Point of intersection of $f(x)$ and $f^{-1}(x)$

Question

Can we say that that if $f(x)$ and $f^{-1}(x)$ intersect, then at least one point of intersection will lie on $y=x$?

Also there are many function e.g. $f(x)=1-x^3$ where point of intersection exists outside $y=x$ There will be $5$(odd) point of intersection of $f(x)=1-x^3$ and $f^{-1}(x)=(1-x)^{1/3}$ out of which one lie on $y=x$. Will there exist a function in which there will be even number of point of intersection but odd number of point of intersection will lie outside $y=x$?

Also it is clear that in case of strictly increasing continuous function , point of intersection if exists will lie on $y=x$ but will also be true for strictly increasing discontinuous function?

Answer 1

It is perfectly possible for the graphs of $f$ and $f^{-1}$ to cross away from the line $y=x$.

We need is $f(p)=q$ and $f^{-1}(p)=q$.

Given that $f^{-1}$ exists, and so $f$ and $f^{-1}$ are one-to-one, the condition that $f^{-1}(p)=q$ becomes $f(f^{-1}(p)) = f(q)$, i.e. $p=f(q)$. Hence:

The graphs of $f$ and $f^{-1}$ cross at the points $(x,y)=(p,q)$ where both $f(p)=q$ and $f(q)=p$.

This is trivially true on the line $y=x$, where $p=q$.

Answer 2

if the graphs of $f$ and $f^{-1}$ intersect at $(x,y)$, then they will also intersect at $(y,x)$.

If the function is additionally assumed to be continuous, then -- since $(x,y)$ and $(y,x)$ are on different sides of the diagonal $x=y$ -- the function graph must cross the line $x=y$. And such a point is necessarily also a crossing of $f$ and $f^{-1}$.