Calculus
posted by Anonymous on .
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.

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?