Consider the function f(x)=3x4−2x3−6x2+6x+1 and differentiate it with respect to x f′(x)=12x3−6x2−12x+6 Equating it to 0 we get, 12x3−6x2−12x+6=0 2x3−x2−2x+1=0 (x−1)(2x2+x−1)=0 (x−1)(2x−1)(x+1)=0 Therefore, the critical points are x=1,x=
1
2
and x=−1. Now double derivative is given by: f"(x)=6x2−2x−2 Now the function will have local maxima when f"(x)<0 At x=1 the value of double derivative is: f"(1)=2 At x=−1 the value of double derivative is: f"(−1)=6 At x=