Let $f(x)=(f(x))^2$ that is continuous for every $x \in\mathbb R$.
Prove using the intermediate value theorem that this function is constant.
I noticed that the $f(x)$ could only be equal to : $1,0$
I know I can evaluate its limit at infinity:
$\lim_{x\to\infty}f(x)=1$
$\lim_{x\to-\infty}f(x)=1$
Can I conclude anything from that that the function is constant?
I know there exist a $c\in R$ such that $1\le f(c) \le 1$, but exists is not proving that for all $c \in R$ :S
Can someone please arrange the mess I've done?:)
Thanks
Best Answer
A continuous function $\mathbb{R} \to \mathbb{R}$ whose image consists of finitely many points is constant.
Indeed, if $f: \mathbb{R} \to \mathbb{R}$ is continuous and not constant, there are $a,b \in \mathbb{R}$ with $f(a) < f(b)$. By the intermediate value theorem, $f$ will take all values in the interval $[f(a),f(b)]$, which are infinitely many values.