f(x) = $\frac{4x+3}{6x-4}$ Let f $(x_1)$ = f $(x_2)$ ⇒ $\frac{4x_1+3}{6x_1-4}$ = $\frac{4x_2+3}{6x_2-4}$ ⇒ 24 $x_1x_2$ - 16 $x_1$ + 18 $x_2$ - 12 = 24 $x_1x_2$ + 18 $x_1$ - 16 $x_2$ - 12 ⇒ 18 $x_2$ + 16 $x_2$ = 18 $x_1$ + 16 $x_1$ ⇒ 34 $x_2$ = 34 $x_1$ Since, $\frac{4x_1+3}{6x_1-4}$ is a real number, therefore, for every y in the co-domain of f, there exists a number x in R - $\left\{ \frac{2}{3} \right\}$ such that f(x) = y = $\frac{4x_1+3}{6x_1-4}$ Therefore, f(x) is onto. Hence, $f^{-1}$ exists. Now, let y = $\frac{4x_1+3}{6x_1-4}$ ⇒ 6xy - 4y = 4x + 3 ⇒ 6xy - 4xx = 4y + 3 ⇒ x (6y - 4) = 4y+ 3 ⇒ x = $\frac{4y+3}{6y-4}$ ⇒ y = $\frac{4x+3}{6x-4}$ int exchanging the variables x and y ⇒ $f^{-1}$ (x) = $\frac{4x+3}{6x-4}$ [put y = $f^{-1}$ x]