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The problem asks which function \( f(x) \) could **not** be an expression for a function that is **continuous** on the interval \( -1 < x < 3 \), but **not continuous** on the closed interval \( -1 \leq x \leq 3 \).

Let's analyze each option:

### Key points:
1. The function is continuous for \( -1 < x < 3 \), which means it has no breaks, holes, or vertical asymptotes in this open interval.
2. The function **is not continuous** on the closed interval \( -1 \leq x \leq 3 \), meaning that there is likely a discontinuity at \( x = -1 \) or \( x = 3 \).

### Analyze each option:
#### A) \( \frac{x+1}{x-3} \)
- This function has a potential discontinuity at \( x = 3 \), where the denominator is zero. However, it is continuous for \( -1 < x < 3 \), since \( x = 3 \) is outside the interval.
- Therefore, this could be a valid option.

#### B) \( \frac{x-3}{x+1} \)
- This function has a potential discontinuity at \( x = -1 \), where the denominator becomes zero.
- Since the function is **not continuous** at \( x = -1 \), this could be a valid option.

#### C) \( \frac{(x+1)}{(x-3)} \)
- This is similar to option A and would have a discontinuity at \( x = 3 \), but it's continuous in the open interval \( -1 < x < 3 \).
- Thus, this could also be valid.

#### D) \( \frac{1}{(x+1)(x-3)} \)
- This function has two potential discontinuities: one at \( x = -1 \) and one at \( x = 3 \).
- Since both \( x = -1 \) and \( x = 3 \) are problematic, this function could be **discontinuous** at both ends of the closed interval \( -1 \leq x \leq 3 \).

### Conclusion:
The function that **could not** be continuous on \( -1 < x < 3 \) while being discontinuous at \( -1 \leq x \leq 3 \) is:

**Option D: \( \frac{1}{(x+1)(x-3)} \)**

This is because it has discontinuities at both \( x = -1 \) and \( x = 3 \), which conflicts with the requirement that it is continuous on the open interval \( -1 < x < 3 \).

Let me know if you need further details or explanations!

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Here are some related questions to explore:

1. What are the general conditions for a rational function to be continuous?
2. How can you determine if a function is discontinuous at a particular point?
3. What is the difference between continuity on an open interval and a closed interval?
4. How do you handle vertical asymptotes when determining the continuity of a function?
5. Can a piecewise function be continuous if its pieces are rational functions?

**Tip**: Always check the denominators of rational functions for potential points of discontinuity by finding when they are zero.

The function f is continuous on the interval -1 < x < 3 and is not continuous on the interval -1 ≤ x ≤ 3. Which of the following could not be an expression for f(x)?

Understand how to identify functions that are continuous on an open interval but discontinuous on a closed interval, focusing on rational functions and their points of discontinuity. Learn about the key principles of function continuity and how to solve related problems. Suitable for Grades 10-12.

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