How to find the solutions of this functional equation 
$$f(\tfrac{1}{2}+x)+f(\tfrac{1}{2}-x)=8xf\big(4(\tfrac{1}{2}+x)(\tfrac{1}{2}-x)\big)\qquad\text{for}\qquad x\in(0,\tfrac{1}{2})$$

I have no idea about how to tackle this equation. The original problem asks me to verify that a function indeed fits in, but I want to know how to tell the solution directly from the equation. Thanks to the comments, I have to add that $f$ could be defined on $(0,1)$. Also if no one knows how to proceed for a while, I will post the answer the book gives (perhaps as well as the motivation for this problem), and maybe some people then know how to tackle this equation with the given answer.
 A: If you have the values of a continuous $f$ given on $(\frac12,1)$, then you get values on $(\frac14,\frac12)$ from the functional equation isolated for the first term,
$$
f(\tfrac12-x)=8xf(1-4x^2)-f(\tfrac12+x),
\\
f(a)=4(1-2a)f(4a(1-a))-f(1-a).
$$ 
To ensure continuity in $a=\frac12$, we need $f(\frac12)=0$.
From the values in $(\frac14,1)$ you can determine the values of $f(a)$ for all $a$ such that
$$
a_1=\tfrac14<4a(1-a)\iff (a-\tfrac12)^2<\tfrac3{16}\iff a_2=\tfrac12-\tfrac{\sqrt3}4<a<\tfrac12+\tfrac{\sqrt3}4
$$
From these you get values for the interval $(\frac12-\frac{\sqrt3}4,\frac14)$ for $x\in (0,\frac14,\frac12)$. Continuity in $a=\frac14$ follows from continuity in $a=\frac34$. With these values we can extend to $(a_3,1)$ etc. with 
$$
a_{k+1}=\frac12(1-\sqrt{1-a_k})=\frac12\frac{a_k}{1+\sqrt{1-a_k}}
$$
This iteration of the lower interval bounds $a_0=\frac12, a_1=\frac14,a_2=\frac1{4(2+\sqrt3)},...$ converges to zero, so that all values of $f$ are determined by fixing the values on $[\frac12,1]$ to an arbitrary continuous function with $f(\frac12)=0$.
