Precalculus Student Solution Chapter 12

Chapter 12 Try It 12.1. a = 5, f (x) = 2x 2 − 4, and L = 46. 12.2. a. 0; b. 2; c. does not exist; d. −2; e. 0; f. does not exist; g. 4; h. 4; i. 4 12.3. ⎛20sin(x) ⎞ =5 4x ⎠ x → 0⎝ lim 295 12.4. does not exist 12.5. 26 12.6. 59 12.7. 10 12.8. −64 12.9. −3 12.10. −1 50 12.11. −1 8 12.12. 2 3 12.13. −1 12.14. a. removable discontinuity at x = 6; b. jump discontinuity at x=4 12.15. yes 12.16. No, the function is not continuous at x = 3. There exists a removable discontinuity at x = 3. 12.17. x=6 12.18. 3 12.19. f ′(a) = 6a + 7 12.20. f ′(a) = −15 (5a + 4) 2 12.21. 3 2 12.22. 0 12.23. −2, 0, 0, −3 12.24. a. b. c. d. e. After zero seconds, she has traveled 0 feet. After 10 seconds, she has traveled 150 feet east. After 10 seconds, she is moving eastward at a rate of 15 ft/sec. After 20 seconds, she is moving westward at a rate of 10 ft/sec. After 40 seconds, she is 100 feet westward of her starting point. 296 12.25. The graph of f is continuous on (−∞, 1) ∪ (1, 3) ∪ (3, ∞). The graph of f is discontinuous at x = 1 and x = 3. The graph of f is differentiable on (−∞, 1) ∪ (1, 3) ∪ (3, ∞). The graph of f is not differentiable at x = 1 and x = 3. 12.26. y = 19x − 16 12.27. –68 ft/sec, it is dropping back to Earth at a rate of 68 ft/s. Section Exercises 1. The value of the function, the output, at x = a is f (a). When the lim f (x) x→a is taken, the values of x get infinitely close to a but never equal a. As the values of x approach a from the left and right, the limit is the value that the function is approaching. 3. –4 5. –4 7. 2 9. does not exist 11. 4 13. does not exist 15. Answers will vary. 16. Answers will vary. 17. Answers will vary. 18. Answers will vary. 19. Answers will vary. 20. Answers will vary. 21. Answers will vary. 23. 7.38906 25. 54.59815 27. e 6 ≈ 403.428794, e 7 ≈ 1096.633158, en 29. lim f (x) = 1 x→ −2 31. ⎛ 2 x − 6⎞ 5 = ≈ 0.83 lim x − 6 x → 3⎝ x 2 − 9 ⎠ 33. ⎛ x2 − 1 ⎞ = − 2.00 x → 1⎝x 2 − 3x + 2 ⎠ lim 297 35. ⎛ 10 − 10x 2 ⎞ x → 1⎝x 2 − 3x + 2 ⎠ lim 37. ⎛ x = 20.00 ⎞ 2 x → −1 ⎝4x + 4x + 1 ⎠ 2 does not exist. Function values decrease without bound as lim x approaches –0.5 from either left or right. 39. lim 7tanx = 7 3x 3 x→0 40. 41. lim 2sinx = 1 2 x → 0 4tanx 298 43. lim e x→0 e − 1 x2 45. lim x → − 1− = 1.0 |x + 1| = −(x + 1) = − 1 (x + 1) x+1 x + 1| (x + 1) lim | = = 1; + x+1 (x + 1) x→ −1 x + 1| lim | x+1 x→ −1 47. and since the right-hand limit does not equal the left-hand limit, does not exist. 1 lim x → − 1 (x + 1) 2 does not exist. The function increases without bound as x approaches −1 from either side. 49. lim x→0 5 2 1 − e x does not exist. Function values approach 5 from the left and approach 0 from the right. 51. Through examination of the postulates and an understanding of relativistic physics, as v → c, m → ∞. Take this one step further to the solution, (12.33) lim m = lim− v → c− 53. If v→c mo / 1 − (v 2 c 2) =∞ f is a polynomial function, the limit of a polynomial function as x approaches a will always be f (a). 55. It could mean either (1) the values of the function increase or decrease without bound as x approaches c, or (2) the left and right-hand limits are not equal. 57. −10 3 59. 6 299 61. 1 2 63. 6 65. does not exist 67. −12 69. − 5 10 71. −108 73. 1 75. 6 77. 1 79. 1 81. does not exist 83. 6+ 5 85. 3 5 87. 0 89. −3 91. does not exist; right-hand limit is not the same as the left-hand limit. 93. 2 95. Limit does not exist; limit approaches infinity. 97. 4x + 2h 99. 2x + h + 4 101. cos(x + h) − cos(x) h 103. −1 x(x + h) 105. −1 x+h+ x 107. 2 f (x) = x + 5x + 6 x+3 109. does not exist 111. 52 113. Informally, if a function is continuous at x = c, then there is no break in the graph of the function at f (c), and f (c) is defined. 115. discontinuous at a = −3; f ( − 3) does not exist 117. removable discontinuity at a = −4; f ( − 4) is not defined 300 119. Discontinuous at a = 3; lim f (x) = 3, x→3 but f (3) = 6, which is not equal to the limit. 121. lim f (x) x→2 does not exist. 123. lim f (x) = 4; lim f (x) = 1 x → 1− lim f (x) x→1 125. x → 1+ . Therefore, does not exist. lim f (x) = 5 ≠ lim f (x) = − 1 x → 1− lim f (x) x→1 127. x → 1+ does not exist. lim f (x) = − 6 lim f (x) = − 1 3 x → − 3− x → − 3+ lim f (x) x→ −3 129. , Therefore, does not exist. f (2) is not defined. 