The functions \( f \) and \( g \) are defined as follows. \[ \begin{array}{l} f(x)=\frac{x^{2}}{x+7} \\ g(x)=\frac{x+9}{x^{2}-81} \end{array} \] For each function, find the domain. Write each answer as an interval or union of intervals. Domain of \( f \) : Domain of \( g \)

To find the domain of the functions \( f(x) \) and \( g(x) \), we need to identify the values of \( x \) for which each function is defined. ### Domain of \( f(x) \) The function \( f(x) = \frac{x^2}{x + 7} \) is a rational function. The domain of a rational function is all real numbers except where the denominator is zero. 1. Set the denominator equal to zero: \[ x + 7 = 0 \] Solving for \( x \): \[ x = -7 \] Thus, the function \( f(x) \) is undefined at \( x = -7 \). Therefore, the domain of \( f \) is all real numbers except \( -7 \). In interval notation, the domain of \( f \) is: \[ (-\infty, -7) \cup (-7, \infty) \] ### Domain of \( g(x) \) The function \( g(x) = \frac{x + 9}{x^2 - 81} \) is also a rational function. We need to find where the denominator is zero. 1. Set the denominator equal to zero: \[ x^2 - 81 = 0 \] This can be factored as: \[ (x - 9)(x + 9) = 0 \] Solving for \( x \): \[ x = 9 \quad \text{or} \quad x = -9 \] Thus, the function \( g(x) \) is undefined at \( x = 9 \) and \( x = -9 \). Therefore, the domain of \( g \) is all real numbers except \( -9 \) and \( 9 \). In interval notation, the domain of \( g \) is: \[ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \] ### Summary of Domains - Domain of \( f \): \((- \infty, -7) \cup (-7, \infty)\) - Domain of \( g \): \((- \infty, -9) \cup (-9, 9) \cup (9, \infty)\)