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Unformatted text preview: obiahu (vo879) – HW12 – Schultz – (56395) 1 This printout should have 15 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 + 3 √ x . 1. g ( x ) = 2 √ x ( 4 x 2 + 5 x + 3 ) + C 2. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x − 3 parenrightbigg + C 3. g ( x ) = √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C 4. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C cor rect 5. g ( x ) = 2 √ x ( 4 x 2 + 5 x − 3 ) + C 6. g ( x ) = √ x ( 4 x 2 + 5 x + 3 ) + C Explanation: After division g ′ ( x ) = 4 x 3 / 2 + 5 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 8 5 x 5 / 2 + 10 3 x 3 / 2 + 6 x 1 / 2 = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Find the value of f (0) when f ′′ ( t ) = 2(3 t + 1) and f ′ (1) = 2 , f (1) = 3 . 1. f (0) = 6 2. f (0) = 5 3. f (0) = 4 correct 4. f (0) = 8 5. f (0) = 7 Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 3 t 2 + 2 t + C where C is an arbitrary constant. But if f ′ (1) = 2, then f ′ (1) = 3 + 2 + C = 2 , i.e., C = − 3 . From this it follows that f ′ ( t ) = 3 t 2 + 2 t − 3 , and the most general antiderivative of the latter is f ( t ) = t 3 + t 2 − 3 t + D , where D is an arbitrary constant. But if f (1) = 3, then f (1) = 1 + 1 − 3 + D = 3 , i.e., D = 4 . Consequently, f ( t ) = t 3 + t 2 − 3 t + 4 . At x = 0, therefore, f (0) = 4 . obiahu (vo879) – HW12 – Schultz – (56395) 2 003 10.0 points Find the value of f (0) when f ′ ( t ) = 5 sin2 t , f parenleftBig π 2 parenrightBig = 1 . 1. f (0) = − 5 2. f (0) = − 3 3. f (0) = − 7 4. f (0) = − 6 5. f (0) = − 4 correct Explanation: Since d dx cos mt = − m sin mt , for all m negationslash = 0, we see that f ( t ) = − 5 2 cos 2 t + C where the arbitrary constant C is determined by the condition f ( π/ 2) = 1. But cos 2 t vextendsingle vextendsingle vextendsingle t = π/ 2 = cos π = − 1 . Thus f parenleftBig π 2 parenrightBig = 5 2 + C = 1 , and so f ( t ) = − 5 2 cos 2 t − 3 2 . Consequently, f (0) = − 4 . 004 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 ?...
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This note was uploaded on 09/06/2010 for the course M 32795 taught by Professor Schultz during the Spring '10 term at University of Texas at Austin.
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