Mathematics - Application of Derivative Question with Solution | TestHub

$y=x+\frac{1}{x}\text{ or }\frac{dy}{dx}=1-\frac{1}{x^2}$ For $\max$. or $\min .,1-\frac{1}{x^2}=0\Rightarrow x=\pm 1$ $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{2}{{\mathrm{x}}^3}\Rightarrow {\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}_{\mathrm{x}=2}=2(+$ ve minima) Therefore $x=1$