Given \( f(x)=\frac{1}{x+1} \) and \( g(x)=\frac{1}{x-2} \), find the domain of \( f(g(x)) \) Domain:

To determine the domain of \( f(g(x)) \), we first need to analyze the inner function \( g(x) = \frac{1}{x-2} \). The function \( g(x) \) is defined for all \( x \) except \( x=2 \), so its domain is \( x \in \mathbb{R} \setminus \{2\} \). Next, we substitute \( g(x) \) into \( f(x) \). The function \( f(x) = \frac{1}{x+1} \) must also be considered. For \( f(g(x)) = f\left(\frac{1}{x-2}\right) \), we need to ensure that \( g(x) + 1 \neq 0 \). This means \( \frac{1}{x-2} + 1 \neq 0 \), simplifying to \( \frac{1}{x-2} \neq -1 \). Solving this inequality, we get: \[ \frac{1}{x-2} \neq -1 \quad \Rightarrow \quad 1 \neq - (x-2) \quad \Rightarrow \quad 1 \neq -x + 2 \quad \Rightarrow \quad x \neq 1 \] So, \( f(g(x)) \) is also undefined at \( x = 1 \). Therefore, the domain of \( f(g(x)) \) will exclude both \( x = 1 \) and \( x = 2 \). Putting it all together, the domain of \( f(g(x)) \) is: Domain: \( x \in \mathbb{R} \setminus \{1, 2\} \)