Posted by Anonymous on Tuesday, March 29, 2011 at 12:14am.
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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