Answer:
The answer to your question is the last option
Explanation:
Here there is a rational function so, we need to find the values where the denominator equals zero.
![f(x) = (x^(2) + 6x + 5)/(x^(2)- 25 )](https://img.qammunity.org/2020/formulas/mathematics/high-school/g0hoyrj72b9b31x2y3oo6pc8gvnm46wg40.png)
![x^(2) - 25 = 0](https://img.qammunity.org/2020/formulas/mathematics/high-school/9pta5lykvy6m69ypa0m4j0lqg8540pxnbn.png)
Factor (x - 5)(x + 5) = 0
x₁ - 5 = 0 x₂ + 5 = 0
The function does not exist in -5 and 5
x₁ = 5 x₂ = -5
Domain
( -∞ , -5) U (-5, 5) U (5, ∞)