If f:[1, ∞) →[2, ∞) is given by f(x)=x+1/x then f-1(x) equals :

$y=x+\frac{1}{x}$ so $yx=x^2+1⇒x^2-yx+1=0$

so $x=\frac{y±\sqrt{y^2-4}}{2}$  $f^{-1}(x)=\frac{x±\sqrt{x^2-4}}{2}$

as $x∈(1,∞)$

so $f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}$