f1(x)=x2−2x+7 f1′(x)=2x−2 , so f(x) is decreasing in [−3,0] and positive also f2(x)=e4x3−12x2−180x+31 f2′(x)=e4x3−12x2−180x+31⋅12x2−24x−180 =12(x−5)(x+3)e4x3−12x2−180x+31 So, f2(x) is also decreasing and positive in {−3,0} ∴ absolute maximum value of f(x) occurs at x=−3