Find the interval in which $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}$ is always increasing, $\displaystyle \mathrm{x} \neq 0$.

$\displaystyle \mathrm{f}^{\prime}(\mathrm{x})=1-\frac{1}{\mathrm{x}^{2}}=\frac{\mathrm{x}^{2}-1}{\mathrm{x}^{2}}$ $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})=0$ gives $\displaystyle \mathrm{x}=1,-1$ f is decreasing in $\displaystyle (-1,0) \mathrm{U}(0,1)$ as $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})<0$ f is increasing in $\displaystyle \mathrm{R}-(-1,1)$ as $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})>0$