HW9 (solutions)

# HW9 (solutions) - myers(nmm698 HW09 Hamrick(54868 This...

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myers (nmm698) – HW09 – Hamrick – (54868) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points IF f is the Function defned 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 | = b x , x 0 , - x , x < 0 , we see that f ( x ) = x + | x |- 4 = b 2 x - 4 , x 0 , - 4 , x < 0 . Thus on [ - 4 , 4] the graph oF f is 2 4 - 2 - 4 2 4 - 2 - 4 Consequently: A. ±alse: by inspection. B. True: f is continuous, but not di²eren- 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

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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 deFnition 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 . 005 10.0 points If the graph of the function deFned on [ - 3 , 3] by f ( x ) = x 2 + ax + b has an absolute minimum at ( - 2 , 2), deter- mine the value of f (1). 1. f (1) = 13 2. f (1) = 12 3. f (1) = 11 correct 4. f (1) = 10 5. f (1) = 9 Explanation: The absolute minimum of f on the inter- val [ - 3 , 3] will occur at a critical point c in ( - 3 , 3), i.e. , at a solution of f ( x ) = 2 x + a = 0 , or at at an endpoint of [ - 3 , 3]. Thus, since this absolute minimum is known to occur at x = - 2 in ( - 3 , 3), it follows that f ( - 2) = 0 , f ( - 2) = 2 .

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

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