The functions \( f \) and \( g \) are defined as follows. \[ f(x)=\frac{x-2}{x^{2}+4 x+4} \] \[ g(x)=\frac{x-5}{x^{2}-4 x-5} \] 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 the functions are defined. This typically involves finding values that make the denominator zero, as these values will be excluded from the domain. ### Finding the Domain of \( f(x) \) The function \( f(x) \) is given by: \[ f(x) = \frac{x-2}{x^2 + 4x + 4} \] First, we need to find when the denominator is zero: \[ x^2 + 4x + 4 = 0 \] This can be factored as: \[ (x + 2)^2 = 0 \] Setting the factor equal to zero gives: \[ x + 2 = 0 \implies x = -2 \] Thus, the function \( f(x) \) is undefined at \( x = -2 \). The domain of \( f(x) \) is all real numbers except \( -2 \): \[ \text{Domain of } f(x) = (-\infty, -2) \cup (-2, \infty) \] ### Finding the Domain of \( g(x) \) The function \( g(x) \) is given by: \[ g(x) = \frac{x-5}{x^2 - 4x - 5} \] Next, we find when the denominator is zero: \[ x^2 - 4x - 5 = 0 \] This can be factored as: \[ (x - 5)(x + 1) = 0 \] Setting each factor equal to zero gives: \[ x - 5 = 0 \implies x = 5 \] \[ x + 1 = 0 \implies x = -1 \] Thus, the function \( g(x) \) is undefined at \( x = 5 \) and \( x = -1 \). The domain of \( g(x) \) is all real numbers except \( -1 \) and \( 5 \): \[ \text{Domain of } g(x) = (-\infty, -1) \cup (-1, 5) \cup (5, \infty) \] ### Summary of Domains - The domain of \( f(x) \) is: \[ (-\infty, -2) \cup (-2, \infty) \] - The domain of \( g(x) \) is: \[ (-\infty, -1) \cup (-1, 5) \cup (5, \infty) \]