Question: Suppose $f(x)$ is a bounded polynomial, in other words, there is an $M$ such that $|f(x)|\le M$ for all $x\in R$. Prove that $f$ must be a constant.
I think the question assumes $M\in R$. I can't think of any polynomial that is bounded by some real number unless the polynomial degree is zero, so $f$ should be constant. But, I am not sure about how to prove this.
My attempt: Prove the statement by contrapositive.
If $f$ is not constant, then $f$ should be unbounded. Let $f(x)=b_dx^d+b_{d-1}x^{d-1}+...+b_1x+b_0$ for $d>0$. Since polynomial is infinitely differentiable (explanation), we can differentiate $f$ for $d$ times. Then, we get $b_d\in R$. That is, $f$ is increasing or decreasing continuously. Therefore, if $f$ is bounded, $f$ should be constant.
Is it okay? If not, could you give some hint?
Thank you in advance!