If $f(x) = \dfrac{2x + 1}{x - 3}$, $x \neq 3$, find $f^{-1}(x)$.

Let $y = \dfrac{2x+1}{x-3}$. Then $y(x - 3) = 2x + 1$, so $xy - 3y = 2x + 1$ and $x(y - 2) = 3y + 1$, giving $x = \dfrac{3y + 1}{y - 2}$. Interchanging the variables, $f^{-1}(x) = \dfrac{3x + 1}{x - 2}$. Simply inverting the fraction is a common error and does not give the inverse function.