HW15 - Jolley Garrett Homework 15 Due Dec 4 2007 3:00 am...

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Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 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 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 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 ) = 2 sin - 1 ( e 3 x ) . 1. f 0 ( x ) = 6 e 3 x 1 + e 6 x 2. f 0 ( x ) = 2 e 3 x 1 - e 6 x 3. f 0 ( x ) = 2 1 + e 6 x 4. f 0 ( x ) = 2 e 3 x 1 + e 6 x 5. f 0 ( x ) = 6 1 - e 6 x 6. f 0 ( x ) = 6 1 + e 6 x 7. f 0 ( x ) = 6 e 3 x 1 - e 6 x correct 8. f 0 ( 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 0 ( x ) = 6 e 3 x 1 - e 6 x . keywords: 003 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = tan - 1 x 5 - x 2 · . ( Hint : first simplify f .) 1. f 0 ( x ) = 1 5 - x 2 correct 2. f 0 ( x ) = 5 5 - x 2
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Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 2 3. f 0 ( x ) = 5 5 + x 2 4. f 0 ( x ) = x x 2 + 5 5. f 0 ( 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 0 ( 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 ) = 5 tan - 1 ( e 2 x ) + 3 e - 2 x . 1. f 0 ( x ) = 1 1 - e 4 x ( 2 e 2 x + 3 e - 2 x ) 2. f 0 ( x ) = 1 1 + e 4 x ( 2 e 2 x + 3 e - 2 x ) 3. f 0 ( x ) = 2 1 + e 4 x ( 2 e 2 x + 3 e - 2 x ) 4. f 0 ( x ) = 2 1 - e 4 x ( 2 e 2 x + 3 e - 2 x ) 5. f 0 ( x ) = 2 1 + e 4 x ( 2 e 2 x - 3 e - 2 x ) cor- rect 6. f 0 ( x ) = 1 1 + e 4 x ( 2 e 2 x - 3 e - 2 x ) Explanation: By the Chain Rule, f 0 ( x ) = 2 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 0 can now be simplified by bringing the right hand side to a common denominator, for then f 0 ( x ) = 2 5 e 2 x - 3 e - 2 x (1 + e 4 x ) 1 + e 4 x = 2 5 e 2 x - 3 e - 2 x - 3 e 2 x 1 + e 4 x . Consequently, f 0 ( 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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Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 3 1. f 0 ( x ) = - 4 x 2 16 - x 4 correct 2. f 0 ( x ) = 2 x 2 16 + x 4 3. f 0 ( x ) = - 4 x 2 16 + x 4 4. f 0 ( x ) = 6 16 - x 4 5. f 0 ( x ) = 2 x 2 16 - x 4 6. f 0 ( x ) = 6 16 + x 4 Explanation: Since ln r 2 + x 2 - x = 1 2 h ln(2 + x ) - ln(2 - x ) i , we see that d dx ˆ ln r 2 + x 2 - x ! = 1 2 1 2 + x + 1 2 - x = 2 4 - x 2 . On the other hand, d dx tan - 1 x 2 · = 1 2 ˆ 1 1 + x 2 4 ! = 2 4 + x 2 . Thus f 0 ( x ) = 2 4 + x 2 - 2 4 - x 2 = 2(4 - x 2 ) - 2(4 + x 2 ) (4 + x 2 )(4 - x 2 ) .
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