f(x) = $\frac{4\mathit{x}+3}{6\mathit{x}-4}$
Let f $\left({\mathit{x}}_1\right)$ = f $\left({\mathit{x}}_2\right)$
⇒ $\frac{4{\mathit{x}}_1+3}{6{\mathit{x}}_1-4}$ = $\frac{4{\mathit{x}}_2+3}{6{\mathit{x}}_2-4}$
⇒ 24 ${\mathit{x}}_1{\mathit{x}}_2$ - 16 ${\mathit{x}}_1$ + 18 ${\mathit{x}}_2$ - 12 = 24 ${\mathit{x}}_1{\mathit{x}}_2$ + 18 ${\mathit{x}}_1$ - 16 ${\mathit{x}}_2$ - 12
⇒ 18 ${\mathit{x}}_2$ + 16 ${\mathit{x}}_2$ = 18 ${\mathit{x}}_1$ + 16 ${\mathit{x}}_1$
⇒ 34 ${\mathit{x}}_2$ = 34 ${\mathit{x}}_1$
Since, $\frac{4{\mathit{x}}_1+3}{6{\mathit{x}}_1-4}$ 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 = $\frac{4{\mathit{x}}_1+3}{6{\mathit{x}}_1-4}$
Therefore, f(x) is onto.
Hence, ${\mathit{f}}^{-1}$ exists.
Now, let y = $\frac{4{\mathit{x}}_1+3}{6{\mathit{x}}_1-4}$
⇒ 6xy - 4y = 4x + 3
⇒ 6xy - 4xx = 4y + 3
⇒ x (6y - 4) = 4y+ 3
⇒ x = $\frac{4\mathit{y}+3}{6\mathit{y}-4}$
⇒ y = $\frac{4\mathit{x}+3}{6\mathit{x}-4}$ int exchanging the variables x and y
⇒ ${\mathit{f}}^{-1}$ (x) = $\frac{4\mathit{x}+3}{6\mathit{x}-4}$ [put y = ${\mathit{f}}^{-1}$ x]