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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.

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    $\begingroup$ Well, you need to prove that there is no upper or lower bound on the values of $f(x)$. $\endgroup$ – Don Thousand Apr 1 at 13:30
  • $\begingroup$ Suppose the lower bound was $f(x)$. Then by the fact that $f$ is strictly increasing, we can take $f(x - a) < f(x)$ for $a > 0$. Does that work? $\endgroup$ – wutv1922 Apr 1 at 13:31
  • $\begingroup$ You have a stronger hypothesis than strictly increasing, and you will have to use it. For example, $\exp(x)$ is strictly increasing but not bijective (but it also does not satisfy the stronger hypothesis.) $\endgroup$ – hunter Apr 1 at 13:34
  • $\begingroup$ How can you assume that lower bound is attained by $f$? $\endgroup$ – Dbchatto67 Apr 1 at 13:35
  • $\begingroup$ @hunter I'm guessing I need to use the fact that it's continuously differentiable. I still can't get anything. $\endgroup$ – wutv1922 Apr 1 at 13:48
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Notice that $f(x)=\int_0^{x}f'(t)dt \implies \lim_{x\to\pm\infty}|f(x)|>\lim_{x\to\pm\infty}|x|c=\infty$.

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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.

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