 Solution

Assume $f\in C^{n+1}[a,b]$ and reorder the interpolation nodes increasingly as $x_0<\cdots <x_n$. The polynomial interpolation error $e=f-p_n$ vanishes at all of them. Rolle's theorem suppliesThese $n$ points are distinct because their containing intervals are disjoint. Thus $q=p_n^{\prime }$ is a polynomial of polynomial degree at most $n-1$ that interpolates the function $f^{\prime }$ at the interpolation nodes ${\eta }_1,\dots ,{\eta }_n$.

Apply the polynomial interpolation error theorem to $f^{\prime }$ with these $n$ interpolation nodes. Since the $n$th derivative of $f^{\prime }$ is $f^{(n+1)}$, for every $x\in [a,b]$ there is a ${\xi }_x\in (a,b)$ withFor $x={\eta }_i$ the equality is zero on both sides; otherwise the same auxiliary-function and repeated-Rolle's theorem proof applies. This also covers $n=1$, where $q$ is a constant polynomial and the proof is the ordinary mean value theorem for $f^{\prime }$.

Writing $M={\max }_{[a,b]}|f^{(n+1)}|$ gives the genuine error boundThe first formula is sharper, but its intermediate interpolation nodes depend on $f$; the last bound eliminates those unknown locations. This is derivative interpolation error from Rolle points.