Section 7. 2 AN53 8. f x x 1 3 2 1( ) ( ) = + − 9. 2− 10. (0, 0) p 4 , 3 p 2 , 0 , 0 p 2 2 p 4 , 2 23 3p 4 , 2 3 3p 4 , 23 P 2 y 5 x P 11. 3 3 3 2 − 12. y 2 3x ( ) = 13. y x 3 cos 6 π( ) = 14. (a) f x x3 3 ( ) = − − ; m 3; 1,0, 0, 3 ( ) ( ) = − − − (22, 3) (21, 0) (0, 23) (1, 26) y 4 x 5 (b) f x x 1 6; 0, 5 2 ( ) ( ) ( ) = − − − , 6 1,0, 6 1,0 ( ) ( ) − + + (22, 3) (21.45, 0) (1, 26) (3.45, 0) y 4 x 5 (c) We have that y 3 = when x 2 = − and y 6 = − when x 1. = Both points satisfy y aex = . Therefore, for 2, 3 ( ) − we have ae 3 2 = − , which implies that a e3 2 = . But for 1, 6 ( ) − we have ae 6 1 − = , which implies that a e6 1 = − − . Therefore, there is no exponential function y aex = that contains 2, 3 ( ) − and 1, 6 ( ) − . 15. (a) f x x x x 1 6 2 3 5 ( ) ( )( )( ) = + − − (3, 0) (5, 0) (0, 5) (22, 0) y 10 x 6 (b) R x x x x x 2 3 5 3 2 ( ) ( )( )( ) ( ) = − + − − − (5, 0) (0, 5) (3, 0) (22, 0) y 10 x 4 CHAPTER 7 Analytic Trigonometry 7. 1 Assess Your Understanding (page 483) 5. = x y sin 6. π ≤ ≤ x 0 7. T 8. T 9. T 10. d 11. 0 13. π − 2 15. 0 17. π 4 19. π 3 21. π5 6 23. π 4 25. π − 6 27. 0. 10 29. 1.37 31. 0.51 33. −0.38 35. −0.12 37. 1.08 39. π4 5 41. π − 3 8 43. π − 8 45. π 3 47. π − 5 49. π 3 51. π 4 53. Not defined 55. 1 4 57. 4 59. Not defined 61. π 63. ( ) = − − − f x x sin 2 5 1 1 Range of = f Domain of [ ] = − −f 3, 7 1 Range of π π = ⎡ − ⎣ ⎢ ⎤ ⎦ ⎥ −f 2 , 2 1 65. ( ) ( ) = − − − f x x 1 3 cos 2 1 1 Range of = f Domain of [ ] = − −f 2, 2 1 Range of π = ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ −f 0, 3 1 67. ( ) ( ) = − + − − − f x x tan 3 1 1 1 Range of = f Domain of f , 1 ( ) = −∞ ∞ − Range of f 1 2 , 2 1 1 π π ( ) = − − − − 69. f x x 1 2 sin 3 1 1 1( ) ( ) = − ⎡ ⎢ ⎣ ⎤ ⎦ ⎥ − − Range of f Domain = of f 3, 3 1 [ ] = − − Range of f 1 2 4 , 1 2 4 1 π π = ⎡ − − − + ⎣ ⎢ ⎤ ⎦ ⎥ − 71. 2 2 { } 73. 1 4 { } − 75. 3 { } 77. 1 { } − 79. (a) 13.92 h or 13 h, 55 min (b) 12 h (c) 13.85 h or 13 h, 51 min 81. (a) 13.3 h or 13 h, 18 min (b) 12 h (c) 13.26 h or 13 h, 15 min 83. (a) 12 h (b) 12 h (c) 12 h (d) It is 12 h. 85. 3.35 min 87. (a) 3 π square units (b) 5 12 π square units 89. 4250 mi 91. 7 4 , 7 4 { } − 93. 2 3 , 2 ⎡ − ⎣ ⎢ ⎤ ⎦ ⎥ 94. The graph passes the horizontal-line test. y 5 x 5 y x (0, 1) (1, 2) 1, 1 2 95. f x x log 1 1 2 ( ) ( ) = − − 96. x x x x 2 1 3 3 1 2 2 3 2 2 ( ) ( ) ( ) + + − − + − − 97. ln3 4 { } 98. 20 mph 99. 3 4 100. sin 7 25 , tan 7 24 , sec 25 24 , csc 25 7 , cot 24 7 θ θ θ θ θ = = = = = 101. Quadrant II 102. 12 4 3 π − 7. 2 Assess Your Understanding (page 490) 4. x y sec ; 1; 0; π = ≥ 5. cosine 6. F 7. T 8. T 9. 6 π 11. 2 π − 13. 6 π 15. 2 3 π 17. 3 4 π 19. 4 π − 21. 1.32 23. 0.46 25. 0.34 − 27. 2.72 29. 0.73 − 31. 2.55 33. 2 2 35. 3 3 − 37. 2 39. 2 41. 2 2 − 43. 2 3 3 45. 3 4 π 47. 3 π − 49. 2 4 51. 5 2 53. 14 2 − 55. 3 10 10 − 57. 5 59. 4 π − 61. u 1 1 2 + 63. u u 1 2 − 65. u u 1 2 − 67. u u 1 2 − 69. u 1 71. 5 13 73. 3 4 π 75. 3 4 − 77. 5 13 79. 5 6 π 81. 15 − 83. (a) 31.89 θ = ° (b) 54.64 ft in diameter (c) 37.96 ft high 85. (a) 22.3 θ = ° (b) = v 2940.23 ft sec 0 87. x 2 2 − 90. i i 5, 5, 2, 2 − − 91. Neither 92. 7 4 π 93. π ≈ 5 2 7.85 in. 94. (a) 5 2 , 31 2 ( ) − (b) Concave down (c) Increasing: (−∞ − ⎤ ⎦ ⎥ , 5 2 ; decreasing: 5 2 , ) ⎡ − ∞ ⎣ ⎢ 95. 3, 3 { } − 96. x x x x 3, 4, 7 or 3,4 4,7 7, { } [ ) ( ) ( ) ≥ ≠ ≠ − ∪ ∪ ∞ 97. ( ) [ ] ( ) = − = − y x y x 4 sin 6 1 or 4 sin 6 6 98. x c x c 1 1 2 2 − + − + − 99. 3 3 6 5π − +
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