f(x)=xex(1−x),x∈Rf′(x)=ex(1−x)⋅[1+x−2x2] =−ex(1−x⋅[2x2−x−1]=−2ex(1−x⋅[(x+21)(x−1])f′(x)=−2ex(1−x)A where A=(x+21)(x−1) Now, exponential function is always + ve and f′(x) will be opposite to the sign of A which is -ve in [−21,1] Hence, f′(x) is + ve in [−21,1] ∴f(x) is increasing on [−21,1]