Let f: R → R be defined as f(x) = \ cases - (4)/(3) x ^ 3 + 2 x ^ 2...

We know that, $\dfrac{d}{dx}\left({x}^{n}\right) = n{x}^{n - 1} and$\dfrac{d}{dx}\left({e}^{x}\right) = {e}^{x}$</p><p>Hence,$\dfrac{d}{dx}\left(- \dfrac{4}{3}{x}^{3} + 2{x}^{2} + 3x\right) = - \dfrac{4}{3} \times 3{x}^{2} + 2 \times 2x + 3 = - 4{x}^{2} + 4x + 3$</p><p>And, using product rule, we get$\dfrac{d}{dx}\left(3x{e}^{x}\right) = 3x\dfrac{d}{dx}\left({e}^{x}\right) + 3{e}^{x}\dfrac{d}{dx}\left(x\right) = 3x{e}^{x} + 3{e}^{x} = 3{e}^{x}\left(x + 1\right)$</p><p>$⇒ {f}^{'}\left(x\right) = \left\{$\begin{cases}- 4{x}^{2} + 4x + 3, & x > 0 \\ 3{e}^{x} ( 1 + x ) , & x \le 0\end{cases}$\right)

For $x > 0, {f}^{'} ( x ) = - 4{x}^{2} + 4x + 3 = - \left(2x + 1\right)\left(2x - 3\right)$

The sign scheme of ${f}^{'}\left(x\right) is</p><p><img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/8fd6e4d4-1a95-441d-b595-ee07edd8cb03-WhatsAppImage2021-12-10at2.16.02PM.jpeg" style="width: 300px; height: 133px;" /></p><p>We know that, if${f}^{'}\left(x\right)$> 0$ then for those value of $x, f \left(x\right)$ is increasing.

$⇒ f ( x )$ is increasing in \left(- \dfrac{1}{2},\dfrac{3}{2}\right) .

For $x \le 0, {f}^{'} ( x ) = 3{e}^{x} ( 1 + x )$

As ${e}^{x} > 0, \forall x \in R$

$⇒ {f}^{'} ( x ) > 0 \forall x \in ( - 1, 0 ) .</p><p>So,$ in complete domain,f ( x )$is increasing in \left(- 1, 0\right)$ \cup \left(- \dfrac{1}{2}, \dfrac{3}{2}\right)$= \left(- 1, \dfrac{3}{2}\right)$ .