Let $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ be defined as $\text{f (x)=}\left\{\begin{array}{ll}-\frac{4}{3}{\mathrm{x}}^3+2{\mathrm{x}}^2+3\mathrm{x},& \mathrm{x}>0\\ 3{\mathrm{xe}}^{\mathrm{x}},& \mathrm{x}\le 0\end{array}\text{.}\right.$ 

Then f is increasing function in the interval.

$$\begin{gathered} case 1 \\ x>0 \\ f\left(x\right)=\frac{-4}{3}{x}^3+2x^2+3x \\ f\text{'}\left(x\right)=-4x^2+4x+3=-(4x^2-4x-3)=-(4x^2-6x+2x-3)=-(2x-3)(2x+1) \\ f\left(x\right) is increa\sin g \Rightarrow f\text{'}\left(x\right)>0 \\ x\in \left(\frac{-1}{2},\frac{3}{2}\right) \\ x\in (0,\frac{3}{2}) \\ case 2: \\ x\le 0 \\ f\left(x\right)=3x{e}^x \\ f\text{'}\left(x\right)=3e^x+3x{e}^x=3e^x(x+1) \\ f\left(x\right) is increa\sin g \Rightarrow f\text{'}\left(x\right)>0 \\ x>-1 \\ x\in \left[-1,0\right] \\ x\in (-1,0]\cup [0,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.) \\ x\in (-1,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.) \end{gathered}$$