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MATHEMATICS 1571 Final Examination Review Problems 1. Let f(x ...

MATHEMATICS 1571 Final Examination Review Problems 1. Let f(x ...

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8i.∫3x 2√ 6x 3 + 1dxl.∫sec x tanxcos(sec x)dx∫j. x √ 2 − xdxk.∫ √ 33 + z-1dxz 2m.n.∫∫sin x 3(cosx3) 3dxtan x(sec x) 2 dx59. Differentiate.a. f(x) =∫ x3( t 2 − 8 ) dt b. g(x) =∫ x 24(t 2 + 6) 5 2dt60. Find the area of the region bounded by y = x 2 and y = x 4 .6<strong>1.</strong> Find the area of the region bounded by y = x 2 and y = 2 − x.62. Find the area of the region bounded by y = x 4 and y = 8x.63. Find the volume of the solid generated by revolving the region bounded by y = x 3 2, thex-axis, and the line x = 4 about the x-axis.64. Find the volume of the solid generated by revolving the region bounded by y 2 = 4x andx 2 = 4y about the x-axis.65. Find the volume of the solid generated by revolving the region in the first quadrantbounded by y = x 2 , y = 4, and the y-axis about each of the following.a. the y-axis b. y = 466. Find the volume of the solid generated by revolving the region bounded by y = x + 4 x ,the x-axis, and the lines x = 1 and x = 3 about the y-axis.67. A solid has as its base the region in the first quadrant bounded by x 2 + y 2 = 25. Everyplane section of the solid taken perpendicular to the x-axis is a square. Find the volume ofthe solid.68. A solid has as its base the region in the xy-plane bounded by the graphs of y = x andy 2 = x. Find the volume of the solid if every cross section by a plane perpendicular to thex-axis is a semicircle with diameter in the xy-plane.69. Find the average value of the function f(x) = x 2 + x + 1 on the interval [-1, 2].70. Find the average value of the function f(x) = sin x on the interval [0, π].


7<strong>1.</strong> Assume that the density of water is 62.5 lb/ft 3 . A cylindrical water tank with a circularbase has radius 3 feet and height 10 feet. Find the work required to empty the tank bypumping the water out of the top for each of the following situations.a. The tank is full. b. The tank is half full.972. A bucket with 24 lb of water is raised 30 feet from the bottom of a well. Find the workdone in each of the following cases.a. The weight of the empty bucket is 4 lb and the weight of the rope is negligible.b. The bucket weighs 4 lb and the rope weighs 4 oz/ft.c. The bucket weighs 4 lb, the rope weighs 4 oz/ft, and water is leaking out of the bucket ata constant rate so that only 18 lb of water remain in the bucket when it reaches the top.


1073. Match each numbered item with a lettered item. (There are more lettered items thannumbered items. Some lettered items do not match any numbered item.)<strong>1.</strong> Definition of limx→af(x) = L.a. limx→af(x) = f(a).2. Definition of limx→af(x) = L.b. For every ε > 0 there is a corresponding number Nsuch that |f(x) − L| < ε whenever x > N.3. Definition of lim f(x) = L.x→∞4. Definition of “f is continuous at a”.5. The Intermediate Value Theorem.6. Definition of the derivative of f at a.7. Definition of a function f being differentiable at a.8. Theorem relating differentiability and continuity.9. The power rule for differentiation.10. Definition of the differential.1<strong>1.</strong> Definition of a function f having an absolute maximumat c.12. Definition of a function f having a local maximumat c.13. The Extreme Value Theorem.14. Definition of a function that is increasing on an intervalI.15. The Mean Value Theorem.16. Definition of an antiderivative of f on an interval I.c. If f is continuous on the closed interval [a, b], and Nis a number strictly between f(a) and f(b), then thereexists a number c in (a, b) such that f(c) = N.f(a + h) − f(a)d. The limit limexists.h→0 he. If f is differentiable at a, then f is continuous at a.f. If f is continuous at a, then f is differentiable at a.g.ddx xn = nx n−1 .h. The function g has the property g ′ (x) = f(x) for allx in I.i. limh→0f(a + h) − f(a)hj. f(c) ≥ f(x) for all x in the domain of f.k. For every ε > 0 there is a corresponding number δ > 0such that |f(x) − L| < ε whenever a < x < a + δ.l. For every ε > 0 there is a corresponding number δ > 0such that |f(x) − L| < ε whenever 0 < |x − a| < δ.m. There is an open interval I containing c such thatf(c) ≥ f(x) for all x in I.n. f ′ (x) > 0 for all x in I.o. If f is continuous on [a, b], then there are numbers cand d in [a, b] such that f(c) is an absolute maximum forf in [a, b] and f(d) is an absolute minimum for f in [a, b].p. If f is differentiable, dy = f ′ (x)dx.q. f(x 1) < f(x 2) whenever x 1< x 2and x 1and x 2arein I.r. If f is continuous on [a, b] and differentiable on(a, b), then there is a number c in (a,b) such thatf ′ (c) = f(b)−f(a)b−a.


