Question 20
For each $x$ in $[0, 1],$ define $$
f(x) = \begin{cases} 2x, & 0 \le x \le \tfrac{1}{2},\\ 2 - 2x, & \tfrac{1}{2} \lt x \le 1. \end{cases}
$$ Let $f^{[2]}(x) = f(f(x)),$ and $f^{[n+1]}(x) = f^{[n]}(f(x))$ for each integer $n \ge 2.$ For how many values of $x$ in $[0, 1]$ is $f^{[2005]}(x) = \tfrac{1}{2}?$
f(x) = \begin{cases} 2x, & 0 \le x \le \tfrac{1}{2},\\ 2 - 2x, & \tfrac{1}{2} \lt x \le 1. \end{cases}
$$ Let $f^{[2]}(x) = f(f(x)),$ and $f^{[n+1]}(x) = f^{[n]}(f(x))$ for each integer $n \ge 2.$ For how many values of $x$ in $[0, 1]$ is $f^{[2005]}(x) = \tfrac{1}{2}?$