When actually $f(g(x))=g(f(x))$ holds? We can see that if $f(x)=g(x)=x$ then $f(g(x))=g(f(x))$.
I would like to see other examples of functions $f(x)$ and $g(x)$ such that $f(g(x))=g(f(x))$.   
P.S. By definition we also must have $D_{f\circ g}= D_{g\circ f}$   
 A: Functions need not be inverses of each other to commute. A simple example. Let $f(x)=x+1$ and $g(x)=x+2$.  Here, of course
$$f(g(x))=g(f(x))=x+3.$$
A: Here is a family of examples: let $h$ be any function from a set to itself, and $n,m$ any positive integers. Then $f(x)=h^m(x),g(x)=h^n(x)$ (here superscripts denote functional power, so for example $h^3(x)=h(h(h(x)))$). Then clearly $f(g(x))=h^{m+n}(x)=g(f(x))$.
A: Take $f(x)=x+1$ and $g(x)=x-1$
$$f(g(x))=g(x)+1\\
=(x-1)+1\\
=x\\
\text{ and }\\
g(f(x))=f(x)-1\\
(x+1)-1\\
=x$$
A: I think if you take two inverse functions, defined over e.g. $\Bbb R$ (or over the same subset of $\Bbb R$),
then you trivially get the desired property, because $x = f(g(x)) = g(f(x))$.
Here is an example where the two functions are not inverse.  
$f(x) = x^n$  
$g(x) = x^m$
where $n$ and $m$ are e.g. integers   
Also:   
$f(x) = a \cdot x$  
$g(x) = b \cdot x$
where $a$ and $b$ are e.g. reals
A: I'm not sure there's much to say.
There's $g= f^{-1}$ so $f (g (x)) = x $
Then there's $f = g $ so $f (g (x))e f (f (x)) $
Then there's $f (g (x)) = g (f (x)) = h (x)$ so $g (x) = f^{-1}(g(f (x))) = f^{-1}(h (x)) $ and/or $f (x)=g^{-1}(h (x))$.
As long as we can find invertible functions we can find these.  Is there anything more that needs to be said?
