Let $ f:(-1,1)\to(-1,1)$ be continuous, $f(x)=f(x^2)$ for every $x$ then the value of $f(\frac 1 4)$

Question

Let $f:(-1,1)\to(-1,1)$ be continuous, $f(x)=f(x^2)$ for every $x$ and $f(0)=\frac 1 2$. Then $f(\frac 1 4)$ is $A)\frac{1}{16}\quad B)\frac{1}{4}\quad C)\frac{1}{2}\quad$ $D)$ can't be determined.

I don't know how to start with this problem? Please give me hints.

Answer 1

$f\left(\cfrac{1}{4}\right) = f\left(\cfrac{1}{16}\right) = f\left(\cfrac{1}{256}\right) = f\left(\cfrac{1}{256^2}\right)=\dots$

Can you see where it is going?

$f\left(\cfrac 14\right) = f(0) = \cfrac 12$