12. Find the range of the function $y=\sqrt{x+1}+\sqrt{4-2 x}$.

 untranslated portion:
 
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。

---

12. Find the range of the function $y=\sqrt{x+1}+\sqrt{4-2 x}$.
12. Let $\sqrt{x+1}=u \geqslant 0, \sqrt{4-2 x}=v \geqslant 0$, eliminating $x$, we get $\frac{u^{2}}{3}+\frac{v^{2}}{6}=1(u \geqslant 0, v \geqslant 0)$, whose graph is the part of the ellipse $\frac{u^{2}}{3}+$ $\frac{v^{2}}{6}=1$ in the first quadrant. Given $y=u+v$, we know that $v=-u+y$ represents a line in the $u o v$ plane, with $y$ being its intercept on the $v$-axis.

Since the line intersects the part of the ellipse, we know that $y$ is maximized when and only when the line is tangent to the part of the ellipse. At this point, using the discriminant method, we find $y_{\text {max }}=3$. Also, when and only when $v=-u+y$ passes through the point $(\sqrt{3}, 0)$, $y_{\text {min }}=\sqrt{3}$. Therefore, the range of $y$ is $[\sqrt{3}, 3]$.