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The problem asks to **discuss the continuity of the function** \( f(x) = \frac{x+2}{x-3} \) on two intervals:

1. **Open interval**: \(-2 < x < 3\)
2. **Closed interval**: \([-2 \leq x \leq 3]\)

Let's analyze the function and its continuity:

### 1. **Identifying the Domain**:
The function \( f(x) = \frac{x+2}{x-3} \) has a denominator \( x-3 \), which is undefined at \( x = 3 \). Therefore, the function is undefined at \( x = 3 \).

### 2. **Continuity on the Open Interval** \((-2 < x < 3)\):
On this interval, the function is defined and continuous everywhere except at \( x = 3 \), where the function becomes undefined. However, for all other values of \( x \) in the open interval \((-2, 3)\), the function is continuous because it is a rational function (the quotient of two polynomials) and rational functions are continuous on their domain.

Therefore:
- The function is continuous on the open interval \((-2, 3)\) except at \( x = 3 \).

### 3. **Continuity on the Closed Interval** \([-2 \leq x \leq 3]\):
For the closed interval \([-2, 3]\), we must check continuity at the boundaries \( x = -2 \) and \( x = 3 \):

- **At \( x = -2 \)**: The function value is:
  \[
  f(-2) = \frac{-2 + 2}{-2 - 3} = 0
  \]
  Since the function value exists and is finite, the function is continuous at \( x = -2 \).

- **At \( x = 3 \)**: The function is undefined at \( x = 3 \) (because the denominator becomes zero). Therefore, the function is discontinuous at \( x = 3 \).

Hence, on the closed interval \([-2, 3]\), the function is continuous at \( x = -2 \), but it is not continuous at \( x = 3 \).

### Conclusion:
- On the **open interval** \((-2, 3)\), \( f(x) \) is continuous everywhere except at \( x = 3 \).
- On the **closed interval** \([-2, 3]\), \( f(x) \) is continuous at \( x = -2 \), but it is **discontinuous** at \( x = 3 \).

Would you like more details or have any questions about this?

Here are 5 related questions:
1. What is the definition of a discontinuity?
2. How do you determine if a rational function is continuous on an interval?
3. What types of discontinuities exist (e.g., jump, removable, infinite)?
4. How would the continuity change if we extended the domain to all real numbers?
5. How is the behavior of the function near a discontinuity like \( x = 3 \)?

**Tip**: Always check the denominator of a rational function to identify possible points of discontinuity (where the denominator equals zero).

Discuss the continuity of the function f(x) = (x + 2) / (x - 3) on the open interval -2 < x < 3 and on the closed interval -2 ≤ x ≤ 3.

Explore the continuity of the rational function f(x) = (x + 2) / (x - 3) on the intervals -2 < x < 3 and -2 ≤ x ≤ 3. Understand the points of discontinuity, especially at x = 3, and learn how to apply the concept of continuity to rational functions.

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