math

what point of f(x)= e^(1-x^2) is concave down? I think it is -3/2

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  1. f' = -2x e^(1-x^2) zero when x = 0
    f" = -2x (-2x e^(1-x^2)) -2 e^(1-x^2)
    = (4 x^2 -2)e^(1-x^2)
    when is that negative ?
    e^whatever always +
    so
    when 2 (2x^2-1) is negative

    x^2 always +
    when x^2 <1/2
    -1/sqrt 2 < x <1/sqrt 2 ?

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