HWK5 - Valenica, Daniel Homework 5 Due: Oct 2 2007, 3:00 am...

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Unformatted text preview: Valenica, Daniel Homework 5 Due: Oct 2 2007, 3:00 am Inst: Cheng 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. 001 (part 1 of 1) 10 points Determine y when y = x { x 2 } . 1. y =- y x 2 (ln x- 1) 2. y = xy (2 ln x + 1) correct 3. y =- y (ln x + 1) 4. y =- xy (2 ln x + 1) 5. y = y x 2 (ln x- 1) 6. y = y (ln x + 1) 7. y =- y (2 ln x- 1) 8. y = xy (2 ln x- 1) Explanation: After taking natural logs we see that ln y = x 2 ln x. Thus by implicit differentiation, 1 y dy dx = 2 x ln x + x. Consequently, y = xy (2 ln x + 1) . keywords: 002 (part 1 of 1) 10 points If g = f , determine d dx ln(2 x 2 + f (7 x 2 )) . 1. 14 xg (7 x 2 ) 2 x 2 + f (7 x 2 ) 2. x n 2 + 7 g (7 x 2 ) 2 x 2 + f (7 x 2 ) o 3. 4 x + 14 xg (7 x 2 ) (2 x 2 + f (7 x 2 )) 2 4. 2 + 7 g (7 x 2 ) 2 x 2 + f (7 x 2 ) 5. 2 x n 2 + 7 g (7 x 2 ) 2 x 2 + f (7 x 2 ) o correct Explanation: By the Chain Rule d dx ln(2 x 2 + f (7 x 2 )) = 4 x + 14 xf (7 x 2 ) 2 x 2 + f (7 x 2 ) . Thus d dx ln(2 x 2 + f (7 x 2 )) = 2 x n 2 + 7 g (7 x 2 ) 2 x 2 + f (7 x 2 ) o . keywords: 003 (part 1 of 1) 10 points Find the absolute minimum of f on the interval [5 , 8] when f ( x ) = 4 n x- 5 ln x- 2 5 o- 20 . 1. abs. min = 10 . 21 2. abs. min = 8 . 35 Valenica, Daniel Homework 5 Due: Oct 2 2007, 3:00 am Inst: Cheng 2 3. abs. min = 7 4. abs. min = 8 correct 5. abs. min = 6 Explanation: As f is differentiable everywhere on (2 , ), its absolute minimum on the interval [5 , 8] will occur at an end-point of [5 , 8] or at a local minimum of f in (5 , 8). Now f ( x ) = 4 n 1- 5 x- 2 o and f 00 ( x ) = 20 ( x- 2) 2 . So x = 7 will be a critical point at which f ( x ) will have a local minimum. But f (5) = 12- 4 n 8 ln 3 5 o = 10 . 21 , f (7) = 8 , f (8) = 4 n 3- 5 ln 6 5 o = 8 . 35 . Thus, on [5 , 8], abs. min = 8 . keywords: 004 (part 1 of 1) 10 points Determine the indefinite integral I = Z 6 x ( x- 3) 2 dx. 1. I = ln( x- 3) 6- 18 x- 3 + C correct 2. I = 3 ln( x- 3) 2 + C 3. I = 18 ( x- 3) 2 + C 4. I =- 6 x- 3 + C 5. I = ln( x- 3) 6 + 18 ( x- 3) 2 + C Explanation: Set u = x- 3 ; then du = dx , so I = 6 Z x ( x- 3)- 2 dx = 6 Z ( u + 3) u- 2 du = 6 Z du u + 18 Z u- 2 du. But 6 Z du u = 6 ln | u | + C = ln u 6 + C, while 18 Z u- 2 du =- 18 u- 1 + C....
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This homework help was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas at Austin.

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HWK5 - Valenica, Daniel Homework 5 Due: Oct 2 2007, 3:00 am...

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