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Unformatted text preview: gutierrez (ig3472) HW02 Neitzke (56585) 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 + 5 x + 2 x . 1. g ( x ) = x ( 4 x 2 + 5 x + 2 ) + C 2. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 5 3 x 2 parenrightbigg + C 3. g ( x ) = x parenleftbigg 4 5 x 2 + 5 3 x + 2 parenrightbigg + C 4. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 5 3 x + 2 parenrightbigg + C cor rect 5. g ( x ) = 2 x ( 4 x 2 + 5 x + 2 ) + C 6. g ( x ) = 2 x ( 4 x 2 + 5 x 2 ) + C Explanation: After division g ( x ) = 4 x 3 / 2 + 5 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 8 5 x 5 / 2 + 10 3 x 3 / 2 + 4 x 1 / 2 = 2 x parenleftbigg 4 5 x 2 + 5 3 x + 2 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 4 5 x 2 + 5 3 x + 2 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ( t ) = 4(3 t + 1) and f (1) = 3 , f (1) = 1 . 1. f ( t ) = 2 t 3 4 t 2 + 7 t 4 2. f ( t ) = 6 t 3 + 4 t 2 7 t 2 3. f ( t ) = 2 t 3 + 2 t 2 7 t + 4 correct 4. f ( t ) = 2 t 3 2 t 2 + 7 t 6 5. f ( t ) = 6 t 3 + 2 t 2 7 t + 0 6. f ( t ) = 6 t 3 4 t 2 + 7 t 8 Explanation: The most general antiderivative of f has the form f ( t ) = 6 t 2 + 4 t + C where C is an arbitrary constant. But if f (1) = 3, then f (1) = 6 + 4 + C = 3 , i.e., C = 7 . From this it follows that f ( t ) = 6 t 2 + 4 t 7 . The most general antiderivative of f is thus f ( t ) = 2 t 3 + 2 t 2 7 t + D , where D is an arbitrary constant. But if f (1) = 1, then f (1) = 2 + 2 7 + D = 1 , i.e., D = 4 . Consequently, f ( t ) = 2 t 3 + 2 t 2 7 t + 4 . gutierrez (ig3472) HW02 Neitzke (56585)...
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 Spring '09
 GOGOLEV
 Calculus

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