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Let $f : mathbbR rightarrow mathbbR$ be a continuously
differentiable function with the property that $exists c > 0$ such
that $f'(x) > c$ for all $xinmathbbR$.

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.

Notice that $f(x)=int_0^xf'(t)dt implies lim_xtopminfty|f(x)|>lim_xtopminfty|x|c=infty$.

Claim one: for any $x in mathbbR$, $f(x + 1) > f(x) + c$.

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
$$
f'(y) = fracf(x + 1) - f(x)(x+1) - 1 = f(x+1) - f(x) leq c,
$$
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.