M408 C Fall 2011 Dr. Jeffrey Danciger Exam 2 November 3, Section time (circle one): 11:00am 1:00pm 2:00pm

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1 M408 C Fall 2011 Dr. Jeffrey Danciger Exam 2 November 3, 2011 NAME EID Section time (circle one): 11:00am 1:00pm 2:00pm No books, notes, or calculators. Show all your work. Do NOT open this exam booklet until instructed to do so. 1

2 Please leave this page blank. Question Points Score (bonus) 5 Total 80 2

3 1. [15 pts.] The following chart contains partial information about the continuous function f(x) (defined for all numbers) and its derivatives. f(x) f (x) f (x) x < 2 + x = 2 10 DNE 2 < x < x = < x < 4 + x = x > 4?? (a): Find all local maxima and local minima. Justify your answer. There are critical points at x = 2 and x = 4. x = 2: f (x) changes from to +. Hence, by the first derivative test, there is a local minimum at x = 2. x = 4: f (4) = 0 and f (4) < 0. Hence, by the second derivative test, there is a local maximum at x = 4. (b): On what intervals is the graph of f(x) concave up? concave down? Locate any inflection points. The graph of f(x) is concave up on the intervals (, 2) and ( 2, 0). The graph of f(x) is concave down on the interval (0, ). There is an inflection point at x = 0 because the concavity changes from positive (concave up) to negative (concave down). There is NOT an inflection point at x = 2. (c): Identify the intervals of increase and decrease for f(x). To figure out what is happening for x > 4, note that f (x) = d dx f (x) is negative for all x > 4. Hence f (x) is decreasing for x > 4. Since f (4) = 0, f (x) must be negative for all x > 4. Intervals of increase: ( 2, 4) Intervals of decrease: (, 2) and (4, ). 3

4 2. [10 pts.] Two cars start moving from the same point. One travels east at 15 mi/h and the other travels north at 20 mi/h. At what rate is the distance between the cars increasing two hours later? Let x(t) be the distance of the first car from the starting point at time t. Let y(t) be the distance of the second car from the starting point at time t. The paths of the two cars form a right triangle, so the distance f between the two is given by the pythagorean theorem: f = x 2 + y 2. We want to know df after two hours. By the chain rule: dx dy df 2x + 2y = 2 x 2 + y. 2 After two hours x = 30mi and y = 40mi. We are given that dx = 20mi/h. Hence after two hours, dy = 15mi/h and that df = (30mi)(15mi/h) + (40mi)(20mi/h) (30mi) 2 + (40mi) = mi/h 50 = 25 mi/h. 4

5 3. [10 pts.] The conditions y = x x 4 define a line segment in the x-y plane. Find the points on the line segment with maximum and minimum distance to the origin. The distance from the point (x, y) to the origin is given by the distance formula distance = x 2 + y 2. We want to extremize the distance, when (x, y) lies on the line y = x + 2 and 1 x 4. Hence we are trying to optimize the function f(x) = x 2 + ( x + 2) 2 on the closed interval [ 1, 4]. to do this, use the closed interval method: Critical points: f (x) = 1 ( x 2 + ( x + 2) 2) 1 2 (2x + 2( x + 2)( 1)) 2 2x 2 = x 2 + (2 x). 2 The denominator of f (x) is never zero, so f (x) is always defined. f (x) = 0 exactly when 2x 2 = 0. So the only critical point is x = 1. At this point: f(1) = = 2. Check endpoints: f( 1) = 10. f(4) = 20. Hence, the absolute minimum occurs when x = 1 (at the point (1, 1) on the line segment) and the absolute maximum occurs when x = 4 (at the point (4, 2) on the line segment). 5

6 4. [10 pts.] Find the limit. You may use any method. Please show all steps clearly. (a): lim θ π 2 1 sin θ 1 + cos 2θ Plugging in θ = π 2 gives the indeterminate form 0 0. So we use L Hospital s rule: lim θ π 2 1 sin θ 1 + cos 2θ = lim θ π 2 cos θ 2 sin 2θ This limit again has indeterminate form 0 0, so we use L Hospital s rule again. lim θ π 2 1 sin θ 1 + cos 2θ = lim θ π 2 = lim θ π 2 = sin π 2 4 cos π = 1 4 cos θ 2 sin 2θ sin θ 4 cos 2θ (b): lim x ln x x 0 + As x 0 +, ln x. Hence the limit is an indeterminate product of form 0. Rewrites as a fraction, and then use L Hospital s rule. ln x lim x ln x = lim x 0 + x 0 + 1/x = lim x 0 + 1/x 1/x 2 = lim x 0 +( x) = 0. by L Hospital s rule 6

7 5. [16 pts] Sketch the graph of y = xe x. Label all important features of the graph. You may use this page and the next one to show your work. Domain: y defined for all numbers. Intercepts: y(0) = 0. And xe x = 0 only when x = 0. (e x > 0 for all x). Symmetry: No symmetry. Asymptotes: No vertical asymptotes because the function is defined for all numbers. Horizontal asymptotes: lim x xe x = lim x x e x = lim x So horizontal asymptote at y = 0 as x. (No h.a. as x ) Intervals of I/D: lim x xe x = y = e x + xe x ( 1) 1 e x = 0. = e x xe x = (1 x)e x So y is increasing when x < 1 and y is decreasing when x > 1. Local Extrema: y switches from + to when x = 1, so y has a local max at x = 1. Intervals of Concavity: y = e x (e x xe x ) = (x 2)e x. So, the graph of y is concave up for x > 2 and concave down for x < 2. The concavity switches at x = 2, so there is an inflection point at x = 2. 7

8 Problem 5 extra page 8

9 6. [10 pts] (a): Let f(x) be continuous on [1, 5] and differentiable on (1, 5). What does the Mean Value Theorem say about f(x)? There exists an x-value c in (1, 5) so that f (c) = f(5) f(1) 5 1 (b): Suppose that f (x) > 3 for all x, and that f(1) = 1. Show that f(5) > 13. Using MVT we have that and so f (c) = f(5) 1 4 f(5) = 4f (c) + 1. Since f (x) > 3 for all x, we have f (c) > 3. Thus f(5) > = 13. 9

10 7. [9 pts.] (a): Find the linearization L(x) of f(x) = x 1/3 at x = 8. L(x) = f(a) + f (a)(x a). In this case, a = 8. f(8) = 2 and f (x) = 1 3 x 2/3, so f (8) = Hence, L(x) = (x 8). 12 (b): Use your answer form part (a) to estimate the value of f(8.2) = L(8.2) = (0.2) = (c): What is the concavity of f(x) near x = 8? Use this to explain whether your answer from part (b) is an over-estimate or an under-estimate. f (x) = 2 9 x 5/3. Hence f (x) < 0 for all positive x. In particular, near x = 8, the graph of f is concave down. Hence the tangent line at x = 8 lies above the graph. This means the linearization gives an overestimate. 10

11 8. [Bonus - 5pts] Let f(x) be de defined for all x. Suppose f(x) is continuous and differentiable whenever x 0 and satisfies f(0) = 1 lim f(x) = a x 0 lim f(x) = 1 x 0 + lim f (x) = b. x 0 + Give conditions on the constants a, b that guarantee f(0) is a local maximum. f(0) is a local maximum if for x near 0, the values of f(x) are all less than or equal to f(0). To guarantee this we need the following two conditions. a < 1: This guarantees that f(x) < f(0) = 1 for nearby x < 0. b < 0: This guarantees that f(x) < f(0) for nearby x > 0. 11