6. The range of the function $f(x)=\sqrt{3 x-6}+\sqrt{3-x}$ is
$\qquad$
6. $[1,2]$.

Notice that the domain of $f(x)=\sqrt{3(x-2)}+\sqrt{3-x}$ is $[2,3]$. Therefore, let $x=2+\sin ^{2} \alpha\left(0 \leqslant \alpha \leqslant \frac{\pi}{2}\right)$.
$$
\begin{array}{l}
\text { Then } f(x)=\sqrt{3 \sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \alpha} \\
=\sqrt{3} \sin \alpha+\cos \alpha=2 \sin \left(\alpha+\frac{\pi}{6}\right) . \\
\text { And } \frac{\pi}{6} \leqslant \alpha+\frac{\pi}{6} \leqslant \frac{2 \pi}{3} \text {, at this time, } \\
\frac{1}{2} \leqslant \sin \left(\alpha+\frac{\pi}{6}\right) \leqslant 1 .
\end{array}
$$

Therefore, $1 \leqslant f(x) \leqslant 2$.