## Calculus 8th Edition

(a)$$(f+g)(x)=f(x)+g(x)= \sqrt{3-x} +\sqrt{x^2-1}$$ $$D_{f+g}=(- \infty, -1] \cup [1,3]$$ (b)$$(f-g)(x)=f(x)-g(x)=\sqrt{3-x}-\sqrt{x^2-1}$$ $$D_{f-g}=(- \infty, -1] \cup [1,3]$$ (c) $$(fg)(x)=f(x)g(x)=\sqrt{3-x}\sqrt{x^2-1}=\sqrt{-x^3+3x^2+x-3}$$ $$D_{fg}=(- \infty, -1] \cup [1,3]$$ (d)$$(f/g)(x)=f(x)/g(x)=\frac{\sqrt{3-x}}{\sqrt{x^2-1}}$$ $$D_{f/g}=(- \infty, -1) \cup (1,3]$$
To find the domains, one should consider the following points. The domain of $f+g$, $D_{f+g}$, is $D_f \cap D_g$ . The domain of $f-g$, $D_{f-g}$, is $D_f \cap D_g$ . The domain of $fg$, $D_{fg}$, is $D_f \cap D_g$ . The domain of $f/g$, $D_{f/g}$, is $(D_f \cap D_g)- \{ \text{roots of } g(x) \}$ . The domain of root function $\sqrt {f}$, $D_\sqrt{f}$ is $D_f - \{ x \mid f(x) < 0 \}$, so we have $$D_f= \mathbb R - \{ x \mid x>3\}= (- \infty , 3]$$ andD_g= \mathbb R - \left ( \{ x \mid -1