Suppose $f:\mathbb{R} \to \mathbb{R}$ is differentiable and $f'(x) \geq c > 0, \forall x.$ Then $f(\mathbb{R}) =\mathbb{R} \ $ Suppose $f:\mathbb{R} \to \mathbb{R}$ is differentiable and there exists $c>0$ such that $f'(x) \geq c, \forall x.$  Could anyone advise me how to prove $f(\mathbb{R}) =\mathbb{R} \ $
I have established that $f(x) \geq f(0)+cx, \forall x \geq 0$ and $f(x) \leq f(0) +cx, \forall x \leq 0.$ Would it help?  Thank you. 
 A: HINT:
Given a $y$ and searching for an $x$ such that $f(x)=y$, you want to find $f(b)\geq y$ and $f(a) \leq y$ and use the intermediate value theorem. You already have two inequalities to use...
A: If $f'(x)\geq c>0$, then $f$ is strictly increasing. Let show that $f$ is unbounded.
Suppose that $f$ is upper bounded. Therefore $$\lim_{x\to\infty }f(x)=\alpha\in\Bbb R$$
(because $f$ is increasing). By the mean value theorem, there is a $\xi_x\in]x,x+1[$
 such that
$$f(x+1)-f(x)=f'(\xi_x).$$
If you let $x\to \infty $, you get $$\displaystyle\lim_{x\to\infty }f'(\xi_x)=0,$$ which is a contradiction. If $f$ is lower bounded, you can do the same reasoning as previous. Finally, we get $$\lim_{x\to\infty }f(x)=\infty \quad\text{and}\quad\lim_{x\to-\infty }f(x)=-\infty, $$
and thus $f(\mathbb R)=\mathbb R$.
A: as pointed by andres caicedo i need to fix my argument. hopefully, this is correct.
we will show that $\lim_{x \to b-} f(x) = \infty$ where $b$ is the maximal such that $f(x)$ is defined for all $x < b.$  there are two possibilities: (i) $b < \infty$ and $f(b-) = \infty$ or $b = \infty.$  the first of these alternatives can be argued by the uniqueness of the initial value problem for differential equation.  
in the case $b = \infty,$ we can use ftc, $f(b) -f(a) = \int_a^b f'(x) dx \ge c(b-a), a \le b.$ keep $a$ fixed and let $b \to \infty$ to get $f(\infty) = \infty.$ likewise $f(-\infty) = -\infty.$ 
