Solutions HW 9

# Solutions HW 9 - mao (tm23477) Hw 09 gualdani (57180) 1...

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Unformatted text preview: mao (tm23477) Hw 09 gualdani (57180) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find all the critical values of f ( x ) = x (1 x ) 2 / 5 . 1. x = 1 2. x = 1 , 5 7 correct 3. x = 1 , 5 7 4. x = 1 , 5 7 5. x = 1 , 5 7 6. x = 1 7. x = 5 7 8. x = 5 7 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 ) = (1 x ) 2 / 5 2 x 5 (1 x ) 3 / 5 = 5 (1 x ) 2 x 5 (1 x ) 3 / 5 = 5 7 x 5 (1 x ) 3 / 5 . Thus f (1) does not exist, while f ( c ) = 0 when 7 x = 5 . Consequently, the critical val- ues of f occur at x = 1 , 5 7 . 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 (4 , 7) 2. local maximum at x = 4 3. lim x 4 f ( x ) = 5 correct 4. lim x 2 + f ( x ) = lim x 2- f ( x ) 5. critical point at x = 2 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 = 4. Inspection of the graph now shows that the only property f does not have is lim x 4 f ( x ) = 5 . 003 10.0 points mao (tm23477) Hw 09 gualdani (57180) 2 Find all the critical points of f when f ( x ) = x x 2 + 16 . 1. x = 0 , 4 2. x = 4 , 16 3. x = 16 , 4 4. x = 4 , 5. x = 4 , 4 correct 6. x = 16 , 16 Explanation: By the Quotient Rule, f ( x ) = ( x 2 + 16) 2 x 2 ( x 2 + 16) 2 = 16 x 2 ( x 2 + 16) 2 . Since f is differentiable everywhere, the only critical points occur at the solutions of f ( x ) = 0, i.e. , at the solutions of 16 x 2 = 0 . Consequently, the only critical points are x = 4 , 4 . 004 (part 1 of 3) 10.0 points Let f be the function defined by f ( x ) = x radicalbig 1 x 2 2 on [ 1 , 1]. (i) Find the derivative of f . 1. f ( x ) = 1 x 2 2 x 2 2. f ( x ) = 1 2 x 2 1 x 2 correct 3. f ( x ) = radicalbig 1 x 2 4. f ( x ) = 2 x radicalbig 1 x 2 5. f ( x ) = 2 x 1 x 2 6. f ( x ) = 2 x 2 1 x 2 Explanation: By the Product and Chain Rules, f ( x ) = radicalbig 1 x 2 x 2 1 x 2 = (1 x 2 ) x 2 1 x 2 ....
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## This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.

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Solutions HW 9 - mao (tm23477) Hw 09 gualdani (57180) 1...

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