# A function whose reciprocal is an inverse

Hi! Can anyone think of any continuous function that satisfies the following? ‘-1’ denotes the ‘inverse’ of the function.

$f^{-1}(x)= \frac{1}{f(x)}$ Note by Inquisitor Math
3 months ago

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

- 2 months, 4 weeks ago

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

- 2 months, 4 weeks ago

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 $\forall x \in (0,e^{2π}]$

- 2 months, 4 weeks ago

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

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

- 2 months, 4 weeks ago

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

- 2 months, 4 weeks ago

There may be no function such that:

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

- 2 months, 4 weeks ago

$f:\{1\} ➝ \{1\} ,f(x)=x$ satisfies

- 2 months, 4 weeks ago

Continuity cannot be defined for such a function.

- 2 weeks, 5 days ago

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.

- 2 weeks, 5 days ago