Find local maximum value and local minimum value (whichever exists) for the function $\displaystyle \mathrm{f}(\mathrm{x})=4 \mathrm{x}^{2}+\frac{1}{\mathrm{x}}(\mathrm{x} \neq 0)$.

$$\begin{aligned} & \mathrm{f}(\mathrm{x})=4 \mathrm{x}^{2}+\frac{1}{\mathrm{x}}(\mathrm{x} \neq 0) \\ & \mathrm{f}^{\prime}(\mathrm{x})=8 \mathrm{x}-\frac{1}{\mathrm{x}^{2}}=0 \\ & \Rightarrow \mathrm{x}^{3}=\frac{1}{8} \Rightarrow \mathrm{x}=\frac{1}{2} \\ & \mathrm{f}^{\prime \prime}(\mathrm{x})=8+\frac{2}{\mathrm{x}^{3}}>0 \text { at } \mathrm{x}=\frac{1}{2} \\ & \therefore \text { Local minimum value }=\mathrm{f}\left(\frac{1}{2}\right)=3 \end{aligned} $$