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

Question

Evelyn Agard · Accepted Answer

#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

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))#

Axel Alvarez · Answer

undefined To find the points of inflection, concavity, and critical points of the function $ y = e^{-x^2} $, we need to take its second derivative and analyze it.

First, find the first derivative of $ y $ with respect to $ x $:
$$ \frac{dy}{dx} = -2xe^{-x^2} $$

Then, find the second derivative:
$$ \frac{d^2y}{dx^2} = -2e^{-x^2} + 4x^2e^{-x^2} $$

For points of inflection, set the second derivative equal to zero and solve for $ x $:
$$ -2e^{-x^2} + 4x^2e^{-x^2} = 0 $$
$$ e^{-x^2}(4x^2 - 2) = 0 $$
$$ 4x^2 - 2 = 0 $$
$$ x^2 = \frac{1}{2} $$
$$ x = \pm\frac{1}{\sqrt{2}} $$

Now, we need to determine the concavity of the function. For concave up, the second derivative is positive; for concave down, it's negative.

Evaluate the second derivative at $ x = \pm\frac{1}{\sqrt{2}} $:
$$ \frac{d^2y}{dx^2}\left(\frac{1}{\sqrt{2}}\right) = -2e^{-\frac{1}{2}} + 4\left(\frac{1}{\sqrt{2}}\right)^2e^{-\frac{1}{2}} = -2e^{-\frac{1}{2}} + 2e^{-\frac{1}{2}} = 0 $$
$$ \frac{d^2y}{dx^2}\left(-\frac{1}{\sqrt{2}}\right) = -2e^{-\frac{1}{2}} + 4\left(-\frac{1}{\sqrt{2}}\right)^2e^{-\frac{1}{2}} = -2e^{-\frac{1}{2}} + 2e^{-\frac{1}{2}} = 0 $$

Therefore, the function $ y = e^{-x^2} $ is concave up and has points of inflection at $ x = \pm\frac{1}{\sqrt{2}} $. To find the critical points, set the first derivative equal to zero and solve for $ x $:

$$ -2xe^{-x^2} = 0 $$
$$ x = 0 $$

Thus, the critical point is at $ x = 0 $.