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Unformatted text preview: Garcia, Ilse – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: Fonken 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 µ 3 + x 1 + √ 3 x ¶ exists, and if it does, find its value. 1. limit = π 3 2. limit = π 2 3. limit = 0 4. limit = π 4 5. limit = π 6 correct 6. limit does not exist Explanation: Since lim x →∞ 3 + x 1 + √ 3 x = 1 √ 3 , we see that lim x →∞ tan 1 µ 3 + x 1 + √ 3 x ¶ exists, and that the limit = tan 1 1 √ 3 = π 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 ) = 2 e x √ 1 e 2 x correct 2. f ( x ) = 2 1 + e 2 x 3. f ( x ) = e x √ 1 e 2 x 4. f ( x ) = e x 1 + e 2 x 5. f ( x ) = 1 √ 1 e 2 x 6. f ( x ) = 2 e x 1 + e 2 x 7. f ( x ) = 1 1 + e 2 x 8. f ( x ) = 2 √ 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 √ 6 x 2 · . ( Hint : first simplify f .) 1. f ( x ) = x x 2 + 6 2. f ( x ) = x √ x 2 6 Garcia, Ilse – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: Fonken 2 3. f ( x ) = √ 6 √ 6 + x 2 4. f ( x ) = √ 6 √ 6 x 2 5. f ( x ) = 1 √ 6 x 2 correct Explanation: If tan θ = x √ 6 x 2 , then by Pythagoras’ theorem applied to the right triangle p 6 x 2 x θ √ 6 we see that sin θ = x √ 6 . Thus f ( x ) = sin 1 ‡ x √ 6 · . Consequently, f ( x ) = 1 √ 6 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 ) = 5tan 1 ( e 3 x ) + 4 e 3 x . 1. f ( x ) = 3 1 + e 6 x ( e 3 x + 4 e 3 x ) 2. f ( x ) = 2 √ 1 e 6 x ( e 3 x + 4 e 3 x ) 3. f ( x ) = 3 1 + e 6 x ( e 3 x 4 e 3 x ) correct 4. f ( x ) = 2 1 + e 6 x ( e 3 x + 4 e 3 x ) 5. f ( x ) = 3 √ 1 e 6 x ( e 3 x + 4 e 3 x ) 6. f ( x ) = 2 1 + e 6 x ( e 3 x 4 e 3 x ) Explanation: By the Chain Rule, f ( x ) = 3 µ 5 e 3 x 1 + e 6 x 4 e 3 x ¶ since d dx tan 1 x = 1 1 + x 2 ( e 3 x ) 2 = e 6 x . The expression for f can now be simplified by bringing the right hand side to a common denominator, for then f ( x ) = 3 • 5 e 3 x 4 e 3 x (1 + e 6 x ) 1 + e 6 x ‚ = 3 µ 5 e 3 x 4 e 3 x 4 e 3 x 1 + e 6 x ¶ . Consequently, f ( x ) = 3 1 + e 6 x ( e 3 x 4 e 3 x ) . keywords: 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 1 2 ˆ tan 1 x 2 ln r 2 + x 2 x !...
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This note was uploaded on 11/05/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.
 Spring '08
 schultz

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