<s>[INST] Orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ converges. I view the orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ as the sequence $(x_{n})$. So 
$$x_n = \frac{1}{2}, f\bigg(\frac{1}{2}\bigg), f\bigg(f\bigg(\frac{1}{2}\bigg)\bigg), \dots$$ 
$(x_n)$ appears to dance around the fixed point $\frac{2}{3}$ while slowly converging to it. 
The odd-numbered iterates are increasing and the even-number iterates are decreasing. Let $(a_{k})$ be the sequence of odd terms of $(x_{n})$ where $k = 2n-1$ and $(b_{k})$ be the sequence of even terms of $(x_{n})$ where $k = 2n$.
Show $(a_k)$ is monotonic & bounded.
1. Increasing:
$a_{k}<a_{k+1} \forall \ k \in \ \mathbb{N}$. In terms of $(x_{n})$, show $x_{2n-2}<x_{2n+1} \forall \ n \in \ \mathbb{N}$.
$$x_{2n+1} - x_{2n-1}>0$$
$$(-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+9x_{2n-1})-(x_{2n-1}) > 0$$
$$-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+8x_{2n-1}> 0$$
$$x_{2n-1}(2-3x_{2n-1})^3 > 0$$
$$0<x_{2n-1}<\frac{2}{3}$$
$$0<a_{k}<\frac{2}{3}$$
Hence, the subsequence $(a_k)$ is strictly increasing and
$(a_k)$ is bounded below by the first point of the subsequence
$a_1=\frac{1}{2}$.
Question:
How do I show $a_{k}$ is bounded above by the fixed point $\frac{2}{3}$?
Maybe, use epsilon delta proof to show convergence, which implies $(x_n)$. Or by induction, assume $x_{2n-1}< \frac{2}{3}$ show $x_{2n+1}=f(f(x_{2n-1}))<\frac{2}{3}$
 [/INST] Note that $f(\frac{1}{2})=\frac{3}{4}$(as commented by bonext), $f(\frac{2}{3})=\frac{2}{3}$(as you have noticed) and $f(\frac{3}{4})=\frac{9}{16}\in(\frac{1}{2},\frac{2}{3})$. Also note that $f$ is strictly decreasing on $[\frac{1}{2},1]$. Then we have
$$f((\frac{1}{2},\frac{2}{3}))\subset(\frac{2}{3},\frac{3}{4})\quad\mbox{and}\quad f((\frac{2}{3},\frac{3}{4}))\subset(\frac{1}{2},\frac{2}{3}).$$
Now by induction, it is easy to see $x_{2n-1}\in(\frac{1}{2},\frac{2}{3})$ and $x_{2n}\in(\frac{2}{3},\frac{3}{4})$ when $n\ge 2$.
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