Example 11 Find the range of the function $f(x)=\sqrt{2 x-1}-\sqrt{x+3}$.
Solve: From $2 x-1 \geqslant 0$ and $x+3 \geqslant 0$, we get the domain as $x \geqslant \frac{1}{2}$. Let $\sqrt{x+3}=t$, then $t \geqslant \frac{\sqrt{14}}{2}$, and $x=t^{2}-3$, thus
$$
f(x)+\sqrt{x+3}=\sqrt{2 x-1} \xlongequal{\text { let } f(x)=y}\left\{\begin{array}{l}
t \geqslant-y \\
t \geqslant \frac{\sqrt{14}}{2} \\
t^{2}-2 y t-y^{2}-7=0
\end{array}\right.
$$
$\Leftrightarrow(\mathrm{I})\left\{\begin{array}{l}t \geqslant \frac{\sqrt{14}}{2} \\ y \geqslant-\frac{\sqrt{14}}{2} \\ t^{2}-2 y t-y^{2}-7=0\end{array} \quad\right.$ or ( II ) $\left\{\begin{array}{l}t \geqslant-y, \\ y \leqslant-\frac{\sqrt{14}}{2}, \\ t^{2}-2 y t-y^{2}-7=0 .\end{array}\right.$
Let $g(t)=t^{2}-2 y t-y^{2}-7$.
$\left\{\begin{array}{l}g\left(\frac{\sqrt{14}}{2}\right) \leqslant 0 \\ y \geqslant-\frac{\sqrt{14}}{2}\end{array}\right.$ or $\quad\left\{\begin{array}{l}\Delta \geqslant 0, \\ -\frac{b}{2 a}=y \geqslant \frac{\sqrt{14}}{2}, \\ y \geqslant-\frac{\sqrt{14}}{2} .\end{array}\right.$

Therefore, $y \geqslant-\frac{\sqrt{14}}{2}$. However, solution ( II ) has no solution.
Thus, the range of the function is $\left[-\frac{\sqrt{14}}{2},+\infty\right)$.