If $f:\left[1,\infty \right)\to \left[2,\infty \right)$  is given by $f\left(x\right)=x+\frac{1}{x}$ , then $f^{-1}\left(x\right)=$

$f:\left[2,\infty \right)\to \left[1,\infty \right)$

$f\left(x\right)=x+\frac{1}{x}$

$y=f\left(x\right)=x+\frac{1}{x}$

$y=f\left(x\right)$

$x=f^{-1}\left(y\right)$                              

$y=x+\frac{1}{x}$

$y=\frac{x^2+1}{x}$

$xy=x^2+1$

$f^{-1}\left(y\right)=\frac{y\pm \sqrt{y^2-4}}{2}$                         $x^2-xy+1=0$

$f^{-1}\left(x\right)=\frac{x\pm \sqrt{x^2-4}}{2}$                        $x=\frac{y\pm \sqrt{y^2-4}}{2}$

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