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Math Help - Maxima and Minima

  1. #1
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    Maxima and Minima

    I need to find the local maxima and minima using the second derivative of X^2/x-2.

    Now I know how to do this with the first derivative which is (x-4)x/(x-2)^2

    (0,0) Local max and (4,8) for local min

    derivative 2 equals 8/(x-2)^3

    I'm lost with this one!
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  2. #2
    Senior Member Twig's Avatar
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    You have the derivatives, both the second and the first.

    Now, if the second derivative is negative for the x-value at which the function attains an extremevalue, then you have a maxima.

    If the second derivative is positive, you have a minima.

    So you evaluate the second derivative for x=0 \mbox{ and } x=4 .

    This is sometimes called the second-derivative test.
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  3. #3
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    Hi!

    Quote Originally Posted by lisa1984wilson View Post
    I need to find the local maxima and minima using the second derivative of X^2/x-2.
    It is f(x) = x^2/(x-2).

    You find the critical points by solving f'(x) = 0

    the second derivative tells you something about min or max.


    Quote Originally Posted by lisa1984wilson View Post
    Now I know how to do this with the first derivative which is (x-4)x/(x-2)^2

    (0,0) Local max and (4,8) for local min

    derivative 2 equals 8/(x-2)^3

    I'm lost with this one!
    Why is that? Everything is correct.

    f''(x) = 8/(x - 2)^3

    f''(0) = -1 < 0 => (0,0) is maximum

    f''(4) = 1 > 0 => (4,8) is minimum

    Yours
    Rapha

    Edit: Someone was faster than me.
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