We have. ‌f(x)=x3+2ax2+3(a+1)x+5 ‌f′(x)=3x2−4ax+3a+3 As the function is increasing in its entire demain ‌f(x)>0 ‌3x2−4x+3x+3>0 As cellicient of x2>0, then ‌D<0 ‌16a2−36a−36<0 ‌4a2−9a−9<0 ‌(4a+3)(a−3)<0 ‌⇒‌‌a∈(‌