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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Step-by-step Solution
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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