Answer :
To determine the domain of the function [tex]\( f(x) = \frac{x^2 + 6x + 5}{x^2 - 25} \)[/tex], we need to find the values of [tex]\( x \)[/tex] for which the function is defined. For a rational function like this, the function is undefined where the denominator is zero because division by zero is undefined.
1. Find where the denominator is zero:
We need to solve the equation:
[tex]\[ x^2 - 25 = 0 \][/tex]
To solve this, we recognize that it is a difference of squares:
[tex]\[ x^2 - 25 = (x - 5)(x + 5) = 0 \][/tex]
This equation is satisfied when:
[tex]\[ x - 5 = 0 \quad \text{or} \quad x + 5 = 0 \][/tex]
Hence,
[tex]\[ x = 5 \quad \text{or} \quad x = -5 \][/tex]
2. Define the domain:
The function [tex]\( f(x) \)[/tex] will be undefined at [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex]. Thus, we need to exclude these points from the domain.
The domain will include all real numbers except [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex]. So, in interval notation, the domain is:
[tex]\[ (-\infty, -5) \cup (-5, 5) \cup (5, \infty) \][/tex]
3. Compare with the provided choices:
Given the options, the one that corresponds to this interval notation is:
[tex]\[ (-\infty,-5) \cup (-5,5) \cup (5, \infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{(-\infty, -5) \cup (-5, 5) \cup (5, \infty)} \][/tex]
1. Find where the denominator is zero:
We need to solve the equation:
[tex]\[ x^2 - 25 = 0 \][/tex]
To solve this, we recognize that it is a difference of squares:
[tex]\[ x^2 - 25 = (x - 5)(x + 5) = 0 \][/tex]
This equation is satisfied when:
[tex]\[ x - 5 = 0 \quad \text{or} \quad x + 5 = 0 \][/tex]
Hence,
[tex]\[ x = 5 \quad \text{or} \quad x = -5 \][/tex]
2. Define the domain:
The function [tex]\( f(x) \)[/tex] will be undefined at [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex]. Thus, we need to exclude these points from the domain.
The domain will include all real numbers except [tex]\( x = 5 \)[/tex] and [tex]\( x = -5 \)[/tex]. So, in interval notation, the domain is:
[tex]\[ (-\infty, -5) \cup (-5, 5) \cup (5, \infty) \][/tex]
3. Compare with the provided choices:
Given the options, the one that corresponds to this interval notation is:
[tex]\[ (-\infty,-5) \cup (-5,5) \cup (5, \infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{(-\infty, -5) \cup (-5, 5) \cup (5, \infty)} \][/tex]
