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Unformatted text preview: Version 037 K Exam 3 Hamrick (54868) 1 This printout should have 18 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 + x + 4 x . 1. g ( x ) = 2 x ( 2 x 2 + x 4 ) + C 2. g ( x ) = 2 x ( 2 x 2 + x + 4 ) + C 3. g ( x ) = 2 x parenleftbigg 2 5 x 2 + 1 3 x 4 parenrightbigg + C 4. g ( x ) = x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg + C 5. g ( x ) = x ( 2 x 2 + x + 4 ) + C 6. g ( x ) = 2 x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg + C cor rect Explanation: After division g ( x ) = 2 x 3 / 2 + x 1 / 2 + 4 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 + 2 3 x 3 / 2 + 8 x 1 / 2 = 2 x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Find the value of f (0) when f ( t ) = sin 2 t , f parenleftBig 2 parenrightBig = 1 . 1. f (0) = 2 2. f (0) = 2 correct 3. f (0) = 1 4. f (0) = 0 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) = 1. But cos 2 t vextendsingle vextendsingle vextendsingle t = / 2 = cos = 1 . Thus f parenleftBig 2 parenrightBig = 1 2 + C = 1 , and so f ( t ) = 1 2 cos 2 t 3 2 . Consequently, f (0) = 2 . 003 10.0 points Determine the inverse, f 1 , of f when f ( x ) = 5 3 x 3 . Version 037 K Exam 3 Hamrick (54868) 2 1. f 1 ( x ) = parenleftBig x 3 5 parenrightBig 1 / 3 2. f 1 ( x ) = parenleftBig 3 x 5 parenrightBig 1 / 3 3. f 1 ( x ) = parenleftBig x 5 3 parenrightBig 1 / 3 4. f 1 ( x ) = parenleftBig 5 x 3 parenrightBig 1 / 3 correct 5. f 1 ( x ) = parenleftBig 3 + x 5 parenrightBig 1 / 3 6. f 1 ( x ) = parenleftBig 5 + x 3 parenrightBig 1 / 3 Explanation: To determine f 1 we solve for x in y = 5 3 x 3 and then interchange x, y . In this case 3 x 3 = 5 y, i.e. x = parenleftBig 5 y 3 parenrightBig 1 / 3 . Consequently, f 1 ( x ) = parenleftBig 5 x 3 parenrightBig 1 / 3 . 004 10.0 points Find the inverse, f 1 , of f ( x ) = 1 e x . 1. f 1 ( x ) = ln parenleftBig 1 x 1 parenrightBig , x > 1 2. f 1 ( x ) = ln( x 1) 3. f 1 ( x ) = ln( x 1) , x > 1 4. f 1 ( x ) = ln(1 x ) , x < 1 correct 5. f 1 ( x ) = ln(1 x ) 6. f 1 ( x ) = ln parenleftBig 1 1 x parenrightBig , x < 1 7. f 1 ( x ) = ln parenleftBig 1 1 x parenrightBig 8. f 1 ( x ) = ln...
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 Fall '08
 Satasivian

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