Example. Find the range of the function $y=4 \sqrt{8-16 x}$ $+3 \sqrt{9 x-\frac{9}{2}}$.
The function's domain is $-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}$. The function can be transformed as follows:
$$
\begin{array}{l}
y=16 \sqrt{\frac{1}{2}-x}+9 \sqrt{x+\frac{1}{2}} . \\
\text { Let } x_{0}=\sqrt{\frac{1}{2}-x} \quad(0 \leqslant x_{0} \leqslant 1), \\
y_{0}=\sqrt{x+\frac{1}{2}} \quad\left(0 \leqslant y_{0} \leqslant 1\right) .
\end{array}
$$

Then $x_{0}^2 + y_{0}^2 = 1$.
Consider $\left(x_{0}, y_{0}\right)$ as a point on the arc of a circle, and the original function can be transformed as:
$$
y=16 x_{0}+9 y_{0} \text {. }
$$

Recall the point-to-line distance formula:
$$
d=\frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}},
$$

Then the original function can be transformed as:
$$
y=\sqrt{16^{2}+9^{2}} \cdot \frac{16 x_{0}+9 y_{0}}{\sqrt{16^{2}+9^{2}}},
$$

where $x_{0}^2 + y_{0}^2 = 1$. Let
$$
d'=\frac{16 x_{0}+9 y_{0}}{\sqrt{16^{2}+9^{2}}}>0 \text {, }
$$
$d'$ represents the distance from a point on the arc to the line $16 x + 9 y = 0$, and we have:
$$
y=\sqrt{16^{2}+9^{2}} \cdot d' \text {. }
$$

As shown in the figure, we have:
$$
\begin{array}{l}
d'_{(A) \text { min }}=\frac{16 \times 0+9 \times 1}{\sqrt{16^{2}+9^{2}}}=\frac{9}{\sqrt{337}}, \\
d'_{(C) \text { max }}=1 .
\end{array}
$$

By the continuity of the curve, we have:
$$
d' \in\left[\frac{9}{\sqrt{337}}, 1\right] \text {. }
$$

Substituting $y=\sqrt{16^{2}+9^{2}} d'$, we can find the range of the function $y \in[9, \sqrt{337}]$.
Therefore, for any function of the form
$$
y=A \sqrt{a x+b}+B \sqrt{c x+d}
$$

the function can be transformed as follows:
(1) When $a c > 0$, the function can be transformed as:
$$
y=m \sqrt{x+p}+n \sqrt{x+q} \text {, }
$$

where $x \geqslant -p, y \geqslant -q$.
Let $x_{0}=\sqrt{x+p}, y_{0}=\sqrt{x+q}$. Then
$$
x_{0}^2 - y_{0}^2 = p - q \text {. }
$$

Then consider the distance from a point on a segment of a hyperbola to the line $m x + n y = 0$, and the range can be found.

In addition, for any function of the form $y=A \sqrt{a x+b}+B x$, it can also be transformed as:
$$
y=m \sqrt{x+p}+B x(a>0),
$$

or $y=m \sqrt{p-x}+B x(a<0)$.
Let $x_{0}=\sqrt{x+p}$ (or $x_{0}=\sqrt{p-x}$ ),
$$
y_{0}=x \text {. }
$$

Then $x_{0}^2 - y_{0} = p$ (or $x_{0}^2 + y_{0} = p$) (considering a segment of a parabola), and the original function can be transformed as:
$$
y=m x_{0}+B y_{0} \text {. }
$$

Then by finding the distance from a point on a segment of a parabola to a line, the range of the original function can be determined.