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

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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 0 when y = x { x 2 } . 1. y 0 = - y x 2 (ln x - 1) 2. y 0 = xy (2 ln x + 1) correct 3. y 0 = - y (ln x + 1) 4. y 0 = - xy (2 ln x + 1) 5. y 0 = y x 2 (ln x - 1) 6. y 0 = y (ln x + 1) 7. y 0 = - y (2 ln x - 1) 8. y 0 = 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 0 = xy (2 ln x + 1) . keywords: 002 (part 1 of 1) 10 points If g = f 0 , 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 0 (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
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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 0 ( 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. Consequently, I = ln( x - 3) 6 - 18 x - 3 + C . keywords: 005 (part 1 of 1) 10 points Evaluate the definite integral I = Z ln 3 0 2 e x e x + 1 dx . Correct answer: 1 . 38629 . Explanation: Set u = e x + 1 . Then du = e x dx, so Z ln 3 0 2 e x e x + 1 dx = Z 4 2 2 u du since e ln 3 = 3 . But Z 4 2 1 u du = h ln u i 4 2 = ln 2 .
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Valenica, Daniel – Homework 5 – Due: Oct 2 2007, 3:00 am – Inst: Cheng 3 Thus I = 2 ln 2 .
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