Exam2_Solutions_F10

# Exam2_Solutions_F10 - Math 115 Second Midterm November 16,...

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page 2 1 . [10 points] Given below is a graph of a function f ( x ) and a table for a function g ( x ). 3 5 7 x M 2 2 f L x R x 0 1 2 3 4 g ( x ) 4 3 1 2 20 3 g ( x ) -2 - 5 2 1 2 3 - 1 3 Give answers for the following or write “Does not exist.” No partial credit will be given. i) d dx f ( g ( x )) at x = 0 By the chain rule d dx f ( g ( x )) = f ( g ( x )) g ( x ). At x = 0 we have f ( g (0)) g (0) = f (4) · ( - 2) = 4 . ii) d dx [ f ( x ) g ( x )] at x = 2 By the product rule d dx [ f ( x ) g ( x )] = f ( x ) g ( x ) + f ( x ) g ( x ). At x = 2 we have f (2) g (2) + f (2) g (2) = (2 / 3)(1) + (4 / 3)(1 / 2) = 4 / 3 . iii) d dx b f ( x ) g ( x ) B at x = 4 By the quotient rule d dx b f ( x ) g ( x ) B = f ( x ) g ( x ) - f ( x ) g ( x ) [ g ( x )] 2 . At x = 4 we have f (4) g (4) - f (4) g (4) [ g (4)] 2 = ( - 2)(20 / 3) - (0)( - 1 / 3) (20 / 3) 2 = - 3 / 10 . iv) d dx [ g ( f ( x ))] at x = 3 We know g ( f (3)) = 1 / 2, so for values of y near f (3), g ( y ) looks like a line with slope 1/2. So g ( f ( x )) “looks like” 1 2 f ( x )+ b for some constant b , for x near 3. Since f ( x ) is “pointy” at x = 3, 1 2 f ( x ) + b looks like a vertically compressed version of this pointy graph (near x = 3), which is still pointy. So g ( f ( x )) is also pointy at x = 3, hence not diFerentiable. v)
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## This note was uploaded on 01/28/2012 for the course MATH 115 taught by Professor Blakelock during the Fall '08 term at University of Michigan.

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Exam2_Solutions_F10 - Math 115 Second Midterm November 16,...

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