Precalculus Student Solution Chapter 6

134 Chapter 6 Try It 6.1. 6π 6.2. 1 2 compressed 6.3. π; 2 right 6.4. 2 units up 6.5. midline: y = 0; amplitude: | A| = 1 ; 2 period: 2π P= = 6π; |B| phase shift: C =π B 6.6. f (x) = sin(x) + 2 6.7. two possibilities: ⎛ ⎞ y = 4sin⎝π x − π ⎠ + 4 5 5 or ⎛π ⎞ 4π y = − 4sin⎝ x + ⎠ + 4 5 5 6.8. midline: y = 0; amplitude: | A| = 0.8; period: P = 2π = π; |B| phase shift: C =0 B or none 6.9. 135 midline: y = 0; amplitude: | A| = 2; period: P = 2π = 6; |B| phase shift: C= −1 B 2 6.10. 7 6.11. y = 3cos(x) − 4 136 6.12. 6.13. It would be reflected across the line y = − 1, becoming an increasing function. 6.14. g(x) = 4tan(2x) 6.15. This is a vertical reflection of the preceding graph because A is negative. 6.16. 137 6.17. 6.18. 138 6.19. arccos(0.8776) ≈ 0.5 6.20. a. − π; 2 b. π − ; 4 c. π; d. π 3 6.21. 1.9823 or 113.578° 6.22. sin −1(0.6) = 36.87° = 0.6435 radians 6.23. π ; 2π 8 9 6.24. 3π 4 6.25. 12 13 6.26. 4 2 9 6.27. 4x 16x 2 + 1 Section Exercises 1. The sine and cosine functions have the property that f (x + P) = f (x) for a certain P. This means that the function values repeat for every P units on the x-axis. 3. The absolute 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. 7. 139 amplitude: 2; 3 period: 2π; midline: y = 0; maximum: y=2 3 occurs at x = 0; minimum: y= −2 3 occurs at x = π; for one period, the graph starts at 0 and ends at 2π 9. amplitude: 4; period: 2π; midline: y = 0; maximum y = 4 occurs at x = π; 2 minimum: y = − 4 occurs at x = 3π ; 2 one full period occurs from x = 0 to x = 2π 11. 140 amplitude: 1; period: π; midline: y = 0; maximum: y = 1 occurs at x = π; minimum: y = − 1 occurs at x = π; 2 one full period is graphed from x = 0 to x=π 13. amplitude: 4; period: 2; midline: y = 0; maximum: y = 4 occurs at x = 0; minimum: y = − 4 occurs at x=1 15. amplitude: 3; period: π; 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; one period occurs from 141 x = 0 to x=π 4 17. amplitude: 5; period: 2π ; 5 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; one full period can be graphed on x = 0 to x = 2π 5 19. amplitude: 1 ; period: 2π; midline: y = 1; maximum: y = 2 occurs at x = 2.09; maximum: y = 2 occurs at t = 2.09; minimum: y = 0 occurs at t = 5.24; phase shift: − π; 3 vertical translation: 1; one full period is from t = 0 to t = 2π 21. 142 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 ; 3 vertical shift: 0 23. amplitude: 2; midline: y = − 3; period: 4; equation: ⎛ ⎞ f (x) = 2sin⎝π x⎠ − 3 2 25. amplitude: 2; period: 5; midline: y = 3; equation: ⎛ ⎞ f (x) = − 2cos⎝2π x⎠ + 3 5 27. amplitude: 4; period: 2; midline: y = 0; equation: ⎛ ⎛ ⎞⎞ f (x) = − 4cos⎝π ⎝x − π ⎠⎠ 2 29. amplitude: 2; period: 2; midline y = 1; equation: f (x) = 2cos(πx) + 1 32. π , 5π 6 6 34. π , 3π 4 4 36. 3π 2 38. π , 3π 2 2 40. π , 3π 2 2 42. π , 11π 6 6 44. The graph appears linear. The linear functions dominate the shape of the graph for large values of x. 143 46. The graph is symmetric with respect to the y-axis and there is no amplitude because the function is not periodic. 48. a. Amplitude: 12.5; period: 10; midline: y = 13.5; b. ⎛ ⎞ h(t) = 12.5sin⎝π (t − 2.5)⎠ + 13.5; 5 c. 26 ft 49. 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. 51. Answers will vary. Using the unit circle, one can show that tan(x + π) = tan x. 53. The period is the same: 2π. 55. IV 144 57. III 59. period: 8; horizontal shift: 1 unit to left 61. 1.5 63. 5 65. −cotxcosx − sinx 67. stretching factor: 2; period: π; 4 asymptotes: ⎛ ⎞ x = 1 ⎝π + πk⎠ + 8, where k is an integer 4 2 69. stretching factor: 6; period: 6; asymptotes: x = 3k, where k is an integer 71. 145 stretching factor: 1; period: π; asymptotes: x = πk, where k is an integer 73. Stretching factor: 1; period: π; asymptotes: x = π + πk, where k is an integer 4 75. stretching factor: 2; period: 2π; asymptotes: x = πk, where k is an integer 77. stretching factor: 4; period: 2π ; 3 asymptotes: x = π k, where k is an odd integer 6 146 79. stretching factor: 7; period: 2π ; 5 asymptotes: x = π k, where k is an odd integer 10 81. stretching factor: 2; period: 2π; asymptotes: x = − π + πk, where k is an integer 4 83. 