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solve this question correctly and get everything correctly:

The function \( f(x) = \frac{x^2 + 5x}{x^3 - 25x} \) has discontinuities where the denominator is zero. 

### Step-by-step solution:
1. The denominator is \( x^3 - 25x \), which factors to \( x(x - 5)(x + 5) \).
2. Setting the denominator equal to zero: 
   \[
   x(x - 5)(x + 5) = 0
   \]
   yields the solutions \( x = 0, x = 5, x = -5 \).

### Answers:
1. **Number of discontinuities:** There are 3 discontinuities.
2. **Values of \( x \) at which \( f(x) \) has discontinuities:** The function is discontinuous at \( x = -5, 0, 5 \).

Would you like more details or have any questions?

Here are 5 related questions for further understanding:
1. How does factoring help in identifying discontinuities in rational functions?
2. Can a discontinuity be removed by simplifying the function?
3. What happens at each discontinuity in terms of limits and function behavior?
4. What type of discontinuities are present in this function? (removable, infinite, or jump)
5. How do discontinuities affect the domain of a function?

**Tip:** Always check the denominator of a rational function to find its discontinuities by identifying where it equals zero.

Consider the function f(x) = (x^2 + 5x) / (x^3 - 25x). Determine the number of discontinuities of f(x). Determine the values of x at which f(x) has a discontinuity. Enter your answer as a comma-separated list of values.

Learn how to determine the number of discontinuities in the rational function f(x) = (x^2 + 5x) / (x^3 - 25x). This problem covers algebraic concepts such as rational functions and their discontinuities. Suitable for students in Grades 11-12.

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