I'm looking for a function $f(x)$ so that $f_{inv}(x) = \frac{1}{f(x)}$. Does such a function exist and if so, how do I find it?
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1What's the domain of $f$? There's always the trivial example $f: {1} \rightarrow {1}, 1 \mapsto 1$. – Jair Taylor Dec 14 '19 at 23:05
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Let us focus on the set of monotonic functions. If $f$ is increasing (decreasing) then $f_{\text{inv}}$ is also increasing (decreasing), however $\frac 1{f(x)}$ is decreasing (increasing). Hence such a function does not exist. – Pavel R. Dec 14 '19 at 23:12
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1Here is an older question which is very similar – pjs36 Dec 14 '19 at 23:25
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2Does this answer your question? Is there a function whose inverse is exactly the reciprocal of the function, that is $f^{-1} = \frac{1}{f}$? – ViHdzP Dec 14 '19 at 23:27
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Yeah, I had searched first but without the term "reciprocal". Thanks for the help – OscarVFE Dec 14 '19 at 23:33
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that would mean $x = f^{-1}(f(x)) = \frac 1{f(x)}$ so $f(x) = \frac 1x$. – fleablood Dec 15 '19 at 00:47