QPaperGen

We are given:
\(f(x) = 2x^2 - 3x\). \[\]


Differentiate:
\[f'(x) = \dfrac{d}{dx}(2x^2 - 3x) = 4x - 3\]

Set \(f'(x) = 0\) to find critical point:
\[4x - 3 = 0 \Rightarrow x = \dfrac{3}{4}\]

Now analyze the sign of \(f'(x)\):

In the interval \(\left(-\infty,\dfrac{3}{4}\right)\), we have \(f'(x) = 4x - 3 < 0\)
\(\Rightarrow f(x)\) is strictly decreasing on \(\left(-\infty,\dfrac{3}{4}\right)\)
\[\]

In the interval \(\left(\dfrac{3}{4},\infty\right)\), we have \(f'(x) = 4x - 3 > 0\)
\(\Rightarrow f(x)\) is strictly increasing on \(\left(\dfrac{3}{4},\infty\right)\)

\[\]Final Answer:

(a) \(f(x)\) is strictly increasing in \(\left(\dfrac{3}{4}, \infty\right)\)

(b) \(f(x)\) is strictly decreasing in \(\left(-\infty, \dfrac{3}{4}\right)\)