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Math Help - Absolute Extrema

  1. #1
    Junior Member Sammyj's Avatar
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    Absolute Extrema

    F(x) = x^4 - 18x^2 + 1; [-4,4}

    F'(x) = 4x^3 -36x = 0

    Critical points x = -3, x=3

    x | f(x)= x^4 - 18x^2 + 1
    -------------------------
    -4 | -31
    -3 | -80
    3 | -80
    4 | -31

    I would had said the absolute max is -4 & 4 and min is -3, 3
    The book says max = o, min = -3

    Where did I go wrong?

    f(x) = (1-x)/(x+3); [0,3]
    I got .33333 for the max, -.3333 for the min. The book says max is 0 and min is 3. What did I miss?
    Last edited by Sammyj; November 6th 2008 at 11:36 AM.
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