M408K Hm. 9 Solutions

# M408K Hm 9 - Mahon Kevin – Homework 9 – Due 3:00 am – Inst Edward Odell 1 This print-out should have 21 questions Multiple-choice questions

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Unformatted text preview: Mahon, Kevin – Homework 9 – Due: Oct 27 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 21 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 If the graph of the function defined on [- 3 , 3] by f ( x ) = x 2 + ax + b has an absolute minimum at (- 1 ,- 2), deter- mine the value of f (1). 1. f (1) = 5 2. f (1) = 4 3. f (1) = 2 correct 4. f (1) = 3 5. f (1) = 1 Explanation: The absolute minimum of f on the inter- val [- 3 , 3] will occur at a critical point c in (- 3 , 3), i.e. , at a solution of f ( x ) = 2 x + a = 0 , or at at an endpoint of [- 3 , 3]. Thus, since this absolute minimum is known to occur at x =- 1 in (- 3 , 3), it follows that f (- 1) = 0 , f (- 1) =- 2 . These equations are enough to determine the values of a and b . Indeed, f (- 1) =- 2 + a = 0 , so a = 2, in which case f (- 1) = 1- 2 + b =- 2 , so b =- 1. Consequently, f (1) = 1 + a + b = 2 . keywords: Stewart5e, absolute minimum, quadratic function 002 (part 1 of 3) 10 points Let f be the function defined by f ( x ) = x p 1- x 2 + 3 on [- 1 , 1]. (i) Find the derivative of f . 1. f ( x ) = p 1- x 2 2. f ( x ) = 2- x 2 √ 1- x 2 3. f ( x ) = 2 x p 1- x 2 4. f ( x ) = 1- 2 x 2 √ 1- x 2 correct 5. f ( x ) = 2 x √ 1- x 2 6. f ( x ) = √ 1- x 2 2 x 2 Explanation: By the Product and Chain Rules, f ( x ) = p 1- x 2- x 2 √ 1- x 2 = (1- x 2 )- x 2 √ 1- x 2 . Consequently, f ( x ) = 1- 2 x 2 √ 1- x 2 . 003 (part 2 of 3) 10 points (ii) Find all the critical points of f in (- 1 , 1). 1. x = 1 √ 2 2. x = 1 2 Mahon, Kevin – Homework 9 – Due: Oct 27 2005, 3:00 am – Inst: Edward Odell 2 3. x = ± 1 2 4. x = 1 4 5. x = ± 1 √ 2 correct 6. x = ± 1 4 Explanation: Since f is differentiable everywhere on (- 1 , 1), the critical points of f in (- 1 , 1) are the solutions of f ( x ) = 1- 2 x 2 √ 1- x 2 = 0 . Consequently, the critical points of f occur at x = ± 1 √ 2 . 004 (part 3 of 3) 10 points (iii) Determine the absolute minimum value of f on [- 1 , 1]. 1. abs. min. value = 4 2. abs. min. value = 2 3. abs. min. value = 7 2 4. abs. min. value = 5 2 correct 5. abs. min. value = 3 6. abs. min. value = 3 2 Explanation: We have to check the values of f at the endpoints of the interval [- 1 , 1] and at the critical points x = ± 1 √ 2 . Now f (- 1) = f (1) = 3 , while f ‡- 1 √ 2 · = 5 2 , f ‡ 1 √ 2 · = 7 2 . Consequently, abs. min. value = 5 2 . keywords: Stewart5e, derivative, critical point absolute minimum, absolute maximum, square root function 005 (part 1 of 1) 10 points Find the absolute minimum value of f ( x ) = 1 3 x 3- 4 x 2 + 12 x + 7 on the interval [0 , 3]....
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## This homework help was uploaded on 04/15/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.

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M408K Hm 9 - Mahon Kevin – Homework 9 – Due 3:00 am – Inst Edward Odell 1 This print-out should have 21 questions Multiple-choice questions

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