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Calculus

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

a) Inflection Values
b) Intervals on which f(x) is concave downward
c) Intervals on which f(x) is concave upward.

  • Calculus -

    If you look at the graph of this function you will see a standard "bell curve"

    f'(x) = -2x(e^(-x^2))
    f''(x) = -2x(-2x)(e^(-x^2)) - 2(e^(-x^2))
    = 4x^2(e^(-x^2)) - 2(e^(-x^2))
    = 2e^(-x^2)(2x^2 - 1)

    (e^(-x^2)) = 0
    1/(e^(x^2))= 0 -----> no solution

    or

    2x^2 - 1 = 0
    x = ± 1/√2

    sub that back into f''(x) to find the values at the points of inflection

    b) f(x) is concave down when f''(x) < 0

    f''(x) = 2(e^(-x^2))(2x^2-1)
    clearly (e^(-x^2)) or 1/(e^(x^2)) is always positive, so all we need to look at is the 2x^2 -1 part

    concave up for -1/√2 < x < 1/√2

    c) mmmmhhh?

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