Example 6 (1996 National High School League Question) 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. $1-\frac{1}{2} \sqrt[3]{4}+\sqrt[3]{4}$
D. None of the above answers is correct
Solution: Choose B. Reason: Since $1 \leqslant x \leqslant 2$, we know that
$g(x)=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant 3 \sqrt[3]{\frac{x}{2} \cdot \frac{x}{2} \cdot \frac{1}{x^{2}}}=\frac{3}{2} \sqrt[3]{2}$, with equality if and only if $\frac{x}{2}=\frac{1}{x^{2}}$, i.e., $x=\sqrt[3]{2}$, at which $g(x)$ has its minimum value $\frac{3}{2} \sqrt[3]{2}$.
Thus, from the given conditions, we have $\left\{\begin{array}{l}-\frac{p}{2}=\sqrt[3]{2}, \\ q-\frac{p^{2}}{4}=\frac{3}{2} \sqrt[3]{2},\end{array}\right.$ solving which yields $\left\{\begin{array}{l}p=-2 \sqrt[3]{2}, \\ q=\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4} .\end{array}\right.$
Therefore, $f(x)=x^{2}-2 \sqrt[3]{2} x+\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4}=(x-\sqrt[3]{2})^{2}+\frac{3}{2} \sqrt[3]{2}$.
Since $2-\sqrt[3]{2}>\sqrt[3]{2}-1$, we know that $f(x)$ is a decreasing function on $x \in[1, \sqrt[3]{2}]$ and an increasing function on $x \in[\sqrt[3]{2}, 2]$, hence $f(x)$ has its maximum value $f(2)=4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$.