CALCL HWK5 - Arwikar, Dev Homework 5 Due: Oct 2 2007, 3:00...

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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, 3:00...

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