HW10.pdf - young(toy68 HW10 villafuerte altu(53785 This...

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young (toy68) – HW10 – villafuerte altu – (53785) 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 10.0points Find all the critical points of f ( x ) = x (4 - x ) 5 . 1. x = - 4 , 2 3 2. x = - 4 , 1 3. x = - 1 4. x = 4 , 2 3 correct 5. x = 1 6. x = 4 , 1 7. x = - 2 3 8. x = 2 3 Explanation: As f is a polynomial, it is differentiable everywhere, so the only critical points of f are those values of c where f ( c ) = 0. Now by the Product and Chain Rules, f ( x ) = (4 - x ) 5 - 5 x (4 - x ) 4 = (4 - x ) 4 (4 - x - 5 x ) = (4 - x ) 4 (4 - 6 x ) . Consequently, the critical points of f all occur at x = 4 , 2 3 . 002 10.0points 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. critical point at x = 2 2. f ( x ) > 0 on ( - 1 , 2) correct 3. local minimum at x = 4 4. lim x 4 + f ( x ) = lim x 4 - f ( x ) 5. lim x 4 f ( x ) = 4 Explanation: The given graph has a removable disconti- nuity at x = 4. On the other hand, recall that f has a localmaximum 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. Inspection of the graph now shows that the only property f does not have is f ( x ) > 0 on ( - 1 , 2) . 003 10.0points
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young (toy68) – HW10 – villafuerte altu – (53785) 2 If the graph of the function defined on [ - 3 , 3] by f ( x ) = x 2 + ax + b has an absolute minimum at (2 , 2), determine the value of f (1). 1. f (1) = 3 correct 2. f (1) = 4 3. f (1) = 7 4. f (1) = 6 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 , so a = - 4, in which case f (2) = 4 - 8 + b = 2 , so b = 6. Consequently, f (1) = 1 + a + b = 3 . 004 10.0points Find all the critical points of f when f ( x ) = x x 2 + 4 . 1. x = - 4 , 2 2. x = - 2 , 2 correct 3. x = - 4 , 4 4. x = - 2 , 4 5. x = - 2 , 0 6. x = 0 , 2 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 10.0points Determine the absolute minimum value of f ( x ) = x radicalbig 1 - x 2 + 2 on [ - 1 , 1]. 1. absolute min. value = 1 2 2. absolute min. value = 3 2 correct 3. absolute min. value = 5 2 4. absolute min. value = 1 5. absolute min. value = 3
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young (toy68) – HW10 – villafuerte altu – (53785) 3 6. absolute min. value = 2 Explanation: Since f is differentiable everywhere in ( - 1 , 1), the absolute minimum value of f ( x ) will occur at a critical point of f in ( - 1 , 1) or at an end-point x = ± 1.
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