Proving an unknown function with some properties is bijective 
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuously
  differentiable function with the property that $\exists c > 0$ such
  that $f'(x) > c$ for all $x\in\mathbb{R}$. 
I want to show $f$ is bijective

Injectivity follows easily because $f$ is strictly increasing. How can I show $f$ is onto? Usually I just compute the inverse function, but this isn't possible here.
 A: Notice that $f(x)=\int_0^{x}f'(t)dt \implies \lim_{x\to\pm\infty}|f(x)|>\lim_{x\to\pm\infty}|x|c=\infty$. 
A: Claim one: for any $x \in \mathbb{R}$, $f(x + 1) > f(x) + c$.
Proof: suppose that $f(x + 1) \leq f(x) + c$. By the mean value theorem, there is a point $y \in [x, x+1]$ with
$$
f'(y) = \frac{f(x + 1) - f(x)}{(x+1) - 1} = f(x+1) - f(x) \leq c,
$$
a contradiction.
Claim two: $f(x)$ is surjective.
Proof: Say we want to show that $b$ is in the range of $f(x)$. We'll break into cases, although the argument in both cases is really the same. First, if $b = f(0)$, we're obviously done.
If $b > f(0)$, then for any positive integer $N$, we have $f(N) > f(0) + Nc$ by iteratively applying claim one, and we can pick $N$ large enough to make the right hand side greater than $b$. Then applying the intermediate value theorem we see that $b$ is in the range of $f$. 
If $b < f(0)$, then for any positive integer $N$, we have $f(-N) < f(0) - Nc$ by applying claim one iteratively again. Now pick $N$ large enough to make the left hand side less than $b$, and apply the intermediate value theorem.
In the two non-trivial cases, we are using that $c \neq 0$.
Claim three: $f(x)$ is injective. As you say, this is immediate, since $f$ is increasing.
