HW 9 Gagan Solutions

HW 9 Gagan Solutions - Nanda Gagan – Homework 9 – Due...

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Unformatted text preview: Nanda, Gagan – Homework 9 – Due: Oct 21 2004, 3:00 am – Inst: R Gompf 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the absolute maximum value of y = 4 3 x 3 + 3 2 x 2- x + 1 on the interval [0 , 2]. 1. max value = 47 3 correct 2. max value = 50 3 3. max value = 83 96 4. max value = 2 5. max value = 1 Explanation: Differentiating y = 4 3 x 3 + 3 2 x 2- x + 1 , we see that y = 4 x 2 + 3 x- 1 . Thus the critical points occur at the solutions of the equation 4 x 2 + 3 x- 1 = (4 x- 1)( x + 1) = 0 , i.e. , at x = 1 4 ,- 1, only the first of which lies in the interval [0 , 2]. But y (0) = 1 , y ‡ 1 4 · = 83 96 , y (2) = 47 3 . Consequently, on [0 , 2] the abs max value = 47 3 . 002 (part 1 of 1) 10 points Determine the absolute maximum value of f ( x ) = 3 + 2 x x 2 + 4 on the interval [- 1 , 2]. 1. abs max = 1 correct 2. abs max = 1 5 3. abs max = 7 8 4. abs max = 3 2 5. none of these Explanation: By the Quotient Rule f ( x ) = 2 x 2 + 8- 6 x- 4 x 2 ( x 2 + 4) 2 = 8- 6 x- 2 x 2 ( x 2 + 4) 2 for all x . Hence the critical points x 1 , x 2 of f are the solutions of the equation 2 x 2 + 6 x- 8 = 0 . But 2 x 2 + 6 x- 8 = 2( x- 1)( x + 4) , so x 1 = 1 which lies inside [- 1 , 2], while x 2 =- 4 which lies outside the interval [- 1 , 2]. Thus the absolute maximum of f on this interval is attained at the point x = 1, x =- 1 or x = 2. Computing the values of f at these points and comparing values we see that the absolute maximum of f on [- 1 , 2] is abs max = 1. 003 (part 1 of 2) 10 points Nanda, Gagan – Homework 9 – Due: Oct 21 2004, 3:00 am – Inst: R Gompf 2 Let f be the function defined on [- 1 , 2] by f ( x ) = 3 x 2 / 3- 2 x. (i) What is its maximum value? Correct answer: 5 . Explanation: The function f is continuous on [- 1 , 2] and differentiable on (- 1 , 0) ∪ (0 , 2) where it has derivative f ( x ) = 2 x 1 / 3- 2 = 2(1- x 1 / 3 ) x 1 / 3 . Since f is not differentiable at x = 0, the critical points of f on (- 1 , 2) thus occur at x = 0 and at the zeros, if any, of f ( x ) = 0, i.e. , when x 1 / 3- 1 = 0, in other words when x = 1. Hence the maximum value of f ( x ) on [- 1 , 2] will occur at x = 0 , 1 or at one of the end-points x =- 1 , 2. But f (- 1) = 5 , f (0) = 0 , f (1) = 1 , f (2) = 3 · 4 1 / 3- 4 . Now 3 · 4 1 / 3- 4 < 5. Consequently, max value = 5 and this occurs at the end-point x =- 1. 004 (part 2 of 2) 10 points (ii) What is its minimum value? Correct answer: 0 . Explanation: The minimum value of f ( x ) will occur at x = 0 , 1 or at one of the end-points x =- 1 , 2. But f (- 1) = 5 , f (0) = 0 , f (1) = 1 , f (2) = 3 · 4 1 / 3- 4 ....
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This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas.

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HW 9 Gagan Solutions - Nanda Gagan – Homework 9 – Due...

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