Define
m=bβaf(b)βf(a)β
and set Ο:IβR by Ο(x)=f(x)βmx. Since f is continuous on [a,b] and differentiable at every point of (a,b) by hypothesis, the same is true for Ο. Also,
Ο(b)βΟ(a)=(f(b)βmb)β(f(a)βma)=f(b)βf(a)βm(bβa)=0,
so Ο(a)=Ο(b). Therefore Rolle's Theorem in One Dimension applies to Ο on [a,b], and there exists cβ(a,b) such that Οβ²(c)=0. Because
Οβ²(c)=fβ²(c)βm,
we obtain
fβ²(c)=m=bβaf(b)βf(a)β.
This is the desired conclusion.