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Unformatted text preview: white (taw933) HW12 benzvi (55600) 1 This printout should have 18 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 ) = 4 x 2 + x + 3 x . 1. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x 3 parenrightbigg + C 2. g ( x ) = x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg + C 3. g ( x ) = x ( 4 x 2 + x + 3 ) + C 4. g ( x ) = 2 x ( 4 x 2 + x + 3 ) + C 5. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg + C cor rect 6. g ( x ) = 2 x ( 4 x 2 + x 3 ) + C Explanation: After division g ( x ) = 4 x 3 / 2 + x 1 / 2 + 3 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 8 5 x 5 / 2 + 2 3 x 3 / 2 + 6 x 1 / 2 = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ( t ) = 6(3 t 1) and f (1) = 6 , f (1) = 5 . 1. f ( t ) = 9 t 3 6 t 2 + 3 t 1 2. f ( t ) = 3 t 3 3 t 2 + 3 t + 2 correct 3. f ( t ) = 3 t 3 + 6 t 2 3 t 1 4. f ( t ) = 9 t 3 + 6 t 2 3 t 7 5. f ( t ) = 9 t 3 3 t 2 + 3 t 4 6. f ( t ) = 3 t 3 + 3 t 2 3 t + 2 Explanation: The most general antiderivative of f has the form f ( t ) = 9 t 2 6 t + C where C is an arbitrary constant. But if f (1) = 6, then f (1) = 9 6 + C = 6 , i.e., C = 3 . From this it follows that f ( t ) = 9 t 2 6 t + 3 . The most general antiderivative of f is thus f ( t ) = 3 t 3 3 t 2 + 3 t + D , where D is an arbitrary constant. But if f (1) = 5, then f (1) = 3 3 + 3 + D = 5 , i.e., D = 2 . Consequently, f ( t ) = 3 t 3 3 t 2 + 3 t + 2 . white (taw933) HW12 benzvi (55600) 2 003 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 2 , ( B ) F 2 ( x ) = cos 2 x 4 , ( C ) F 3 ( x ) = sin 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 1 only 2. F 3 only correct 3. F 2 only 4. none of them 5. F 1 and F 3 only 6. F 1 and F 2 only 7. all 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 ) Not antiderivative. ( B ) Not antiderivative. ( C ) Antiderivative. 004 10.0 points Find the most general antiderivative of the function f ( x ) = 7 cos x 3 sin x ....
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This note was uploaded on 12/05/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.
 Spring '08
 schultz
 Differential Calculus

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