Find the interval in which the function f ( x )=2 x^3-3 x is strictly increasing.

Here, $\displaystyle \mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-3 \mathrm{x}$ $$\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2}-3=3\left(2 \mathrm{x}^{2}-1\right) $$ $$\text { Now, } \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}= \pm \frac{1}{\sqrt{2}} $$ $$\mathrm{f}(\mathrm{x}) \text { is strictly increasing in }\left(-\infty,-\frac{1}{\sqrt{2}}\right) \cup\left(\frac{1}{\sqrt{2}}, \infty\right) $$