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Calculus

posted by on .

Find the absolute extrema of the function. (Round your answer to three decimal places.)
f(x) = xe-x2 on [0,2]
Absolute maximum value:
at x =

Absolute minimum value:
at x =

  • Calculus - ,

    f = xe^-x^2
    f' = (1-2x^2) e^-x^2

    so, where is f' = 0?

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