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

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.

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

Note by Ralph James
6 months ago

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

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

then \(f^{-1}(x)=x^{\frac{1}{i}}=x^{-i}=\frac{1}{x^i}=\frac{1}{f(x)}\) Janardhanan Sivaramakrishnan · 6 months ago

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

EDIT: Nope, again \( f^{-1} \) doesn't exist. Take \( x = e^{2\pi n} \). Ameya Daigavane · 6 months ago

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@Ameya Daigavane That makes sense. \(f^{-1}(x)\) would turn out be a multi-valued function in this case. Janardhanan Sivaramakrishnan · 6 months ago

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Does y=x^i work? Oliver Daugherty-Long · 6 months ago

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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) Aditya Lalbondre · 5 months, 3 weeks ago

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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 \(f(1) = -1, f(-1) = 1\) Aloysius Ng · 5 months, 4 weeks ago

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@Aloysius Ng Adding on, this gives, \[ f(f(f(f(y)))) = y \] Ameya Daigavane · 5 months, 4 weeks ago

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I take the function as \(f(x)=x^i\). It satisfy the given condition Rishabh Deep Singh · 6 months ago

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@Rishabh Deep Singh Then \( f^{-1} \) will not be defined.

EDIT: Original function mentioned was \( f(x) = 1 \) New function still does not have an inverse defined. Ameya Daigavane · 6 months ago

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@Ameya Daigavane OOps Sorry! Rishabh Deep Singh · 6 months ago

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

What can we say about \( f(5) , f ( \frac{1}{4} ), f ( \frac{1}{5} ) \)? Calvin Lin Staff · 6 months ago

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Comment deleted 6 months ago

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@Akhil Bansal How?
\( f^{-1} (5) = 4 = \dfrac{1}{f(5)} \Rightarrow f(5) = \dfrac{1}{4} \)
\( f^{-1} (x) = \dfrac{1}{f(x)} \Rightarrow x = f\left(\dfrac{1}{f(x)}\right) \)
\( \Rightarrow 4= f\left(\dfrac{1}{5}\right) \) and \( \dfrac{1}{5} = f\left(\dfrac{1}{4}\right) \) Ameya Daigavane · 6 months ago

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One I can think of is 1/x, but it wasn't defined for \(x= 0\) Kay Xspre · 6 months ago

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@Kay Xspre \( f^{-1}(x) = f(x) \) then, sadly not what we wanted. Ameya Daigavane · 6 months ago

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