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Unformatted text preview: silva (jrs4378) – Hw 09 – gualdani – (56455) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 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 (2 , 4) , B. differentiable at x = 2 , C. local maximum at x = 4 . 1. B and C only 2. A only 3. C only 4. B only 5. all of them 6. none of them 7. A and B only 8. A and C only correct 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 has , B. f does not have , C. f has . 002 10.0 points If the graph of the function defined 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) = 8 2. f (1) = 7 correct 3. f (1) = 6 4. f (1) = 9 5. f (1) = 5 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 . These equations are enough to determine the values of a and b . Indeed, f ′ ( − 2) = − 4 + a = 0 , silva (jrs4378) – Hw 09 – gualdani – (56455) 2 so a = 4, in which case f ( − 2) = 4 − 8 + b = − 2 , so b = 2. Consequently, f (1) = 1 + a + b = 7 . 003 10.0 points If f is a continuous function on [0 , 6] having (1) a local maximum at 2, (2) an absolute minimum at 4 , and (3) an absolute maximum at 6, 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 4. 2 4 6 2 4 x y correct 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 silva (jrs4378) – Hw 09 – gualdani – (56455) 3 can be the graph of f . keywords: absolute minimum, absolute maxi mum, local maximum, local minimum 004 10.0 points Find all the critical points of f when f ( x ) = x x 2 + 4 . 1. x = − 4 , 4 2. x = − 4 , 2 3. x = 0 , 2 4. x = − 2 , 5. x = − 2 , 4 6. x = − 2 , 2 correct Explanation: By the Quotient Rule, f ′ ( x ) = ( x 2 + 4) − 2 x 2 ( x 2 + 4) 2 = 4 − x 2 ( x 2 + 4) 2 ....
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This note was uploaded on 08/29/2010 for the course M 46455 taught by Professor Gualdani during the Spring '10 term at University of Texas.
 Spring '10
 Gualdani

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