If $f(x)=\large\frac{1}{1-x}$ is a function defined in $R-\{1\}$ then $\;fofof(x)$ = ?

$\begin{array}{1 1} 1-x \\ x-1 \\ x \\ \large\frac{1}{1-x} \end{array}$

$fof(x)=\large\frac{1}{1-\big(\frac{1}{1-x}\big)}=\large-\frac{1-x}{x}$
$=\large\frac{x-1}{x}$
$fofof(x)=\large\frac{1}{1-\big(\frac{x-1}{x}\big)}$$=x$
answered May 12, 2013
edited Mar 21, 2014