We have, f(x)=xex(1−x) On differentiating both sides w.r.t. x, we get f′(x)=ex(1−x)+x(1−2x)ex(1−x) ⇒f′(x)=(1+x−2x2)ex(1−x) ⇒f′(x)=−(x−1)(2x+1)ex(1−x) since, ex(1−x)>0 for all x. Therefore, signs of f′(x) for different values of x are as shown below.