 Solution

If $x$ is a node, both sides vanish and any $\xi \in (a,b)$ works. Otherwise put $\omega (t)={\prod }_i(t-x_i)$ and $C=(f(x)-p_n(x))/\omega (x)$. The function $H(t)=f(t)-p_n(t)-C\omega (t)$ has $n+2$ distinct zeros, at the nodes and at $x$. Repeated application of Rolle's theorem gives an interior point $\xi$ where $H^{(n+1)}(\xi )=0$. As $p_n^{(n+1)}=0$ and ${\omega }^{(n+1)}=(n+1)!$, this gives $C=f^{(n+1)}(\xi )/(n+1)!$. ThereforeThe distinct nodes and $n\ge 1$ ensure $a<b$, so the interior point is meaningful.