Find the value of $f(\frac14)$

Question

$$f:(-1,1)\mapsto(-1,1)\in C^1\quad f(x)=f(x^2)\quad f(0)=\frac12$$ Then $f(\frac14)$ is?

All I could deduce is that $f(x)=f(-x)$. A hint would be great.

Answer 1

Using $f(x)=f(x^2)$ You'll get

$$f(x)=f(x^{2^n}) \forall n \in \mathbb N $$

Now apply $$\lim_{n \to \infty} f(x^{2^n})$$

Since $|x|<1$, $$\lim_{n \to \infty} x^{2^n}=0$$

Hence you get $$f(x)=\lim_{n \to \infty} f(x^{2^n})=f(0)$$

Therefore $ f(x)$ is a constant function.

Hence, finally $$f\left(\frac 14\right)=f(0)=\frac 12$$

Answer 2

Consider this: $$ f(\frac14)=f(\frac1{16})=f(\frac1{256})=...$$

You can form a sequence of $f( \frac1{4 ^{2^n}} )$ and then use continuity of $f $ to show that this tends to $f(0)$