Calculus
posted by Anonymous .
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
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^21)
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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