latestmid - MATH 1300D-MIDTERM2-2003 Solutions Version A...

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MATH 1300D-MIDTERM2-2003 Solutions Version A Multiple Choice Section Question 1 - Let f ( x ) = x 2 - 1 x +2 , then f 0 ( x ) = Solution: Use quotient rule. f 0 ( x ) = 2 x · ( x + 2) - 1 · ( x 2 - 1) ( x + 2) 2 = x 2 + 4 x + 1 ( x + 2) 2 Answer: A. Question 2 - Find dy dx for the equation xy 3 - 3 y = 2 x . Solution: Implicit differentiation. y 3 + x · ± 3 y 2 dy dx - 3 dy dx = 2 ( 3 xy 2 - 3 ) dy dx = 2 - y 3 dy dx = 2 - y 3 3 xy 2 - 3 Answer: B. Question 3 - What are the critical points of f ( x ) = 3 - 1 x + 1 x 2 ? Solution: We have: f 0 ( x ) = 1 x 2 - 2 x 3 = 1 x 3 ( x - 2) . Thus, f 0 (0) is undefined and f 0 (2) = 0. However, x = 0 is not in the domain of the original function f ( x ). Thus, by definition, it is not a critical point. Answer: B. 1
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Question 4 - Let f ( x ) = x 4 - 2 x 2 +3 and let I be the interval [ - 2 , 3]. Which of the following is true for the function f on I? Solution: f is continuous and the domain I is a closed interval. Thus, we only need to plug in the critical points of f and endpoints of I into f and pick the largest and smallest values. Since f 0 ( x ) = 4 x 3 - 4 x = 4 x ( x - 1)( x + 1), the critical points are x = 0 , ± 1. The endpoints of I are 3 and -2. We have: x -2 -1 0 1 3 f ( x ) 11 2 3 2 66 Answer: C. Question 5
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latestmid - MATH 1300D-MIDTERM2-2003 Solutions Version A...

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