Waste less time on Facebook — follow Brilliant.
×

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
1 year, 7 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 - 1 year, 7 months ago

Log in to reply

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 - 1 year, 7 months ago

Log in to reply

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

Janardhanan Sivaramakrishnan - 1 year, 7 months ago

Log in to reply

Does y=x^i work?

Oliver Daugherty-Long - 1 year, 7 months ago

Log in to reply

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 - 1 year, 7 months ago

Log in to reply

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

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 - 1 year, 7 months ago

Log in to reply

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

Ameya Daigavane - 1 year, 7 months ago

Log in to reply

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

Rishabh Deep Singh - 1 year, 7 months ago

Log in to reply

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 - 1 year, 7 months ago

Log in to reply

OOps Sorry!

Rishabh Deep Singh - 1 year, 7 months ago

Log in to reply

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

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

Calvin Lin Staff - 1 year, 7 months ago

Log in to reply

Comment deleted Mar 26, 2016

Log in to reply

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 - 1 year, 7 months ago

Log in to reply

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

Kay Xspre - 1 year, 7 months ago

Log in to reply

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

Ameya Daigavane - 1 year, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...