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
