$\text{ Let }f\left(x\right)=(x-3{)}^5(x+1{)}^4\text{ then }$

detailed solution

Correct option is C

f′(x)=(x−3)4(x+1)3(9x−7) and f′′(x)=8(x−3)3(x+1)29x2+14x+1f′′(x)=24(x−3)2(x+1)21x3−49x2++7x+13fiv(x)=24(x−3)(3x−1)21x3−49x2+7x+13+168(x+3)2(x+1)9x2−14x+1         fv(x)=48(3x−5)21x3−49x2+7x+13+336(x−3)9x2−14x+1(9x−7)          f(3)=f′(−1)=f′(7/9)=0 At x=−1,fiv(x)<0 so x=−1 is a point of maxima  At x=3,f′′(x)=f′′′(x)=fiv(x)=0 but fv(x)≠0 So _f has neither maximum nor minimum at x=3 At x=7/9,f′′(x)>0 and therefore is a point of minima