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Unformatted text preview: jiwani (amj566) Homework12 Fouli (58395) 1 This printout should have 22 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 + 4 x + 1 x . 1. g ( x ) = 2 x parenleftbigg 2 5 x 2 + 4 3 x + 1 parenrightbigg + C cor rect 2. g ( x ) = 2 x ( 2 x 2 + 4 x 1 ) + C 3. g ( x ) = 2 x ( 2 x 2 + 4 x + 1 ) + C 4. g ( x ) = x parenleftbigg 2 5 x 2 + 4 3 x + 1 parenrightbigg + C 5. g ( x ) = 2 x parenleftbigg 2 5 x 2 + 4 3 x 1 parenrightbigg + C 6. g ( x ) = x ( 2 x 2 + 4 x + 1 ) + C Explanation: After division g ( x ) = 2 x 3 / 2 + 4 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 + 8 3 x 3 / 2 + 2 x 1 / 2 = 2 x parenleftbigg 2 5 x 2 + 4 3 x + 1 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 2 5 x 2 + 4 3 x + 1 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ( t ) = 2(3 t 2) and f (1) = 1 , f (1) = 2 . 1. f ( t ) = t 3 2 t 2 + 2 t + 1 correct 2. f ( t ) = t 3 + 2 t 2 2 t + 1 3. f ( t ) = t 3 + 4 t 2 2 t 1 4. f ( t ) = 3 t 3 2 t 2 + 2 t 1 5. f ( t ) = 3 t 3 4 t 2 + 2 t + 1 6. f ( t ) = 3 t 3 + 4 t 2 2 t 3 Explanation: The most general antiderivative of f has the form f ( t ) = 3 t 2 4 t + C where C is an arbitrary constant. But if f (1) = 1, then f (1) = 3 4 + C = 1 , i.e., C = 2 . From this it follows that f ( t ) = 3 t 2 4 t + 2 . The most general antiderivative of f is thus f ( t ) = t 3 2 t 2 + 2 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 1 2 + 2 + D = 2 , i.e., D = 1 . Consequently, f ( t ) = t 3 2 t 2 + 2 t + 1 . jiwani (amj566) Homework12 Fouli (58395) 2 003 10.0 points Find the value of f (0) when f ( t ) = sin 2 t, f parenleftBig 2 parenrightBig = 3 . 1. f (0) = 2 correct 2. f (0) = 0 3. f (0) = 1 4. f (0) = 3 5. f (0) = 1 Explanation: Since d dx cos mt = m sin mt, for all m negationslash = 0, we see that f ( t ) = 1 2 cos 2 t + C where the arbitrary constant C is determined by the condition f ( / 2) = 3. But cos 2 t vextendsingle vextendsingle vextendsingle t = / 2 = cos = 1 . Thus f parenleftBig 2 parenrightBig = 1 2 + C = 3 , and so f ( t ) = 1 2 cos 2 t + 5 2 . Consequently, f (0) = 2 . 004 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 2 , ( B ) F 2 ( x ) = sin 2 x, ( C ) F 3 ( x ) = cos 2 x 4 ....
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This note was uploaded on 07/10/2011 for the course KIN 321M taught by Professor Jensen during the Spring '11 term at University of Texas at Austin.
 Spring '11
 Jensen

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