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The problem asks which of the given functions satisfy the conditions of the **Intermediate Value Theorem (IVT)** for the existence of roots in the indicated intervals.

**IVT Conditions**:
1. The function must be continuous on the closed interval.
2. The function must take values at the endpoints of the interval that lead to a sign change, indicating the existence of at least one root within the interval.

Let’s analyze each function:

### I. \( f(x) = \frac{x - 2}{x} \), \( [-3, 3] \)
- **Discontinuity**: The function has a discontinuity at \( x = 0 \) because the denominator becomes zero.
- **Conclusion**: Since \( f(x) \) is not continuous on \( [-3, 3] \), the IVT cannot be applied to this function.

### II. \( f(x) = x^2 + 3 \), \( [-3, 3] \)
- **Continuity**: This is a polynomial function, which is continuous everywhere, including on \( [-3, 3] \).
- **Sign Change**: The function does not change sign within the interval because \( f(x) = x^2 + 3 \) is always positive (its minimum value is 3).
- **Conclusion**: No sign change occurs, so the IVT cannot be used to prove the existence of a root.

### III. \( f(x) = 2x^3 - 3x \), \( [-3, 3] \)
- **Continuity**: This is a polynomial function, and it is continuous on \( [-3, 3] \).
- **Sign Change**: Evaluate \( f(x) \) at the endpoints:
  - \( f(-3) = 2(-3)^3 - 3(-3) = -54 + 9 = -45 \)
  - \( f(3) = 2(3)^3 - 3(3) = 54 - 9 = 45 \)
- There is a sign change from negative to positive, indicating the existence of a root in the interval.
- **Conclusion**: The IVT can be applied to this function.

### Final Answer:
The Intermediate Value Theorem can be used for function **III only**. Therefore, the correct answer is:

**III only.**

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Do you need further clarification or details? Here are five related questions you can explore:

1. What is the formal definition of the Intermediate Value Theorem?
2. How do you check for discontinuities in rational functions like in function I?
3. Why are polynomial functions always continuous?
4. How does the sign of the function at the endpoints guarantee a root inside the interval?
5. Can the IVT be used if a function is only partially continuous on the interval?

**Tip**: Always check for both continuity and a sign change when applying the IVT.

For which of the following functions can we use the Intermediate Value Theorem to prove the existence of roots in the indicated interval?

Explore how the Intermediate Value Theorem can be applied to different functions within specific intervals. This detailed guide covers the conditions for continuity and sign change, with a focus on polynomial and rational functions.

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