For each $x$ in $[0,1]$, define
\[\begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\ f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array}\]
Let $f^{[2]}(x) = f(f(x))$, and $f^{[n + 1]}(x) = f^{[n]}(f(x))$ for each integer $n \geq 2$. For how many values of $x$ in $[0,1]$ is $f^{[2005]}(x) = \frac {1}{2}$?
\[(\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}\]
For the two functions $f(x)=2x,0\le x\le \frac{1}{2}$ and $f(x)=2-2x,\frac{1}{2}\le x\le 1$,as long as $f(x)$ is between $0$ and $1$, $x$ will be in the right domain, so we don't need to worry about the domain of $x$.  

Also, every time we change $f(x)$, the expression for the final answer in terms of $x$ will be in a different form(although they'll all satisfy the final equation), so we get a different starting value of $x$.  Every time we have two choices for $f(x$) and altogether we have to choose $2005$ times. Thus, $2^{2005}\Rightarrow\boxed{E}$.
Note: the values of x that satisfy $f^{[n]}(x) = \frac {1}{2}$ are $\frac{1}{2^{n+1}}$, $\frac{3}{2^{n+1}}$, $\frac{5}{2^{n+1}}$, $\cdots$ ,$\frac{2^{n+1}-1}{2^{n+1}}$.