f(x)=x3,f′(x)=3x2 and
f′(c)=3c2Let
a=−1,b=1,f(a)=f(−1)=(−1)3=−1f(b)=f(1)=(1)3.=+1.By Lagrange's mean value theorem, we have
f(b)=f(a)+(b−a)f′(c)
f(1)=f(−1)+(1+1)f′(c)
1=−1+2f′(c)
⇒1=−1+2(3c2)
⇒1=−1+6c22=6c2
2=6c2⇒c2=13,c=1√3ϵ[−1,1]
Hence, Lagrange's mean theorem is verified.