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Unformatted text preview: Horstman (mdh995) – Homework01 – Radin – (58505) 1 This printout should have 16 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 + 3 x + 1 √ x . 1. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg + C cor rect 2. g ( x ) = √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg + C 3. g ( x ) = √ x ( 4 x 2 + 3 x + 1 ) + C 4. g ( x ) = 2 √ x ( 4 x 2 + 3 x − 1 ) + C 5. g ( x ) = 2 √ x ( 4 x 2 + 3 x + 1 ) + C 6. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + x − 1 parenrightbigg + C Explanation: After division g ′ ( x ) = 4 x 3 / 2 + 3 x 1 / 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 + 2 x 3 / 2 + 2 x 1 / 2 = 2 √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ′′ ( t ) = 4(3 t + 1) and f ′ (1) = 2 , f (1) = 5 . 1. f ( t ) = 6 t 3 − 4 t 2 + 8 t − 5 2. f ( t ) = 6 t 3 + 2 t 2 − 8 t + 5 3. f ( t ) = 2 t 3 + 2 t 2 − 8 t + 9 correct 4. f ( t ) = 6 t 3 + 4 t 2 − 8 t + 3 5. f ( t ) = 2 t 3 − 2 t 2 + 8 t − 3 6. f ( t ) = 2 t 3 − 4 t 2 + 8 t − 1 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) = 2, then f ′ (1) = 6 + 4 + C = 2 , i.e., C = − 8 . From this it follows that f ′ ( t ) = 6 t 2 + 4 t − 8 . The most general antiderivative of f is thus f ( t ) = 2 t 3 + 2 t 2 − 8 t + D , where D is an arbitrary constant. But if f (1) = 5, then f (1) = 2 + 2 − 8 + D = 5 , Horstman (mdh995) – Homework01 – Radin – (58505) 2 i.e., D = 9 . Consequently, f ( t ) = 2 t 3 + 2 t 2 − 8 t + 9 . 003 10.0 points Consider the following functions: ( A ) F 1 ( x ) = sin 2 x 2 , ( B ) F 2 ( x ) = − cos 2 x 4 , ( C ) F 3 ( x ) = − cos 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 2 only 2. F 1 only 3. F 2 and F 3 only 4. F 1 and F 3 only 5. all of them correct 6. none of them 7. F 1 and F 2 only 8. 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 ) Antiderivative. ( B ) Antiderivative. ( C ) Antiderivative. 004 10.0 points Find the value of f ( π ) when f ′ ( t ) = 2 3 cos 1 3 t − 8 sin 2 3 t and f ( π 2 ) = 6....
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This note was uploaded on 11/30/2008 for the course M 58510 taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin
 Calculus

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