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Unformatted text preview: Tran, Tony Homework 15 Due: Dec 4 2007, 3:00 am Inst: Samuels 1 This printout should have 18 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 Determine if lim x tan 1 1 + x 3 + 3 x exists, and if it does, find its value. 1. limit = 4 2. limit = 0 3. limit does not exist 4. limit = 3 5. limit = 2 6. limit = 6 correct Explanation: Since lim x 1 + x 3 + 3 x = 1 3 , we see that lim x tan 1 1 + x 3 + 3 x exists, and that the limit = tan 1 1 3 = 6 . keywords: limit, limit at infinity, arctan, 002 (part 1 of 1) 10 points Find the derivative of f ( x ) = 2sin 1 ( e x ) . 1. f ( x ) = e x 1 e 2 x 2. f ( x ) = e x 1 + e 2 x 3. f ( x ) = 1 1 e 2 x 4. f ( x ) = 2 1 e 2 x 5. f ( x ) = 2 e x 1 e 2 x correct 6. f ( x ) = 1 1 + e 2 x 7. f ( x ) = 2 1 + e 2 x 8. f ( x ) = 2 e x 1 + e 2 x Explanation: Since d dx sin 1 x = 1 1 x 2 , d dx e ax = ae ax , the Chain Rule ensures that f ( x ) = 2 e x 1 e 2 x . keywords: 003 (part 1 of 1) 10 points Determine f ( x ) when f ( x ) = tan 1 x 7 x 2 . ( Hint : first simplify f .) 1. f ( x ) = 7 7 + x 2 2. f ( x ) = x x 2 7 Tran, Tony Homework 15 Due: Dec 4 2007, 3:00 am Inst: Samuels 2 3. f ( x ) = 1 7 x 2 correct 4. f ( x ) = 7 7 x 2 5. f ( x ) = x x 2 + 7 Explanation: If tan = x 7 x 2 , then by Pythagoras theorem applied to the right triangle p 7 x 2 x 7 we see that sin = x 7 . Thus f ( x ) = sin 1 x 7 . Consequently, f ( x ) = 1 7 x 2 . Alternatively, we can differentiate f using the Chain Rule and the fact that d dx tan 1 x = 1 1 + x 2 . . keywords: 004 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 6tan 1 ( e 2 x ) + 3 e 2 x . 1. f ( x ) = 1 1 + e 4 x ( 3 e 2 x + 3 e 2 x ) 2. f ( x ) = 1 1 e 4 x ( 3 e 2 x + 3 e 2 x ) 3. f ( x ) = 2 1 + e 4 x ( 3 e 2 x + 3 e 2 x ) 4. f ( x ) = 1 1 + e 4 x ( 3 e 2 x 3 e 2 x ) 5. f ( x ) = 2 1 e 4 x ( 3 e 2 x + 3 e 2 x ) 6. f ( x ) = 2 1 + e 4 x ( 3 e 2 x 3 e 2 x ) cor rect Explanation: By the Chain Rule, f ( x ) = 2 6 e 2 x 1 + e 4 x 3 e 2 x since d dx tan 1 x = 1 1 + x 2 ( e 2 x ) 2 = e 4 x . The expression for f can now be simplified by bringing the right hand side to a common denominator, for then f ( x ) = 2 6 e 2 x 3 e 2 x (1 + e 4 x ) 1 + e 4 x = 2 6 e 2 x 3 e 2 x 3 e 2 x 1 + e 4 x . Consequently, f ( x ) = 2 1 + e 4 x ( 3 e 2 x 3 e 2 x ) . keywords: 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = tan 1 x 3 ln r 3 + x 3 x !...
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This note was uploaded on 04/27/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz

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