What are the values and types of the critical points, if any, of #f(x)=3e^(-2x^2))#?

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Answer 1 · 2017-01-08T13:52:03

#x=0# is a local maximum.

We can find critical points by equating the first derivative to zero:

#d/(dx) (3e^(-2x^2)) = -12xe^(-2x^2)#

So #f'(x) = 0 => x=0# and #x=0#is the only critical point.

We can now look at the sign of #f'(x)# and see:

#x < 0 => f'(x) > 0 #
#x > 0 => f'(x) < 0 #

So the point #x=0# is a local maximum.

graph{3e^(-2x^2) [-10, 10, -5, 5]}