f(x)=2x3−9x2+12x+1 f′(x)=6x2−18x+12 f′′(x)=12x−18 For critical points, f′(x)=0 6x2−18x+12=0 ⇒‌‌x2−3x+2=0 ⇒(x−1)(x−2)=0 ⇒‌‌x=1,2 Now, f′′(l)=12(l)−18=−6<0 ∴x=1 is print of maxima, and maximum value f(1)=2−9+12+1=6