HW9 (solutions) - myers (nmm698) HW09 Hamrick (54868) 1...

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Unformatted text preview: myers (nmm698) HW09 Hamrick (54868) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points If f is the function defined on [- 4 , 4] by f ( x ) = x + | x |- 4 , which of the following properties does f have? A. Absolute maximum at x = 0 . B. Continuous at x = 0 . 1. both of them 2. B only correct 3. neither of them 4. A only Explanation: Since | x | = braceleftbigg x, x ,- x, x < , we see that f ( x ) = x + | x |- 4 = braceleftbigg 2 x- 4 , x ,- 4 , x < . Thus on [- 4 , 4] the graph of f is 2 4- 2- 4 2 4- 2- 4 Consequently: A. False: by inspection. B. True: f is continuous, but not differen- tiable at x = 0 . 002 10.0 points If f is a continuous function on [0 , 6] having (1) an absolute maximum at 2 , and (2) an absolute minimum at 4, which one of the following could be the graph of f ? 1. 2 4 6 2 4 x y 2. 2 4 6 2 4 x y 3. 2 4 6 2 4 x y correct myers (nmm698) HW09 Hamrick (54868) 2 4. 2 4 6 2 4 x y 5. 2 4 6 2 4 x y 6. 2 4 6 2 4 x y Explanation: By inspection, only 2 4 6 2 4 x y can be the graph of f . keywords: absolute minimum, absolute maxi- mum, local maximum, local minimum 003 10.0 points Find all the critical values of f ( x ) = x (2- x ) 1 / 5 . 1. x =- 2 , 5 3 2. x = 2 , 5 3 correct 3. x = 2 4. x =- 2 ,- 5 3 5. x = 5 3 6. x =- 2 7. x = 2 ,- 5 3 8. x =- 5 3 Explanation: The critical values of f are those values of c where f ( c ) does not exist or f ( c ) = 0 . Now by the Product and Chain Rules, f ( x ) = (2- x ) 1 / 5- 1 x 5 (2- x ) 4 / 5 = 5 (2- x )- 1 x 5 (2- x ) 4 / 5 = 10- 6 x 5 (2- x ) 4 / 5 . Thus f (2) does not exist, while f ( c ) = 0 when 6 x = 10 . Consequently, the critical values of f occur at x = 2 , 5 3 . 004 10.0 points myers (nmm698) HW09 Hamrick (54868) 3 If f is the function whose graph is given by 2 4 6 2 4 6 which of the following properties does f have? A. f ( x ) > 0 on (- 1 , 2) , B. critical point at x = 2 , C. local minimum at x = 4 . 1. none of them 2. all of them 3. B and C only 4. C only 5. A and C only 6. A only 7. B only correct 8. A and B only Explanation: The given graph has a removable disconti- nuity at x = 4 and a critical point at x = 2. On the other hand, recall that f has a local maximum at a point c when f ( x ) f ( c ) for all x near c . Thus f could have a local max- imum even if the graph of f has a removable discontinuity at c ; similarly, the definition of local minimum allows the graph of f to have a local minimum at a removable disconitu- ity. So it makes sense to ask if f has a local extremum at x = 4. Inspection of the graph now shows of the three properties A. f does not have , B. f has , C. f does not have ....
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HW9 (solutions) - myers (nmm698) HW09 Hamrick (54868) 1...

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