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Answer:
To determine the values of \( x \) that are not in the domain of \( f(x) = \frac{x^2 + 6x + 5}{x^2 - 9} \), we identify where the denominator is zero because division by zero is undefined.
The denominator of \( f(x) \) is \( x^2 - 9 \). We set the denominator equal to zero to find the values where \( f(x) \) is undefined:
\[ x^2 - 9 = 0 \]
Solving for \( x \):
\[ x^2 = 9 \]
\[ x = \pm 3 \]
Therefore, \( x = 3 \) and \( x = -3 \) are the values where \( f(x) \) is not defined because they make the denominator zero.
Step-by-step explanation:
- The function \( f(x) \) is undefined where the denominator \( x^2 - 9 \) equals zero because division by zero is not allowed in mathematics.
- By solving \( x^2 - 9 = 0 \), we find that \( x = \pm 3 \).
- Hence, \( x = 3 \) and \( x = -3 \) are the values where \( f(x) \) is undefined, indicating these points are outside the domain of the function \( f(x) \).