7. The minimum value of the function $f(x)=\sqrt{9(x-3)^{2}+\left(x^{2}-12\right)^{2}}+$ $\sqrt{9 x^{2}+\left(x^{2}-\frac{9}{4}\right)^{2}}$ is $\qquad$
7. $\frac{57}{4}$.

Obviously, $\frac{1}{3} f(x)$
$$
=\sqrt{(x-3)^{2}+\left(\frac{x^{2}}{3}-4\right)^{2}}+\sqrt{x^{2}+\left(\frac{x^{2}}{3}-\frac{3}{4}\right)^{2}}
$$

Let $y=\frac{x^{2}}{3}$. This is the equation of a parabola, with its focus at $F\left(0, \frac{3}{4}\right)$, and the directrix
equation is $y=-\frac{3}{4}$.
Let point $A(3,4)$, as
shown in Figure 3. Then equation (1) can
be rewritten as
$$
\begin{array}{l}
\frac{1}{3} f(x) \\
=\sqrt{(x-3)^{2}+(y-4)^{2}}+\sqrt{x^{2}+\left(y-\frac{3}{4}\right)^{2}} .
\end{array}
$$

It represents the sum of the distances from a point $M(x, y)$ on the parabola to point $A$ and to the focus $F$, i.e., $\frac{1}{3} f = M A + M F$.

Noting that point $A$ is above the parabola, the distance from point $M$ to the focus is equal to its distance to the directrix, i.e., $M F = M H$. Therefore, when point $M$ moves to $M_{1}$ on the perpendicular line $A H_{1}$, the value of $M A + M H$ is minimized as
$$
A M_{1} + M H_{1} = A H_{1} = 4 + \frac{3}{4} = \frac{19}{4},
$$

i.e., $\frac{1}{3} f \geqslant \frac{19}{4} \Rightarrow f \geqslant \frac{57}{4}$.