 4 ___ __ + 29. x 2 x x + 1 _________ + x 2 + x + 3 2x − 1 __________ + x 2 + 6x + 1 2 _________ + x 2 −3x + 9 1 __ + 43. x x + 1 _____ x + 2 1 ___ 8x 3 ________ (4x + 5) 2 7 ________ 2(x + 1) 2 7 _________ 2(3x + 2) 2 4 − 3x __________ + x 2 + 3x + 8 1 _________ + x 2 + x + 1 1 ___________ + 4x 2 + 6x + 9 3 ______ 3x + 2 3 _____ x + 2 2 _____ x + 3 3 _____ x + 3 1
_____ x − 1 4 _____ x − 1 1 ___________ − x 2 + 3x + 25 4x __ x 2 − 6x + 36 2 _______ (x − 7) 2 x + 6 ______ x 2 + 1 1 _____ x + 6 41. − 49. 45. 33. 37. 47. 51. 39. 35. 31. + − + − − − + 1 ______ 2x − 3 4x + 3 _______ (x 2 + 1) 2 3x _____________ (x 2 + 3x + 25) 2 2x + 3 _______ (x + 2) 2 x ________ + 8(x 2 + 4) 9 ___ x 2 + 16 _____ x − 1 53. − − 55. − 57. − 16 ___ x 1 _____ x + 1 5 _____ x − 2 5 ___ 4x 59. − − 2 _______ + (x + 1) 2 3 _________ + 10(x + 2) 5 ________ + 2(x + 2) − 10 − x _________ 2(x 2 + 4) 2 7 _______ (x − 1) 2 5 _______ (x + 1)3 7 _____ x + 8 11 ________ + 2(x + 4) − 7 _________ 10(x − 8) 5 _______ 4(x − 4) Section 9.5 1. No, they must have the same dimensions. An example would include two matrices of different dimensions. One cannot add the following two matrices because the first is a 2×2 matrix and the  has no sum. second is a 2×3 matrix. 3. Yes, if the dimensions of A are m × n and the dimensions of B are n × m, both products will be defined 5. Not necessarily. To find AB, we multiply the first row of A by the first column of B to get the first entry of AB. To find BA, we multiply the first row of B by the first column of A to get the first entry of BA. Thus, if those are unequal, then the matrix multiplication does not commute. ODD ANSWERS C-36 7.  11 19 15 94 17 67  9.  −4 2  8 1 11. Undefined; dimensions do not match 13.  9 27 63 36 0 192  15.  −64 −12 −
28 −72 −360 −20 −12 −116  17.  21.  1,800 1,200 1,300 800 1,400 600 700 400 2,100 60 41 2 −16 120 −216   23.  19.  20 102  28 28 −68 24 136 −54 −12 64 −57 30 128  25. Undefined; dimensions do not match.  29.  −8 41 −3 40 −15 −14 4 27 42 27.  31.  33. Undefined; inner dimensions do not match. −350 1,050 350 350  −840 650 −530 330 360 250 −10 900 110  35.  39.  43.  47.  51.  55.  37.  1,400 700  −1,400 700 490,000 0 41.   0 490,000 45.   49.  53.  −4 29 21  −27 −3 1 1 −18 −9 −198 505 369 −72 126 91 2 24 −4.5 12 32 −9 −8 64 61    332,500 927,500 −227,500 87,500 −2 3 4  −7 9 −7 −3 −2 −2  −28 59 46 −4 16 7 0 1.6 9 −1   0.5 3 0.5 2 1 2 10 7 10  57.  , n even, , n odd. 59. Bn =    Section 9.6 1. Yes. For each row, the coefficients of the variables are written across the corresponding row, and a vertical bar is placed; then the constants are placed to the right of the vertical bar. 3. No, there are numerous correct methods of using row operations on a matrix. Two possible ways are the following: (1) Interchange rows 1 and 2. Then R 2 = R 2 − 9R 1. (2) R 2 = R 1 −9R 2. Then divide row 1 by 9. 5
