f(x)=x³+3x²-9x+2

f’(x)=3x²+6x-9

f’(x)=0

3x²+6x-9=0

by solving the quadratic we get roots as follows

x=-3 and x=1

Hence by second derivative test

f’’(x)>0 or f”(x)<0 so it’s a point of minimum or maximum respecctively.

f”(x)=6x+6

f”(-3)=6(-3)+6

=-12<0

so At x=-3 maximum value.

f”(1)=6+6

=12>0

so At x=1 minimum value.

Option=(B)