3. The range of the function $y=\sqrt{7-x}+\sqrt{9+x}$ is the interval $\qquad$ .
3. $[4,4 \sqrt{2}]$.

Obviously, the domain of the function is $x \in[-9,7]$, within this interval $y>0$.
Since $(7-x)+(9+x)=16$, that is
$$
\frac{7-x}{16}+\frac{9+x}{16}=1 \text {, }
$$

there exists $\alpha \in\left[0, \frac{\pi}{2}\right]$, such that
$$
\sqrt{\frac{7-x}{16}}=\sin \alpha, \sqrt{\frac{9+x}{16}}=\cos \alpha .
$$

Then $\frac{y}{4}=\sqrt{\frac{7-x}{16}}+\sqrt{\frac{9+x}{16}}$
$=\sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)$.
Since $0 \leqslant \alpha \leqslant \frac{\pi}{2}$, then $\frac{\pi}{4} \leqslant \alpha+\frac{\pi}{4} \leqslant \frac{3 \pi}{4}$.
Within this range $\frac{\sqrt{2}}{2} \leqslant \sin \left(\alpha+\frac{\pi}{4}\right) \leqslant 1$, hence
$1 \leqslant \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \leqslant \sqrt{2}$.
Therefore, $4 \leqslant y \leqslant 4 \sqrt{2}$ (when $x=7$, $y_{\text {min }}=4$; when $x=-1$, $y_{\max }=4 \sqrt{2}$ ).