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Unformatted text preview: momin (rrm497) – Homework 01 – cheng – (58520) 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 ) = 2 x 2 + 5 x + 1 √ x . 1. g ( x ) = 2 √ x ( 2 x 2 + 5 x + 1 ) + C 2. g ( x ) = 2 √ x parenleftbigg 2 5 x 2 + 5 3 x − 1 parenrightbigg + C 3. g ( x ) = 2 √ x ( 2 x 2 + 5 x − 1 ) + C 4. g ( x ) = 2 √ x parenleftbigg 2 5 x 2 + 5 3 x + 1 parenrightbigg + C cor rect 5. g ( x ) = √ x ( 2 x 2 + 5 x + 1 ) + C 6. g ( x ) = √ x parenleftbigg 2 5 x 2 + 5 3 x + 1 parenrightbigg + C Explanation: After division g ′ ( x ) = 2 x 3 / 2 + 5 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 4 5 x 5 / 2 + 10 3 x 3 / 2 + 2 x 1 / 2 = 2 √ x parenleftbigg 2 5 x 2 + 5 3 x + 1 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 2 5 x 2 + 5 3 x + 1 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ′′ ( t ) = 4(3 t − 2) and f ′ (1) = 3 , f (1) = 1 . 1. f ( t ) = 2 t 3 + 8 t 2 − 5 t − 4 2. f ( t ) = 2 t 3 + 4 t 2 − 5 t + 0 3. f ( t ) = 6 t 3 + 8 t 2 − 5 t − 8 4. f ( t ) = 6 t 3 − 8 t 2 + 5 t − 2 5. f ( t ) = 2 t 3 − 4 t 2 + 5 t − 2 correct 6. f ( t ) = 6 t 3 − 4 t 2 + 5 t − 6 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 , momin (rrm497) – Homework 01 – cheng – (58520) 2 i.e., D = − 2 . Consequently, f ( t ) = 2 t 3 − 4 t 2 + 5 t − 2 . 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 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. none of them 2. all of them 3. F 3 only 4. F 2 and F 3 only 5. F 1 only 6. F 2 only 7. F 1 and F 2 only 8. F 1 and F 3 only correct 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 ) Not antiderivative. ( C ) Antiderivative. 004 10.0 points Find the value of f ( π ) when f ′ ( t ) = 4 3 cos 1 3 t − 4 sin 2 3 t and f ( π 2 ) = 2....
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This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas.
 Fall '08
 RAdin
 Calculus

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