= Solution
Use the corrected <Hermite interpolating polynomial> from part (a). The auxiliary function $\varphi$ vanishes at all $n+1$ nodes and at the additional point $x$. Its <derivative> vanishes at all nodes, because both $e'(x_i)$ and the <derivative> of $\pi^2$ vanish there. <Rolle's theorem> between the $n+2$ ordered zeros of $\varphi$ gives $n+1$ further, distinct zeros of $\varphi'$. Thus $\varphi'$ 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 $\varphi$, and $\varphi'(x)$ need not vanish.
Repeated application of <Rolle's theorem> gives $\varphi^{(2n+2)}(\xi)=0$ for some interior $\xi$. Since $h$ has degree at most $2n+1$ and $\pi^2$ is <monic> of degree $2n+2$,
$$
0=f^{(2n+2)}(\xi)-\frac{e(x)}{\pi(x)^2}(2n+2)!.
$$
Rearranging gives
$$
\boxed{e(x)=\frac{f^{(2n+2)}(\xi)}{(2n+2)!}\prod_{i=0}^n(x-x_i)^2.}
$$
At 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.
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