 Solution

Use the corrected Hermite interpolating polynomial from part (a). The auxiliary function $\phi$ vanishes at all $n+1$ nodes and at the additional point $x$. Its derivative vanishes at all nodes, because both $e^{\prime }(x_i)$ and the derivative of ${\pi }^2$ vanish there. Rolle's theorem between the $n+2$ ordered zeros of $\phi$ gives $n+1$ further, distinct zeros of ${\phi }^{\prime }$. Thus ${\phi }^{\prime }$ has $2n+2$ distinct zeros on the closed interval. The printed phrase that these include $x$ is not generally true; the extra zeros lie in intervals between zeros of $\phi$, and ${\phi }^{\prime }(x)$ need not vanish.

Repeated application of Rolle's theorem gives ${\phi }^{(2n+2)}(\xi )=0$ for some interior $\xi$. Since $h$ has degree at most $2n+1$ and ${\pi }^2$ is monic of degree $2n+2$,Rearranging givesAt an interpolation node, both sides are zero, so the error formula extends to the whole closed interval. The false location claim about one zero is unnecessary for this result.