soal yang jika yaaa .... pertanyaanya maka f^-1 =

[tex]\displaystyle f(x)=\frac{2}{x+2} \\ y=\frac{2}{x+2} \\ (x+2)y=2 \\ x+2=\frac{2}{y} \\ x=\frac{2}{y}-2 \\ x=\frac{2-2y}{y} \\ f^{-1}(x)=\frac{2-2x}{x} [/tex]