Exam 3. MA Exam 3 Fall Solutions. 1. (10 points) Find where the following functions are increasing and decreasing.

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1 Exam 3 Solutions. (0 points) Find where the following functions are increasing and decreasing. (a) (5 points) f(x) = x2 + 2x 3, x 3 2. x2 f(x) = + 2x 3 f (x) = 2x( x3 ) ( + 2x 3 ) 2 Critical points: x = 0, x =. The function is increasing for 0 < x < and decreasing elsewhere. Derivative pt Critical points 2 pt Increasing interval 2 pt Should indicate that derivative pos/neg function incr/decr (b) (5 points) g(x) = x 3 3x 2 9x + 4. g(x) = x 3 3x 2 9x + 4 g (x) = 3x 2 6x 9 = 3 (x 3) (x + ) Critical points: x =, x = 3 Decreasing for < x < 3 and increasing elsewhere Derivative pt Critical points 2 pt Increasing interval 2 pt Should indicate that derivative pos/neg function incr/decr Solutions

2 2. (0 points) Let f(x) = x 2 5x + on [0, 2]. Find a value, c, from the Mean Value Theorem so that f f(b) f(a) (c) =. b a f(b) f(a) b a = 5 2 = 5 2 f (x) = 2x 5 Slope of secant line 2 pt Derivative 3 pt Solve 2c 5 = 5 2 for c c = 5 4 Set up correct equation 3 pt Correct solution 2 pt Solutions Page 2 of

3 3. (2 points) Let f(x) = 2x 5 5x 4, x [, 4]. (a) (4 points) Locate the critical points of f and determine the intervals on which f is increasing and the intervals on which f is decreasing. f(x) = 2x 5 5x 4 f (x) = 0x (x 2) Critical points: x = 0, x = 2. Decreasing for 0 < x < 2 and increasing elsewhere Derivative pt Critical points 2 pt Increasing interval 2 pt Should indicate that derivative pos/neg function incr/decr (b) (4 points) Locate the possible inflection points for f and determine the intervals on which f is concave up and the intervals on which f is concave down. f (x) = 40x 3 60x 2 = 20x 2 (2x 3) Possible inflection points at: x = 0, x = 3 2 Function concave down for x < 3 2 concave up for x > 3 2 has an inflection point at x = 3 2 Possible inflection points 2 pt Concavity pt Inflection point pt (c) (4 points) Evaluate f at the critical points and endpoints. f( ) = 7 f(0) = 0 f(2) = 6 f(4) = 768 Each value pt (d) (4 points) What are the global maximum values and global minimum values of f(x)? Absolute max: f(4) = 768 Absolute min: f(2) = 6 Each value 2 pt Solutions Page 3 of

4 4. (0 points) The measles pathogenesis function f(t) = t(t 2)(t + ) is used to model the development of the disease, where t is measured in days and f(t) represents the number of infected cells per milliliter of plasma. What is the peak infection time for the measles virus? f (t) = 3t t + 2 f (t) = 40 6t Set f (t) = 0 and solve for t Critical point: t = ( ) f (3.839) < 0 concave down Function has a maximum at t = Peak infection time is 4 days. Derivative pt 2nd derivative pt Find critical point 2 pt Correct critical point 2pt Test for maximum 3 pt Answer question pt Solutions Page 4 of

5 5. (8 points) Mezzadra et al. estimate the average individual daily milk consumption for Charolais, Angus, and Hereford calves to be the function M(t) = 6.28t e 0.025t, t 26, where M(t) is the milk consumption (in kg) and t is the age of the calf (in weeks). (a) (5 points) Find the age of a calf at which maximum daily consumption occurs. M (t) = 6.28 ( 0.242t e 0.025t) ( 0.025t e 0.025t) = 6.28t e 0.025t ( 0.242t ) Set M (t) = 0 Critical points: t = 0, t = 9.68 First Derivative Test easiest to check Maximum consumption at t = Derivative pt Find critical point pt Correct critical point pt Test for maximum pt Answer question pt (b) (3 points) How much milk is consumed on this day? M(9.68) = 8.54 kg M(9.68) 2 pt Correct value pt Solutions Page 5 of

