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?

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!