Derivatives, bounds and surjectivity Let $f: \mathbb{R} \to \mathbb{R}$ be a derivable function.
We know that:
$\lim_{x\to-\infty}\ f'(x) = \lim_{x\to+\infty}\ f'(x) = \frac1 2$
Prove that such function is surjective.
Ok my thoughts so far are:
Let's suppose that such a function has an upper bound. Then let $s=\sup(f)$ be such upper bound. Since it is continuous, there has to be a neighbourhood of $+\infty$ in which the derivative remains positive, and we can take it big enough to include $s$ (?). Then we can conclude that in such neighbourhood the function is increasing, which goes against the fact that s is an upper bound.
Therefore the function has to not have an upper bound. A similar reasoning can be done for the lower bound and we can conclude that the function in surjective.
Pretty sure it's completely wrong, but still a try.

We may choose $M>0$ and $N<0$ such that if $x\in (-\infty, N)\cup (M,\infty),$ then  $f'(x)>0. $
Now, consider separately the intervals $(-\infty, N),\ [N,M],\ (M,\infty). $
On the first and third of these, $f$ is increasing and continuous, hence surjective. 
Wlog $f(M)\ge f(N)$. Then, the IVT implies that $f([N,M])$ is onto $[f(N),f(M)],\ $ 
so $f$ is surjective, as required.
