TITLE: Fixed Point Proof
QUESTION [5 upvotes]: *I would only like a hint! Not a full proof.
Prove: 
If a function $f(x)$ is differentiable on $\mathbb{R}$ and $f'(x) < 1$ for all $x\in\mathbb{R}$, then $f(x)$ has at most one fixed point.
So far, I have applied the Mean Value Theorem and developed the following inequality, which may or may not be of any help:
$$f(b)-b < f(a) - a$$
For all  $a,b\in\mathbb{R},\ a<b$. I have a feeling this is much easier than I am making it.

REPLY [5 votes]: Ok guys, all of your answers have helped, so I wanted to post the proof I developed for critique:
Let $f$ be differentiable on $\mathbb{R}$, and $f'(x)<1$ for all $x\in\mathbb{R}$. Suppose $a,b\in\mathbb{R}$ are fixed points of the function $f$ such that $a<b$. Then, by the Mean Value Theorem, there exists a $c\in(a,b)$ such that:
        $$f'(c) = \frac{f(b)-f(a)}{b-a}.$$
    Using the fact that $a$ and $b$ are fixed point gives:
        $$f'(c) = \frac{f(b)-f(a)}{b-a} = \frac{b-a}{b-a} = 1.$$
    Which is a contradiction. So, $f$ cannot have more than 1 fixed point.