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 $x \leq 0$
 A: From the mean value theorem, 


*

*$f(x) \le cx + f(0)$ for $x<0$.

*$f(x)\ge cx + f(0)$ for $x>0$.
It follows that:
$$\lim_{x \to \pm \infty} f(x) = \pm \infty$$
Hence by the IVT, $f$ hits every $y\in \mathbb R$.
A: From $c>0$ and $f(x) \geq f'(0)+ cx$ for $x \ge 0$ we get
$ \lim_{x \to \infty}f(x)= \infty$.
From $c>0$ and $f(x) \le  f'(0)+ cx$ for $x \le 0$ we get
$ \lim_{x \to -\infty}f(x)= -\infty$.
Now we derive , by the intermediate value theorem: $f( \mathbb R)= \mathbb R$
A: 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 $\mathbb{R}$.