131. f (−3) is not defined. 133. f (0) is not defined. 135. Continuous on ( − ∞, ∞) 137. Continuous on ( − ∞, ∞) 139. Discontinuous at x = 0 and x=2 141. Discontinuous at x=0 143. Continuous on (0, ∞) 145. Continuous on [4, ∞) 147. Continuous on ( − ∞, ∞) . 149. 1, but not 2 or 3 151. 1 and 2, but not 3 153. f (0) is undefined. 155. ( − ∞, 0) ∪ (0, ∞) 157. At x = − 1, the limit does not exist. At x = 1, . Thus 301 f (1) does not exist. At x = 2, there appears to be a vertical asymptote, and the limit does not exist. 159. x 3 + 6x 2 − 7x (x + 7)(x − 1) 161. The function is discontinuous at x = 1 because the limit as x approaches 1 is 5 and f (1) = 2. 163. The slope of a linear function stays the same. The derivative of a general function varies according to x. Both the slope of a line and the derivative at a point measure the rate of change of the function. 165. Average velocity is 55 miles per hour. The instantaneous velocity at 2:30 p.m. is 62 miles per hour. The instantaneous velocity measures the velocity of the car at an instant of time whereas the average velocity gives the velocity of the car over an interval. 167. The average rate of change of the amount of water in the tank is 45 gallons per minute. If f (x) is the function giving the amount of water in the tank at any time t, then the average rate of change of f (x) between t = a and t = b is f (a) + 45(b − a). 169. f ′(x) = − 2 171. f ′(x) = 4x + 1 173. f ′(x) = 1 (x − 2) 2 175. −16 (3 + 2x) 2 177. f ′(x) = 9x 2 − 2x + 2 179. f ′(x) = 0 181. −1 3 183. undefined 185. f ′(x) = − 6x − 7 187. f ′(x) = 9x 2 + 4x + 1 189. y = 12x − 15 191. k = − 10 or k=2 193. Discontinuous at x = − 2 and x = 0. Not differentiable at –2, 0, 2. 195. Discontinuous at x = 5. Not differentiable at -4, –2, 0, 1, 3, 4, 5. 302 197. f (0) = − 2 199. f (2) = − 6 201. f ′ (−1) = 9 203. f ′ (1) = − 3 205. f ′ (3) = 9 207. Answers vary. The slope of the tangent line near x = 1 is 2. 209. At 12:30 p.m., the rate of change of the number of gallons in the tank is –20 gallons per minute. That is, the tank is losing 20 gallons per minute. 211. At 200 minutes after noon, the volume of gallons in the tank is changing at the rate of 30 gallons per minute. 213. The height of the projectile after 2 seconds is 96 feet. 215. The height of the projectile at t = 3 seconds is 96 feet. 217. The height of the projectile is zero at t = 0 and again at t = 5. In other words, the projectile starts on the ground and falls to earth again after 5 seconds. 219. 36π 221. $50.00 per unit, which is the instantaneous rate of change of revenue when exactly 10 units are sold. 223. $21 per unit 225. $36 227. f '(x) = 10a − 1 229. 4 (3 − x) 2 Review Exercises 230. 2 232. does not exist 234. f (x) does not exist), x = 3 (jump discontinuity), Discontinuous at x = − 1(x lim → a and x = 7 (x lim f (x) does not exist). → a 235. 3 237. lim x→ −2 f (x) = 1 239. 2 241. −15 243. 3 245. 12 247. −10 249. −1 9 251. At x = 4, the function has a vertical asymptote. 303 253. removable discontinuity at a= −5 2 255. continuous on ( − ∞, ∞) 257. removable discontinuity at x = 2. f (2) is not defined, but limits exist. 259. discontinuity at x = 0 and x = 2. Both f (0) and f (2) are not defined. 261. removable discontinuity at x = – 2. f ( – 2) is not defined. 263. 0 265. ln(x + h) − ln(x) h 267. =4 269. y = − 8x + 16 271. 12π Practice Test 272. 3 274. 0 276. −1 278. lim f (x) = − 5 a 2 and lim f (x) = 9 x → 2− x → 2+ Thus, the limit of the function as x approaches 2 does not exist. 279. −1 50 281. 1 283. removable discontinuity at x=3 285. f '(x) = − 3 3 2a 2 287. discontinuous at –2,0, not differentiable at –2,0, 2. 289. not differentiable at x = 0 (no limit) 291. the height of the projectile at t = 2 seconds 293. the average velocity from t = 1 to t = 2 295. 1 3 297. 0 299. 2 300. x=1 302. y = − 14x − 18 304. The graph is not differentiable at x = 1 (cusp). 306. f '(x) = 8x 308. f '(x) = − 1 (2 + x) 2 310. f '(x) = − 3x 2 312. f '(x) = 1 2 x−1 This content is available for free at http://cnx.org/content/col11667/1.5