The correct option is D x = –5 is point of local maxima.
f'(x)=−3x3−24x2−45x=−3x(x2+8x+15)⇒f'(x)=−3x(x+3)(x+5)⇒f'(x)=0⇒x=0, −3, −5f''(x)=−9x2−48x−45=−3(3x2+16x+15)f''(0)=−45<0.Therefore, x = 0 is point of local maximaf''(−3)=18>0Therefore, x = −3 is point of local minimaf''(−5)=−30<0Therefore, x = −5 is point of local maxima.