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# HW 15 - Tran Tony – Homework 15 – Due Dec 4 2007 3:00...

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Unformatted text preview: Tran, Tony – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: Samuels 1 This print-out should have 18 questions. Multiple-choice 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 + x 3 + √ 3 x ¶ exists, and if it does, find its value. 1. limit = π 4 2. limit = 0 3. limit does not exist 4. limit = π 3 5. limit = π 2 6. limit = π 6 correct Explanation: Since lim x →∞ 1 + x 3 + √ 3 x = 1 √ 3 , we see that lim x →∞ tan- 1 µ 1 + x 3 + √ 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 ) = 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 θ = x √ 7- x 2 , then by Pythagoras’ theorem applied to the right triangle p 7- x 2 x θ √ 7 we see that sin θ = 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 µ 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 µ 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 ) = ˆ tan- 1 x 3- ln r 3 + x 3- x !...
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HW 15 - Tran Tony – Homework 15 – Due Dec 4 2007 3:00...

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