3. Given $f(x)=a x^{3}+b x^{2}+c x+d$, if $1 \leqslant f(-1) \leqslant 2,1 \leqslant f(1) \leqslant 3,2 \leqslant f(2) \leqslant 4$, $-1 \leqslant f(3) \leqslant 1$, find the range of values for $f(4)$.
3. Taking $x_{k}=-1,1,2,3$, by the Lagrange formula we have
$$
\begin{array}{l}
f(4)=\frac{(4-1)(4-2)(4-3)}{(-1-1)(-1-2)(-1-3)} \cdot f(-1)+\frac{(4+1)(4-2)(4-3)}{(1+1)(1-2)(1-3)} f(1)+ \\
\frac{(4+1)(4-1)(4-3)}{(2+1)(2-1)(2-3)} f(2)+\frac{(4+1)(4-1)(4-2)}{(3+1)(3-1)(3-2)} f(3) \\
=-\frac{1}{4} f(-1)+\frac{5}{2} f(1)+(-5) f(2)+\frac{5}{4} f(3) \text {. } \\
\text { Since }-\frac{1}{2} \leqslant-\frac{1}{4} f(-1) \leqslant-\frac{1}{4}, \frac{5}{2} \leqslant \frac{5}{2} f(1) \leqslant \frac{15}{2},-20 \leqslant-5 f(2) \leqslant-10,-\frac{15}{4} \leqslant \frac{5}{4} f(3) \leqslant \\
\frac{15}{4} \text {, thus }-21 \frac{3}{4} \leqslant f(4) \leqslant 1 \text {, hence the maximum value of } f(4) \text { is } 1 \text {, and the minimum value is }-21 \frac{3}{4} \text {. Therefore, the range of } f(4) \text { is } \\
{\left[-21 \frac{3}{4}, 1\right] \text {. }} \\
\end{array}
$$