Show $(\forall x\in I)\quad f(x)=1 \quad \mbox{or} \quad (\forall x\in I)\quad f(x)=-1$

Question

Let $I$ be an interval of $\mathbb{R}$ and $f$ a continuous function defined from I to $\mathbb{R}$ such that: $$(\forall x\in I)\quad (f(x))^{2}=1$$

Show that : $(\forall x\in I)\quad f(x)=1 \quad \mbox{or} \quad (\forall x\in I)\quad f(x)=-1$

Answer 1

I think that $f^2(x)$ means $(f(x))^2$ and not $f(f(x))$.

Suppose to the contrary, that there are $u,v \in I$ such that $f(u)=-1$ and $f(v)=1$. The intermediate value theorem gives now a point $w \in I$ with $f(w)=0$, hence

$$0 =f^2(w)=1,$$

which is absurd !