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Unformatted text preview: choi (kc24547) HW02 BERG (56525) 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 ) = 3 x 2 + x + 2 x . 1. g ( x ) = x parenleftbigg 3 5 x 2 + 1 3 x + 2 parenrightbigg + C 2. g ( x ) = 2 x parenleftbigg 3 5 x 2 + 1 3 x + 2 parenrightbigg + C cor rect 3. g ( x ) = 2 x ( 3 x 2 + x 2 ) + C 4. g ( x ) = 2 x parenleftbigg 3 5 x 2 + 1 3 x 2 parenrightbigg + C 5. g ( x ) = 2 x ( 3 x 2 + x + 2 ) + C 6. g ( x ) = x ( 3 x 2 + x + 2 ) + C Explanation: After division g ( x ) = 3 x 3 / 2 + x 1 / 2 + 2 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 6 5 x 5 / 2 + 2 3 x 3 / 2 + 4 x 1 / 2 = 2 x parenleftbigg 3 5 x 2 + 1 3 x + 2 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 3 5 x 2 + 1 3 x + 2 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 6 t 2 + 7 t + 0 4. f ( t ) = 3 t 3 + 3 t 2 7 t + 5 5. f ( t ) = t 3 6 t 2 + 7 t + 2 6. f ( t ) = 3 t 3 + 6 t 2 7 t + 2 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 . choi (kc24547) HW02 BERG (56525)...
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 Spring '10
 ZHENG
 Calculus

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