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Homework 11

# Homework 11 - Garcia Ilse Homework 11 Due Nov 6 2007 3:00...

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Garcia, Ilse – Homework 11 – Due: Nov 6 2007, 3:00 am – Inst: Fonken 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points f is a continuous ±unction on ( - 5 , 3) whose graph is 2 - 2 - 4 2 4 which o± the ±ollowing properties are satisfed? A. f 00 ( x ) > 0 on ( - 2 , 1), B. f has exactly 2 local extrema, C. f has exactly 4 critical points. 1. none o± them 2. A and B only 3. B and C only correct 4. all o± them 5. A and C only 6. A only 7. B only 8. C only Explanation: A. False: the graph changes concavity at the point x = - 1 in the interval ( - 2 , 1); in ±act, f 00 ( x ) > 0 on ( - 2 , - 1), while f 00 ( x ) < 0 on ( - 1 , 1). B. True: f has a local minimum at x = - 2 and a local maximum at x = 1; the graph f does have a horizontal tangent at ( - 3 , 1), but this is an in²ection point. C. True: f 0 ( x ) = 0 at x = - 3 , 1, while f 0 ( x ) does not exist at x = - 2; in addition, the graph o± f has a vertical tangent at x = - 1. keywords: critical point, local extreme, True/False 002 (part 1 o± 1) 10 points In drawing the graph P f the x -axis and y -axis have been omitted, but the point P = (1 , 1) on the graph has been included. Use calculus to determine which o± the ±ol- lowing f could be. 1. f ( x ) = - 2 3 + 3 x - x 2 - 1 3 x 3 2. f ( x ) = - 4 3 + 5 x - 3 x 2 + 1 3 x 3 3. f ( x ) = 10 3 - 5 x + 3 x 2 - 1 3 x 3 correct 4. f ( x ) = - 8 3 + 3 x + x 2 - 1 3 x 3

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Garcia, Ilse – Homework 11 – Due: Nov 6 2007, 3:00 am – Inst: Fonken 2 5. f ( x ) = 8 3 - 3 x + x 2 + 1 3 x 3 6. f ( x ) = 14 3 - 3 x - x 2 + 1 3 x 3 Explanation: From the graph, f ( x ) → -∞ as x → ∞ . Of the six given choices, therefore, f must have the form f ( x ) = a + bx + cx 2 - 1 3 x 3 with { b, c } being one of { 3 , - 1 } , { 3 , 1 } , {- 5 , 3 } . Now f 0 ( x ) = b + 2 cx - x 2 , f 00 ( x ) = 2( c - x ) . Since the graph has a local miminum at x = 1, it follows that b + 2 c - 1 = 0 , c - 1 > 0 , i.e. , b = - 5 and c = 3. On the other hand, to determine a we use the fact that f (1) = a + b + c - 1 3 = 1 . Consequently, f ( x ) = 10 3 - 5 x + 3 x 2 - 1 3 x 3 . keywords: polynomial, local maximum 003 (part 1 of 1) 10 points If f is a function on ( - 4 , 4) having exactly two critical points and the sign of f 0 , f 00 are given in - 2 0 2 f 0 > 0 f 0 > 0 f 0 > 0 f 0 < 0 f 00 < 0 f 00 > 0 f 00 < 0 decide which of the following could be the graph of f . 1. 2 4 - 2 - 4 2 4 - 2 - 4 2. 2 4 - 2 - 4 2 4 - 2 - 4 3. 2 4 - 2 - 4 2 4 - 2 - 4 4. 2 4 - 2 - 4 2 4 - 2 - 4
Garcia, Ilse – Homework 11 – Due: Nov 6 2007, 3:00 am – Inst: Fonken 3 5. 2 4 - 2 - 4 2 4 - 2 - 4 6. 2 4 - 2 - 4 2 4 - 2 - 4 correct Explanation: For the given sign chart - 2 0 2 f 0 > 0 f 0 > 0 f 0 > 0 f 0 < 0 f 00 < 0 f 00 > 0 f 00 < 0 an inspection of the graphs shows that two of them fail to have exactly two critical points, leaving just four possible graphs for f . To distinguish among these we use the fact that (i) if f 0 ( x ) > 0 on ( a, b ), then f ( x ) is in- creasing on ( a, b ), while (ii) if f 0 ( x ) < 0 on ( a, b ), then f ( x ) is decreasing on ( a, b ), and that (iii) if f 00 ( x ) > 0 on ( a, b ), then the graph is concave UP on ( a, b ), while (iv) if f 00 ( x ) < 0 on ( a, b ), then the graph is concave DOWN on ( a, b ).

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Homework 11 - Garcia Ilse Homework 11 Due Nov 6 2007 3:00...

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