Find the intervals in which the function

The given function is $f(x)=2 x^{2}-3 x$.

$f^{\prime}(x)=4 x-3$

$\therefore f^{\prime}(x)=0 \Rightarrow x=\frac{3}{4}$

Now, the point $\frac{3}{4}$ divides the real line into two disjoint intervals i.e., $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$.

In interval $\left(-\infty, \frac{3}{4}\right), f^{\prime}(x)=4 x-3<0$.

Hence, the given function $(f)$ is strictly decreasing in interval $\left(-\infty, \frac{3}{4}\right)$.

In interval $\left(\frac{3}{4}, \infty\right), f^{\prime}(x)=4 x-3>0$.

Hence, the given function $(f)$ is strictly increasing in interval $\left(\frac{3}{4}, \infty\right)$.