Consider the function, f(x)=2x3−3x2−12x+5 So, f′(x)=6x2−6x−12=0 Now, x2−x−2=0 (x+1)(x−2)=0 x=−1,2∈[−2,4] Therefore, f(−2)=−16−12+24+5=1 f(−1)=−2−3+12+5=12 f(2)=16−12−24+5=−15 f(4)=128−48−48+5=37 So, the absolute maxima occurs at x = 4 .