2、Function $f(x)=\sqrt{5 x+15}+\sqrt{12-x}$ has the range of
Solution: $[\sqrt{15}, 3 \sqrt{10}]$
Let $u=\sqrt{x+3}, v=\sqrt{12-x}$, then $u^{2}+v^{2}=15(u \geq 0, v \geq 0)$
Let $u=\sqrt{15} \cos \theta, v=\sqrt{15} \sin \theta$, where $\theta \in\left[0, \frac{\pi}{2}\right]$
Then $f(x)=\sqrt{5 x+15}+\sqrt{12-x}=\sqrt{5} u+v=5 \sqrt{3} \cos \theta+\sqrt{15} \sin \theta=3 \sqrt{10} \sin (\theta+\varphi)$

where $\tan \varphi=\sqrt{5}$
Therefore, when $\sin (\theta+\varphi)=1$, $f(x)_{\text {max }}=3 \sqrt{10}$.
When $\theta=\frac{\pi}{2}, \cos \varphi=\frac{\sqrt{6}}{6}$, $f(x)_{\min }=\sqrt{15}$.