We have, f(x)=(x−1)2(x+1)3f′(x)=2(x−1)(x+1)3+3(x−1)2(x+1)2 ⇒ f′(x)=(x−1)(x+1)2[2x+2+3x−3] ⇒ f′(x)=(x−1)(x+1)2(5x−1) For maxima or minima put f′(x)=0∴(x−1)(x+1)2(5x−1)=0 ⇒ x=−1,51,1f′′(x)=(x+1)2(5x−1)+2(x−1)(x+1)(5x−1)+5(x−1)(x+1)2f′′(−1)=0f′′(1)=16>0f′′(51)=25−144<0∴f(x) has local maxima at x=51