[Math] Show $f$ has exactly one fixed point if $f'(x) <1$ for all $x$

Question

I'm trying to use the mean value theorem to show that for differentiable $f:\mathbb{R} \to \mathbb{R}$, if $f'(x) < 1 \forall x\in\mathbb{R}$, then $f$ has exactly one fixed point. I know how to show that if $f'(x)\neq 1$ then $f$ can have at most one fixed point. I think I need to use that $f'(x) \leq 1$ to see that $f$ is decreasing, but I'm not sure how to use that.

Answer 1

Let $f(x) = x+1$ for $x \in \mathbb{R}$. Then $f'(x) = 1$ for all $x$ and $f$ has no fixed points.