The functions \( f \) and \( g \) are defined as follows. \[ f(x)=\frac{x-7}{x^{2}+14 x+49} \] \( g(x)=\frac{x-5}{x^{2}-2 x-15} \) 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 \( x \) that make the denominators of each function non-zero. ### For the function \( f(x) \): The function is defined as: \[ f(x) = \frac{x-7}{x^2 + 14x + 49} \] First, we need to find when the denominator is zero: \[ x^2 + 14x + 49 = 0 \] This can be factored as: \[ (x + 7)^2 = 0 \] Thus, the only solution is: \[ x + 7 = 0 \implies x = -7 \] Therefore, the function \( f(x) \) is undefined at \( x = -7 \). The domain of \( f(x) \) is all real numbers except \( -7 \): \[ \text{Domain of } f(x) = (-\infty, -7) \cup (-7, \infty) \] ### For the function \( g(x) \): The function is defined as: \[ g(x) = \frac{x-5}{x^2 - 2x - 15} \] Next, we find when the denominator is zero: \[ x^2 - 2x - 15 = 0 \] We can factor this quadratic: \[ (x - 5)(x + 3) = 0 \] Thus, the solutions are: \[ x - 5 = 0 \implies x = 5 \] \[ x + 3 = 0 \implies x = -3 \] Therefore, the function \( g(x) \) is undefined at \( x = 5 \) and \( x = -3 \). The domain of \( g(x) \) is all real numbers except \( 5 \) and \( -3 \): \[ \text{Domain of } g(x) = (-\infty, -3) \cup (-3, 5) \cup (5, \infty) \] ### Summary of Domains: - Domain of \( f(x) \): \( (-\infty, -7) \cup (-7, \infty) \) - Domain of \( g(x) \): \( (-\infty, -3) \cup (-3, 5) \cup (5, \infty) \)