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

1 Answer
Andrea S.
Jan 13, 2017

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

Explanation:

Given:

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

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))#

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