. No. A matrix with 0 entries for an entire row would have either zero or infinitely many solutions. 7.  0 16 9 −1 \| 4 2  9.  1 5 8 12 3 0 3 4 9 \| 19 4 −7  11. −2x + 5y = 5 6x − 18y = 26 15. 4x + 5y − 2z = 12 y + 58z = 2 3x + 2y = 13 13. −x − 9y + 4z = 53 8x + 5y + 7z = 80 17. No solutions 19. (−1, −2) 8x + 7y − 3z = −5 21. (6, 7) 23. (3, 2) 25.   27.  x, 1 4 1 ___ __ __, 5 15 2 5 196 ____ ___  31. , − 33. (31, −42, 87) 39 13 15 15 18 ___ ___ ___ , 13 13 13 x __ 41.  x, −, −1  2 47. (1, 2, 3) (5x + 1)  29. (3, 4)  35.  9 1 21 __ ___ ___,, 8 20 40 3 1 y  39.  x, y, _ _ − x − 2 2 45. (8, 1, −2) 10 __, z  7 53. 860 red velvet, 1,340 chocolate 51. No solutions exist. 55. 4% for account 1, 6% for account 2 59. Banana was 3%, pumpkin was 7%, and rocky road was 2% 61. 100 almonds, 200 cashews, 600 pistachios 49.  −4z + 17 __ 7 43. (125, −25, 0) 57. $126 37. , 3z −, − Section 9.7 1. If A−1 is the inverse of A, then AA−1 = I, the identity matrix. Since A is also the inverse of A−1, A−1 A = I. You can also check by proving this for a 2 × 2 matrix. 3. No, because ad and bc are both 0, so ad − bc = 0, which requires us to divide by 0 in the �
�. The inverse is formula. found with the following calculation: 1 0 0 1 0 1 0 −1 1  __________ −1 0 0(0) − 1(1)  = I 1 0 5. Yes. Consider the matrix   =  . 9. AB = BA =   = I 13. 1 0 17. There is no inverse A−1 = 21. 15. 1 ___ 17. AB = BA =  11. AB = BA =  −2 7 1   ___ 9 3 69 −5 5 −3  20 −3 12 1 −1 4 18 60 −168  −56 −140 448 40 80 −280 2 __ 33.  −, − 3 39.  − 35 __ 34 229 _ , − 690 25.   31.  5 1 __ __, − 3 2 37. (5, 0, −1) 41.  568 ___ 345 13 _ 138 , − 23. 1 ____ 209    ___ −1 3 29 4  __ 19. 7 47 −57 69 10 19 −12 −24 38 −13  0.5 1.5 1 −0.5  27. (−5, 6) 29. (2, 0), −, −  35.  7, 1 1 11 __ __ ___ , 6 5 2 77 97 __ __  17 34 43.  −  8 37 ___ ___, 15 30 2 1 −1 −1  0 1 1 −1 0 −1 1 1 0 1 −1 1   45.  10 ____ 123  2 __, −1, 5 1 __ 47. 2 3 2 1 −7 18 −53 32 10 24 −36 21 9 −9 46 −16 −5  1 ___ 39 49. ODD ANSWERS 511 −1 −1 −1 −1 1  53. Infinite solutions 55. 50% oranges, 25% bananas, 20% apples 57. 10 straw hats, 50 beanies, 40 cowboy hats 59. Tom ate 6, Joe ate 3,
and Albert ate 3 61. 124 oranges, 10 lemons, 8 pomegranates Section 9.8 51. x − 3z = 7 y + 2z = − 5 with infinite solutions C-37 53.  −2 2 1 7 2 −8 5 0 19 −10 22 3 \|  55.  1 0 3 −1 4 0 0 1 2 \| 12 0 −7  59. No solutions exist 57. No solutions exist  63. No inverse exists 1 2 7  __ 61. 6 1 8 67. (−1, 0.2, 0.3) 1 __  77. ( x, 5x + 3) 71. 0 73. 6 75.  6, 2 69. 17% oranges, 34% bananas, 39% apples  79.  0, 0, − 1 __ 2 65. (−20, 40) 1. A determinant is the sum and products of the entries in the matrix, so you can always evaluate that product—even if it does end up being 0. 7. 7 9. −4 11. 0 19. 224 3. The inverse does not exist. 13. −7, 990.7 5. −2 15. 3 29. (2, 5) 37. (−1, 0, 3) 21. 15 23. −17.03 25. (1, 1) 1 __  31.  −1, − 3, 1, 2  39.  1 __ 2 45. 24 33. (15, 12) 41. (2, 1, 4) 17. −1  27.  1 1 __ __, 3 2 35. (1, 3, 2) 49. Yes; 18, 38 55. 120 children, 1,080 adult 47. 1 53. $7,000 in first account, $3,000 in 43. Infinite solutions 51. Yes; 33, 36, 37 second account yellow, 6 gal blue 61. Strawberries 18%, oranges 9%, kiwi 10% first movie, 230 for the second movie, 312 for the third movie 65. 20–29: 2,100, 30–39: 2,600, 40–49: 825 400 cranberries, 300 cashews 59. 13 green tomatoes, 17 red tomatoes 57. 4 gal 63. 100 for the 67.
