A MATH 1225 Practice Test 3 (38 pts) NAME: SOLUTIONS CRN:

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1 A MATH 15 Practice Test (8 pts) NAME: SOLUTIONS CRN: Multiple Choice (1 pt each) No partial credit will be given. Clearl circle one answer. No calculator! 1. The concentration of a Drug A in the bloodstream t hours after being administered is given b the equation C(t) = 10 ln(t t + 1). Suppose that a patient were given Drug A two hours ago. Using differentials, what is the approximate change in C over the next fifteen minutes? (A) 0 (B) 1 (C) 1 (D) 1 C. Which of the following statements must be true for some c in the interval [ 1, 5]? (A) f (c) = { f(5) f( 1), where f(x) = (B) c ln() = 1 ( 1) x, x < x +, x (C) f (c) = 0 given f() = f(7) (D) c 1 c = ( ) ( ) ( 1) B. The graph of = f (x), the first derivative of the function f, with domain (, ) is given below. = f (x) 8 10 x At which of the following x-values does f have a local maximum? (A) x = 0 (B) x = (C) x = (D) x = 7 C 1

2 Free Response Show reasoning that is complete and correct b the standards of this course. Whenever using theorems, ou should explicitl check that all hpotheses are satisfied. Improper use of (or the absence of) proper notation will be penalized. No calculator!. (8 pts) Sketch the graph of a function, f, that satisfies all of the following conditions: (a) the domain of f is (, ) (, ) (b) f is continuous on its domain (c) lim x (d) f(x) = lim x (e) f(0) =, f() = 1 (f) x = is a vertical asmptote (g) f (x) > 0 on (, 0) (h) f (x) < 0 on (1, ) (, ) (, ) (i) f is not differentiable at x = 0 (j) f () = 0 (k) f (x) > 0 on (, 0) (0, ) (, ). One possible solution is: = f(x) 8 10 x

3 5. (1 pts) Define g(x) = f(x) + e x ln(7π) on [0, ]. Suppose that both f and g are defined for all x-values on [0, ]. Suppose that the onl local extrema that f has on the interval [0, ] is a local maximum at x = 1 (a) ( pts) Find g (1). Justif our answer. First note that f has a local extrema at x = 1. Note that f is defined on [0, ], so f is continuous. Therefore, x = 1 is a critical number. We conclude that f (1) = 0 (since f (1) must exist). Thus g (1) = e. g (x) = f (x) + xe x + 0 = f (x) xe x, g (1) = f (1) + e g (1) = 0 + e = e (b) ( pts) Determine whether g is increasing or decreasing on (0, 1). Justif our answer. Note that f has a local maximum at x = 1. Since this is the onl extremum on [0, ], f is increasing on (0, 1). We conclude that f (x) > 0 on (0, 1). Note that f (x) = g (x) + e x x. Thus, g (x) = f (x) + xe x. Observe e x > 0 for all x. Also, x 0, x 0, so xe x 0 for x 0. Therefore, g (x) = f (x) + xe x > 0 on (0, 1). Thus, we conclude that g is increasing on (0, 1). (c) ( pts) Given f(0) = 5 and f(1) = 1, find the absolute maximum and absolute minimum values of g on [0, 1]. We know that g exists for all x-values on the interval [0, 1]. So g is continuous on [0, 1]. Therefore, we can appl the Closed Interval Method. Since g > 0 on (0, 1) (b part (b)), g (x) 0 for all x in (0, 1). Thus, g has no critical numbers on the interval (0, 1). Thus, we must onl compare the values of g at the endpoints on the interval [0, 1]. Note that g(1) = f(1) + e (1) ln(7π) = 1 + e 1 ln(7π) is the absolute maximum and is the absolute minimum. g(0) = f(0) + e 0 + ln(7π) = ln(7π) = ln(7π) (d) ( pts) Using a theorem from class, show that g has at most one root on the interval (0, 1). We want to show that g has at most one root on (0, 1). Suppose not. That is, suppose that g has at least two roots on (0, 1). Suppose g has roots at x = a and x = b, where 0 < a < b < 1. We know that g is defined on [0, ] (given above), Since g is differentiable on [0, ], g must be continuous on [0, ]. Thus, g is continuous on [a, b] g is differentiable on (a, b) g(a) = 0 = g(b) (since g has a root at x = a and x = b). Thus, we can appl Rolle s Theorem. That is, g (c) = 0 for some c on (a, b). But we proved in part (b) that g (x) > 0 for all x in (0, 1). Since 0 < a < b < 1, we know that c is in (0, 1). Thus, g (c) > 0 b part (b). But g (c) > 0 contradicts our assumption that g (c) = 0. Since Rolle s Theorem is alwas true, it must be that one of our hpotheses is erroneous. But we know g must be continuous and differentiable b the given information. Therefore, it must be our assumption that g has at least two roots on (0, 1). We conclude that g has at most one root on the interval (0, 1).

