When f1(x)=[f(x)]1 f^{-1}(x) = [f(x)]^{-1}

Hello, members of the Brilliant community. Does anyone know how to find a function f(x)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.

f1(x)=[f(x)]1=1f(x)f^{-1}(x) = [f(x)]^{-1} = \frac{1}{f(x)}

Note by Ralph James
3 years, 6 months ago

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

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

then f1(x)=x1i=xi=1xi=1f(x)f^{-1}(x)=x^{\frac{1}{i}}=x^{-i}=\frac{1}{x^i}=\frac{1}{f(x)}

Janardhanan Sivaramakrishnan - 3 years, 6 months ago

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

EDIT: Nope, again f1 f^{-1} doesn't exist. Take x=e2πn x = e^{2\pi n} .

Ameya Daigavane - 3 years, 6 months ago

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That makes sense. f1(x)f^{-1}(x) would turn out be a multi-valued function in this case.

Janardhanan Sivaramakrishnan - 3 years, 6 months ago

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Does y=x^i work?

Oliver Daugherty-Long - 3 years, 6 months ago

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

What can we say about f(5),f(14),f(15) f(5) , f ( \frac{1}{4} ), f ( \frac{1}{5} ) ?

Calvin Lin Staff - 3 years, 6 months ago

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

Rishabh Deep Singh - 3 years, 6 months ago

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Then f1 f^{-1} will not be defined.

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

Ameya Daigavane - 3 years, 6 months ago

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OOps Sorry!

Rishabh Deep Singh - 3 years, 6 months ago

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f1(x)=1f(x)f^{-1}(x) = \frac{1}{f(x)}

Let y=f1(x)y = f^{-1}(x), i.e. x=f(y)x= f(y).

f1(f(y))=1f(f(y))f^{-1}(f(y)) = \frac{1}{f(f(y))}

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

f(f(y))=1yf(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)f(x) = f(y), x=yx = y, and vice versa.

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

f(f(1y))=y=f(f1(y))f(f(\frac{1}{y})) = y = f(f^{-1}(y))

f(1y)=f1(y)=1f(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)=1f(1)=1, f(-1)=-1 or f(1)=1,f(1)=1f(1) = -1, f(-1) = 1

Aloysius Ng - 3 years, 6 months ago

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Adding on, this gives, f(f(f(f(y))))=y f(f(f(f(y)))) = y

Ameya Daigavane - 3 years, 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 - 3 years, 6 months ago

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

Kay Xspre - 3 years, 6 months ago

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f1(x)=f(x) f^{-1}(x) = f(x) then, sadly not what we wanted.

Ameya Daigavane - 3 years, 6 months ago

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