300 almonds, Chapter 9 Review exercises 3. (−2, 3) 1. No 9. (300, 60) 5. (4, −1) 11. (10, −10, 10) 17.  x, 7. No solutions exist 13. No solutions exist 14x 8x ____ ___ , 5 5 23. No solution 25. No solution 19. 11, 17, 33 15. (−1, − 2, 3) 21. (2, − 3), (3, 2) 272−3−4−5 −1 −1 −2 −3 −4 −5 29. y 5 4 3 2 1 −2−3 −1 −1 −2 −3 −4 −5 1 2 3 4 5 6 7 x 31. − 10 33. 7 _ x + 5 − 15 _ (x + 5)2 35. 37. + −4x + 1 3 __ _ x − 5 x 2 + 5x + 25 −16 8  −4 −12 39.  43. Undefined; inner dimensions do not match x − 4 _ x 2 − 2 + 5x + 3 _ (x2 − 2)2 41. Undefined; dimensions do not match 45.  113 28 10 44 81 −41 84 98 −42  47.  −127 −74 176 −2 11 40 28 77 38  49. Undefined; inner dimensions do not match Chapter 9 practice test, − 3. No solutions exist 1. Yes 13x 16x ____ ____ 7.  x,  5 5 — — 17 ), (−2 √ 9. (−2 √ 2, − √ 112−3−4−5 −1 −1 −2 −3 −4 − — 17 ), (2 √ — 2, − √ — 17 ), (2 √ — 2, √ — 17 ) − 13. 17 51 −8 11 5 ______ 3x + 1 15.  17.  2x + 3 ________ (3x + 1)2  12 −20  −15 30 1 __ 19. − 8 14 −2 13 −2 3 −6 1 −5 12 21.  \| 25. (100, 90) 27.  140 −1 11 23. No solutions exist. , 0 
29. 32 or more cell phones per day 1 ____ 100 ChapteR 10 Section 10.1 1. An ellipse is the set of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant. 3. This special case would be a circle. 5. It is symmetric about the x-axis, y-axis, and the origin. 7. Yes; + x 2 ___ 32 = 1 y 2 ___ 22 9. Yes; x 2 ______ 2   1 __ 2 = 1; endpoints of major axis: (0, 7) and (0, −7); y 2 ______ 2   1 __ 3 = 1 + x 2 ___ 22 y 2 ___ 72 — — + 5 ), (0, −3 √ 11. endpoints of minor axis: (2, 0) and (−2, 0); foci: (0, 3 √ 13; endpoints of major axis: (1, 0) and (−1, 0); (1)2 + 2   1 __ 3  ; foci:  1 1 ,  0, − endpoints of minor axis:  0, _ _ 3 3 (x − 2)2  − _ = 1; endpoints of major + 72 axis: (9, 4), (−5, 4); endpoints of minor axis: (2, 9), (2, − 1); foci: (2 + 2 √ (y − 4)2 _ 52 ), (2 − 2 √ 15. 6, 4) — — — — ODD ANSWERS C-38 (x + 5)2 _ + 22 (y − 7)2 _ 32 = 1; endpoints of major axis: (−5, 10), 17. (−5, 4); endpoints of minor axis: (−3, 7), (−7, 7); foci: (−5, 7 + √ (x − 1)2 (−5, 7 − √ _ = 1; endpoints of + 32 major axis: (4, 4), (−2, 4); endpoints of minor axis: (1, 6), (1, 2); 5, 4) 21. foci: (1 + √ (y − 4)2 _ 22 5, 4), (1 − √
