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Unformatted text preview: mao (tm23477) Hw 12 gualdani (57180) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find all functions g such that g ( x ) = x 2 + 3 x + 4 x . 1. g ( x ) = 2 x parenleftbigg 1 5 x 2 + x 4 parenrightbigg + C 2. g ( x ) = 2 x ( x 2 + 3 x + 4 ) + C 3. g ( x ) = 2 x parenleftbigg 1 5 x 2 + x + 4 parenrightbigg + C cor rect 4. g ( x ) = x parenleftbigg 1 5 x 2 + x + 4 parenrightbigg + C 5. g ( x ) = 2 x ( x 2 + 3 x 4 ) + C 6. g ( x ) = x ( x 2 + 3 x + 4 ) + C Explanation: After division g ( x ) = x 3 / 2 + 3 x 1 / 2 + 4 x 1 / 2 , so we can now find an antiderivative of each term separately. But d dx parenleftbigg ax r r parenrightbigg = ax r 1 for all a and all r negationslash = 0. Thus 2 5 x 5 / 2 + 2 x 3 / 2 + 8 x 1 / 2 = 2 x parenleftbigg 1 5 x 2 + x + 4 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 1 5 x 2 + x + 4 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ( t ) = 6( t + 1) and f (1) = 2 , f (1) = 4 . 1. f ( t ) = t 3 3 t 2 + 7 t 1 2. f ( t ) = t 3 + 3 t 2 7 t + 7 correct 3. f ( t ) = 3 t 3 + 3 t 2 7 t + 5 4. f ( t ) = t 3 6 t 2 + 7 t + 2 5. f ( t ) = 3 t 3 + 6 t 2 7 t + 2 6. f ( t ) = 3 t 3 6 t 2 + 7 t + 0 Explanation: The most general antiderivative of f has the form f ( t ) = 3 t 2 + 6 t + C where C is an arbitrary constant. But if f (1) = 2, then f (1) = 3 + 6 + C = 2 , i.e., C = 7 . From this it follows that f ( t ) = 3 t 2 + 6 t 7 . The most general antiderivative of f is thus f ( t ) = t 3 + 3 t 2 7 t + D , where D is an arbitrary constant. But if f (1) = 4, then f (1) = 1 + 3 7 + D = 4 , i.e., D = 7 . Consequently, f ( t ) = t 3 + 3 t 2 7 t + 7 . mao (tm23477) Hw 12 gualdani (57180) 2 003 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 4 , ( B ) F 2 ( x ) = sin 2 x , ( C ) F 3 ( x ) = cos 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 3 only 2. F 2 only 3. F 1 only correct 4. all of them 5. F 1 and F 3 only 6. F 1 and F 2 only 7. none of them 8. F 2 and F 3 only Explanation: By trig identities, cos 2 x = 2 cos 2 x 1 = 1 2 sin 2 x , while sin 2 x = 2 sin x cos x . But d dx sin x = cos x, d dx cos x = sin x . Consequently, by the Chain Rule, ( A ) Antiderivative. ( B ) Not antiderivative. ( C ) Not antiderivative. 004 10.0 points Find f ( t ) when f ( t ) = 2 cos 1 3 t 4 sin 2 3 t and f ( 2 ) = 7. 1. f ( t ) = 10 cos 1 3 t 6 sin 2 3 t + 5 2. f ( t ) = 6 cos 1 3 t + 6 sin 2 3 t + 1 3. f ( t ) = 6 sin 1 3 t + 6 cos 2 3 t + 1 correct 4. f ( t ) = 6 sin 1 3 t + 2 cos 2 3 t + 3 5. f ( t ) = 10 sin 1 3 t 6 cos 2 3 t + 5 6. f ( t ) = 6 cos 1 3 t + 2 sin 2 3 t + 3 Explanation: The function f must have the form...
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This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz

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