Given $$ f(x)=\frac{1}{x-1} $$ and $$ g(x)=\frac{1}{x+3} $$, find the domain of $$ f(g(x)) $$ Domain:

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1. **Find the domain of $$ g(x) $$:**  
   Since $$ g(x)=\frac{1}{x+3} $$, the denominator cannot be zero. That is,  
   $$
   x+3 \neq 0 \quad \Longrightarrow \quad x \neq -3.
   $$

2. **Determine when $$ f(g(x)) $$ is defined:**  
   The composite function is given by  
   $$
   f(g(x)) = f\left(\frac{1}{x+3}\right)=\frac{1}{\frac{1}{x+3}-1}.
   $$  
   For $$ f(x)=\frac{1}{x-1} $$, the input $$ g(x) $$ must not be $$ 1 $$ (since $$ 1-1=0 $$). Therefore,  
   $$
   \frac{1}{x+3} \neq 1.
   $$
   Solve for $$ x $$ by setting  
   $$
   \frac{1}{x+3} = 1 \quad \Longrightarrow \quad x+3=1 \quad \Longrightarrow \quad x=-2.
   $$
   Thus, $$ x \neq -2 $$.

3. **Combine the restrictions:**  
   The composite function $$ f(g(x)) $$ is defined when  
   - $$ x \neq -3 $$ (from the domain of $$ g(x) $$)  
   - $$ x \neq -2 $$ (to ensure $$ f(g(x)) $$ is defined)

Therefore, the domain of $$ f(g(x)) $$ is  
   $$
   \{x \in \mathbb{R} \mid x \neq -3 \text{ and } x \neq -2\}.
   $$
The domain of $$ f(g(x)) $$ is all real numbers except $$ x = -3 $$ and $$ x = -2 $$.

Given and , find the domain of
Domain:

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