Is f(x) = x smooth? It may sound too basic to even be a question, but I couldn't find a straight answer in Wolfram Alpha, Wolfram Mathworld or Wikipedia. Several other examples of more complicated functions are given.
In Wolfram Mathworld it is written that 

A smooth function is a function that has continuous derivatives up to
  some desired order over some domain. (...) The number of continuous
  derivatives necessary for a function to be considered smooth depends
  on the problem at hand, and may vary from two to infinity.

$f(x) = x$ has derivative 1 of the first order and 0 of second order, so I would say based on this it has at least 2 derivatives. I think it also  has an infinite number of derivatives which are also 0.
Another page on Wolfram Mathworld says the following:

A $C^{\infty}$ function is a function that is differentiable for all degrees of differentiation. (...) All polynomials are $C^{\infty}$. (...) $C^{\infty}$ functions are also called "smooth" (...).

Since $f(x) = x$ is a polynomial, I'm concluding that the paragraphs above mean it is also smooth. 
 A: You may be having issues with the difference between existence and triviality.
If $f(x)=x$ then
$f(x)=x$ is continuouss
$f'(x)=1$ is continuous
$f''(x)=0$ is continuous
$f'''(x)=0$ is continuous
etc...
So all its derivatives are continuous. On the other hand, take $g(x)=x\times|x|$
$g(x)=x\times|x|$ is continuous
$g'(x)=\frac{|x|}2$ is continuous
$g''(x) = \frac12$ if $x>0$, $g''(x)=-\frac12$ if $x<0$ $g''(0)$ is undefined
Clearly $g''$ is not continuous because of $g''(0)$ not existing, and so $g$ only has two derivatives.
Intuitively, all of $f$'s derivatives have no breaks in their graph ($y=0$ is simply a nice line), while $g''$ graph has a gaping hole in it at $x=0$.  
A: A function is smooth is it has derivatives of infinite order. $f(x) = x$ is smooth because it has infinitely many derivatives which are all 0, except for the first one. Polynomials are smooth because eventually their derivatives are 0.
A: Yes, the identity function has derivatives of every finite order, and is therefore smooth. It doesn't matter that most of the derivatives are $0$ everywhere -- being $0$ is a perfectly cromulent way to exist.
