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HW09-solutions

# HW09-solutions - hinojosa(jah4698 HW09 seckin(57195 This...

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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

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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 , 0 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 . Consequently, f ( x ) = 1 2 x 2 1 x 2 . 005 (part 2 of 3) 10.0 points (ii) Find all the critical points of f in ( 1 , 1). 1. x = ± 1 2 2. x = 1 2 3. x = ± 1 2 correct 4. x = 1 2 5. x = 1 4 6. x = ± 1 4 Explanation:
hinojosa (jah4698) – HW09 – seckin – (57195) 3 Since f is differentiable everywhere on ( 1 , 1), the critical points of f in ( 1 , 1) are the solutions of f ( x ) = 1 2 x 2 1 x 2 = 0 .

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