Example 3 Given the function $f(x)=a x^{2}-c$, satisfying: $-4 \leqslant f(1) \leqslant-1,-1 \leqslant f(2) \leqslant 5$, then, $f(3)$ should satisfy ( ).
(A) $-7 \leqslant f(3) \leqslant 26$
(B) $-4 \leqslant f(3) \leqslant 15$
(C) $-1 \leqslant f(3) \leqslant 20$
(D) $-\frac{28}{3} \leqslant f(3) \leqslant \frac{35}{3}$ (1983 National High School League Question)
Notice that $f(x)$ is an even function, so $f(-1)=f(1)$. Taking $x_{1}, x_{2}, x_{3}$ as $-1, 1, 2$ respectively, then by the Lagrange formula, we have
$$
\begin{aligned}
f(x) & =\frac{(x-1)(x-2)}{(-1-1)(-1-2)} \cdot f(-1)+\frac{(x+1)(x-2)}{(1+1)(1-2)} \cdot f(1)+\frac{(x+1)(x-1)}{(2+1)(2-1)} \cdot f(2) \\
& =-\frac{1}{3}\left(x^{2}-4\right) \cdot f(-1)+\frac{1}{3}\left(x^{2}-1\right) \cdot f(2),
\end{aligned}
$$

Therefore, $f(3)=-\frac{5}{3} f(1)+\frac{8}{3} f(2)$. Given the conditions $\frac{5}{3} \leqslant-\frac{5}{3} f(1) \leqslant \frac{20}{3},-\frac{8}{3} \leqslant \frac{8}{3} f(2) \leqslant \frac{40}{3}$, we have $-1 \leqslant f(3) \leqslant 2$. When $f(1)=-1, f(2)=-1$, we can find $a=0, c=1$, such that $f(3)=-1$. When $f(1)=-4, f(2)=5$, we can find $a=3, c=7$, such that $f(3)=20$. Therefore, the correct choice is C.

Note: Using the Lagrange formula, we can find a general term formula for any finite sequence: $a_{1}, a_{2}, \cdots, a_{m}$:
$$
a_{n}=f(n)=b_{1} \cdot n^{m-1}+b_{2} \cdot n^{m-2}+\cdots+b_{m}.
$$

This formula indicates that the general term of the sequence $a_{1}, a_{2}, \cdots, a_{m}$ is the value of the polynomial function $g(x)$ when $x=n$, where $b_{1}, b_{2}, \cdots, b_{m}$ are undetermined constants.
For example, let $a_{1}=1, a_{2}=5, a_{3}=11$. We need to find a general term formula for this sequence.
We can set $a_{n}=f(n)=b_{1} n^{2}+b_{2} n+b_{3}$.
Here, $x=n, x_{1}=1, x_{2}=2, x_{3}=3$, and $f(1)=1, f(2)=5, f(3)=11$.
By the Lagrange formula, we get
$$
\frac{(n-2)(n-3)}{(1-2)(1-3)} \cdot f(1)+\frac{(n-1)(n-3)}{(2-1)(2-3)} \cdot f(2)+\frac{(n-1)(n-2)}{(3-1) \cdot(3-2)} \cdot f(3)=f(n),
$$

Simplifying, we get $n^{2}+n-1=f(n)$.
Thus, the general term of the sequence is $a_{n}=n^{2}+n-1 (n=1,2,3)$.
Using the Lagrange formula, we can also find the remainder of polynomial division.
By the polynomial division theorem, for polynomials $f(x)$ and $g(x) (g(x) \neq 0)$, there must exist polynomials $q(x)$ and $r(x) (\operatorname{deg} r(x)<\operatorname{deg} g(x)$ or $r(x)=0)$, such that $f(x)=g(x) \cdot q(x)+r(x)$.

Let $\operatorname{deg} f(x) \geqslant n, \operatorname{deg} g(x)=n$, and $g(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)$. By the Lagrange formula, we have $r(x)=\sum_{i=1}^{n} f\left(x_{i}\right) \cdot R_{i}(x), R_{i}(x)=\prod_{j=1}^{n} \frac{x-x_{i}}{x_{i}-x_{j}}$, and $r\left(x_{i}\right)=f\left(x_{i}\right)$. Thus, by the factor theorem, we have $\left(x-x_{i}\right) \mid[f(x)-r(x)]$, so $g(x) \mid [f(x)-r(x)]$.

Since $\operatorname{deg} g(x)=n$, then $\operatorname{deg} r(x) \leqslant n-1<n$, so $r(x)$ is the remainder when $f(x)$ is divided by $g(x)$.