Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function. $x\in\mathbb{R}$ is a fixed point of $f$ if $f(x)=x$. Show that if $f'(t)\neq 1\;\forall\;t\in\mathbb{R}$, then $f$ has at most one fixed point.
My biggest problem with this is that it doesn't seem to be true. For example, consider $f(x)=x^2$. Then certainly $f(0)=0$ and $f(1)=1 \Rightarrow 0$ and $1$ are fixed points. But $f'(x)=2x\neq 1 \;\forall\;x\in\mathbb{R}$. Is there some sort of formulation that makes this statement correct? Am I missing something obvious? This is a problem from an old exam, so I'm assuming that maybe there's some sort of typo or missing condition.
Best Answer
This problem is straight out of baby Rudin.
Assume by contradiction that $f$ has more than one fixed point. Select any two distinct fixed points, say, $x$ and $y$.
Then, $f(x) = x$ and $f(y) = y$. By the Mean Value Theorem, there exists some $\alpha \in (x,y)$ such that $f'(\alpha) = \frac{f(x)-f(y)}{x-y} = \frac{x-y}{x-y} = 1$, contradicting the hypothesis.