Prove that $f(x)$ is a constant function. [duplicate]

Here is the question: Let f be a real valued continuous function on $[0, ∞)$. Suppose $f (x) = f (x^2)$ for all x ≥ 0, prove that f (x) is a constant function.

My attempt: Since f(x) is continuous, and $f(x^2)$ is continuous, then $f(x)-f(x^2)$ is continuous.

I will try to prove the contrapositive. If f(x) is not a constant function, then $f(x)$ does not equal $f(x^2)$ for all x.

Let $f(x)$ be of the form $a_n*x^n + .... +a_1*x + a_0$ Let $f(x^2)$ be of the form $a_n*x^(2n) + .... +a_1*x^2 + a_0$ In both cases, $a_j$ does not equal 0 for all j in ${1,...,n}$

So $g(x)=f(x)-f(x^2)$ doesnt equal 0 for some x. Also, $g(x)$ is continuous. Now I'm stuck, and your help would be appreciated.

Also, we haven't yet covered derivatives in our class.

Hint: By assumption, we have $$f(x)=f\left(x^2\right)=f\left(x^4\right)=\cdots=f\left(\lim_{n\to\infty}x^{2^n}\right)=f(0)$$ for any $0\leq x<1$, and $$f(x)=f\left(x^\frac{1}{2}\right)=f\left(x^\frac{1}{4}\right)=\cdots=f\left(\lim_{n\to\infty}x^\frac{1}{2^n}\right)=f(1)$$ for any $x\geq 1$.