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Unformatted text preview: syed (sms3768) Quest HW 8 seckin (56425) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the linearization of f ( x ) = 1 7 + x at x = 0. 1. L ( x ) = 1 7 parenleftBig 1 + 1 14 x parenrightBig 2. L ( x ) = 1 7 parenleftBig 1 1 7 x parenrightBig 3. L ( x ) = 1 7 parenleftBig 1 1 14 x parenrightBig correct 4. L ( x ) = 1 7 parenleftBig 1 + 1 14 x parenrightBig 5. L ( x ) = 1 7 1 7 x 6. L ( x ) = 1 7 + 1 7 x Explanation: The linearization of f is the function L ( x ) = f (0) + f (0) x . But for the function f ( x ) = 1 7 + x = (7 + x ) 1 / 2 , the Chain Rule ensures that f ( x ) = 1 2 (7 + x ) 3 / 2 . Consequently, f (0) = 1 7 , f (0) = 1 14 7 , and so L ( x ) = 1 7 parenleftBig 1 1 14 x parenrightBig . 002 10.0 points Use linear appproximation to estimate the value of 15 1 / 4 . ( Hint : (16) 1 / 4 = 2.) 1. 15 1 / 4 15 8 2. 15 1 / 4 31 16 3. 15 1 / 4 2 4. 15 1 / 4 63 32 correct 5. 15 1 / 4 61 32 Explanation: Set f ( x ) = x 1 / 4 , so that f (16) = 2 as the hint indicates. Then df dx = 1 4 x 3 / 4 . By differentials, therefore, we see that f ( a + x ) f ( a ) df dx vextendsingle vextendsingle vextendsingle x = a x = x 4 a 3 / 4 . Thus, with a = 16 and x = 1, 15 1 / 4 2 = 1 32 . Consequently, 15 1 / 4 63 32 . 003 10.0 points If f is the function defined on [ 4 , 4] by f ( x ) = x +  x   4 , which of the following properties does f have? A. Differentiable at x = 0 . B. Absolute maximum at x = 0 . syed (sms3768) Quest HW 8 seckin (56425) 2 1. neither of them correct 2. B only 3. both of them 4. A only Explanation: Since  x  = braceleftbigg x , x , x , x < , we see that f ( x ) = x +  x  4 = braceleftbigg 2 x 4 , x , 4 , x < . Thus on [ 4 , 4] the graph of f is 2 4 2 4 2 4 2 4 Consequently: A. False: f is continuous, but not differen tiable at x = 0 ....
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 Spring '10
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