review 2 ans - Lescure, Etienne Review 2 Due: Dec 10 2007,...

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Lescure, Etienne – Review 2 – Due: Dec 10 2007, 10:00 pm – Inst: Diane Radin 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points ±ind all the critical values oF f ( x ) = x (2 - x ) 4 . 1. x = 2 5 2. x = 2 3 3. x = - 2 3 4. x = - 2 5 5. x = 2 , 2 3 6. x = 2 , 2 5 correct 7. x = - 2 , 2 3 8. x = - 2 , 2 5 Explanation: As f is a polynomial, it is di²erentiable everywhere, so the only critical values oF f are those values oF c where f 0 ( c ) = 0. Now by the Product and Chain Rules, f 0 ( x ) = (2 - x ) 4 - 4 x (2 - x ) 3 = (2 - x ) 3 (2 - x - 4 x ) = (2 - x ) 3 (2 - 5 x ) . Consequently, the critical values oF f all occur at x = 2 , 2 5 . 002 (part 1 oF 1) 10 points ±ind the absolute minimum value oF f on the interval £ 1 2 , 2 / when f ( x ) = 2 x 2 + 4 x - 3 . 1. abs. min. value = 11 2 2. abs. min. value = 3 correct 3. abs. min. value = 7 4. abs. min. value = 4 5. none oF these 6. abs. min. value = 7 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 0 ( c ) = 0 For some c , 1 2 < c < 2. But f 0 ( x ) = 4 x - 4 x 2 = 4 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 · = 11 2 , f (1) = 3 , f (2) = 7 , we thus see that f has abs. min. value = 3 on the interval £ 1 2 , 2 / . keywords: absolute minimum, critical point 003 (part 1 oF 1) 10 points Let f be the Function defned by f ( x ) = 5 - x 2 / 3 .
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Lescure, Etienne – Review 2 – Due: Dec 10 2007, 10:00 pm – Inst: Diane Radin 2 Consider the following properties: A. has local maximum at x = 0 B. concave down on ( -∞ , 0) (0 , ) Which does f have? 1. both of them 2. neither of them 3. A only correct 4. B only Explanation: The graph of f is 2 4 - 2 - 4 2 4 On the other hand, after diFerentiation, f 0 ( x ) = - 2 3 x 1 / 3 , f 00 ( x ) = 2 9 x 4 / 3 . Consequently, A. TRUE: see graph B. ±ALSE: f 00 ( x ) > 0 , x 6 = 0 . keywords: concavity, local maximum, True/±alse, graph 004 (part 1 of 1) 10 points The derivative, f 0 , of f has graph a b c graph of f 0 Use it to locate the critical point(s) x 0 at which f has a local minimum? 1. x 0 = a, b, c 2. x 0 = c 3. x 0 = b 4. x 0 = b, c 5. x 0 = a correct 6. x 0 = c, a 7. x 0 = a, b 8. none of Explanation: Since the graph of f 0 ( x ) has no ‘holes’, the only critical points of f occur at the x - intercepts of the graph of f 0 , i.e. , at x 0 = a, b, and c . Now by the ²rst derivative test, f will have (i) a local maximum at x 0 if f 0 ( x ) changes from positive to negative as x passes through x 0 ; (ii) a local minimum at x 0 if f 0 ( x ) changes from negative to positive as x passes through x 0 . Consequently, by looking at the sign of f 0 ( x ) near each of x 0 = a, b, and c we see that f has a local minimum only at x 0 = a .
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Lescure, Etienne – Review 2 – Due: Dec 10 2007, 10:00 pm – Inst: Diane Radin 3 keywords: graph, derivative, critical points, local extrema 005 (part 1 of 1) 10 points Determine the interval(s) where f ( x ) = x - sin x is increasing on [0 , 2 π ].
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review 2 ans - Lescure, Etienne Review 2 Due: Dec 10 2007,...

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