H09-solutions - white (taw933) – H09 – ben-zvi –...

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Unformatted text preview: white (taw933) – H09 – ben-zvi – (55600) 1 This print-out should have 19 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. Differentiable at x = 0 . B. Absolute maximum at x = 0 . 1. neither of them correct 2. B only 3. both 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: f is continuous, but not differen- tiable at x = 0 . B. False: by inspection. 002 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 NOT have? 1. f ′ ( x ) > 0 on (2 , 4) 2. lim x → 4 f ( x ) = 4 3. lim x → 2 + f ( x ) = lim x → 2- f ( x ) 4. local minimum at x = 4 5. local maximum at x = 2 correct Explanation: The given graph has a removable disconti- nuity at x = 4. 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 maximum even if the graph of f has a removable discontinuity at c ; simi- larly, the definition of local minimum allows the graph of f to have a local minimum at a removable disconituity. So it makes sense to ask if f has a local extremum at x = 2 , 4. white (taw933) – H09 – ben-zvi – (55600) 2 Inspection of the graph now shows that the only property f does not have is local maximum at x = 2 . 003 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 ,- 1), deter- mine the value of f (1). 1. f (1) = 11 2. f (1) = 9 3. f (1) = 8 correct 4. f (1) = 12 5. f (1) = 10 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) =- 1 . These equations are enough to determine the values of a and b . Indeed, f ′ (- 2) =- 4 + a = 0 , so a = 4, in which case f (- 2) = 4- 8 + b =- 1 , so b = 3. Consequently, f (1) = 1 + a + b = 8 . 004 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 white (taw933) – H09 – ben-zvi – (55600) 3 4. 2 4 6 2 4 x y 5....
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This note was uploaded on 12/05/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.

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H09-solutions - white (taw933) – H09 – ben-zvi –...

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