Prove that exists $x_{0} \in ]0,1[$ such that $[f(x_{0})]^2=x_{0}$?. Let $f$ $\in C[0,1]$ such that $-1 \leq f(x) \leq 1$ $\forall_{x} \in [0,1]$.Prove that exists $x_{0} \in ]0,1[$  such that $[f(x_{0})]^2=x_{0}$.
I thought I can apply some theorem related to Continuity, be it Bolzano's theorem or the Intermediate Value. But I think it should be done in cases, but I don't know what they would be. Any suggestion or suggestion will be welcome. Thanks in advance.

The question as stated here is false as the counterexample $f(x) = 0 $ shows.  In fact there are plenty of other counterexamples with no solution to $f(x)^2 = x$ in the open interval $(0,1)$, such as $f(x) = 1, f(x) = x^n, n>0, \neq \frac{1}{2}$.  However, if the closed interval $[0,1]$ is intended, you can proceed as follows.
Let $g(x) = f(x)^2-x$.  Then, if $g(x)$ is non zero throughout $[0,1]$ the intermediate value theorem implies $g(x)$ cannot change sign n $[0,1]$.  It follows, either $g(x) > 0 $ or $g(x) < 0$ throughout the interval.  In the former case, take $x=1$. It implies $f(1)^2-1>0$ and that contradicts the requirement $-1 \leqslant f(x) \leqslant 1$.  In the latter case, set $x=0$, implying $f(0)^2 < 0$ which is also impossible.  Therefore, by contradiction, $g(x)$ must take the value zero in $[0,1]$ and there is a solution $x_0$ such that $f(x_0)^2=x_0$.
