How do you find the critical points for #f(x)=3e^(-2x(^2))#?

1 Answer
Jun 10, 2018

Answer:

#(0,3)# is a maximum critical point

Explanation:

First, you need to differentiate your equation

#f(x)=3e^(-2x^2)#
#f'(x)=-4xtimes3e^(-2x^2)#
#f'(x)=-12xe^(-2x^2)#
#f''(x)=-12xtimes-4xe^(-2x^2)+e^(-2x^2)times-12#
#f''(x)=48x^2e^(-2x^2)-12e^(-2x^2)#

For stationary points/critical points, #f'(x)=0#

#-12xe^(-2x^2)=0#

#-12x=0# or #e^(-2x^2)=0#

There is no solution for #e^(-2x^2)=0# since the graph NEVER goes to 0. It only APPROACHES 0.

#-12x=0#
#x=0#

To find whether it is maximum or minimum, you sub #x=0# into #f''(x)#. If the answer is greater than zero ie #>0#, then it is a minimum. If the answer is smaller than zero ie #<0#, then it is a maximum.

#f''(0)=-12e^0 = -12 <0#
Therefore, at #x=0#, it is a maximum
To find the coordinate, sub #x=0# back into #f(x)# and you will get #y=3# --> #(0,3)# is a maximum