Suppose that $x_{0}, x_{1}, \ldots, x_{n} \in[-1,1]$ are distinct points. Let $f$ be an infinitely differentiable real-valued function on an open interval containing $[-1,1]$. Let $p$ be the unique polynomial of degree at most $n$ such that $f\left(x_{r}\right)=p\left(x_{r}\right)$ for $r=0,1, \ldots, n$. Show that for each $x \in[-1,1]$ there is some $\xi \in(-1,1)$ such that

$$f(x)-p(x)=\frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_{0}\right) \ldots\left(x-x_{n}\right)$$

Now take $x_{r}=\cos \frac{2 r+1}{2 n+2} \pi$. Show that

$$|f(x)-p(x)| \leqslant \frac{1}{2^{n}(n+1) !} \sup _{\xi \in[-1,1]}\left|f^{(n+1)}(\xi)\right|$$

for all $x \in[-1,1]$. Deduce that there is a polynomial $p$ of degree at most $n$ such that

$$\left|\frac{1}{3+x}-p(x)\right| \leqslant \frac{1}{4^{n+1}}$$

for all $x \in[-1,1]$.