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A close solution is f(x) = (1+x)/(1-x) which results in f(f(x)) = -1/x. Looking at the function by substituting x = tan θ, we can easily see that f simply adds π/4 to the argument θ, which when done twice in a row gives -cot θ or -1/x.

Trying to find a function in the reals which satisfies f(f(x)) = 1/x may ,thus , be impossible as simply adding some φ to the argument θ can never result in cot θ , although I don't have a proof for this.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestI think I heard of one... note that $\dfrac{1}{f(f(x))}=x.$

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Bingo! $f(x)=x^{i }$

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Wow, I was thinking it would be hard to get a real function but you have taken yours to complex

And btw replace $\sqrt{-1}$ with iota because $\sqrt{-1}$ is equal to iota and -iota

And one more you have to restrict the domain of the function so that the inverse exists, one restriction could be $\forall x \in (0,e^{2π}]$

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$f(x) = x^{-1}$

$f^{-1}(x) = x^{-1}$

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Here $f(x)=f^{-1}(x)\cancel{=}\frac{1}{f(x)}$

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There may be no function such that:

$f^{-1}(x) = \frac{1}{f(x)}$

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$f:\{1\} ➝ \{1\} ,f(x)=x$ satisfies

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A close solution is f(x) = (1+x)/(1-x) which results in f(f(x)) = -1/x. Looking at the function by substituting x = tan θ, we can easily see that f simply adds π/4 to the argument θ, which when done twice in a row gives -cot θ or -1/x.

Trying to find a function in the reals which satisfies f(f(x)) = 1/x may ,thus , be impossible as simply adding some φ to the argument θ can never result in cot θ , although I don't have a proof for this.

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