Functions validity. Why does writing a function differently make it valid for a originally invalid input? 
$e.g:$
$$f(x) = \frac{1} {(\frac1x+2)(\frac1x-3)} \implies x≠0$$
Which  may alternatively be written as:
$$f(x) =\frac{x^2}{(1+2x)(1-3x)}$$
Which is valid for $x=0$?
Both graphically represent the same function. 
Thanks. 

let $$g(x)=\frac{x^2}{(1+2x)(1-3x)}$$ and $$f(x) = \frac{1} {(\frac1x+2)(\frac1x-3)}$$
we have 
$D_f=R-\{0,\frac{1}{3},\frac{-1}{2}\}$ and $D_g=R-\{\frac{1}{3},\frac{-1}{2}\}$, thus $g\ne f$