6 6. (0 points) A rectangle has its base on the x-axis, its lower left corner at (0, 0), and its upper right corner on the curve y = 2. What is the smallest perimeter the rectangle x can have? Dimensions: x and y Constraint: y = 2 x Objective function: P = 2x + 2y = 2x + 4 x,x > 0 P (x) = 2x + 4 x P (x) = 2 4 x 2 Set P (x) = 0 x = 2 P (t) = 8/x 3 > 0 for all x > 0 Function is always concave up P ( 2 ) a minimum x = 2, y = 2/x = 2 Minimum perimeter is 4 2 Constraint eqn: pt Objective function: pt Derivative: 2 pt Set P = 0 pt pt 2nd derivative: pt Concavity: pt Find x and y: pt Answer question: pt Solutions Page 6 of

7 7. (0 points) Find the following limits (a) (2 points) lim x 0 x cos(x) sin(x). x cos(x) lim x 0 sin(x) cos(x) x sin x = lim x 0 cos(x) = 0 = Correct derivatives pt Correct answer pt ln (x) (b) (2 points) lim x 3x. ln (x) lim x 3x = lim x x 3 = lim x 3x = 0 Correct derivatives pt Correct answer pt ln (x) (c) (2 points) lim x x 2. lim x ln (x) x 2 = lim x = 2. x 2x Correct derivatives pt Correct answer pt ln x (d) (2 points) lim. x x ln x lim = lim x x x x 2 x 2 = lim x = 0 x Correct derivatives pt Correct answer pt Solutions Page 7 of

8 (e) (2 points) If f is continuous, f(2) = 0, and f f(2 + 3x) + f(2 + 5x) (2) = 7, find lim. x 0 x f(2 + 3x) + f(2 + 5x) lim = x 0 x 3f (2 + 3x) + 5f (2 + 5x) = lim x 0 = lim (7f (2 + 3x) + 5f (2 + 5x)) x 0 = = 56. Correct derivatives pt Correct answer pt Solutions Page 8 of

9 8. (0 points) A state game commission introduces 50 deer into newly acquired state game lands. The population N of the herd can be modeled by where t is the time in years. N = 0(5 + 3t) t (a) (5 points) Find the linear approximation to approximate the population of the herd size at t = 5. 0(5 + 3t) N(t) = t N(5) = N 28 (t) = ( t) 2 N (5) = L(t) = (t 5) 9 N(5) pt Correct derivative pt N (5) pt L(t) 2 pt (b) (5 points) Use your approximation to approximate N(6). N(6) L(6) L(t) = (t 5) 9 L(6) = Knowing how to use linearization pt Use their L(t) pt Correct evaluation pt Solutions Page 9 of

10 9. Use Newton s method for five steps to find the root of x 4 2x 2 = 0 starting with x 0 =. (a) ( point) What is f(x) f(x) = x 4 2x 2 pt (b) ( point) What is f (x) f (x) = 4x 3 2 pt The algorithm is x n+ = x n f(x n) f (x n ) and x 0 =. (c) (2 points) x =? x = pt (d) (2 points) x 2 =? x 2 = pt (e) (2 points) x 3 =? x 3 = pt (f) (2 points) x 4 =? x 4 = pt (g) (2 points) x 5 =? x 5 = pt Solutions Page 0 of

11 0. (8 points) If g(x) = ln (a) (4 points) Find g (x). ( ) 3 x. 3 + x ( ) 3 x g(x) = ln 3 + x g (x) = 3 x 3 + x 6 = (3 x)(3 + x) Allocate points equitably depending on approach. Answer = pt 4 pt (b) (4 points) Find g (x). ( ) 3 x g(x) = ln 3 + x g (x) = (3 x) + 2 (3 + x) 2 = 2x (9 x 2 ) 2 Allocate points equitably depending on approach. Answer = pt 4 pt Solutions Page of