Show that if f is differentiable on R and |f ' (x)| < 1 for all x in R, then f has at most one fixed point.
Assume that there are two fixed points $\displaystyle x<y$. Then $\displaystyle f$ on $\displaystyle [a,b]$ satisfies the continous of the Mean Value Theorem and therefore $\displaystyle f(y) - f(x) = f'(c)(y-x)$ where $\displaystyle x<c<y$. But that gives us $\displaystyle (y-x)=f'(c)(y-x)\implies f'(c) = 1$. This is a contradiction because $\displaystyle |f'| < 1$.