19 ), (x − 3) ), (3 − 3 √ (y − 5)); endpoints of = 1; — — 2 ); foci: (7, 5), (−1, 5) endpoints of major axis: (3 + 3 √ minor axis: (3, 5 + √ 2 ), (3, 5 − √ — — (x + 5)2 _______ 52 (y − 2)2 _______ 22 — — + 25. 21, 2) = 1; endpoints of major axis: (0, 2), (−10, 2); 23. endpoints of minor axis: (−5, 4), (−5, 0); foci: (−5 + √ (x + 3)2 (−5 − √ _______ 52 major axis (2, −4), (−8, −4); endpoints of minor axis (−3, −2), 21, −4), (−3 − √ (−3, − 6); foci: (−3 + √ 27. Foci: (−3, −1 + √ 11 ), (−3, −1 − √ 31. Foci: (−10, 30), (−10, −30) 21, 2), = 1; endpoints of 21, −4). 11 ) 29. Focus: (0, 0) (y + 4)2 _______ 22 + — — — — 33. Center: (0, 0); vertices: (4, 0), (−4, 0), (0, 3), (0, −3); 7, 0), (− √ foci: ( √ y 7, 0) — — 1 __ 35. Center (0, 0); vertices, ,  0, − 1 1 1 __ __ __  ;  −, − foci  0, 4 √ 4 √ 2 2 _____ _____ 63 63 5 4 3 2 1 −2−3−4−5 −1 −1 −2 −3 −4 −5 1 2 3 4 5 x y.3.2.1.1.2.3 x −.3 −.2 −.1 −.1 −.2 −.3 37. Center (−3, 3); vertices (0, 3), (−6, 3), (−3, 0), (−3, 6); focus: (−3, 3). Note that this ellipse is a circle
. The circle has only one focus, which coincides with the center. y 10 7.5 5 2.5 2.5 5 7.5 10 x −10 −7.5 −5 −2.5 −2.5 −5 −7.5 −10 41. Center: (−4, 5); vertices: (−2, 5), (−6, 5), (−4, 6), (−4, 4); 3, 5) foci: (−4 + √ y 3, 5), (−4 − √ — — 10 7.5 5 2.5 2.5 5 7.5 x −10 −5−7.5 −2.5 −2.5 −5 −7.5 39. Center: (1, 1); vertices: (5, 1), (−3, 1), (1, 3), (1, −1); foci: 3 ) (1, 1 + 4 √ 3 ), (12−3−4−5 −1 −1 −2 −3 −4 −5 43. Center: (−2, 1); vertices: (0, 1), (−4, 1), (−2, 5), (−2, −3); 3 ) foci: (−2, 1 + 2 √ 3 ), (−2, 1 − 2 √ — — y.75.5.25.25.5.75 x −.75 −.5 −.25 −.25 −.5 −.75 45. Center: (−2, − 2); vertices: (0, − 2), (−4, −2), (−2, 0), (−2, −4); focus: (−2, −2) y 2 1 x 1 2 −2−3−4−5 −1 −1 −2 −3 −4 −5 = 1 47. 49. 51. 53. 55. + y 2 x 2 ___ ___ 29 25 (x − 4)2 _______ 25 (x + 3)2 _______ 16 + y 2 x 2 ___ ___ 81 9 (x + 2)2 _______ 4 + + (y − 2)2 _______ 1 (y − 4)2 _______ 4 = 1 = 1 = 1 + (y − 2)2 _______ 9 = 1 — 5 π square units 59. Area = 2 √ x 2 ____ 4h 2 = 1, distance: 17.32 feet y 2 _ + 1 __ h 2 4

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