The correct option is
A local maxima at
x=−15, local
minima at
x=−1Given f(x)=−(x−1)3(x+1)2
finding the derivative and equating to zero to find the critical points.
f′(x)=−3(x−1)2(x+1)2+(−(x−1)3)∗2∗(x+1)
⟹f′(x)=−3(x−1)2(x+1)2+(−2)(x−1)3(x+1)
⟹f′(x)=(x−1)2(x+1)(−3(x+1)+(−2)(x−1))
⟹f′(x)=(x−1)2(x+1)(−3x−3−2x+2)
⟹f′(x)=(x−1)2(x+1)(5x+1)(−1)
⟹(x−1)2(x+1)(5x+1)(−1)=0
⟹(x−1)2=0 and (x+1)=0 and (5x+1)=0
Therefore, x=1,−1,−15 are the critical points.
Now substituting the critical points in f(x) to get the local maximum and the local minimum.
Therefore, f(1)=−(1−1)3(1+1)2=0
f(−1)=−(−1−1)3(−1+1)2=0
f(−15)=−(−15−1)3(−15+1)2=(−65)3(−45)2=−34563125=−1.1
Hence, the local maximum occurs at x=−15 and local minimum occurs at x=1 or x=−1