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Math Help - absolute extrema

  1. #1
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    absolute extrema

    Hi

    My problem tonight is I don't have a firm enough grasp of e

    The question is to find all local extreme values and determine if any are absolute

    f(x) = x^2 e^-x^2

    So I think f'(x) = x^2 (-2xe^-x^2) + 2xe^x^2
    = (1-x^2)(-2xe^x^2)

    Now the critical point of (1-x^2) is x= 1 or x=-1 but I don't know what the critical point of the other part is

    I am also not sure how to test for absoluteness and how do I know if it is a local min or max - how can I tell if the sign changes???

    sorry lots of questions - but thanks for reading - hopefully you can help

    thanks

    a calculus beginner
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  2. #2
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    f(x) = x^2 e^{-x^2}

    f'(x) = x^2(-2xe^{-x^2}) + 2xe^{-x^2}

    f'(x) = 2xe^{-x^2}(1 - x^2)

    f'(x) = 0 at x = 0 , x = \pm 1

    for x < -1 , f'(x) > 0 ... f(x) is increasing

    for -1 < x < 0 , f'(x) < 0 ... f(x) is decreasing

    for 0 < x < 1 , f'(x) > 0 ... f(x) is increasing

    for x > 1 , f'(x) < 0 ... f(x) is decreasing

    the above analysis indicates maximums at x = -1 and x = 1

    since f(x) is an even function, the absolute max will occur at both values ...

    f(1) = f(-1) = \frac{1}{e}

    f(0) = 0 is an absolute minimum since f(x) \geq 0 for all x
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