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

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!