The functions \( f \) and \( g \) are defined as follows. \[ f(x)=\frac{x-9}{x^{2}+18 x+81} \] \[ g(x)=\frac{x+3}{x^{2}-5 x-24} \] For each function, find the domain. Write each answer as an interval or union of intervals. \begin{tabular}{|l|l|l|l|}\hline\( \frac{\square}{\square} \) & \( \infty \) & \( -\infty \) \\ \hline Domain of \( f: \square \) \\ \hline\( \square \square, \square) \) & {\( [\square, \square] \)} & \( (\square, \square] \) \\ \hline\end{tabular}

To find the domains of the functions \( f(x) \) and \( g(x) \), we need to determine where each function is defined, which generally involves identifying where the denominators do not equal zero. For \( f(x) = \frac{x-9}{x^{2} + 18x + 81} \), we first analyze the denominator: \[ x^{2} + 18x + 81 = (x + 9)^2. \] This means that the denominator equals zero when \( x + 9 = 0 \) or \( x = -9 \). Therefore, the domain of \( f \) excludes \( -9 \): \[ \text{Domain of } f: (-\infty, -9) \cup (-9, \infty). \] For \( g(x) = \frac{x+3}{x^{2}-5x-24} \), we factor the denominator: \[ x^{2} - 5x - 24 = (x - 8)(x + 3). \] This shows that the denominator equals zero when \( x - 8 = 0 \) or \( x + 3 = 0 \), giving us \( x = 8 \) and \( x = -3 \). Therefore, the domain of \( g \) excludes both \( 8 \) and \( -3 \): \[ \text{Domain of } g: (-\infty, -3) \cup (-3, 8) \cup (8, \infty). \] Thus, filling in the table will give us: \[ \begin{tabular}{|l|l|l|l|}\hline \( \frac{-9}{\infty} \) & \( \infty \) & \( -\infty \) \\ \hline \text{Domain of } f: (-\infty, -9) \cup (-9, \infty) \\ \hline \( (-\infty, -3) \) & \( (8, \infty) \) & \( [-3, 8) \) \\ \hline\end{tabular} \] So the final domains are: - Domain of \( f: (-\infty, -9) \cup (-9, \infty) \) - Domain of \( g: (-\infty, -3) \cup (-3, 8) \cup (8, \infty) \)