# 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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