3. 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 $(\quad)$.
(Hint: $g(x)=x+\frac{1}{x^{2}}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \geqslant \frac{3}{2} \sqrt[3]{2}$. Also, $f(x)$ and $g(x)$ achieve the same minimum value at the same point, so $-\frac{p}{2}=$ $\sqrt[3]{2}, \frac{4 q-p^{2}}{4}=\frac{3}{2} \sqrt[3]{2}$. Solving these, we get $p=-2 \sqrt[3]{2}, q=\frac{3}{2} \sqrt[3]{2}+\sqrt[3]{4}$. Therefore, the maximum value of $f(x)$ is $f(2)=4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$. )