How to find critical points for an equation

In calculus, a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0. The value of the function at a critical is a critical value of the function. These definitions admit generalizations to functions of several variables, differentiable maps between R^m and Rⁿ, and differentiable maps between differentiable manifolds.

Lets see a problem from calculus help with probability problems,which explains us
more about this .

Question:-

f(x)=3x⁴-4x³-12x²+6

Differentiating with respect to x

Solution:-

f'(x)=12x³-12x²-24x

we know f'(x)=0 ,so

12x³-12x²-24x=0

12x(x²-x-2)=0

12x(x²-2x+x-2)=0

12x(x-2)(x+1) = 0

x(x-2)(x+1) = 0

Now we use the zero product property ,which states that
if ab=0 ,then a=0 and b=0,this property is from geometry terms and definitions .

so x=0 (x-2)=0 (x+1)=0

x=0 x=2 x=-1

Therefore,x=0,-1,2 are the critical points for y=f(x)