The functions \( f \) and \( g \) are defined as follows. \[ f(x)=\frac{x}{x^{2}+4} \] \( g(x)=\frac{x+9}{x^{2}-81} \) For each function, find the domain. Write each answer as an interval or union of intervals.

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 \( f(x) = \frac{x}{x^2 + 4} \) is a rational function. The denominator \( x^2 + 4 \) must not be equal to zero for \( f(x) \) to be defined. 1. Set the denominator equal to zero: \[ x^2 + 4 = 0 \] This equation has no real solutions because \( x^2 + 4 \) is always positive (since \( x^2 \geq 0 \) for all real \( x \), and adding 4 keeps it positive). Since the denominator is never zero, the domain of \( f(x) \) is all real numbers: \[ \text{Domain of } f(x) = (-\infty, \infty) \] ### Domain of \( g(x) \) The function \( g(x) = \frac{x + 9}{x^2 - 81} \) is also a rational function. The denominator \( x^2 - 81 \) must not be equal to zero. 1. Set the denominator equal to zero: \[ x^2 - 81 = 0 \] Factor the equation: \[ (x - 9)(x + 9) = 0 \] This gives the solutions: \[ x = 9 \quad \text{and} \quad x = -9 \] Thus, \( g(x) \) is undefined at \( x = 9 \) and \( x = -9 \). Therefore, we exclude these points from the domain. The domain of \( g(x) \) is all real numbers except \( -9 \) and \( 9 \): \[ \text{Domain of } g(x) = (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \] ### Summary of Domains - Domain of \( f(x) \): \( (-\infty, \infty) \) - Domain of \( g(x) \): \( (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \)