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If [tex]f(x)[/tex] and [tex]f^{-1}(x)[/tex] are inverses of each other, then we have
[tex]f\left(f^{-1}(x)\right) = x[/tex]
Since [tex]f(x) = \frac1{x+2}[/tex], we have
[tex]f\left(f^{-1}(x)\right) = \dfrac1{f^{-1}(x)+2} = x[/tex]
Solve for [tex]f^{-1}(x)[/tex] :
[tex]\dfrac1{f^{-1}(x)+2} = x \\\\ 1 = x\left(f^{-1}(x)+2\right) \\\\ 1 = x f^{-1}(x) + 2x \\\\ x f^{-1}(x) = 1 - 2x \\\\ f^{-1}(x) = \dfrac{1-2x}x \\\\ \boxed{f^{-1}(x) = \dfrac1x - 2}[/tex]
(provided that x ≠ 0 and x ≠ -2)