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Define a function $y=f(x)$ on $\mathbb{R^{2}}$. Can the graph of this function and that of $y=f^{-1}(x)$ have intersection(s) that cannot be represented in the form $(k,k)$, where $k$ is a real number? How about other domains?

oseen8202
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    Sure. Consider $f(x)=-x$. – Karl Aug 12 '22 at 16:47
  • Y=$e^x$ and y=log x have empty intersection – Lawrence Mano Aug 12 '22 at 16:48
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    The graph of $f^{-1}$ is the reflection across the line $y=x$ of the graph of $f$. So if you want them to intersect at the point $(a,b)$, just pick $f$ so that $f(a)=b$ and $f(b)=a$ (and $f$ is invertible). – Karl Aug 12 '22 at 16:50
  • Continuing from @Karl ‘s comment, if you are only visualising monotonic functions, that won’t work because: WLOG if $f$ is increasing and $a>b$, then $f(a)=b>a=f(b)$ which is a contradiction. Same goes if $f$ is decreasing. – insipidintegrator Aug 12 '22 at 17:53
  • Is the domain $\mathbb R$ or $\mathbb R^2?$ Anyway, $y=\frac1x.$ – ryang Aug 12 '22 at 19:31

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