Calculus 408K Exam 2 Review

Calculus 408K Exam 2 Review - Barkley, Thane Review 2 Due:...

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Unformatted text preview: Barkley, Thane Review 2 Due: Dec 11 2007, 3:00 pm Inst: Louiza Fouli 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find all the critical values of f ( x ) = x (2- x ) 3 . 1. x =- 1 2 2. x = 1 3. x =- 1 4. x = 2 , 1 2 correct 5. x = 1 2 6. x = 2 , 1 7. x =- 2 , 1 8. x =- 2 , 1 2 Explanation: As f is a polynomial, it is differentiable everywhere, so the only critical values of f are those values of c where f ( c ) = 0. Now by the Product and Chain Rules, f ( x ) = (2- x ) 3- 3 x (2- x ) 2 = (2- x ) 2 (2- x- 3 x ) = (2- x ) 2 (2- 4 x ) . Consequently, the critical values of f all occur at x = 2 , 1 2 . 002 (part 1 of 1) 10 points Find the absolute minimum value of f on the interval 1 2 , 2 / when f ( x ) = x 2 + 2 x- 1 . 1. abs. min. value = 13 4 2. abs. min. value = 5 2 3. abs. min. value = 1 4. none of these 5. abs. min. value = 2 correct 6. abs. min. value = 4 Explanation: The absolute minimum value of f on the interval 1 2 , 2 / occurs at an endpoint x = 1 2 , 2 or at a critical point of f in ( 1 2 , 2 ) , i.e. , when f ( c ) = 0 for some c , 1 2 < c < 2. But f ( x ) = 2 x- 2 x 2 = 2 x 3- 1 x 2 . Now the only solution of x 3- 1 = 0 in 1 2 , 2 / is at c = 1. Since f 1 2 = 13 4 , f (1) = 2 , f (2) = 4 , we thus see that f has abs. min. value = 2 on the interval 1 2 , 2 / . keywords: absolute minimum, critical point 003 (part 1 of 1) 10 points Let f be the function defined by f ( x ) = 5- x 2 / 3 . Consider the following properties: Barkley, Thane Review 2 Due: Dec 11 2007, 3:00 pm Inst: Louiza Fouli 2 A. concave up on (- , 0) (0 , ) B. has local minimum at x = 0 Which does f have? 1. both of them 2. A only correct 3. B only 4. neither of them Explanation: The graph of f is 2 4- 2- 4 2 4 On the other hand, after differentiation, f ( x ) =- 2 3 x 1 / 3 , f 00 ( x ) = 2 9 x 4 / 3 . Consequently, A. TRUE: f 00 ( x ) > , x 6 = 0 B. FALSE: see graph. keywords: concavity, local maximum, True/False, graph 004 (part 1 of 1) 10 points The derivative, f , of f has graph a b c graph of f Use it to locate the critical point(s) x at which f has a local maximum? 1. x = a 2. x = b correct 3. x = c, a 4. x = a, b 5. none of a, b, c 6. x = b, c 7. x = a, b, c 8. x = c Explanation: Since the graph of f ( x ) has no holes, the only critical points of f occur at the x- intercepts of the graph of f , i.e. , at x = a, b, and c . Now by the first derivative test, f will have (i) a local maximum at x if f ( x ) changes from positive to negative as x passes through x ; (ii) a local minimum at x if f ( x ) changes from negative to positive as x passes through x ....
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Calculus 408K Exam 2 Review - Barkley, Thane Review 2 Due:...

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