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Math Help - point of max curvature of f(x) = 8e^3x

  1. #1
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    point of max curvature of f(x) = 8e^3x

    f(x) = 8e^3x
    f'(x) = 24e^3x
    f''(x) = 72e^3x

    k(x) = f''(x) / (1 + [f'(x)]^2) ^ (3/2)

    k(x) = 72e^3x / (1 + 576e^6x) ^(3/2)

    now I know i need to find k'(x) and set it equal to 0 and solve for x, but I just cant seem to come up with an appropriate answer. =/

    k'(x) = long complicated quotient rule expression whose value of x escapes me.

    please help!

    thanks!
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  2. #2
    A Plied Mathematician
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    Here's a little trick you can use, if you like: write the denominator in the numerator with opposite sign of the exponent, and use product rule instead. It'll get you the same result, but sometimes it's easier.
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  3. #3
    MHF Contributor

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    Another useful point: a fraction is equal to 0 if and only if its numerator is 0. You can just ignore the denominator.
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