How to prove that 2/9 is repelling point of period 3 to the Tent function?

Question

Let $$ T(x)=\begin{cases} 2x & 0\leq x\leq 1/2 \\ 2-2x & 1/2\leq x\leq 1 \end{cases} $$ How to prove that 2/9 is repelling point of period 3 to the $T(x)$?

I know that $2/9$ has period of $3$ because $T(T(T(2/9)))=2/9$, but how to prove that it is a repelling point and to do that we need to show that $|(T^{[3]})^{\prime}(2/9)|>1$. But, $T(x)$ is piece wise function so am I supposed to take the top or bottom function then differentiate?

How to find what $T^{3}(x)$ is defined as for the range $0\leq x\leq 1$?

Answer 1

Every power $T^k$ is piecewise linear where the subdivision is by the multiples of $\frac1{2^k}$. One finds that $T^3(\frac{2k+1}8)=1$ and $T^3(\frac k4)=0$ so that on every subinterval the slope is either $+8$ or $-8$. As $\frac18<\frac29<\frac14$, $$T^3(x)=2-8x$$ on that interval, again giving evidence to the fixed point at $\frac29$ and that the slope there is $-8$, which is repelling.

Answer 2

One has $T\bigl({2\over9}\bigr)={4\over 9}<{1\over2}$, and $T'\bigl({2\over9}\bigr)=2$. Then $T^{\circ2}\bigl({2\over9}\big)=T\bigl({4\over9}\bigr)={8\over9}$, and $T'\bigl({4\over9}\bigr)=2$. Finally $T^{\circ3}\bigl({2\over9}\bigr)=T\bigl({8\over9}\bigr)={2\over9}$, and $T'\bigl({8\over9}\bigr)=-2$. It follows that ${2\over9}$ is a point of period $3$ of $T$, and that $$\left(T^{\circ3}\right)'\left({2\over9}\right)=T'\left({2\over9}\right)\cdot T'\left({4\over9}\right)\cdot T'\left({8\over9}\right)=-8\ .$$ As $|-8|>1$ the orbit of period $3$ containing ${2\over9}$ is repelling.