19. Find the range of the function $y=\sqrt{a+x}+\sqrt{b-x}\left(a, b \in \mathbf{R}^{+}\right)$.
19. From $(\sqrt{a+x})^{2}+(\sqrt{b-x})^{2}=(\sqrt{a+b})^{2}$, in a circle with diameter $AB=\sqrt{a+b}$, take point $C$ on the semicircle such that $AC=\sqrt{a+x}, BC=\sqrt{b-x}$; take the midpoint $D$ on the other semicircle, then $AD=BD=\sqrt{\frac{a+b}{2}}$. Connect $CD$, knowing that $CD \leqslant AB$, by Ptolemy's theorem, we have $\sqrt{\frac{a+b}{2}} \cdot \sqrt{a+x}+\sqrt{\frac{a+b}{2}} \cdot \sqrt{b-x}=AB \cdot CD \leqslant(\sqrt{a+b})^{2}$, which means $y=\sqrt{a+x}+$ $\sqrt{b-x} \leqslant \sqrt{2(a+b)}$. Also, in $\triangle ABC$, $AC+BC \geqslant AB$ (equality holds when $C$ coincides with $A$ or $B$), hence $\sqrt{a+b} \leqslant$ $y \leqslant \sqrt{2(a+b)}$.