If a function and its inverse intersect at some point(s), then should there be at least one such point lying on the line y=x?

Question

Like in the function $y=1-x^5$ and its inverse $y=(1-x)^{1/5}$. They have 5 points of intersection and one of them lie on $y=x$. Is it like an isolated case or is it true for every bijective function and it's inverse? The graph mentioned above

Answer 1

If $f$ is increasing then Then the graphs of $y=f(x)$ and $y=f^{-1}(x)$ will intersect only on the line $y=x$.

If $f$ is decreasing then then the graphs of $y=f(x)$ and $y=f^{-1}(x)$ will intersect at most once on the line $y=x$. There may be other solutions as illustrated by OP.

Interestingly in such case the number of point(s) of intersection will be ODD and the points of intersection not lying on line $y=x$ can be paired such that they are reflections of each other in the line $y=x$ (which can again be observed on the graph provided by the OP)

If $f$ is its own inverse e.g. $f(x)=-\frac{x}{1+x}$ then every point in the domain is point of intersection of $y=f(x)$ and $y=f^{-1}(x)$

Answer 2

Assumption: $f : \mathbb{R} \mapsto \mathbb{R}$ is continuous and bijective.

If $(x, f(x))$ is a point on the graph of $f$, then $(f(x), x)$ is a point on the graph of $f^{-1}$. That is, the graphs are mirrors of each other about the line $y=x$.

For them to intersect, thus, $f$ must cross the line $y=x$ which means it must intersect with $f^{-1}$ at this point.