Math 11 (Calculus I) Midterm Exam KEY Multiple Choice. Fill in the answer to each problem on your computer scored answer sheet. Make sure your name, section and instructor are on that sheet. 1. Which of the following are the critical numbers of the function f(x) = x x 7x+? a) 1, 1 b) 1, 7 c) 1, 7 d) 1, 7 e) 1, 7 f) 7, 7 g) None of the above. d). If we let f(x) = 6x 4x+18 on the interval [0, 4], then the number c guaranteed by Rolle s theorem is a) less than 0. b) 0. c) strictly between 0 and 1. d) 1. e) strictly between 1 and. f). g) strictly between and 4. h) 4. i) greater than 4. j) not guaranteed to exist since f(x) doesn t satisfy the hypotheses of the theorem. f). If f, f, and f are continuous and f() = 0, f () =, and f () =, what can we say about the function f(x) at x =? a) f has a local minimum at x =. b) f has a local maximum at x =. c) f is increasing, concave up near x =. d) f is increasing, concave down near x =. e) f is decreasing, concave up near x =. f) f is decreasing, concave down near x =. g) None of these. d)
xe x 4. Find lim x 0 e x 1. a) 0 b) 1 c) 1 d) 1 b) e) f) Limit does not exist. 5. If f(x) = x and the initial approximation in Newton s method is x 1 =, the second approximation x to the root of f(x) in the interval [1, ] is a) b) 7/4 c) 7/4 d) /4 e) 1/4 f) 1 g) None of the above. b) 6. The most general antiderivative of r(θ) = csc θ cot θ + e θ is a) csc θ + e θ b) csc θ + e θ + C c) csc θ + e θ d) cotθ + e θ + C e) cot θ + e θ + C f) csc θ + e θ + C g) None of the above. f) 7. Over which of the following intervals is the function f(x) = x 1/ x 4/ decreasing? a) (0, ) b) (/4, ) c) (1/4, ) d) (, 1/4) e) (0, /4) f) (, /4) g) None of these. c)
8. The graph of the first derivative f of a function f is shown above. On what intervals is the function f increasing? a) (, ) and (1, ) b) (, 0) c) (, 0) and (, ) d) (, 1) e) (, ) and (0, ) f) None of the above. c) 9. The graph of the first derivative f of a function f is shown above. Which of the following is true? a) f is concave up on (, 1). b) f has a point of inflection at x = 0. c) f has a local maximum at approximately x =. d) f has an absolute minimum at x = 0. e) f is concave up on (, ). f) None of the above. e) END OF EXAM
Math 11 (Calculus I) Midterm Exam KEY 1. (7 points) Let f(x) = 4x + 7x on [0, 1]. (a) State the hypotheses of the Mean Value Theorem. Does f(x) satisfy these hypotheses on [0, 1]? Hypotheses of the Mean Value Theorem: Let f : [a, b] R be a function. i. f is continuous on the closed interval [a, b]. ii. f is differentiable on the open interval (a, b). The given f satisfies these hypotheses on [0, 1]. (b) Find any values of c that satisfy the conclusion of the Mean Value Theorem for f(x) on [0, 1]. If there are no such values, explain why not. Find c: Since we require c (0, 1), c = 1. f f(1) f(0) (c) = 1 0 1c + 7 = c = 1 c = ± 1. (9 points) Find the absolute maximum and minimum values of f(x) = 1 sin(x) on the interval [ 0, π ]. Justify your answers. Closed interval method: First find the critical points of f: Since x f (x) = 0 cos(x) = 0 x = π + kπ, k Z x = π 4 [ 0, π ], x = π 4 is the only admissible critical point. + kπ, k Z. Next, make a table: x 0 f(x) 0 π 4 1 π 4 Conclusion: On the given interval, the absolute maximum value of of f is 1, the absolute minimum value of f is 4.
