If y=e^(−x^2)y=ex2, what are the points of inflection, concavity and critical points?

1 Answer
Jan 13, 2017

f(x) = e^(-x^2)f(x)=ex2 has a relative maximum for x=0x=0, two inflection points in x= +-1/sqrt(2)x=±12, is concave down in the interval between the two inflection points and concave up outside

Explanation:

Given:

f(x) = e^(-x^2)f(x)=ex2

we can calculate the first and second order derivatives:

f'(x) = -2xe^(-x^2)

f''(x) = -2e^(-x^2) +4x^2e^(-x^2)= 2(2x^2-1)e^(-x^2)

we can therefore determine that:

(1) By solving the equation:

f'(x) = 0 => -2xe^(-x^2) = 0

we can see that f(x) has a single critical point for x=0, this point is a relative maximum since f''(0) = -2 < 0

Looking at the second derivative, we can see that 2e^(-x^2) is always positive and non null, so that inflection points and concavity are determined by the factor (2x^2-1), so:

(2) f(x) has two inflection points for:

2x^2-1 = 0 => x=+-1/sqrt(2)

(3) As (2x^2-1) is a second order polynomial with leading positive coefficient, we know that is is negative in the inteval between the roots, and positive outside, so:

f(x) is concave up in (-oo,-1/sqrt(2)) and in (1/sqrt(2),+oo)

f(x) is concave down in (-1/sqrt(2),1/sqrt(2))