$f : \mathbb R \to \mathbb R $ be a differentiable and $f'(x) $ is bounded function then $f$ is unbounded. $f : \mathbb R \to \mathbb R $ be a differentiable function with $f(0) =0$. If for all $x \in \mathbb R , 1< f'(x) < 2$ then $f$ is unbounded.
We know that when $f'(x) > 0$ then $f$ is increasing and when $f'(x)$ is bounded then $f$ is uniformly continuous. From there how can we conclude that $f$ is unbounded??

If $f'(x) \ge \epsilon >0$ for all $x$, then $f$ is unbounded above.
The mean value theorem gives $f(x)-f(0) = f'(\xi)(x-0) \ge \epsilon x$, hence
$\lim_{x \to \infty} f(x) = \infty$.
