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CALCL HWK5

# CALCL HWK5 - Arwikar Dev – Homework 5 – Due Oct 2 2007...

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Unformatted text preview: Arwikar, Dev – Homework 5 – Due: Oct 2 2007, 3:00 am – Inst: James Rath 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Is five lucky? Silly numerology? No, it is not. 001 (part 1 of 1) 10 points Determine y when y = x {- 1 /x } . 1. y = xy (2 ln x + 1) 2. y = y (ln x + 1) 3. y = xy (2 ln x- 1) 4. y =- y (2 ln x- 1) 5. y =- y (ln x + 1) 6. y = y x 2 (ln x- 1) correct 7. y =- y x 2 (ln x- 1) 8. y =- xy (2 ln x + 1) Explanation: After taking natural logs we see that ln y =- 1 x ln x. Thus by implicit differentiation, 1 y dy dx = 1 x 2 ln x- 1 x 2 . Consequently, y = y x 2 (ln x- 1) . keywords: 002 (part 1 of 1) 10 points If g = f , determine d dx ln(4 x 2 + f (7 x 2 )) . 1. 4 + 7 g (7 x 2 ) 4 x 2 + f (7 x 2 ) 2. x n 4 + 7 g (7 x 2 ) 4 x 2 + f (7 x 2 ) o 3. 8 x + 14 xg (7 x 2 ) (4 x 2 + f (7 x 2 )) 2 4. 2 x n 4 + 7 g (7 x 2 ) 4 x 2 + f (7 x 2 ) o correct 5. 14 xg (7 x 2 ) 4 x 2 + f (7 x 2 ) Explanation: By the Chain Rule d dx ln(4 x 2 + f (7 x 2 )) = 8 x + 14 xf (7 x 2 ) 4 x 2 + f (7 x 2 ) . Thus d dx ln(4 x 2 + f (7 x 2 )) = 2 x n 4 + 7 g (7 x 2 ) 4 x 2 + f (7 x 2 ) o . keywords: 003 (part 1 of 1) 10 points Find the absolute minimum of f on the interval [3 , 6] when f ( x ) = 2 n x- 4 ln ‡ x- 1 4 ·o- 8 . Arwikar, Dev – Homework 5 – Due: Oct 2 2007, 3:00 am – Inst: James Rath 2 1. abs. min = 1 2. abs. min = 0 3. abs. min = 2 correct 4. abs. min = 2 . 21 5. abs. min = 3 . 54 Explanation: As f is differentiable everywhere on (1 , ∞ ), its absolute minimum on the interval [3 , 6] will occur at an end-point of [3 , 6] or at a local minimum of f in (3 , 6). Now f ( x ) = 2 n 1- 4 x- 1 o and f 00 ( x ) = 8 ( x- 1) 2 . So x = 5 will be a critical point at which f ( x ) will have a local minimum. But f (3) = 4- 2 n 7 ln ‡ 2 4 ·o = 3 . 54 , f (5) = 2 , f (6) = 2 n 2- 4 ln ‡ 5 4 ·o = 2 . 21 . Thus, on [3 , 6], abs. min = 2 . keywords: 004 (part 1 of 1) 10 points Determine the indefinite integral I = Z 4 x ( x- 2) 2 dx. 1. I = ln( x- 2) 4 + 8 ( x- 2) 2 + C 2. I =- 4 x- 2 + C 3. I = 8 ( x- 2) 2 + C 4. I = 2 ln( x- 2) 2 + C 5. I = ln( x- 2) 4- 8 x- 2 + C correct Explanation: Set u = x- 2 ; then du = dx , so I = 4 Z x ( x- 2)- 2 dx = 4 Z ( u + 2) u- 2 du = 4 Z du u + 8 Z u- 2 du. But 4 Z du u = 4 ln | u | + C = ln u 4 + C, while 8 Z u- 2 du =- 8 u- 1 + C....
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CALCL HWK5 - Arwikar Dev – Homework 5 – Due Oct 2 2007...

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