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

Question

Samantha Adkison · Accepted Answer

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

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

Ezra Adams · Answer

undefined To find the critical points of $ f(x) = 3e^{-2x^2} $, you need to find where the derivative of the function is equal to zero or undefined.

First, find the derivative of the function:

$$ f'(x) = \frac{d}{dx} (3e^{-2x^2}) $$

$$ f'(x) = -12xe^{-2x^2} $$

Set $ f'(x) $ equal to zero:

$$ -12xe^{-2x^2} = 0 $$

Solve for $ x $:

$$ -12x = 0 $$

$$ x = 0 $$

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