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Unformatted text preview: silva (jrs4378) HW 12 gualdani (56455) 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 ) = 3 x 2 + 2 x + 1 x . 1. g ( x ) = 2 x ( 3 x 2 + 2 x 1 ) + C 2. g ( x ) = 2 x ( 3 x 2 + 2 x + 1 ) + C 3. g ( x ) = x ( 3 x 2 + 2 x + 1 ) + C 4. g ( x ) = 2 x parenleftbigg 3 5 x 2 + 2 3 x 1 parenrightbigg + C 5. g ( x ) = x parenleftbigg 3 5 x 2 + 2 3 x + 1 parenrightbigg + C 6. g ( x ) = 2 x parenleftbigg 3 5 x 2 + 2 3 x + 1 parenrightbigg + C cor rect Explanation: After division g ( x ) = 3 x 3 / 2 + 2 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 6 5 x 5 / 2 + 4 3 x 3 / 2 + 2 x 1 / 2 = 2 x parenleftbigg 3 5 x 2 + 2 3 x + 1 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 3 5 x 2 + 2 3 x + 1 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Find the value of f (0) when f ( t ) = 6( t + 1) and f (1) = 6 , f (1) = 4 . 1. f (0) = 1 2. f (0) = 2 3. f (0) = 3 correct 4. f (0) = 1 5. f (0) = 0 Explanation: The most general antiderivative of f has the form f ( t ) = 3 t 2 + 6 t + C where C is an arbitrary constant. But if f (1) = 6, then f (1) = 3 + 6 + C = 6 , i.e., C = 3 . From this it follows that f ( t ) = 3 t 2 + 6 t 3 , and the most general antiderivative of the latter is f ( t ) = t 3 + 3 t 2 3 t + D , where D is an arbitrary constant. But if f (1) = 4, then f (1) = 1 + 3 3 + D = 4 , i.e., D = 3 . Consequently, f ( t ) = t 3 + 3 t 2 3 t + 3 . At x = 0, therefore, f (0) = 3 . silva (jrs4378) HW 12 gualdani (56455) 2 003 10.0 points Find the value of f (0) when f ( t ) = 3 sin2 t , f parenleftBig 2 parenrightBig = 5 . 1. f (0) = 5 2. f (0) = 3 3. f (0) = 4 4. f (0) = 6 5. f (0) = 2 correct Explanation: Since d dx cos mt = m sin mt , for all m negationslash = 0, we see that f ( t ) = 3 2 cos 2 t + C where the arbitrary constant C is determined by the condition f ( / 2) = 5. But cos 2 t vextendsingle vextendsingle vextendsingle t = / 2 = cos = 1 . Thus f parenleftBig 2 parenrightBig = 3 2 + C = 5 , and so f ( t ) = 3 2 cos 2 t + 7 2 . Consequently, f (0) = 2 . 004 10.0 points Consider the following functions: ( A ) F 1 ( x ) = sin 2 x , ( B ) F 2 ( x ) = cos 2 x 2 , ( C ) F 3 ( x ) = cos 2 x 4 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 3 only 2. F 2 only correct 3. F 1 and F 3 only 4. F 2 and F 3 only 5. all of them 6. F 1 only 7. none of them 8. F 1 and F 2 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 ....
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This note was uploaded on 08/29/2010 for the course M 46455 taught by Professor Gualdani during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Gualdani

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