Di ketahui f(x) = √x² + 1 dan ( f ₀ g ) (x) = [tex] \frac{1}{x - 2} [/tex][tex] \sqrt{ x^{2} -4x + 5} [/tex] Tentukanlah ( f ₀ g ) ⁻¹ (x)

[tex](f\circ g)(x)=\frac{1}{x-2}\sqrt{x^2-4x+5} \\ y=\sqrt{\frac{x^2-4x+5}{(x-2)^2}} \\ y=\sqrt{\frac{x^2-4x+5}{x^2-4x+4}} \\ y^2=\frac{x^2-4x+5}{x^2-4x+4} \\ y^2=1+\frac{1}{(x-2)^2} \\ \frac{1}{(x-2)^2}=y^2-1 \\ (x-2)^2=\frac{1}{y^2-1} \\ x-2=\pm\sqrt{\frac{1}{y^2-1}} \\ x=2\pm\sqrt{\frac{1}{y^2-1}} \\ \\ (f\circ g)^{-1}(x)=2\pm\sqrt{\frac{1}{x^2-1}}[/tex]