value of parameter a such that $\frac{1}{2}$ is a 3-periodic point

Question

Let

\begin{cases} T_a(x) = ax & \text{ if} & x \in [0; \frac{1}{2}],\\ T_a(x) = a(1 − x) & \text{ if} & x \in [\frac{1}{2}; 1]. \end{cases}

Find a parameter value of $a$ for which $\frac{1}{2}$ is a $3$-periodic point.

I have tried to distinguish several cases.

1. If $0 \leq x \leq \frac{1}{2a}$. Then $0 \leq ax \leq \frac{1}{2}$ and $T_a^2(x)= T(ax)=a^2x$.
2. If $\frac{1}{2a} \leq x \leq \frac{1}{2}$. Then $\frac{1}{2} \leq ax \leq \frac{a}{2}$ and $T_a^2(x)= T(a(1-x))=a(1-ax)$.
3. If $\frac{1}{2} \leq x \leq \frac{a+1}{2}$. Then $\frac{a(a-1)}{2} \leq a(1-x) \leq \frac{a}{2}$ and $T_a^2(x)= T(a(1-x))=a^2(1-x)$.

I am confused about the manner to mimic the response as shown on this post. $$ f^{n}(x)=\left\{ \begin{array}{ccc} a^nx & \text{if} & x\in\big[0,\tfrac{1}{2a^n}\big], \\ 1-a^nx\big(x-\tfrac{1}{2a^n}\big) & \text{if} & x\in\big[\tfrac{1}{2a^n},\tfrac{2}{2a^n}\big], \\ a^n\big(x-\tfrac{2}{2a^n}\big) & \text{if} & x\in\big[\tfrac{2}{2a^n},\tfrac{3}{2a^n}\big], \\ & \vdots & \\ 1-a^n\big(x-\tfrac{2a^n-1}{2a^n}\big) & \text{if} & x\in\big[\tfrac{2a^n-1}{2a^n},1\big], \end{array} \right. $$

Answer 1

(Fill in the gaps as needed. If you're stuck, explain what you've tried and why it doesn't seem to work.)

I'm assuming you're restricted to $ a \in [ 0, 2 ]$ for the function to make sense. In particular, we do not allow $a =-1.$

Hint: Deal with cases based on the supposed value of $ T_ a ( 0.5), T_a ( T_a (0.5) ) $.
IE Case 1: $T_ a ( 0.5) \leq 0.5 , T_a ( T_a (0.5) ) \leq 0.5$.

Show that in all cases, we require $ a = 1$.