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The problem you're looking at involves analyzing a differentiable function \( f \) on the open interval \( (1,10) \), with specific values given for \( f(2) = -5 \), \( f(5) = 5 \), and \( f(9) = -5 \). We need to determine which statements must be true based on this information.

Let's evaluate each statement:

### I. \( f \) has at least 2 zeros.
- The function takes the value \( f(2) = -5 \), \( f(5) = 5 \), and \( f(9) = -5 \). By the Intermediate Value Theorem (IVT), because the function moves from a negative value at \( f(2) \) to a positive value at \( f(5) \), and then back to a negative value at \( f(9) \), there must be at least two points where the function crosses the x-axis, meaning there are at least two zeros.

### II. The graph of \( f \) has at least one horizontal tangent.
- The Mean Value Theorem (MVT) states that if a function is continuous and differentiable on an interval, then there exists at least one point where the slope of the tangent (the derivative) is zero between any two points where the function changes values. Since \( f(2) = -5 \), \( f(5) = 5 \), and \( f(9) = -5 \), there must be at least one point where the derivative is zero, indicating a horizontal tangent.

### III. For some \( c \), \( 2 < c < 5 \), \( f(c) = 3 \).
- The Intermediate Value Theorem (IVT) again applies here. Since \( f(2) = -5 \) and \( f(5) = 5 \), and \( f \) is continuous, there must be some value \( c \) between 2 and 5 where \( f(c) = 3 \), as the function transitions from -5 to 5.

### Conclusion:
All three statements (I, II, and III) are true based on the given function and the theorems involved. Thus, the correct answer is **(E) I, II, and III**.

Would you like further details on any of these concepts?

Here are 5 related questions:
1. What is the Intermediate Value Theorem and how does it apply to continuous functions?
2. Can you provide more examples where the Mean Value Theorem guarantees a horizontal tangent?
3. What other theorems are applicable to differentiable functions on closed intervals?
4. How does the concept of zeros relate to the general behavior of polynomial functions?
5. How can you extend the reasoning here to a non-differentiable function?

**Tip:** The Mean Value Theorem only applies if the function is differentiable on an open interval and continuous on a closed interval!

Let f be a function that is differentiable on the open interval (1,10). If f(2) = -5, f(5) = 5, and f(9) = -5, which of the following must be true?
I. f has at least 2 zeros.
II. The graph of f has at least one horizontal tangent.
III. For some c, 2 < c < 5, f(c) = 3.

Explore a problem involving a differentiable function on the interval (1,10), where key theorems like the Intermediate Value Theorem and Mean Value Theorem help determine zeros, horizontal tangents, and specific function values. This is a college-level calculus problem involving advanced function analysis.

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