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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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This note was uploaded on 04/13/2010 for the course MATH 408L taught by Professor Gogolev during the Spring '09 term at University of Texas.
 Spring '09
 GOGOLEV
 Calculus

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