11<strong>MATHEMATICS</strong> <strong>1571</strong><strong>1.</strong> a. 2a 2 + 4ab + 2b 2 − 5a − 5bb. 4x 22. a. Dom: (-∞, -1] ∪ [4, ∞)Answers to <strong>Final</strong> <strong>Examination</strong> <strong>Review</strong> <strong>Problems</strong>6. a. DNEb. 37. a. -1b. Dom: (-∞, -1) ∪ (-1, 0) ∪ (0, 1) ∪ (1, ∞)3. Left 3 units and down 4 units.4. a. (f ◦ g)(x) = 1x − 1Dom: [0, 1) ∪ (1, ∞)(g ◦ f)(x) =1√(x + 1)(x − 1)Dom: (-∞, -1) ∪ (1, ∞)b. (f ◦ g)(x) =Dom: (-∞, 1) ∪ (1, ∞)(g ◦ f)(x) = x2 + 1x 2Dom: (-∞, 0) ∪ (0, ∞)x 2(x − 1) 2 + 15. a. y-intercept: (0, 0)x-intercept: (0, 0)Symmetry: noneb. y-intercept: nonex-intercepts: noneSymmetry: originc. y-intercept: (0, 2)x-intercepts: (-2, 0), (2, 0)Symmetry: y-axisb. - 1 2c. 4d. -∞e. ∞f. DNEg. 1h. -1i. DNEj. 0k. 1 3l. DNEm. 0n. 1o. 1 3p. 4q. √ 32r. - sin x8. a. 15b. 9 29. a. (-∞, ∞)b. (-∞, -3], (-3, ∞)


1210. a. HA: y = 1VA: x = 1OA: noneb. HA: y = 1VA: x = -2OA: nonec. HA: y = -1, y = 1VA: x = 0OA: noned. HA: y = 1VA: x = 2OA: none1<strong>1.</strong> f(x) = x2 − 9x 2 − 5x + 6a = 3There are infinitely many correct answers.12. -11213. 4 914. 732415. - 8 316. y = 3x17. (2, 6)18. y = 22x− 10721 2119.-13 √ 32<strong>1.</strong> -22. 1 223. - 8 33√41224. 3√ 3425. √ 226. - √ 327. - 9 228. Inc: [ 34 , ∞)29. Dec: ( - 3 2 , 1]30. CU: (-4, 0), (3, ∞)3<strong>1.</strong> CD: (-∞, -2), (0, 2)32. IP: ( -4, - 256333. IP: none) ( ), 3, -171434. Local max: 13 + 32 3√2 at -3 − 2√2Local min: 13 − 323√2 at -3 + 2√235. Local max: 3 at 0Local min: - 13 4 at -√ 5, - 134 at √ 536. a. Max: 0Min: -16b. Max: 16Min: 037. a. 192 ftb. 6 sec20.34 √ 2c. -128 ft/sec


1338. a. 64 ft/secb. 288 ft39. 5000 ft 240. 804<strong>1.</strong> Dimensions: 5 × 5 × 542. Overland 5 mile 8Underwater 5 mile 843. 2 × 2 squareVolume of 128 in 344. 33 √ 5 mph45. 1 ft/sec46. 39 ft/min47. ∆y = .0702dy = .0748. .06 ft 349. a. 7 + 1 14b. 7 + 1 1450. dI = .04M5<strong>1.</strong> 15 meters52. a. Yesb. Yes55. a. 4b. 3c. Dec: [0, 1], [4, 5]Inc: [1, 4]d. IP: (2, 2)56. Dec: [2, 5]Inc: [0, 2]57. a. 12 secondsb. 16 metersc. 4 metersd. 20 meterse. ForwardAcceleratorf. BackwardBrake58. a.130 (3x2 + 4) 5 + Cb.116 (2x4 + 4x) 4 + Cc. 19564d. - 3 x − 3x 2 − 23x 3 + Ce. 38 3f. 1 3√(3x2 − 2x + 1) 3 + Cc. Yesd. No53. f(x) = 2 3 x3 + 3 2 x2 − 8x + 47 6g.h.-13(x + 5) 3 + C-112(10x 2 + 2x + 40) 3 + C54. f(x) = 1 3 x3 + 3 2 x2 + 2xi. 1 9√(6x3 + 1) 3 + C


14j. 2 5√(2 − x)5 − 4 3√(2 − x)3 + Ck. - 3 4√3 (3 + z-1) 4 + Cl. sin(sec x) + Cm. - 3 4(cosx3) 4+ Cn. 1 2 tan2 x + C59. a. f ′ (x) = x 2 − 8b. g ′ (x) = (x 4 + 6) 5 2(2x)60.4156<strong>1.</strong> 9 262. 48 563. 64π64. 96π565. a. 8πb. 256π1566. 100π367. 250368.π24069. 5 270.2π7<strong>1.</strong> a. 28125π foot-poundsb. 21093.75π foot-pounds72. a. 840 foot-poundsb. 952.5 foot-poundsc. 862.5 foot-pounds73.<strong>1.</strong> l 2. k 3. b 4. a 5. c 6. i 7. d 8. e9. g 10. p 1<strong>1.</strong> j 12. m 13. o 14. q 15. r 16. h

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