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

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 -
**Reiny**, Tuesday, March 29, 2011 at 8:44am
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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