Example 1 If on the interval $[1,2]$, the function $f(x)=x^{2}+p x+q$ and $g(x)=x+\frac{1}{x^{2}}$ take the same minimum value at the same point, then the maximum value of $f(x)$ on this interval is ( ).
(A) $4+\frac{11}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(B) $4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(C) $1-\frac{1}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(D) None of the above answers is correct
Solution: $\because 1 \leqslant x \leqslant 2$,
$$
\begin{array}{l}
g(x)=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant 3 \sqrt[3]{\frac{x}{2} \cdot \frac{x}{2} \cdot \frac{1}{x^{2}}} \\
=\frac{3}{2} \sqrt[3]{2} .
\end{array}
$$

When and only when $\frac{x}{2}=\frac{1}{x^{2}}$, i.e., $x=\sqrt[3]{2}$, $g(x)$ has the minimum value $\frac{3}{2} \sqrt[3]{2}$.
$\therefore$ According to the problem, we have $\left\{\begin{array}{l}-\frac{p}{2}=\sqrt[3]{2}, \\ q-\frac{p^{2}}{4}=\frac{3}{2} \cdot \sqrt[3]{2} .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}p=-2 \sqrt[3]{2}, \\ q=\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4} .\end{array}\right.$
Thus, $f(x)=x^{2}-2 \sqrt[3]{2} x+\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4}$ $=(x-\sqrt[3]{2})^{2}+\frac{3}{2} \sqrt[3]{2}$.
Also, $\because 2-\sqrt[3]{2}>\sqrt[3]{2}-1$,
$\therefore f(x)$ is a decreasing function on $x \in[1, \sqrt[3]{2}]$ and an increasing function on $x \in[\sqrt[3]{2}, 2]$.
Therefore, $f(x)$ has a maximum value $f(2)=4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$. Hence, the correct choice is (B).