Let $\lambda\in[0,4]$ and $$\tau(x):=\lambda x(1-x)\;\;\;\text{for }x\in[0,4].$$ Noting that $$\tau'(x)=\lambda(1-2x)=0\Leftrightarrow\lambda=0\vee x=\frac12\tag1$$ and $$\tau'(x)=-2\lambda\tag2,$$ it's easy to conclude that $\tau([0,1])\subseteq[0,1]$. Let $x_0\in\mathbb R$ and $$x_n:=\tau(x_{n-1})\;\;\;\text{for }n\in\mathbb N.$$
Note that $0$ and, if $\lambda\ne0$, $1-1/\lambda$ are the unique fixed points of $\tau$. In general, a sufficient$^1$ condition for a fixed point $x^\ast$ of a map $f\in C^1(\mathbb R)$ to be stable is $|f(x^\ast)|<1$. From this we can easily conclude that
- if $\lambda<1$, the unique fixed point $0$ is stable;
- if $\lambda\in(1,3)$, then $0$ is "probably unstable" (see footnote 1) and $1-1\lambda$ is stable;
- if $\lambda\in(3,4)$, then both $0$ and $1-1/\lambda$ are ("probably") unstable.
Now take a look at the evolution of the system in the picture from the Wikipedia article below. While for $\lambda=3.2$, $x^\ast:=1-1/\lambda$ is a fixed point, the set $\{x^\ast\}$ is not "attracting" (which is caused by the fact that it is not stable, I guess).
Now there is the notion of the (local) stable and unstable "manifolds" $M_s$ and $M_u$ in a neighborhood $O$ (which can be taken to be $O=(0,1)$ here, I guess) of $x^\ast$: $$M_s:=\left\{x\in O:\operatorname{orb}x\text{ maps into }O\text{ and }(\operatorname{orb}x)(n)\xrightarrow{n\to\infty}x^\ast\right\}$$ and $$M_u:=\left\{x\in O:\operatorname{orb}x\text{ maps into }O\text{ and }(\operatorname{orb}x)(-n)\xrightarrow{n\to\infty}x^\ast\right\},$$ where $\operatorname{orb}x:\mathbb N_0\to\mathbb R,x\mapsto\tau^n(x)$.
For large $n$, $(\operatorname{orb}x)(n)$ seems to jump between $a:=0.513$ and $b:=0.799$.
Question: How can we determine $M_s$ and $M_u$ precisely and what's the relation of the points $a$ and $b$ to $M_s$ and $M_u$?
$^1$ But does $|f(x^\ast)|>1$ imply that $x^\ast$ is unstable? Please take note of the separate question I've asked for that: Is a fixed point $x^\ast$ of a map $\tau$ with $|\tau'(x^\ast)|<1$ asymptotically unstable?.