Prove that $f(x)=x$ can have at most one solution if $f'(x)\ne1$

Question

Prove that $f(x)=x$ can have at most one solution if $f'(x)\ne1$

What I did : Use $g(x) = f(x)-x$, then $g'(x) = f'(x)-1\ne0$

I suspect I have to use Rolle's theorem now, But I am having difficulty understanding how that will help

Answer 1

It's exactly as you say. If $g$ has two zeroes, then by Rolle there is a point where $g'=0$. This of course depends on $f$ being differentiable.

Answer 2

Assume that $f(x)=x$ has two solutions $x\neq y$ so that $f(x) = x$ and $f(y) = y$. But then for some $\xi \in \left(x,\,y\right)$ the Mean Value Theorem gives $$f'(\xi)=\frac{f(x)-f(y)}{x-y}=\frac{x-y}{x-y}=1,$$

a contradiction.

I'm assuming that you came across this question in Baby Rudin with the requirement that $f$ is differentiable. Both Rolle's and MVT require differentiability on an interval.