Finding super attractive orbits

Let $f(x)=3 x^2-6 x+3.415$. Find all attractive orbits of $f$.

To find super attractive orbits of f, we must find where $F'(x)=0$ where $F=f^n$.

So, if $F(x)=3x^2−6x+3.415$ then $F'(x)=6x-6=0$ and $x=1$.

Therefore, $x=1$ and it's orbit are super attractive.

@Rick I think you're on the right track and starting at $x=1$ is the right idea. An orbit, however, consists of more than one point. Also, I mistakenly wrote super-attractive originally when I meant only attractive. That's my bad but definitely an important distinction.