Calculus 1 Midterm 3: Function Analysis, Exams of Analytical Geometry and Calculus

Download Calculus 1 Midterm 3: Function Analysis and more Exams Analytical Geometry and Calculus in PDF only on Docsity! MATH 1300: CALCULUS 1 November 15, 2006 MIDTERM 3 I have neither given nor received aid on this exam. Name: © 001 L. Mayhew . . . . . . . . . . . . (8am) © 002 J. Kish . . . . . . . . . . . . . . . . (8am) © 003 P. Newberry . . . . . . . . . . (8am) © 004 A. Spina . . . . . . . . . . . . . . (8am) © 005 M. Hedges . . . . . . . . . . . . (9am) © 006 M. Stackpole . . . . . . . . . (9am) © 007 E. Mankin . . . . . . . . . . . . (9am) © 008 N. Flores . . . . . . . . . . . . . (9am) © 009 A. Szendrei . . . . . . . . . . . (9am) © 010 A. Szendrei . . . . . . . . . . (10am) © 012 J. Fuhrmann . . . . . . . . . (11am) © 013 B. Wang . . . . . . . . . . . . . (11am) © 014 D. Ernst . . . . . . . . . . . . . (12pm) © 015 C. Moody . . . . . . . . . . . . . (1pm) © 016 J. Pearson . . . . . . . . . . . . (2pm) © 017 J. Boisvert . . . . . . . . . . . (2pm) © 018 R. Chestnut . . . . . . . . . . (4pm) © 019 G. Surman . . . . . . . . . . . . (4pm) © 020 N. Sagullo . . . . . . . . . . . .(8am) Box your answers. If you have a question raise your hand and remain seated. In order to receive full credit your answer must be complete, legible and correct. Show all of your work, and give adequate explanations. No calculators, no books, no notes are allowed on this exam. DO NOT WRITE IN THIS BOX! Problem Points Score 1 35 pts 2 20 pts 3 30 pts 4 15 pts 5 10 pts 6 25 pts 7 15 pts TOTAL 150 pts 1. You are given the following information about the function f(x) = (x− 2)3 x2 − 1 : • The vertical asymptotes of f are the lines x = −1 and x = 1. • The line y = x− 6 is an oblique asymptote of f . • The critical points in the domain of f are c1 = −2− √ 7 ≈ −4.65, c2 = −2 + √ 7 ≈ 0.65, and c3 = 2; f is undefined at −1 and 1. • f is increasing on (−∞, c1], [c2, 1], [1, 2], [2,∞), and is decreasing on [c1,−1], [−1, c2]. • f is concave up on (−1, 1), (2,∞), and is concave down on (−∞,−1), (1, 2). • f(c1) ≈ −14.26 and f(c2) ≈ 4.26. Use this information when you complete parts (a)–(d) below. (a) Find the x-intercepts and the y-intercept of f . x-intercept: x = 2, the only solutions of the equation (x− 2)3 x2 − 1 = 0. y-intercept: y = f(0) = −8 −1 = 8. (b) Find all x-values at which f has a relative extremum, and state whether it is a relative maximum or minimum. At x = c1 f has a relative maximum, since f is increasing on (−∞, c1] and decreasing on [c1,−1]. At x = c2 f has a relative minimum, since f is decreasing on [−1, c2] and increasing on [c2, 1]. At x = 2 f does not have a relative extremum, since f is increasing on [1, 2] and [2,∞). (c) Find all x-values at which f has an inflection point. f has one inflection point, namely at x = 2; this is the only number in the domain of f where f changes its direction of concavity. 4. Use L’Hôpital’s Rule to find lim x→0+ √ x ln(x). (Verify that the assumptions of L’Hôpital’s Rule are satisfied.) lim x→0+ √ x ln(x) = lim x→0+ ln(x) x−1/2 . The assumptions of L’Hôpital’s Rule are satisfied for the second limit, since • f(x) = ln(x) and g(x) = x−1/2 are differentiable on (0,∞); • lim x→0+ ln(x) = −∞ and lim x→0+ x−1/2 = ∞; • the limit lim x→0+ d dx [ln(x)] d dx [x−1/2] exists, since lim x→0+ d dx [ln(x)] d dx [x−1/2] = lim x→0+ x−1 −1 2 x−3/2 = lim x→0+ (−2x1/2) = 0. Therefore by L’Hôpital’s Rule, lim x→0+ ln(x) x−1/2 = lim x→0+ d dx [ln(x)] d dx [x−1/2] = 0. Using the first displayed equality we get that lim x→0+ √ x ln(x) = 0. 5. A truck driver handed in a ticket at a toll booth showing that in 2 hours he had covered 158 mi on a toll road with speed limit 70 mph. The driver was cited for speeding. Use the Mean-Value Theorem to explain why. State the assumptions that we have to make about the position function of the truck to be able to apply the Mean-Value Theorem. The average velocity of the truck in this 2 hour time interval was 158 2 = 79 mph. By the Mean-Value Theorem there was a moment in this time interval when the instantaneous velocity of the truck was 79 mph, which is over the speed limit. To be able to apply the Mean-Value Theorem we have to assume that the position function of the truck is continuous on the given closed 2 hour time interval, and is differentiable on the open 2 hour time interval. 6. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 inches. What dimensions will give a box with a square end the largest possible volume? XXXXXXXXXXXX XXXXXXXXXXXX XXXXXXXXXXXX XX XX XX XX XX 6 ?  length girth end If each side of the square end is x inches and the length of the box is y inches, then a box with largest volume will satisfy y + 4x = 108 where x, y ≥ 0, so 0 ≤ x ≤ 108 4 = 27. The volume of such a box is V (x) = x2y = x2(108− 4x) = 108x2 − 4x3. We have to find the absolute maximum of V (x) on [0, 27]. V (x) is continuous on [0, 27] and differentiable on (0, 27) with V ′(x) = 216x−12x2 = 12x(18− x). Hence the only critical point of V (x) in (0, 27) is x = 18. Since V (0) = 0, V (18) = 182(108− 4 · 18) = 182 · 36 > 0, and V (27) = 0, the absolute maximum of V (x) in [0, 27] occurs at x = 18. Thus the dimensions that will give a box with a square end the largest possible volume are: length of box: 36 in, side of square end: 18 in. 7. TRUE or FALSE? Justify your answer. (a) lim x→3 x− 3 x2 − 3 = lim x→3 d dx [x− 3] d dx [x2 − 3] = lim x→3 1 2x = 1 6 . TRUE FALSE (circle one) Justify: Since lim x→3 (x2 − 3) = 6 6= 0, lim x→3 x− 3 x2 − 3 is not an indeterminate form. lim x→3 x− 3 x2 − 3 = limx→3(x− 3) limx→3(x2 − 3) = 0 6 = 0. (b) If f ′(c) = f ′′(c) = 0, then f does not have a relative extremum at x = c. TRUE FALSE (circle one) Justify: For example, if f(x) = x4 and c = 0, then f ′(0) = f ′′(0) = 0, but f has a relative minimum at x = 0. (c) ∫ 2 −2 √ 4− x2 dx = 2π. TRUE FALSE (circle one) Justify: f(x) = √ 4− x2 is continuous and nonnegative on [−2, 2]. Therefore ∫ 2 −2 √ 4− x2 dx is equal to the area of the region between the graph of f and the interval [−2, 2]. Since this region is the upper half of a disc of radius 2, its area is 1 2 22π = 2π.