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Math Help - Calculus

  1. #1
    Junior Member
    Joined
    Sep 2008
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    63

    Calculus

    Find the coordinates of the point(s) on the graph of:

    <br /> <br />
y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x + 1<br />

    where there are horizontal tangents






    Find the x coordinate of the point(s) where:

    <br /> <br />
f(x) = 2x\sqrt{2-x^2}<br />

    has vertical tangents
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  2. #2
    MHF Contributor
    Joined
    Apr 2008
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    1,092
    For both questions, you need to calculate the derivative.

    The derivative of \frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x + 1 is x^2 + x - 2, and so the points at which the original function have horizontal tangents are where the derivative is equal to 0. Hence:

    x^2 + x - 2 = 0

    (x + 2)(x - 1) = 0

    x = -2, 1

    To calculate the derivative of 2x\sqrt{2 - x^2}, use the product rule: h(x) = f(x)\cdot g(x) \implies h'(x) = f(x)g'(x) + g(x)f'(x)

    Then, to answer the question, determine where the derivative is undefined.
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  3. #3
    Junior Member
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    thanks!!
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