## Calculus, 10th Edition (Anton)

$$(f+g)(x)= 3 \sqrt{x-1}$$ $$(f-g)(x)= \sqrt{x-1}$$ $$(fg)(x)= 2x-2$$ $$(f\div g)(x)= 2$$ $$D_{f+g}=D_{f-g}=D_{fg} =[1, + \infty )$$ $$D_{f/g} =(1, + \infty )$$
$$(f+g)(x)=f(x)+g(x)=2\sqrt{x-1}+\sqrt{x-1}=3\sqrt{x-1}$$ $$(f-g)(x)=f(x)-g(x)=2\sqrt{x-1}-\sqrt{x-1}=\sqrt{x-1}$$ $$(fg)(x)=f(x)g(x)=2\sqrt{x-1}\sqrt{x-1}=2(x-1)=2x-2$$ $$(f/g)(x)=f(x)/g(x)=2\sqrt{x-1}/\sqrt{x-1}=2$$ The domains of $f+g$, $f-g$, and $fg$ are the intersection of the domains of $f$ and $g$, and the domain of $f/g$ is the intersection of the domains of $f$ and $g$ minus those points vanishing $g(x)$. So we have $$D_{f+g}=D_{f-g}=D_{fg}=[1, + \infty ) \cap [1, + \infty ) =[1, + \infty )$$ $$D_{f/g}=([1, + \infty ) \cap [1, + \infty ) )-\{ 1 \}=(1, + \infty ).$$