How do you use the second derivative test to find the local extrema for #f(x)=e^(x^2)#?

1 Answer
Apr 4, 2018

The function #f(x) = e^(x^2)# has a local minimum in #x=0# and no local maximum.

Explanation:

Evaluate the first and second derivatives of the function:

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

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

Solving the equation:

#f'(x) = 0#

#2xe^(x^2) = 0#

as the exponential is never null we can see that the only critical point for the function is #x=0#.

Then we see that;

#f''(0) > 0#

so that the critical point is a local minimum.

graph{e^(x^2) [-2, 2, -1, 10]}