Stephens Vaughn
04/23/2023 · Primary School

The functions \( f \) and \( g \) are defined as follows. \[ f(x)=\frac{x+9}{x^{2}+17 x+72} \] \[ g(x)=\frac{x^{2}}{x-3} \] For each function, find the domain. Write each answer as an interval or union of intervals.

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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 \( f(x) \) is given by: \[ f(x) = \frac{x + 9}{x^2 + 17x + 72} \] The function is defined as long as the denominator is not zero. Therefore, we need to find the values of \( x \) that make the denominator zero: \[ x^2 + 17x + 72 = 0 \] To solve this quadratic equation, we can factor it: \[ x^2 + 17x + 72 = (x + 8)(x + 9) = 0 \] Setting each factor to zero gives us: \[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \] \[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \] Thus, the function \( f(x) \) is undefined at \( x = -8 \) and \( x = -9 \). The domain of \( f(x) \) is all real numbers except \( -8 \) and \( -9 \). In interval notation, this is: \[ (-\infty, -9) \cup (-9, -8) \cup (-8, \infty) \] ### Domain of \( g(x) \) The function \( g(x) \) is given by: \[ g(x) = \frac{x^2}{x - 3} \] This function is defined as long as the denominator is not zero. Therefore, we need to find the values of \( x \) that make the denominator zero: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] Thus, the function \( g(x) \) is undefined at \( x = 3 \). The domain of \( g(x) \) is all real numbers except \( 3 \). In interval notation, this is: \[ (-\infty, 3) \cup (3, \infty) \] ### Summary of Domains - The domain of \( f(x) \) is: \[ (-\infty, -9) \cup (-9, -8) \cup (-8, \infty) \] - The domain of \( g(x) \) is: \[ (-\infty, 3) \cup (3, \infty) \]

Quick Answer

- Domain of \( f(x) \): \( (-\infty, -9) \cup (-9, -8) \cup (-8, \infty) \) - Domain of \( g(x) \): \( (-\infty, 3) \cup (3, \infty) \)
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