Find the intervals in which the function $f$ given by $f(x)=2 x^{2}-3 x$ is
$(a)$ increasing
$(b)$ decreasing

(N/A) The given function is $f(x)=2 x^{2}-3 x$.
First,we find the derivative of the function:
$f^{\prime}(x) = \frac{d}{dx}(2x^2 - 3x) = 4x - 3$.
To find the critical points,we set $f^{\prime}(x) = 0$:
$4x - 3 = 0 \Rightarrow x = \frac{3}{4}$.
The point $x = \frac{3}{4}$ divides the real line into two disjoint intervals: $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$.
$(a)$ For the interval $\left(\frac{3}{4}, \infty\right)$,choose a test point $x = 1$. Then $f^{\prime}(1) = 4(1) - 3 = 1 > 0$. Since $f^{\prime}(x) > 0$,the function $f$ is strictly increasing in the interval $\left(\frac{3}{4}, \infty\right)$.
$(b)$ For the interval $\left(-\infty, \frac{3}{4}\right)$,choose a test point $x = 0$. Then $f^{\prime}(0) = 4(0) - 3 = -3 < 0$. Since $f^{\prime}(x) < 0$,the function $f$ is strictly decreasing in the interval $\left(-\infty, \frac{3}{4}\right)$.