The functions \( f \) and \( g \) are defined as follows. \[ f(x)=\frac{x+8}{x^{2}-64} \] \[ g(x)=\frac{x}{x^{2}+1} \] For each function, find the domain. Write each answer as an interval or union of intervals. \begin{tabular}{|l|l|l|l|}\hline Domain of \( f: \square \) & {\( [\square, \square) \)} & \( \square \) & \( \square \cup \square \) \\ \hline & \( (\square, \square) \) & {\( [\square, \square] \)} & \( (\square, \square] \) \\ \hline\end{tabular}

To find the domain of the functions \( f(x) \) and \( g(x) \), we need to identify any values of \( x \) that would make the functions undefined. For \( f(x) = \frac{x+8}{x^2-64} \): 1. The denominator \( x^2 - 64 \) must not be zero. Solving \( x^2 - 64 = 0 \) gives \( x^2 = 64 \) or \( x = \pm 8 \). 2. Thus, the domain excludes \( x = 8 \) and \( x = -8\). The domain of \( f \) in interval notation is: \[ (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \] For \( g(x) = \frac{x}{x^2+1} \): 1. The denominator \( x^2 + 1 \) is always positive since \( x^2 \) is non-negative for all \( x \), and thus \( x^2 + 1 \) is never zero. So the domain of \( g \) is all real numbers: \[ (-\infty, \infty) \] Putting the answers in the specified format, we have: Domain of \( f: (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \) & \( [\text{Interval: } -8, -8] \) & \( \infty \) & \( (-\infty, \infty) \) \\ \hline \( \) & \( (-\infty, -8) \) & \( [\infty] \) & \( \infty \) \\ \hline