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The problem in the image asks why the equation \( f(x) = 0 \) must have at least one solution between \( x = a \) and \( x = b \), given that a continuous function \( f(x) \) is negative at \( x = a \) and positive at \( x = b \).

### This is a typical application of the **Intermediate Value Theorem (IVT)**, which states:

**If \( f(x) \) is continuous on the closed interval \([a, b]\) and takes different signs at \( a \) and \( b \), then there exists at least one \( c \in (a, b) \) such that \( f(c) = 0 \).**

In this case:
- \( f(a) < 0 \) (negative at \( x = a \)).
- \( f(b) > 0 \) (positive at \( x = b \)).

According to the IVT, since the function is continuous and the values of \( f(x) \) at \( a \) and \( b \) are of opposite signs, there must be some \( c \in (a, b) \) where \( f(c) = 0 \).

### Analysis of answer options:

1. **Option A**: 
   - States that \( f(x) = 0 \) has at least one solution because \( f(x) \) is continuous and \( f(a) \) is negative while \( f(b) \) is positive. 
   - This is the correct interpretation of the **Intermediate Value Theorem (IVT)**.

2. **Option B**:
   - This suggests that the continuity of the function is irrelevant, which is incorrect. IVT only applies if the function is continuous.

3. **Option C**:
   - This focuses on a "generic" property of continuous functions, which isn’t directly addressing the IVT principle. It's somewhat vague and incorrect in this context.

4. **Option D**:
   - Suggests that the existence of a root is based on a condition unrelated to the Intermediate Value Theorem. This is incorrect since IVT explicitly requires the function to be continuous.

### Final Answer: **Option A** correctly applies the Intermediate Value Theorem.

Would you like any further clarification or details on the Intermediate Value Theorem?

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Here are five related questions to expand your understanding:
1. What are the conditions for applying the Intermediate Value Theorem?
2. Can the IVT be applied if \( f(a) \) and \( f(b) \) are both negative or both positive?
3. How does the continuity of a function affect its properties, such as having a zero crossing?
4. Could there be more than one solution in the interval \( (a, b) \)? Why or why not?
5. How can you geometrically interpret the Intermediate Value Theorem?

**Tip**: Continuity is a crucial property in many calculus theorems like IVT, Mean Value Theorem, and Extreme Value Theorem. Always check if a function is continuous on the given interval before applying these theorems!

A continuous function f(x) is known to be negative at x = a and positive at x = b. Why does the equation f(x) = 0 have at least one solution between x = a and x = b?

Explore how the Intermediate Value Theorem ensures that a continuous function crossing from negative to positive must have at least one root between two points. This problem emphasizes the importance of continuity and provides insights into finding solutions to equations like f(x) = 0.

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