If [tex]f^{-1}(x)[/tex] is the inverse of [tex]f(x)[/tex], then
[tex]f\left(f^{-1}(x)\right) = \dfrac{2f^{-1}(x)}{f^{-1}(x)+3} = x[/tex]
Solve for [tex]f^{-1}(x)[/tex]. Since [tex]f(x)[/tex] is only defined for [tex]x\neq-3[/tex], this means that [tex]f\left(f^{-1}(x)\right)[/tex] is only defined for [tex]f^{-1}(x)\neq-3[/tex]. Then the denominator is never zero, so we can multiply both sides by it:
[tex]\dfrac{2f^{-1}(x)}{f^{-1}(x)+3} = x \\\\ \dfrac{2f^{-1}(x)}{f^{-1}(x)+3} \cdot (f^{-1}(x)+3) = x(f^{-1}(x)+3) \\\\ 2f^{-1}(x) = x(f^{-1}(x)+3) \\\\ 2f^{-1}(x) = xf^{-1}(x) + 3x \\\\ 2f^{-1}(x) - xf^{-1}(x) = 3x \\\\ (2-x)f^{-1}(x) = 3x \\\\ \boxed{f^{-1}(x) = \dfrac{3x}{2-x}}[/tex]
provided that x ≠ 2.
