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Unformatted text preview: Su, Yung – Homework 5 – Due: Oct 2 2007, 3:00 am – Inst: Shinko Harper 1 This printout should have 17 questions. Multiplechoice 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 { 1 /x } . 1. y = y (ln x + 1) 2. y = y x 2 (ln x 1) 3. y = xy (2 ln x + 1) 4. y = y (2 ln x 1) 5. y = xy (2 ln x 1) 6. y = y (ln x + 1) 7. y = xy (2 ln x + 1) 8. y = y x 2 (ln x 1) correct 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(3 x 2 + f (5 x 2 )) . 1. 2 x n 3 + 5 g (5 x 2 ) 3 x 2 + f (5 x 2 ) o correct 2. 6 x + 10 xg (5 x 2 ) (3 x 2 + f (5 x 2 )) 2 3. x n 3 + 5 g (5 x 2 ) 3 x 2 + f (5 x 2 ) o 4. 10 xg (5 x 2 ) 3 x 2 + f (5 x 2 ) 5. 3 + 5 g (5 x 2 ) 3 x 2 + f (5 x 2 ) Explanation: By the Chain Rule d dx ln(3 x 2 + f (5 x 2 )) = 6 x + 10 xf (5 x 2 ) 3 x 2 + f (5 x 2 ) . Thus d dx ln(3 x 2 + f (5 x 2 )) = 2 x n 3 + 5 g (5 x 2 ) 3 x 2 + f (5 x 2 ) o . keywords: 003 (part 1 of 1) 10 points Find the absolute minimum of f on the interval [6 , 9] when f ( x ) = 2 n x 5 ln ‡ x 3 5 ·o 10 . 1. abs. min = 5 2. abs. min = 7 . 1 Su, Yung – Homework 5 – Due: Oct 2 2007, 3:00 am – Inst: Shinko Harper 2 3. abs. min = 6 . 17 4. abs. min = 6 correct 5. abs. min = 4 Explanation: As f is differentiable everywhere on (3 , ∞ ), its absolute minimum on the interval [6 , 9] will occur at an endpoint of [6 , 9] or at a local minimum of f in (6 , 9). Now f ( x ) = 2 n 1 5 x 3 o and f 00 ( x ) = 10 ( x 3) 2 . So x = 8 will be a critical point at which f ( x ) will have a local minimum. But f (6) = 8 2 n 8 ln ‡ 3 5 ·o = 7 . 1 , f (8) = 6 , f (9) = 2 n 4 5 ln ‡ 6 5 ·o = 6 . 17 . Thus, on [6 , 9], abs. min = 6 . keywords: 004 (part 1 of 1) 10 points Determine the indefinite integral I = Z 4 x ( x 2) 2 dx. 1. I = 2 ln( x 2) 2 + C 2. I = 4 x 2 + C 3. I = ln( x 2) 4 + 8 ( x 2) 2 + C 4. I = 8 ( 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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This note was uploaded on 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas.
 Spring '09
 Harper
 Calculus

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