How to prove that 2/9 is repelling point of period 3 to the Tent function? 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 $0\leq x\leq 1$?

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.