An interesting equation involving many iterations. Let $$f(x) = 1 -|1- 2x |.$$ Find the number of solutions of the equation 
$$f ( f ( f ( f ( f ( f ( f ( f ( f ( f (x))))))))))=x,$$
i.e., $f^{(10)}(x)=x$.
And what about if there is an arbitrary number of iterations?
 A: Draw the graph of $f$. The solutions of $f(x)=x$ are the interseccions with the diagonal of the first quadrant.
Next draw the graph of $f(f(x))$. It looks like two mountains (or tents). Hoy many times does it cross the diagonal?
Repeat for $f(f(f(x)))$. Can you see a pattern?
A: The main question here is what solutions exist for $f(x)=x$, as I commented, an obvious solution is $x=0$, but this may not be unique, so let us see if there are more.
So we consider $1-|1-2x|=x$, if $-\infty\lt x\le \frac{1}{2}$, then we have:
$1-(1-2x)=x\Rightarrow 2x=x$ so $x=0$.
If $1/2\lt x\lt\infty$, then we have:
$1-(2x-1)=x\Rightarrow 2-2x=x\Rightarrow x=2/3$
So we have $x=0,2/3$ as solutions.
You may be interested to know, that on $[0,1]$, this map exhibits behaviour a lot like the tent map, which is a chaotic map.
A: Well $f^{10}(x)=x$ implies that $f$ has to be one-on-one for some $x$ since for a composition to be injective at least one function has to be injective since all functions are $f$ it implies $f$ is injective.From this we know that $f(f(x))=x$ for $$x\leq\frac{1}{2}\\f(x)=2x\\4x=x\implies x=0\\x>\frac{1}{2}\\f(x)=2-2x\\2-2(2-2x)=x\\2-4+4x=x\\4x-2=x\\3x=2\\x=\frac{2}{3}$$
