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.

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

#Calculus

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Jeff Giff - 4 months 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 $\forall x \in (0,e^{2π}]$

Jason Gomez - 4 months ago

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

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

Yajat Shamji - 4 months, 1 week ago

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

Wasi Husain - 4 months ago

There may be no function such that:

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

Yajat Shamji - 4 months ago

@Yajat Shamji – $f:\{1\} ➝ \{1\} ,f(x)=x$ satisfies

Jason Gomez - 4 months ago

@Jason Gomez – Continuity cannot be defined for such a function.

A Former Brilliant Member - 1 month, 3 weeks ago

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

Jeff Giff - 4 months ago

Just think about this, xf(f(x))=1, if we put x=0, we have 0=1, which is absurd, thus f(f(x)) is discontinuous at x=0, but then that means f(x) must also be discontinuous at some point. Thus we cannot find a continuous function, whose reciprocal = inverse.

Kushal Dey - 1 month ago

You can restrict the domain

Jason Gomez - 1 month 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.

A Former Brilliant Member - 1 month, 3 weeks ago

If you try to find functions of the form of a Möbius transformation, we get a^2 + b^2 =0, a = d, b = c. So, we get two solutions

        f(x) = ( ix + 1)/(x + i )        and       f(x) = (-ix + 1)/(x - i )

where i is the square root of -1.

A Former Brilliant Member - 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$