If x is a node, both sides vanish and any ξ∈(a,b)works. Otherwise put ω(t)=∏i(t−xi) and C=(f(x)−pn(x))/ω(x). The function H(t)=f(t)−pn(t)−Cω(t) has n+2 distinct zeros, at the nodes and at x. Repeated application of Rolle's theorem gives an interior point ξ where H(n+1)(ξ)=0. As pn(n+1)=0 and ω(n+1)=(n+1)!, this gives C=f(n+1)(ξ)/(n+1)!. Therefore