3. The range of the function $y=\sqrt{7-x}+\sqrt{9+x}$ is
Answer $[4,4 \sqrt{2}]$.
Analysis Since
$$
y^{\prime}=\frac{1}{2 \sqrt{x+9}}-\frac{1}{2 \sqrt{7-x}}=\frac{1}{2} \frac{\sqrt{7-x}-\sqrt{9+x}}{\sqrt{(x+9)(7-x)}}
$$

the function is decreasing on $(-9,-1)$ and increasing on $[-1,7)$, so
$$
y_{\max }=y(-1)=4 \sqrt{2}, \quad y_{\text {min }}=y(-9)=y(7)=4
$$

Therefore, $4 \leq y \leq 4 \sqrt{2}$.