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Unformatted text preview: mao (tm23477) HW02 Radin (56570) 1 This printout should have 10 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 + 5 x . 1. g ( x ) = 2 x ( 4 x 2 + x + 5 ) + C 2. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x 5 parenrightbigg + C 3. g ( x ) = x ( 4 x 2 + x + 5 ) + C 4. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 5 parenrightbigg + C cor rect 5. g ( x ) = 2 x ( 4 x 2 + x 5 ) + C 6. g ( x ) = x parenleftbigg 4 5 x 2 + 1 3 x + 5 parenrightbigg + C Explanation: After division g ( x ) = 4 x 3 / 2 + x 1 / 2 + 5 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 + 10 x 1 / 2 = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 5 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 5 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ( t ) = 6(2 t 1) and f (1) = 6 , f (1) = 4 . 1. f ( t ) = 6 t 3 + 6 t 2 6 t 2 2. f ( t ) = 2 t 3 + 6 t 2 6 t + 2 3. f ( t ) = 2 t 3 3 t 2 + 6 t 1 correct 4. f ( t ) = 6 t 3 6 t 2 + 6 t 2 5. f ( t ) = 2 t 3 + 3 t 2 6 t + 5 6. f ( t ) = 6 t 3 3 t 2 + 6 t 5 Explanation: The most general antiderivative of f has the form f ( t ) = 6 t 2 6 t + C where C is an arbitrary constant. But if f (1) = 6, then f (1) = 6 6 + C = 6 , i.e., C = 6 . From this it follows that f ( t ) = 6 t 2 6 t + 6 . The most general antiderivative of f is thus f ( t ) = 2 t 3 3 t 2 + 6 t + D , where D is an arbitrary constant. But if f (1) = 4, then f (1) = 2 3 + 6 + D = 4 , mao (tm23477) HW02 Radin (56570) 2 i.e., D = 1 . Consequently, f...
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This note was uploaded on 05/12/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.
 Spring '08
 RAdin

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