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Unformatted text preview: badruddin (ssb776) – HW02 – Radin – (57410) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine f ( t ) when f ′′ ( t ) = 2(3 t − 4) and f ′ (1) = 3 , f (1) = 2 . 1. f ( t ) = t 3 + 4 t 2 − 8 t + 5 2. f ( t ) = 3 t 3 − 4 t 2 + 8 t − 5 3. f ( t ) = t 3 − 4 t 2 + 8 t − 3 correct 4. f ( t ) = 3 t 3 + 8 t 2 − 8 t − 1 5. f ( t ) = 3 t 3 − 8 t 2 + 8 t − 1 6. f ( t ) = t 3 + 8 t 2 − 8 t + 1 Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 3 t 2 − 8 t + C where C is an arbitrary constant. But if f ′ (1) = 3, then f ′ (1) = 3 − 8 + C = 3 , i.e., C = 8 . From this it follows that f ′ ( t ) = 3 t 2 − 8 t + 8 . The most general antiderivative of f is thus f ( t ) = t 3 − 4 t 2 + 8 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 1 − 4 + 8 + D = 2 , i.e., D = − 3 . Consequently, f ( t ) = t 3 − 4 t 2 + 8 t − 3 . 002 10.0 points Find all functions g such that g ′ ( x ) = x 2 + 2 x + 3 √ x . 1. g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x + 3 parenrightbigg + C cor rect 2. g ( x ) = √ x ( x 2 + 2 x + 3 ) + C 3. g ( x ) = √ x parenleftbigg 1 5 x 2 + 2 3 x + 3 parenrightbigg + C 4. g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x − 3 parenrightbigg + C 5. g ( x ) = 2 √ x ( x 2 + 2 x + 3 ) + C 6. g ( x ) = 2 √ x ( x 2 + 2 x − 3 ) + C Explanation: After division g ′ ( x ) = x 3 / 2 + 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 2 5 x 5 / 2 + 4 3 x 3 / 2 + 6 x 1 / 2 = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x + 3 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x + 3 parenrightbigg + C badruddin (ssb776) – HW02 – Radin – (57410) 2 with C an arbitrary constant. 003 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 4 , ( B ) F 2 ( x ) = cos 2 x 2 , ( C ) F 3 ( x ) = sin 2 x . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 1 and F 3 only 2. F 2 only 3. F 2 and F 3 only 4. F 3 only 5. all of them 6. none of them correct 7. F 1 and F 2 only 8. F 1 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 ....
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This note was uploaded on 09/23/2009 for the course M 57420 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.
 Fall '09
 RAdin
 Calculus

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