Let f(x)=xex(1−x), then f(x) is
increasing on [−1/2,1]
decreasing on R
increasing on R
decreasing on [−1/2,1]
f′(x)=ex(1−x)+x(1−2x)ex(1−x)=1+x−2x2ex(1−x)=−(x−1)(2x+1)ex(1−x)
Since ex(1−x)>0 for all x, so f′(x)>0 if and only if (x−1)(2x+1)<0 i.e. −1/2=min(1,−1/2)<x<max(1,−1/2)=1
Thus f increases on [−1/2,1]