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Midterm2ReviewSolutions

# Midterm2ReviewSolutions - Math 016 Review Solutions for...

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Math 016 Review Solutions for Midterm II Thursday, April 17, 2008, 8-10PM at C20 PC. You need # 2 pencil and a picture ID (covers material up to (and including) section 5.5) The exam will be in the form of multiple choices as most of the university’s large classes. 1. Find the derivative of f ( x ) = - 3 - sin x. Since f ( x ) = - (3 - sin x ) 1 2 , f 0 ( x ) = - 1 2 (3 - sin x ) - 1 2 ( - cos x ) = cos x 2 3 - sin x 2. Find the inflection points of 1 x 2 + 1 . By the quotient rule, the second derivative is - 2( x 2 + 1) 2 - ( - 2 x )(2)( x 2 + 1)(2 x ) ( x 2 + 1) 4 = - 2( x 2 + 1)( x 2 + 1 - 4 x 2 ) ( x 2 + 1) 4 = - 2(1 - 3 x 2 ) ( x 2 + 1) 3 = 0 when x = ± 1 3 Since the second derative changes sign at each of these x -values, we have inflection points ( ± 1 3 , 3 4 ). 3. What is the minimum value of x 3 + y 3 if x + y = 1 and both x and y are nonnegative? Substitute y = 1 - x into x 3 + y 3 to get x 3 + (1 - x ) 3 . The derivative 3 x 2 + 3(1 - x ) 2 ( - 1) = 3 x 2 - 3(1 - 2 x + x 2 ) = 3 x 2 - 3 + 6 x - 3 x 2 = 6 x - 3 = 0 when x = . 5 . The first derivative or second derivative test shows there is a local minimum at x = . 5. Then, y = 1 - x = 1 - . 5 = . 5. The minimum value of x 3 + y 3 is . 5 3 + . 5 3 = . 25. 4. The second derivative of the function y = 1 1 + 2 x is The first derivative is - (1 + 2 x ) - 2 (2) = - 2(1 + 2 x ) - 2 . The second deriative is ( - 2)( - 2)(1 + 2 x ) - 3 (2) = 8(1 + 2 x ) - 3 = 8 (1 + 2 x ) 3 .

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5. The derivative of f ( x ) = 9 e - 2 x is f 0 ( x ) = 9 e - 2 x ( - 2) = - 18 e - 2 x . 6. The derivative of f ( x ) = ( x 2 - 5 x + 10) e x is By the product rule and using ( e x ) 0 = e x , f 0 ( x ) = (2 x - 5) e x + ( x 2 - 5 x + 10) e x = ( x 2 - 3 x + 5) e x .
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