4 . (8 pts) Consider the graph of f(x) = xe x x 1 given below x (a) ( pts) Find the linearization to f at x = 1, and plot the linearization on the graph above. Observe f (x) = xe x + e x x. Thus, f(1) = (1)e (1) (1) 1 = e f (1) = (1)e (1) + e (1) (1) = e The linearization of f at center x = a = 1 is given b L(x) = f (a)(x a) + f(a) = f (1)(x 1) + f(1) = (e )(x 1) + (e ). (b) ( pts) Find x using Newton s method to approximate a solution to the equation f(x) = 0 with initial guess x 1 = 1. Draw our approximation on the graph above. x = x 1 f(x 1) f (x 1 ) = x 1 x 1e x1 x 1 1 = 1 e e x1 + x 1 e x1 x 1 e = e e (c) ( pts) What is the relationship between the following two quantities: The solution to the equation L(x) = 0, where L is the linearization of f at x = a. x, the second iteration of Newton s Method used to approximation the root of f(x) = 0 with initial guess x 1 = a. Justif our answer. Consider the general formula for the linearization of a function centered at x = a: L(x) = f(a) + f (a)(x a). If we set L(x) = 0 and solve for x, we obtain the following equation: x = a f(a) f (a) The zero of the linearization centered at x = a is equal to the second iteration of Newton s Method where x = a was our first initial approximation, x 1 of f(x) = 0.

5 7. (7 pts) A cargo ship is carring cargo miles offshore and wishes to reach a warehouse located on the shore 5 miles from the current position of the ship. It can deliver the cargo to a transport vehicle that is waiting onshore. The ship can move at a rate of 1 mph and the vehicle can drive onshore at a rate of mph. Where should the ship land in order to reach the warehouse in the least amount of time? Where should the ship land to reach the warehouse in the greatest amount of time? If the ship is mi. from the shore and 5 mi. from the warehouse, then the point on the shore closest to the ship is 5 = mi. from the warehouse. Cargo Ship Shore Transport Warehouse Let be the location on the shore where the boat lands. The is the distance awa from the warehouse that the truck needs to travel. If we assume that the time to transfer the cargo from the ship to the truck is negligible, the time it takes for the cargo to reach the warehouse is Now, time traveled = distance traveled, so rate T = T boat + T truck T = = Note that If the ship sails directl to the shore, = 0 miles. If the ship sails directl the the warehouse, = miles. If the ship lands between the closest point on the share and the warehouse, then 0 < < miles. Therefore, is restricted to the interval [0, ]. Note that is a polnomial and thus continuous and differentiable everwhere. Also, 9 + is defined everwhere since 9 + > 0, and thus, 9 + is continuous and differentiable everwhere. So the function T is the sum of two functions that are continuous and differentiable on [0, ]. Thus T is also continuous and differentiable on [0, ]. B Extreme Value Theorem, there will be an absolute maximum value and an absolute minimum value for T in [0, ], which will occur at either the critical numbers of T or the endpoints of the interval. We take the derivative of T : dt d = = and solve for when dt d = 0: 5

6 0 = = 1 = + 9 = + 9 = = + 9 = 9 = 0 = = = ± 8 = ± However, since must be in the interval [0, ], the onl critical number is = miles. We now check the values of T for = 0, =, and =. With extreme value theorem, the smallest of the three values will provide the absolute minimum value, and the largest of the three values will provide the absolute maximum value. So, we have T (0) = 9 + = 1. hours ( ) ( ) T = hours T () = = 5 hours Therefore, it takes the least amount of time to reach the warehouse when the ship lands miles up the shoreline, and it takes the most amount of time to reach the warehouse when the ship sails directl to the warehouse.

Math 141: Section 4.1 Extreme Values of Functions - Notes

Math 141: Section 4.1 Extreme Values of Functions - Notes Definition: Let f be a function with domain D. Thenf has an absolute (global) maximum value on D at a point c if f(x) apple f(c) for all x in D