An attractive fixed point that attracts no critical point?

Find an example of a continuously differentiable function $f:\mathbb R\to \mathbb R$ with an attractive fixed point that attracts no critical point.

After laboriously building up a function that satisfies these properties I found that $x^3/3000 + x(1/1000) + 1/1000$ satisfies the given conditions. It has a fixed point at approximately $.0010101$, its derivative is $x^2/1000 + 1/1000$ which is less than $1$ whenever $x < 3\sqrt{111}$. Thus this fixed point is in fact an attractive fixed point. Further $f'(x) > 0$ for all $x$ and thus the point attracts no critical points.

@Levente This is most awesome! I am totally curious how you found such a complicated example, though? I wonder if anyone can find a simpler example?

What about choosing $f(x)=ln(x)+2$, which has a critical point at approximately $x_o=3.14619$, and because $f'(x)=\frac{1}{x}$, we end up with $f(x_o)<1$, while at the same time, the only place where $f'(x)=0$ is...oh wait...

By specifying $f'(x)>0$, we eliminate the possibility of a super attractive orbit. I know that a super attractive orbit must contain a critical point, but is it also true that a critical point must belong to a super attractive orbit? If this is true, then our condition makes it impossible for the critical point to be a part of any orbit, de facto preventing it from being attracted to our attractive fixed point. Is this correct?

@RedCrayon I agree with your first sentence. But I don't see you say that the critical point can't be part of any orbit. I mean, if there's a critical point - then it's certainly part of it's own orbit.