Question

Given $f\left(x\right)=\frac{1}{(1-x)},g\left(x\right)=f\left\{f\right(x\left)\right\}$ and $h\left(x\right)=f\left\{f\right\{f\left(x\right)\left\}\right\}$, then find the value of $f\left(x\right)g\left(x\right)h\left(x\right)$.

Answer 1

$f\left(x\right)=\frac{1}{(1-x)},g\left(x\right)=f\left\{f\right(x\left)\right\}$ and $h\left(x\right)=f\left\{f\right\{f\left(x\right)\left\}\right\}$
Now, $g\left(x\right)=f\left\{f\right(x\left)\right\}$
$=f\left(\frac{1}{(1-x)}\right)$

$=\frac{1}{1-\frac{1}{(1-x)}}$

$\Rightarrow g\left(x\right)=\frac{x-1}{x}$
Now, $h\left(x\right)=f\left\{f\right\{f\left(x\right)\left\}\right\}$

$\Rightarrow h\left(x\right)=f\left\{g\right(x\left)\right\}$

$=f\left(\frac{x-1}{x}\right)$

$=\frac{1}{1-\frac{x-1}{x}}$

$\Rightarrow h\left(x\right)=x$
Consider, $f\left(x\right)g\left(x\right)h\left(x\right)=\frac{1}{(1-x)}\frac{(x-1)}{x}x$
$\Rightarrow f\left(x\right)g\left(x\right)h\left(x\right)=-1$