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

 

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=e^{x(1-x)}+x(1-2x)e^{x(1-x)}\\ =\left(1+x-2x^2\right)e^{x(1-x)}=-(x-1)(2x+1)e^{x(1-x)}\end{array}$

Since $e^{x(1-x)}>0$ for all $x,\text{ so }{f}^{\mathrm{\prime }}\left(x\right)>0$ if and only if

$\begin{array}{r}(x-1)(1+2x)<0\text{ i.e }-1/2=min(1,-1/2)<x<\\ max(1,-1/2)=1\end{array}$

Thus $f$increases on $[– 1/2, 1].$