The correct option is C −23
The given function is a cubic polynomial and, hence, it is continuous and differentiable everywhere. Therefore the LMVT is applicable to this function.
f(−2)=(−2)3−2⋅(−2)2−(−2)+1
f(−2)=−8−8+2+1=−13
f(2)=23−2⋅22−2+1
f(2)=8−8−2+1=−1
f′(x)=3x2−4x−1
According to LMVT, we have:
f′(c)=f(b)−f(a)b−a
⇒3c2−4c−1=−1−(−13)2−(−2),
⇒3c2−4c−1=3,
⇒3c2−4c−4=0
⇒(3c+2)(c−2)=0⇒c=−23,2
Only one root c1=−23 belongs to the open interval (−2,2)