If $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{xe}}^{\mathrm{x}(\mathrm{x}-1)}$, then f(x) is 

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{xe}}^{\mathrm{x}(1-\mathrm{x})}$
or     $$\begin{gathered} {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}(1-\mathrm{x})}+(1-2\mathrm{x}){\mathrm{xe}}^{\mathrm{x}(1-\mathrm{x})} \\ \begin{array}{r}=-{\mathrm{e}}^{\mathrm{x}(1-\mathrm{x})}\left(2{\mathrm{x}}^2-\mathrm{x}-1\right)\\ =-{\mathrm{e}}^{\mathrm{x}(1-\mathrm{x})}(2\mathrm{x}+1)(\mathrm{x}-1)\end{array} \end{gathered}$$
Sign scheme of f '(x) is as follows: 

Thus, g(x) is increasing in [-½,1].