= Solution
{c}
Let $r=f-p_n$ be the <continuous function> describing the error. The preceding <orthogonality> equations say
$$
\int_{-1}^1w(x)r(x)q(x)\,dx=0\qquad(q\in\mathcal P_n).
$$
If $r$ vanishes identically, any $n+1$ distinct points work. If $r$ already has at least $n+1$ distinct zeros, the assertion also follows. Suppose instead that it has at most $n$ distinct zeros. Among its interior zeros, list those at which its sign changes as $t_1<\cdots<t_m$, where $m\leq n$, and define the <polynomial>
$$
q(x)=\prod_{j=1}^m(x-t_j),
$$
with $q=1$ if $m=0$. Between consecutive zeros of $r$, its sign is constant by the <intermediate value theorem>. At each $t_j$, both $r$ and $q$ change sign; at every other zero of $r$, neither sign changes. Hence $rq$ has one fixed sign throughout $[-1,1]$ wherever it is nonzero. Replacing $q$ by $-q$ if necessary makes $rq\geq0$ everywhere.
Since $r$ is not identically zero and $q$ has only finitely many zeros, $rq$ is strictly positive on some open interval. Positivity of $w$ gives
$$
\int_{-1}^1w(x)r(x)q(x)\,dx>0,
$$
contradicting <orthogonality> because $q\in\mathcal P_n$. Thus
$$
\boxed{f(x_i)=p_n(x_i)\text{ at at least }n+1\text{ distinct points in }[-1,1].}
$$
This is the <interpolation property of weighted least-squares polynomial approximation>. It supplies <interpolation nodes> depending on $f$ and $w$, rather than nodes prescribed in advance.
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