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

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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=x+\frac{1}{x} \ & 	ext { let } y=x+\frac{1}{x} \ 	herefore \quad & x y=x^2+1 \ 	herefore \quad & x^2-x y+1=0\end{aligned}$ $\begin{array}{ll}	herefore & x=\frac{y \pm \sqrt{y^2-4}}{2} \ 	herefore & x=\frac{y+\sqrt{y^2-4}}{2} \ 	herefore & \mathrm{f}^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}\end{array} \quad \ldots[\because x \in[1, \infty)]$

If $\mathrm{f}:[1, \infty) \rightarrow[2, \infty)$ is given by $\mathrm{f}(x)=x+\frac{1}{x}$ then…

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