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Unformatted text preview: Version 066 – L FINAL EXAM – Henry – (54974) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x → parenleftBig 3 x 9 e 3 x 1 parenrightBig exists, and if it does, find its value. 1. limit does not exist 2. limit = 9 2 correct 3. limit = 9 4 4. limit = 9 5. limit = 0 6. limit = 3 Explanation: Now 3 x 9 e 3 x 1 = 3 parenleftBig e 3 x 1 3 x x ( e 3 x 1) parenrightBig = f ( x ) g ( x ) where f, g are everywhere differentiable func tions such that lim x → f ( x ) = 0 , lim x → g ( x ) = 0 . Thus L’Hospital’s rule can be applied. But f ′ ( x ) = 9 e 3 x 9 , while g ′ ( x ) = ( e 3 x 1) + 3 xe 3 x , so lim x → f ( x ) g ( x ) = lim x → f ′ ( x ) g ′ ( x ) = lim x → parenleftBig 9( e 3 x 1) e 3 x + 3 xe 3 x 1 parenrightBig . As f ′ and g ′ are differentiable functions such that lim x → f ′ ( x ) = 0 , lim x → g ′ ( x ) = 0 we have to apply L’Hospital’s rule again. But f ′′ ( x ) = 27 e 3 x , g ′′ ( x ) = 6 e 3 x + 9 xe 3 x , from which it follows that lim x → f ′′ ( x ) = 27 , lim x → g ′′ ( x ) = 6 . Consequently, the limit exists and limit = 9 2 . 002 10.0 points Find the value of f x and f y at (1 , 1) when f ( x, y ) = 5 xy 6 x 2 + y 2 . 1. f x vextendsingle vextendsingle vextendsingle (1 , − 1) = 7 , f y vextendsingle vextendsingle vextendsingle (1 , − 1) = 7 cor rect 2. f x vextendsingle vextendsingle vextendsingle (1 , − 1) = 1 , f y vextendsingle vextendsingle vextendsingle (1 , − 1) = 6 3. f x vextendsingle vextendsingle vextendsingle (1 , − 1) = 17 , f y vextendsingle vextendsingle vextendsingle (1 , − 1) = 7 4. f x vextendsingle vextendsingle vextendsingle (1 , − 1) = 17 , f y vextendsingle vextendsingle vextendsingle (1 , − 1) = 3 5. f x vextendsingle vextendsingle vextendsingle (1 , − 1) = 7 , f y vextendsingle vextendsingle vextendsingle (1 , − 1) = 3 Explanation: Differentiating f first with respect to x and then with respect to y we obtain ∂f ∂x = 5 x 2 y 12 x, ∂f ∂y = 5 xy 2 + 2 y. Thus at (1 , 1), f x vextendsingle vextendsingle vextendsingle (1 , − 1) = 7 , f y vextendsingle vextendsingle vextendsingle (1 , − 1) = 7 . 003 10.0 points Version 066 – L FINAL EXAM – Henry – (54974) 2 Evaluate the double integral I = integraldisplay integraldisplay D ( x + 2) dA when D is the bounded region enclosed by y = x and y = x 2 . 1. I = 5 12 correct 2. I = 3 4 3. I = 1 4 4. I = 7 12 5. I = 1 12 Explanation: The area of integration D is the shaded region in the figure To determine the limits of integration, there fore, we have first to find the points of inter section of the line y = x and the parabola y = x 2 . These occur when x 2 = x , i.e. , when x = 0 and x = 1. Thus the double integral can be written as a repeated integral I = integraldisplay 1 bracketleftbiggintegraldisplay x x 2 (...
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This note was uploaded on 02/06/2012 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas.
 Fall '09
 GOGOLEV
 Calculus

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