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

Mar 13, 2017

$x = 2$ is the only critical point.

#### Explanation:

Start by differentiating.

$f ' \left(x\right) = 3 {x}^{2} - 12 x + 12$

$f ' \left(x\right) = 3 \left({x}^{2} - 4 x + 4\right)$

$f ' \left(x\right) = {\left(x - 2\right)}^{2}$

Set this to $0$ and solve for $x$.

$0 = {\left(x - 2\right)}^{2}$

$x - 2 = 0$

$x = 2$

Therefore, there will be one critical value at $x = 2$. Since the slope of the graph at that point is $0$, we call this a stationary point.

Hopefully this helps!