I can't seem to find any function that satisfy these criteria. Let's say $f$ is a function (be in the reals or elsewhere) that's a quotient of two other functions $g$ and $h$ means $f(x)=\frac{g(x)}{h(x)}$, and its inverse $f^{-1}$ is the quotient of the respective inverse functions $g^{-1}$ and $h^{-1}$ means $f^{-1}(x)=\frac{g^{-1}(x)}{h^{-1}(x)}.$ If such functions exists, is there a name for this property?
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- You should use MathJax to write math in your questions. 2. Are you satisfied with the identity function?
– Moishe Kohan Feb 04 '24 at 19:32 -
Thank you for answering, yea the identity function is one of them, but are there any more you can think of? – simogne Feb 04 '24 at 19:36
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You should revise your question accordingly. – Moishe Kohan Feb 04 '24 at 19:38
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How is the identity function a solution? What are $g$ and $h$? – Karl Feb 04 '24 at 19:47
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Are we allowing the solution where $h(x) = 1$ and $g(x) = f(x)$? – John Barber Feb 04 '24 at 20:17
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2$h(x)=1$ is not invertible. People seem to be confusing the identity function with the constant $1$ function, or maybe confusing function inverses with multiplicative inverses. – Karl Feb 04 '24 at 20:34
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1No. There is no name for that property. I don't think anyone considered it significant. It'd be interesting to try to find some. I'm sure there are some. We could try $g(x)=x^r; h(x)=x^s; f(x)=x^{r-s}; g^{-1}=x^{\frac 1r},h^{-1}=x^{\frac 1s}; f^{-1}= x^{\frac 1{r-s}}$ and try to solve for $\frac 1{r-s}= \frac 1r -\frac 1s$ (i.e. $\frac 1{r-s}=\frac{s-r}{rs}; rs =-(r-s)^2; r^2 + 3rs +s^2 = 0$ and so on....) – fleablood Feb 04 '24 at 21:06
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Yes, $f(x)=e^{\frac{2}{\log x}}=\frac{e^{\frac{1}{\log x}}}{e^{\frac{-1}{\log x}}}$ for $x\ne 0, 1$. I'm not aware of a name for this property.
John Zimmerman
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An amusing example is $f(x) = \frac{x}{x-1}, g(x) = -x, h(x) = 1-x$, for $x \ne 1$. Here each of $f$, $g$ and $h$ is its own functional inverse, so $f^{-1} = g^{-1}/h^{-1}$ because $f = g/h$.
FredH
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