Question

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

Answer 1

`x= 2` is the only critical point.

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` and solve for `x`.

`0= (x - 2)^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!