Show that $f$ has at most one fixed point 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.
 A: HINT: Let $g(t)=f(t)-t$. If $f\,'(t)\ne 1$ for each $t\in\Bbb R$, then $g'(t)\ne 0$ for all $t\in\Bbb R$. It follows that either $g'(t)>0$ for all $t\in\Bbb R$, or $g'(t)<0$ for all $t\in\Bbb R$; see Darboux’s theorem.
A: 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.
A: Well, basically, if $f'(t)>1$ then, as this is the slope of $f$ at $x=t$, it means that $f$ grows faster than the $y=x$ function. Similarly for the case $f'(t)<1$, then it grows slower.
The basic thing, is that the range of the derivative, though not necessarily continuous, is still satisfies: if $f'(t_1)>1$ and $f'(t_2)<1$ then there will be a $t$ between with $f'(t)=1$. 
So, suppopse, $x_0$ is a fixed point, then either $f$ grows faster, or slower..
