what is the critical points in this equation F(x) = x^3-3x^2+3x+1?

what is the critical points in this equation F(x) = x^3-3x^2+3x+1?

Mr Google tells me that critical point is what Americans call stationary points.

Ok well that is when the first derivative =0

F(x) = x^3-3x^2+3x+1

F'(x) = 3x^2-6x+3

3x^2-6x+3=0

x^2-2x+1=0

(x-1)^2=0

x=1   

F(1) = 1-3+3+1=2

so the only critical point is (1,2)

F''(x) = 6x-6

F''(1) = 6-6=0 Mmm

As x tends to 1 from above, f"(x)>0    concave up

As x tends to 1 from below, f"(x)<0    concave down

So at (1,2) there is a horizontal point of inflection.