Proving a certain continuously differentiable function is one-to-one and onto Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuously differentiable. Suppose $\exists c > 0$ such that $f'(x) > c$ for all $x \in \mathbb{R}$. Show $f$ is onto and one-to-one.

Clearly, $f$ is one-to-one because it's strictly increasing. But how can I show it's onto? 
I've tried for a while but can't get it.
I think it will use the Intermediate Value Theorem.
 A: By intermediate value theorem, $f$ is surjective if $\displaystyle\sup_{r\in\mathbb R}f(r)=\infty$ and $\displaystyle\inf_{r\in\mathbb R}f(r)=-\infty$.
We prove this by contradiction. Suppose $f(r)<\alpha,\,\forall r\in\mathbb R$ for some $\alpha>0$. Then by mean value theorem, there exists some $0<a<\frac{2\alpha}c$ such that $f(\frac{2\alpha}c)-f(0)=f'(a)\times\frac{2\alpha}c>2\alpha$. But this is impossible since $|f(\frac{2\alpha}c)-f(0)|<\alpha+\alpha=2\alpha$.
The proof that $\displaystyle\inf_{r\in\mathbb R}f(r)=-\infty$ is similar.

Hope this helps.
A: By continuity, $f$ maps intervals to intervals, and from the condition we get for $a\le b$
$$f(b)-f(a)\ge c(b-a)$$
Then consider any number $s>f(0)$.
Then with $x=\frac{s-f(0)}c$, we have $f(x)-f(0)\ge cx=s-f(0)$, thus $s\in [f(0), f(x)]$, so it has a preimage in $[0, x] $.
The case for $s<f(0)$ can be similarly done. 
A: Use the mean value theorem. For any $b>a$ we have for some $d\in(a,b)$
$$f(b)=(b-a)f'(d)+f(a).$$
Take $a=0$, then by the fact that $f'(x)>c>0$ we can see that $f(b)$ increases at least linearly with $b$, so any number greater than $f(0)$ can be attained by $f$. Similar reasoning can be used to show any value smaller than $f(0)$ can be attained by by $f$.
