Given \( f(x)=\frac{1}{x+4} \) and \( g(x)=\frac{16}{x+2} \), find the domain of \( f(g(x)) \) Domain:

To find the domain of \( f(g(x)) \), we start by determining the outputs for \( g(x) \). Here, \( g(x) = \frac{16}{x+2} \), which is defined for all \( x \) except \( x = -2 \). Next, we need to consider when \( f(x) = \frac{1}{x+4} \) is defined. \( f(x) \) is undefined at \( x = -4 \). Thus, we need \( g(x) \neq -4 \): Setting \( g(x) = -4 \): \[ \frac{16}{x+2} = -4 \] Cross-multiplying gives: \[ 16 = -4(x + 2) \implies 16 = -4x - 8 \implies 4x = -24 \implies x = -6 \] So, the domain of \( f(g(x)) \) is all real numbers except \( x = -2 \) and \( x = -6 \). This can be expressed as: \[ (-\infty, -6) \cup (-6, -2) \cup (-2, \infty) \]