HW09-solutions - hinojosa(jah4698 – HW09 – seckin –(57195 1 This print-out should have 21 questions Multiple-choice questions may continue on

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Unformatted text preview: hinojosa (jah4698) – HW09 – seckin – (57195) 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 (2 − x ) 1 / 5 . 1. x = − 5 3 2. x = − 2 3. x = − 2 , 5 3 4. x = − 2 , − 5 3 5. x = 2 , 5 3 correct 6. x = 2 , − 5 3 7. x = 2 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 . 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. lim x → 2 + f ( x ) = lim x → 2- f ( x ) 2. lim x → 4 f ( x ) = 5 correct 3. local maximum at x = 4 4. f ′ ( x ) > 0 on (4 , 7) 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 hinojosa (jah4698) – HW09 – seckin – (57195) 2 Find all the critical points of f when f ( x ) = x x 2 + 16 . 1. x = 0 , 4 2. x = − 4 , 4 correct 3. x = − 4 , 16 4. x = − 16 , 4 5. x = − 16 , 16 6. x = − 4 , 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 − 3 on [ − 1 , 1]. (i) Find the derivative of f . 1. f ′ ( x ) = 2 − x 2 √ 1 − x 2 2. f ′ ( x ) = 2 x radicalbig 1 − x 2 3. f ′ ( x ) = √ 1 − x 2 2 x 2 4. f ′ ( x ) = 1 − 2 x 2 √ 1 − x 2 correct 5. f ′ ( x ) = radicalbig 1 − x 2 6. f ′ ( x ) = 2 x √ 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 09/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.

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HW09-solutions - hinojosa(jah4698 – HW09 – seckin –(57195 1 This print-out should have 21 questions Multiple-choice questions may continue on

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