proving a differentiable function $f: \Bbb R \to \Bbb R$ and a constant $c>0$ with $f'(x) \geq c$ for all $x \in \Bbb R$. is a bijection We have a differentiable function $f: \Bbb R \to \Bbb R$ and a constant $c>0$ with $f'(x) \geq c$ for all $x \in \Bbb R$.
Show that $f$ is a bijection from $\Bbb R$ to $\Bbb R$.
From Rolle's theorem follows that if $f'(x) \neq 0$  $f$ is injective. Which is the case here so we know that $f$ is injective.
I'm stuck on proving it's surjective.
In the questions before this one I already proved with the mean value theorem that $f(x) \geq f(0)+ cx$ if $x \geq 0$, and that  $f(x) \leq f(0)+ cx$ for all $x \leq 0$

Since $\mathbb{R}$ is connected, the image of $f$ is connected. So, if $f$ is not surjective, then the image of $f$ is either bounded above or bounded below. WLOG, suppose that the image of $f$ is bounded above. Then let $y$ be the least upper bound. Now given $c$ such that $f^\prime(x) \geq c$, prove that there must be some $y^\prime \in \mathbb{R}$ such that $f^\prime(y^\prime) < c$.
Hint: Since $f^\prime(x) > 0$ for all $x$, $f$ has to be monotonically increasing. Thus finding such a $y^\prime$ is just about selecting an element large enough in $\mathbb{R}$.
