Let f : R → R be defined as f x = - 4 3 x 3 + 2 x 2 + 3 x ,   x > 0      …

We know that, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ and $\frac{d}{dx}\left(e^x\right)=e^x$

Hence, $\frac{d}{dx}\left(-\frac{4}{3}{x}^3+2x^2+3x\right)=-\frac{4}{3}\times 3x^2+2\times 2x+3=-4x^2+4x+3$

And, using product rule, we get$\frac{d}{dx}\left(3x{e}^x\right)=3x\frac{d}{dx}\left(e^x\right)+3e^x\frac{d}{dx}\left(x\right)=3x{e}^x+3e^x=3e^x\left(x+1\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\left\{\begin{array}{rl}-4x^2+4x+3,& x>0\\ 3e^x(1+x),& x\le 0\end{array}\right.$

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

The sign scheme of $f^{\text{'}}\left(x\right)$ is

We know that, if $f^{\text{'}}\left(x\right)>0$ then for those value of $x, f\left(x\right)$ is increasing.

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

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

As $e^x>0, \forall x\in R$

$\Rightarrow f^{\text{'}}\left(x\right)>0 \forall x\in (-1, 0).$

So, in complete domain, $f\left(x\right)$ is increasing in $\left(-1, 0\right)\cup \left(-\frac{1}{2}, \frac{3}{2}\right)=\left(-1, \frac{3}{2}\right).$