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If $f(x)$ is differentiable and $f'(x) \neq 1$ for all $x \in \mathbb{R}$, then $f(x)$ can have at most one fixed point.

The standard answer using the mean value theorem is here: If $f'(x)\not = 1$ for all real numbers $x$, then $f$ has at most one fixed point

Can somebody help me to understand intuitively why this is true? Thank you

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    It may help to think about the Darboux property here. If $f'(x) > 1$ somewhere then it cannot dip below $1$, so the function is always growing "above" the line $y=x$, where you would expect to see fixed points. Vice versa argument if $f'(x) <1$. – Randall Oct 05 '21 at 18:11
  • In the end, I’d argue you need some intuition about the mean value theorem, because it is sort of intuitive. – Thomas Andrews Oct 05 '21 at 18:14
  • Alternatively, $g(x)=f(x)-x$ would have two zeros, $x_1<x_2$, and then between them somewhere there must be a maximum of minimum. Then $g’(c)=0$ at the maximum or minimum. – Thomas Andrews Oct 05 '21 at 18:19
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    Agreeing with @Thomas Andrews here that simply getting comfortable with the MVT is the best way to go. (1) It's incredibly important and (2) the explanation it lends to this is very simple. If $f$ has more than one fixed point then two points lie on the line $y=x$, which has slope $1$. Thus the derivative must equal $1$ somewhere. Take the contrapositive, and you're done. What could be nicer than this? – Randall Oct 05 '21 at 18:37
  • @Randall I’m not sure I’d call Darboux intuitive, given that it is by nature a work-around to the non-intuitive case on discontinuous derivatives. – Thomas Andrews Oct 05 '21 at 19:56
  • @ThomasAndrews I agree, actually. – Randall Oct 05 '21 at 19:59

1 Answers1

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Perhaps you prefer the more intuitive but equivalent Rolle’s theorem?

Let $f$ be a continuous function on $[a,b],$ differentiable on $(a,b)$ with $f(a)=a,f(b)=b.$

Let $g(x)=f(x)-x.$ Then $g(a)=g(b)=0,$ and $g$ is continuous and differentiable. By Rolle’s theorem, $g’(c)=0$ for some $c\in (a,b).$ But $g’(x)=f’(x)-1.$ So $f’(c)=1.$

Rolle’s theorem follows intuitively from the fact that a differentiable function has derivative zero at a local maximum/minimum inside an open interval.

Of course, Rolle and the Mean Value Theorem are pretty much equivalent, so perhaps that isn’t much help.

Thomas Andrews
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