# What are the values and types of the critical points, if any, of #f(x)=x^3-6x^2+12x-6#?

##### 1 Answer

Mar 13, 2017

#### Explanation:

Start by differentiating.

#f'(x) = 3x^2 - 12x + 12#

#f'(x) = 3(x^2 - 4x + 4)#

#f'(x) = (x- 2)^2#

Set this to

#0= (x - 2)^2#

#x - 2 = 0#

#x = 2#

Therefore, there will be one critical value at *stationary point*.

Hopefully this helps!