# 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
2 years, 1 month 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 $$f^{-1}(x)=x^{\frac{1}{i}}=x^{-i}=\frac{1}{x^i}=\frac{1}{f(x)}$$

- 2 years, 1 month ago

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

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

- 2 years, 1 month ago

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

- 2 years, 1 month ago

Does y=x^i work?

- 2 years, 1 month ago

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)

- 2 years, 1 month ago

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

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 $$f(1) = -1, f(-1) = 1$$

- 2 years, 1 month ago

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

- 2 years, 1 month ago

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

- 2 years, 1 month ago

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

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

- 2 years, 1 month ago

OOps Sorry!

- 2 years, 1 month ago

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

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

Staff - 2 years, 1 month ago

Comment deleted Mar 26, 2016

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 $$\dfrac{1}{5} = f\left(\dfrac{1}{4}\right)$$

- 2 years, 1 month ago

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

- 2 years, 1 month ago

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

- 2 years, 1 month ago