6. Find the range of the function $f(x)=\sqrt{x+1}+\sqrt{4-2 x}$.

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6. It can also be solved as in Question 5. Here is another solution: Let $\sqrt{1+x}=u, \sqrt{4-2 x}=v$, eliminating $x$ gives $\frac{u^{2}}{3}+\frac{v^{2}}{6}=1(u \geqslant 0$, $v \geqslant 0$ ), its graph is the part of the ellipse $\frac{u^{2}}{3}+\frac{v^{2}}{6}=1$ in the first quadrant. The original problem becomes finding $f(x)=u+v$, i.e., the intercept $f(x)$ of the line $v=-u+f(x)$ when the point is in the first quadrant of the $u O v$ plane.

Drawing the graph (omitted), by the line intersecting the part of the ellipse, we get that when and only when the line $v=-u+f(x)$ passes through point $A(\sqrt{3}, 0)$, the minimum intercept is $\sqrt{3}$, i.e., the minimum value of $f(x)$ is $\sqrt{3}$; when and only when the line $v=-u+f(x)$ is tangent to the part of the ellipse, the intercept is maximum, i.e., the maximum value of $f(x)$ is 3. Therefore, the range of $f(x)$ is $[\sqrt{3}, 3]$.