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Unformatted text preview: Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 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 1 + 2 x 2 5 + 4 x ¶ exists, and if it does, find its value. 1. limit = π 6 2. limit = 0 3. limit = π 3 4. limit does not exist 5. limit = π 4 6. limit = π 2 correct Explanation: Since 1 + 2 x 2 5 + 4 x→ ∞ as x → ∞ , we see that lim x →∞ tan 1 µ 1 + 2 x 2 5 + 4 x ¶ exists, and that the limit = tan 1 ∞ = π 2 . keywords: limit, limit at infinity, arctan, 002 (part 1 of 1) 10 points Find the derivative of f ( x ) = 2sin 1 ( e 3 x ) . 1. f ( x ) = 6 e 3 x 1 + e 6 x 2. f ( x ) = 2 e 3 x √ 1 e 6 x 3. f ( x ) = 2 1 + e 6 x 4. f ( x ) = 2 e 3 x 1 + e 6 x 5. f ( x ) = 6 √ 1 e 6 x 6. f ( x ) = 6 1 + e 6 x 7. f ( x ) = 6 e 3 x √ 1 e 6 x correct 8. f ( x ) = 2 √ 1 e 6 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 ) = 6 e 3 x √ 1 e 6 x . keywords: 003 (part 1 of 1) 10 points Determine f ( x ) when f ( x ) = tan 1 ‡ x √ 5 x 2 · . ( Hint : first simplify f .) 1. f ( x ) = 1 √ 5 x 2 correct 2. f ( x ) = √ 5 √ 5 x 2 Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 2 3. f ( x ) = √ 5 √ 5 + x 2 4. f ( x ) = x x 2 + 5 5. f ( x ) = x √ x 2 5 Explanation: If tan θ = x √ 5 x 2 , then by Pythagoras’ theorem applied to the right triangle p 5 x 2 x θ √ 5 we see that sin θ = x √ 5 . Thus f ( x ) = sin 1 ‡ x √ 5 · . Consequently, f ( x ) = 1 √ 5 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 2 x ) + 3 e 2 x . 1. f ( x ) = 1 √ 1 e 4 x ( 2 e 2 x + 3 e 2 x ) 2. f ( x ) = 1 1 + e 4 x ( 2 e 2 x + 3 e 2 x ) 3. f ( x ) = 2 1 + e 4 x ( 2 e 2 x + 3 e 2 x ) 4. f ( x ) = 2 √ 1 e 4 x ( 2 e 2 x + 3 e 2 x ) 5. f ( x ) = 2 1 + e 4 x ( 2 e 2 x 3 e 2 x ) cor rect 6. f ( x ) = 1 1 + e 4 x ( 2 e 2 x 3 e 2 x ) Explanation: By the Chain Rule, f ( x ) = 2 µ 5 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 • 5 e 2 x 3 e 2 x (1 + e 4 x ) 1 + e 4 x ‚ = 2 µ 5 e 2 x 3 e 2 x 3 e 2 x 1 + e 4 x ¶ . Consequently, f ( x ) = 2 1 + e 4 x ( 2 e 2 x 3 e 2 x ) . keywords: 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = ˆ tan 1 x 2 ln r 2 + x 2 x !...
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This note was uploaded on 10/29/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.
 Spring '08
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