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

Question

4497 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T09:54:53

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

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]}