147 stretching factor: 7; 5 period: 2π; asymptotes: x = π + πk, where k is an integer 4 85. ⎛ ⎛ ⎞⎞ y = tan⎝3⎝x − π ⎠⎠ + 2 4 87. f (x) = csc(2x) 89. f (x) = csc(4x) 91. f (x) = 2cscx 93. f (x) = 1 tan(100πx) 2 95. 148 97. 99. 101. 103. 149 a. b. c. ⎛ π π⎞ ⎝− 2 , 2 ⎠; x= −π 2 and π x= ; 2 the distance grows without bound as |x| approaches π 2 —i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it; d. 3; when x = − π, 3 the boat is 3 km away; e. 1.73; when x = π, 6 the boat is about 1.73 km away; f. 1.5 km; when 105. a. b. c. x=0 ⎛ ⎞ h(x) = 2tan⎝ π x⎠; 120 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; 150 d. As x approaches 60 seconds, the values of h(x) grow increasingly large. The distance to the rocket is growing so large that the camera can no longer track it. 106. The function y = sinx is one-to-one on ⎡ π π⎤ ⎣− 2 , 2 ⎦; y = sinx, thus, this interval is the range of the inverse function of f (x) = sin −1 x. The function y = cosx is one-to-one on [0, π]; thus, this interval is the range of the inverse function of y = cosx, f (x) = cos −1 x. 108. π 6 is the radian measure of an angle between −π 2 and π 2 whose sine is 0.5. 110. In order for 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⎤ ⎣− 2 , 2 ⎦ so that it is one-to-one and possesses an inverse. 112. True . The angle, θ 1 that equals arccos( − x) , x > 0 , will be a second quadrant angle with reference angle, θ 2 , where θ 2 equals arccosx , x > 0 . Since θ 2 is the reference angle for θ1 , θ 2 = π − θ 1 and arccos( − x) = π − arccosx - 114. −π 6 116. 3π 4 118. −π 3 120. π 3 122. 1.98 124. 0.93 126. 1.41 128. 0.56 radians 130. 0 132. 0.71 151 134. -0.71 136. −π 4 138. 0.8 140. 5 13 142. x−1 −x 2 + 2x 144. x2 − 1 x 146. x + 0.5 −x 2 − x + 34 148. 2x + 1 x+1 150. 2x + 1 x 152. t 154. domain [−1, 1]; range [0, π] 156. approximately x = 0.00 158. 0.395 radians 160. 1.11 radians 162. 1.25 radians 164. 0.405 radians 166. No. The angle the ladder makes with the horizontal is 60 degrees. Review Exercises 168. amplitude: 3; period: 2π; midline: y = 3; no asymptotes 152 170. amplitude: 3; period: 2π; midline: y = 0; no asymptotes 172. amplitude: 3; period: 2π; midline: y = − 4; no asymptotes 174. amplitude: 6; period: 2π ; 3 midline: y = − 1; no asymptotes 153 176. stretching factor: none; period: π; midline: y = − 4; asymptotes: x = π + πk, 2 where k is an integer 178. stretching factor: 3; period: π; 4 midline: y = − 2; asymptotes: x = π + π k, 8 4 where k is an integer 154 180. amplitude: none; period: 2π; no phase shift; asymptotes: x = π k, 2 where k is an odd integer 182. amplitude: none; period: 2π ; 5 no phase shift; asymptotes: x = π k, 5 where k is an integer 155 184. amplitude: none; period: 4π; no phase shift; asymptotes: x = 2πk, where k is an integer 186. largest: 20,000; smallest: 4,000 188. amplitude: 8,000; period: 10; phase shift: 0 190. In 2007, the predicted population is 4,413. In 2010, the population will be 11,924. 192. 5 in. 194. 10 seconds 196. π 6 198. π 4 200. π 3 202. No solution 204. 12 5 206. The graphs are not symmetrical with respect to the line y = x. They are symmetrical with respect to the y -axis. 156 208. The graphs appear to be identical. Practice Test 209. amplitude: 0.5; period: 2π; midline y=0 157 211. amplitude: 5; period: 2π; midline: y=0 213. amplitude: 1; period: 2π; midline: y=1 215. amplitude: 3; period: 6π; midline: y=0 158 217. amplitude: none; period: π; midline: y = 0, asymptotes: x = 2π + πk, 3 where k is an integer 219. amplitude: none; period: 2π ; 3 midline: y = 0, asymptotes: x = π k, 3 where k is an integer 221. amplitude: none; period: 2π; midline: y= −3 159 223. amplitude: 2; period: 2; midline: y = 0; f (x) = 2sin⎛⎝π(x − 1)⎞⎠ 225. amplitude: 1; period: 12; phase shift: −6; midline y = −3 227. ⎛ ⎞ D(t) = 68 − 12sin⎝ π x⎠ 12 229. period: π; 6 horizontal shift: −7 231. f (x) = sec(πx); period: 2; phase shift: 0 233. 4 235. The views are different because the period of the wave is 1. 25 Over a bigger domain, there will be more cycles of the graph. 160 237. 3 5 239. On the approximate intervals (0.5, 1), (1.6, 2.1), (2.6, 3.1), (3.7, 4.2), (4.7, 5.2), (5.6, 6.28) 241. ⎛ ⎛ ⎞⎞ f (x) = 2cos⎝12⎝x + π ⎠⎠ + 3 4 243. This graph is periodic with a period of 2π. 161 245. π 3 247. π 2 249. 1 − (1 − 2x) 2 251. 1 1 + x4 253. x+1 x 255. False 257. approximately 0.07 radians