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Unformatted text preview: hyun (hh7953) – HW02 – gogolev – (57440) 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 ) = 4(3 t − 2) and f ′ (1) = 3 , f (1) = 1 . 1. f ( t ) = 6 t 3 − 4 t 2 + 5 t − 6 2. f ( t ) = 2 t 3 − 4 t 2 + 5 t − 2 correct 3. f ( t ) = 2 t 3 + 4 t 2 − 5 t + 0 4. f ( t ) = 2 t 3 + 8 t 2 − 5 t − 4 5. f ( t ) = 6 t 3 + 8 t 2 − 5 t − 8 6. f ( t ) = 6 t 3 − 8 t 2 + 5 t − 2 Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 6 t 2 − 8 t + C where C is an arbitrary constant. But if f ′ (1) = 3, then f ′ (1) = 6 − 8 + C = 3 , i.e., C = 5 . From this it follows that f ′ ( t ) = 6 t 2 − 8 t + 5 . The most general antiderivative of f is thus f ( t ) = 2 t 3 − 4 t 2 + 5 t + D , where D is an arbitrary constant. But if f (1) = 1, then f (1) = 2 − 4 + 5 + D = 1 , i.e., D = − 2 . Consequently, f ( t ) = 2 t 3 − 4 t 2 + 5 t − 2 . 002 10.0 points Find all functions g such that g ′ ( x ) = 5 x 2 + 3 x + 5 √ x . 1. g ( x ) = √ x ( 5 x 2 + 3 x + 5 ) + C 2. g ( x ) = 2 √ x ( x 2 + x − 5 ) + C 3. g ( x ) = √ x ( x 2 + x + 5 ) + C 4. g ( x ) = 2 √ x ( x 2 + x + 5 ) + C correct 5. g ( x ) = 2 √ x ( 5 x 2 + 3 x + 5 ) + C 6. g ( x ) = 2 √ x ( 5 x 2 + 3 x − 5 ) + C Explanation: After division g ′ ( x ) = 5 x 3 / 2 + 3 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 2 x 5 / 2 + 2 x 3 / 2 + 10 x 1 / 2 = 2 √ x ( x 2 + x + 5 ) is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x ( x 2 + x + 5 ) + C with C an arbitrary constant. 003 10.0 points Consider the following functions: hyun (hh7953) – HW02 – gogolev – (57440) 2 ( 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 only 2. F 2 only 3. none of them correct 4. all of them 5. F 3 only 6. F 1 and F 2 only 7. F 1 and F 3 only 8. F 2 and F 3 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 . But d dx sin x = cos x, d dx cos x = − sin x . Consequently, by the Chain Rule, ( A ) Not antiderivative. ( B ) Not antiderivative....
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This note was uploaded on 01/18/2010 for the course M 58460 taught by Professor Gogolev during the Fall '09 term at University of Texas.
 Fall '09
 GOGOLEV

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