# Math Help - Real Analysis

1. ## Real Analysis

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.

2. Originally Posted by dhhnerd
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 $x. Then $f$ on $[a,b]$ satisfies the continous of the Mean Value Theorem and therefore $f(y) - f(x) = f'(c)(y-x)$ where $x. But that gives us $(y-x)=f'(c)(y-x)\implies f'(c) = 1$. This is a contradiction because $|f'| < 1$.