x. (8 points) Find: lim (tan x)cos Let y = (tan x) cos x. Then ln(y) = cos x ln(tan x). We have: lim ln(y) cos x ln(tan x) ln(tan x) sec x sec x tan x sec x tan x sec x tan x cos x sin x = 0. (L Hopital Rule) Hence, lim (tan x)cos x = e 0 = 1. ( 1 4. (6 points) Find: lim x 0 + x 1 e x 1 Apply L Hopital twice, we have: ( 1 lim x 0 + x 1 ) e x 1 ) x 0 + e x 1 x x(e x 1) x 0 + e x 1 x 0 + e x 1 + xe x e x e x + e x + xe x 1 x 0 + + x = 1. 5. (6 points) The velocity of a falling ball is given by v(t) = 9.8t 5 m/s, and its initial height is 1000 m. How high is the ball after 10 seconds? Antidifferentiate v, we get: By assumption, s(0) = s 0 = 1000, so s(t) = 4.9t 5t + s 0. s(t) = 4.9t 5t + 1000. At t = 10s, s(10) = 4.9 10 5 10 + 1000 = 60 m 6. (18 points) For the function f(x) = x x + 1 : (a) (4 points) Find x- and y- intercepts, horizontal and vertical asymptotes, and symmetries. x-intercept: Solve f(x) = 0, get x = 0, so the x-intercept is (0, 0). y-intercept: Set x = 0, get f(x) = 0, so the y-intercept is also (0, 0).
x Horizontal asymptote(s): lim = 0, so y = 0 is the horizontal asymptote. x ± x + 1 Vertical asymptote(s): Since the domain of f is R, there are no vertical asymptotes. Symmetries: f( x) = x = f(x), so f is odd and hence, the graph of f is x + 1 symmetric over the origin. (b) (6 points) Find points where f has local maxima and minima (both x and y values), and intervals on which f is increasing and decreasing. f (x) = (x + 1) x(x) = x + x)(1 + x) = (1. (x + 1) (x + 1) (x + 1) So, f (x) = 0 iff (1 x)(1 + x) = 0, iff x = 1 or x = 1. x 1 1 f (x) 0 + 0 f(x) Decreasing Min Increasing Max Decreasing From the above table: f has a local min at x = 1, f( 1) = 1; f has a local max at x = 1, f(1) = 1. The function f is decreasing on (, 1) (1, ). It is increasing on ( 1, 1). (c) (6 points) Find points of inflection (both x and y values), and intervals on which f is concave up and down. f (x) =... = 4 x(x ) (x + 1). So, f (x) = 0 iff x(x ) = 0, iff x = 0 or x = ±. x 0 f (x) 0 + 0 0 + f(x) concave down concave up concave down concave up From the above table: f has inflection points at x = 0 and x = ±, f(0) = 0, f( ) =, f( ) =. The function f is concave down on (, ) (0, ). It is concave up on (, 0) (, ). (d) ( points) Sketch the graph of f(x). 7. (10 points) A rectangular box with a square bottom and an open top has a volume of 10 cubic inches. Material for the base costs 5 cents per square inch and material for the sides costs cents per square inch. Find the cost of the cheapest such box.
Let x denote the the side of the square bottom and y denote the height of the box. We have: Volume = x y = 10. We want to minimize: 5x + 8xy. From the first equation y = 10 x. So, 5x + 8xy = 5x + 8x 10 x = 5x + 80 := C(x). x C (x) = 10x 80 x. C (x) = 0 10x 80 = 0 x = 8 x = 8 =. x C (x) = 10 + 160 x. C () = 10 + 160 = 0 > 0. So, by the Second Derivative Test, C has a 8 local min at x =. Thus, the cost of the cheapest such box is C() = 5() + 80 = 60. [Second way] Since x > 0, by the Arithmetic-Geometric Means Inequality: C(x) = 5x + 80 x = 5x + 40 x + 40 x 5x 40 x 40 x = 8000 = 60. This immediately gives a global min. The equality happens iff 5x = 40 x, iff x = 8, iff x =. END OF EXAM