If $f(x)=f^{-1}(x)$ Does it mean $f(x)=x$ or $f(x)=-x$

Question

If $f(x)=f^{-1}(x)$ $\forall$ $x \in Dom(f)$, Then Does it mean $f(x)=x$ or $f(x)=-x$

My question: Are there any other functions whose inverse is itself?

Answer 1

Sure. Take $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by$$f(x)=\begin{cases}x&\text{ if }x\in\mathbb{Q}\\-x&\text{ otherwise.}\end{cases}$$

Answer 2

Or take $\;f(x):=\cfrac1x\;,\;\;x\in (0,\infty)\;$ .

Answer 3

Actually it means that domain and codomain of $f$ coincide and that $f\circ f$ equals the identity function on this set.

The function $f$ is a so-called involution.

See here for some examples.

Answer 4

You know that the graph of $y=f^{-1}(x)$ is the graph of $y=f(x)$ flipped about the line $y=x$, right?

So any function whose graph is symmetric around the line $y=x$ will be its own inverse. So things like $y=1/x$ and $y= 3-x$ all work.

Answer 5

$${\displaystyle f(x)=\ln \left({\frac {e^{x}+1}{e^{x}-1}}\right).}$$

check that $$f\circ f(x)=x$$

Answer 6

Let $A,B$ be two disjoint subsets of $\Bbb R$ of equal cardinality, which means that there exists a bijection $g\colon A\to B$. Define $$f(x)=\begin{cases}g(x)&x\in A\\g^{-1}(x)&x\in B\\x&\text{otherwise}\end{cases} $$ Then $f$ is its own inverse. On the other hand, every involution of $\Bbb R$ is of this form (we can let $A=\{\,x\in \Bbb R\mid f(x)<x\,\}$, $B=\{\,x\in \Bbb R\mid f(x)>x\,\}$).

Answer 7

Simplest examples are in the form $$ \text{a symmetric expression in }x,y=\text{const}. $$ The expressions $x+y$, $xy$ and $e^x+y^y-e^{x+y}$ has already been mentioned. You can try to figure out other, for example, $x^2+y^2$ gives the function $$ x^2+y^2=C\quad\Rightarrow\quad y=f(x)=\sqrt{C-x^2}. $$ Do not forget to describe the domain of the function.