Let f(x) = sqrt{frac{x+1}{x+3}} and g(x) = sq | vedclass

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(B) The domain of $f/g$ is the set of all $x$ such that $f(x)$ is defined,$g(x)$ is defined,and $g(x) 
eq 0$.For $f(x) = \sqrt{\frac{x+1}{x+3}}$ to b

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$[-1, 2)$

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(B) The domain of $f/g$ is the set of all $x$ such that $f(x)$ is defined,$g(x)$ is defined,and $g(x) 
eq 0$.For $f(x) = \sqrt{\frac{x+1}{x+3}}$ to be defined,we need $\frac{x+1}{x+3} \geq 0$. This holds for $x \in (-\infty, -3) \cup [-1, \infty)$.For $g(x) = \sqrt{\frac{2-x}{x+3}}$ to be defined and $g(x) 
eq 0$,we need $\frac{2-x}{x+3} > 0$. Multiplying by $-1$,we get $\frac{x-2}{x+3} The domain of $f/g$ is the intersection of these two sets: $((-\infty, -3) \cup [-1, \infty)) \cap (-3, 2) = [-1, 2)$.

Let $f(x) = \sqrt{\frac{x+1}{x+3}}$ and $g(x) = \sqrt{\frac{2-x}{x+3}}$ be two real-valued functions. Then the domain of $f/g$ is

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