 Solution

If $x$ is a node the error and the product both vanish, and any interior $\xi$ works. Otherwise put $\omega (t)={\prod }_{i=0}^n(t-x_i)$ and $C=(f(x)-p_n(x))/\omega (x)$. The functionhas $n+2$ distinct zeros, at the nodes and at $x$. Repeated Rolle's theorem gives a point $\xi$ between the extreme zeros, and therefore in $(a,b)$, where $R^{(n+1)}(\xi )=0$. Since $p_n^{(n+1)}=0$ and ${\omega }^{(n+1)}=(n+1)!$, one gets $C=f^{(n+1)}(\xi )/(n+1)!$. Substitution gives