Given:
\begin{equation} f_1(x)= 4x(1-x)\\ \end{equation}
\begin{equation} f_2(x)=\begin{cases} 1-2x , & \text{if $ 0 \leq x \leq 1/2$}.\\ 2(x-1/2)/3, & \text{if $ 1/2< x \leq 1$}.\\ \end{cases} \end{equation}
How do I find their corresponding Markov partitions? Note, I may be wrong with my terminology here. I am trying to determine which intervals should I include when checking if they are Markov partitions.
My attempt( and confusion):
So by sketching both functions, I can guess $f_1(x)$ has two intervals $[0,a]$ and $[a,1]$. I proceed to find $a$ by taking its first derivative, giving me $1/2$. Okay.
For $f_2(x)$, again from the sketch I can guess two intervals and also because how of the function is given.
This is where I realize my approach is wrong. Apparently there are three intervals for $f_2(x)$, $[0,1/3], [1/3,1/2], [1/2,1]$. So from here, I tried plotting $y=x$ onto $f_2(x)$ which reveals to me that at $0\leq x \leq 1/2$, there is an intersection. So I equate $x=1-2x$ and gives me $x=1/3$ as the intersection point. So this must be what I had to do.
BUT, if I applied the same method again back to $f_1(x)$, the intersection point becomes $3/4$ which isn't really a point of interest (contradicts with $1/2$).
What am I doing wrong here? If you are explaining this graphically, would appreciate it if you are able to include the sketches.