6. Find the range of the function $f(x, y)=\sqrt{x+\frac{1}{2}}+\sqrt{y+\frac{1}{2}}$, where $x+y=1$.
6. Since $\left(\sqrt{x+\frac{1}{2}}\right)^{2}+\left(\sqrt{y+\frac{1}{2}}\right)^{2}=2=(\sqrt{2})^{2}$, we can construct a circle $\odot O$ with diameter $AC=\sqrt{2}$, and the chords $AB=\sqrt{x+\frac{1}{2}}, BC=\sqrt{y+\frac{1}{2}}, CD=1, DA=1$, then $f(x, y)=AB+BC=AB \cdot CD + BC \cdot DA = AC \cdot BD = \sqrt{2} BD$.
When $x=-\frac{1}{2}$, $AB=0$, at this time $y=\frac{3}{2}$, then $BC=\sqrt{2}$, thus $BD=1$.
When $x=\frac{1}{2}$, $AB=1=CD$, at this time, $y=\frac{1}{2}$, then $AB=BC=1$, thus $BD=\sqrt{2}$.
Therefore, $1 \leqslant BD \leqslant \sqrt{2}$. Hence $1 \cdot \sqrt{2} \leqslant f(x) \leqslant \sqrt{2} \cdot \sqrt{2}=2$, so the range of values is $[\sqrt{2}, 2]$.