f(x)=2x3−3x2−36x+9 ‌x∈[−3,3] ‌f(x)=2x3−3x2−36x+9 On differentiating w.r.t. x, we get f′(x)=6x2−6x−36 On putting f′(x)=0 for critical numbers. ‌⇒‌‌6x2−6x−36=0 ‌⇒‌‌x2−x−6=0 ‌⇒‌‌x2−3x+2x−6=0 ‌⇒‌‌x(x−3)+2(x−3)=0 ‌⇒‌‌(x−3)(x+2)=0 ‌‌‌x=3,−2 ‌f(3)=2×27−3×9−36×3+9 ‌‌‌=54−27−108+9=−72(min) f(−2)=2(−8)−3(4)−36(−2)+9 =−16−12+72+9=53(max.) f(−3)‌=2(−27)−3(9)−36(−3)+9 ‌=−54−27+108+9=36 The absolute maximum value is 53.