Estimating $n_0$ such that $f^n (x) \in [1+2\sigma, \infty)$, $\forall\ n>n_0$. Where $f(x) = x^3 + \frac{2}{3 \sqrt{3}} + \varepsilon$.

Question

Let $\varepsilon>0$ be a real number and $f_{\varepsilon}:\mathbb{R}\to \mathbb{R}$ be the smooth map $$f_\varepsilon(x) = x^3 + \frac{2}{3\sqrt 3} + \varepsilon. $$ Besides that consider the dynamical system $$\Phi_{\varepsilon}(n,x) = f^{n}_{\varepsilon}(x). $$

It is easy to see that the function $f_{\varepsilon}$ has only one fixed point $x^*<0$ and such fixed point is repelling. Moreover, for every $x>0$ it is also easy to see that $\Phi_\varepsilon (n,x)\to \infty$, when $n\to \infty$.

My question Let $\varepsilon,\sigma,x>0$ be real numbers. Is it possible to exhibit an estimate $n(\sigma,x,\varepsilon)$, such that $\forall$ $n>n(\sigma,x,\varepsilon)$, $$\Phi_\varepsilon (n,x) \in [1+2\sigma,\infty]\ \ \ \ \ \ \ \ \ \ ? $$

Answer 1

For $x>0$, the minimum of $g_\epsilon(x) = f_\epsilon(x)-x$ is attained at $x=\frac{1}{\sqrt3}$ and its value is equal to $\epsilon$. Therefore,

$$f_\epsilon(x)\ge x +\epsilon$$ for $x >0$.

Based on this inequality, you get by induction $$\Phi_\epsilon(n,x) \ge n\epsilon$$

And you’re done by picking $$n(\sigma, x, \epsilon)=\frac{1+2\sigma}{\epsilon}$$

Note: this is a very coarse value. It can be refined with deeper analysis of the problem.