Hello, members of the Brilliant community. Does anyone know how to find a function \(f(x) \) if its inverse is equal to its reciprocal? I have been trying to come up with ways to do so, but to no avail.

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

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## Comments

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TopNewestI think the solution suggested by Oliver Daugherty-Long is correct.

If \(f(x)=x^{i}\), where \(i=\sqrt{-1}\),

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

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Okay, but are there any \( f:\mathbb{R}\to \mathbb{R}\)?

EDIT: Nope, again \( f^{-1} \) doesn't exist. Take \( x = e^{2\pi n} \).

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That makes sense. \(f^{-1}(x)\) would turn out be a multi-valued function in this case.

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Does y=x^i work?

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HInt: Suppose that \( f(4) = 5 \).

What can we say about \( f(5) , f ( \frac{1}{4} ), f ( \frac{1}{5} ) \)?

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I take the function as \(f(x)=x^i\). It satisfy the given condition

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Then \( f^{-1} \) will not be defined.

EDIT: Original function mentioned was \( f(x) = 1 \) New function still does not have an inverse defined.

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OOps Sorry!

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

Let \(y = f^{-1}(x)\), i.e. \(x= f(y)\).

\(f^{-1}(f(y)) = \frac{1}{f(f(y))}\)

\(y = \frac{1}{f(f(y))}\)

\(f(f(y)) = \frac{1}{y}\)

Maybe that helps...?

Edit 1: I just realised that the existence of an inverse proves that f(x) is bijective.

Therefore, if \(f(x) = f(y)\), \(x = y\), and vice versa.

\(f(f(y)) = \frac{1}{y}\)

\(f(f(\frac{1}{y})) = y = f(f^{-1}(y))\)

\(f(\frac{1}{y}) =f^{-1}(y) = \frac{1}{f(y)}\)

And we get that f(1) and f(-1) equals 1 or -1, f(1) not equal to f(-1).

i.e. \(f(1)=1, f(-1)=-1\) or \(f(1) = -1, f(-1) = 1\)

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Adding on, this gives, \[ f(f(f(f(y)))) = y \]

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In general Consider f(x) =y then replace all the x with y and y with x Then find y interms of x if u have one value for y,then that itself is inverse of f(x)

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One I can think of is 1/x, but it wasn't defined for \(x= 0\)

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\( f^{-1}(x) = f(x) \) then, sadly not what we wanted.

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