f(x)=3x3−9x2−27x+15 f′(x)=9x2−18x−27 For maxima or minima: f′(x)=0⇒9x2−18x−27=0 ⇒x2−2x−3=0 ⇒x2−3x+x−3=0 ⇒x(x−3)+1(x−3)=0 ⇒x=−1,3 f′′(x)=18x−18 f′′(−1)=−18−18=−36<0 f′′(3)=18(3)−18=36>0 ∴ f′(x) has maximum value at x=−1. & max. value =3(−1)3−9(−1)2−27(−1)+15 =−3−9+27+15=30