If $f(x) = \cfrac{{(4x + 3)}}{{(6x - 4)}},\;x \ne \cfrac{2}{3},$ show that $fof(x) = x,$ for all $x\not = \cfrac{2}{3}.$ What is the inverse of f?

$f(x) = \cfrac{{4x + 3}}{{6x - 4}},\;x \ne \cfrac{2}{3}$

L.H.S. $= fof(x) = f(f(x))$

$= f\left[ {\cfrac{{4x + 3}}{{6x - 4}}} \right] = \cfrac{{4\left( {\cfrac{{4x + 3}}{{6x - 4}}} \right) + 3}}{{6\left( {\cfrac{{4x + 3}}{{6x - 4}}} \right) - 4}}$

$= \cfrac{{16x + 12 + 18x - 12}}{{24x + 18 - 24x + 16}} = \cfrac{{34x}}{{34}} = x = R.H.S.$

Now, $y = \cfrac{{4x + 3}}{{6x - 4}} \Rightarrow 6xy - 4y = 4x + 3$

$\Rightarrow$ $6xy - 4x = 4y + 3 \Rightarrow x(6y - 4) = 4y + 3$

$\Rightarrow$ $x = \cfrac{{4y + 3}}{{6y - 4}} \Rightarrow y = \cfrac{{4x + 3}}{{6x - 4}}$

Hence, inverse of f is f.