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hw 9 - silva(jrs4378 Hw 09 gualdani(56455 This print-out...

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silva (jrs4378) – Hw 09 – gualdani – (56455) 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 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 ,
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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
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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 , 0 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 . Since f is differentiable everywhere, the only critical points occur at the solutions of f ( x ) = 0, i.e. , at the solutions of 4 x 2 = 0 . Consequently, the only critical points are x = 2 , 2 . 005 10.0 points Find all the critical points of f when f ( x ) = x 4 / 5 ( x 3) 2 . 1. x = 0 , 6 7 , 3 correct 2. x = 0 , 6 7 3. x = 3 7 , 3 4. x = 0 , 3 7 5. x = 6 7 , 3 6. x = 0 , 3 7 , 3 Explanation: Since x 1 / 5 is defined for all x , the function f is defined for all x , but it will not be differ- entiable at x = 0. Thus x = 0 is one critical point. Now by the Chain and Product Rules f ( x ) = 4 5 x 1 / 5 ( x 3) 2 + 2 x 4 / 5 ( x 3) = 2( x 3) 5 braceleftBig
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