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Unformatted text preview: Granillo, Yvette Homework 9 Due: Oct 27 2005, 3:00 am Inst: Edward Odell 1 This printout should have 21 questions. Multiplechoice 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 (2 , 1), deter mine the value of f (1). 1. f (1) = 1 2. f (1) = 2 3. f (1) = 0 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 = 2 in ( 3 , 3), it follows that f (2) = 0 , f (2) = 1 . These equations are enough to determine the values of a and b . Indeed, f (2) = 4 + a = 0 , so a = 4, in which case f (2) = 4 8 + b = 1 , so b = 3. Consequently, f (1) = 1 + a + b = 0 . 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 + 2 on [ 1 , 1]. (i) Find the derivative of f . 1. f ( x ) = 2 x 1 x 2 2. f ( x ) = 2 x 2 1 x 2 3. f ( x ) = 1 x 2 2 x 2 4. f ( x ) = p 1 x 2 5. f ( x ) = 2 x p 1 x 2 6. f ( x ) = 1 2 x 2 1 x 2 correct 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 Granillo, Yvette Homework 9 Due: Oct 27 2005, 3:00 am Inst: Edward Odell 2 2. x = 1 2 3. x = 1 2 correct 4. x = 1 4 5. x = 1 2 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 = 2 2. abs. min. value = 3 3. abs. min. value = 1 4. abs. min. value = 1 2 5. abs. min. value = 3 2 correct 6. abs. min. value = 5 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) = 2 , while f  1 2 = 3 2 , f 1 2 = 5 2 . Consequently, abs. min. value = 3 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 3 x 2 + 5 x + 4 on the interval [0 , 3]....
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This test prep was uploaded on 04/16/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.
 Spring '08
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