3. Let $f(x)=\sqrt{2-x}+\sqrt{3 x+12}, x \in[-4,2]$, then the range of $f(x)$ is $(\quad)$.
A. $[\sqrt{6},+\infty)$
B. $[\sqrt{6}, 2 \sqrt{3}]$
C. $[-\sqrt{6}, 3 \sqrt{2}]$
D. $[\sqrt{6}, 2 \sqrt{6}]$
3. D
3. Let $a=\sqrt{2-x} \quad b=\sqrt{3 x+12}$

Then $3 a^{2}+b^{2}=18$
$$
\begin{array}{l}
a+b \leqslant \sqrt{\left(\frac{1}{3}+1\right)\left(3 a^{2}+b^{2}\right)}=\sqrt{\frac{4}{3} \cdot 18}=2 \sqrt{16} \\
a+b \geqslant \sqrt{\frac{1}{3}}\left(3 a^{2}+b^{2}\right)=\sqrt{\frac{1}{3}} \cdot 18=\sqrt{6}
\end{array}
$$