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Let's break down this problem.

### Part 1: Showing that \( x^3 - 15x + 2 = 0 \) has three solutions in the interval \([-4, 4]\)
The given equation is:
\[
f(x) = x^3 - 15x + 2
\]
We need to prove that this equation has three solutions in the interval \([-4, 4]\). This can be done using the **Intermediate Value Theorem (IVT)**, which applies because \( f(x) \) is a continuous function (a polynomial).

The IVT states that if a function is continuous on an interval \([a, b]\), and \( f(a) \) and \( f(b) \) have opposite signs, then there is at least one root in the interval \((a, b)\).

To find the number of solutions, we should evaluate \( f(x) \) at key points in the interval \([-4, 4]\) and observe sign changes.

### Evaluating \( f(x) \) at key points:
1. \( f(-4) = (-4)^3 - 15(-4) + 2 = -64 + 60 + 2 = -2 \)
2. \( f(-2) = (-2)^3 - 15(-2) + 2 = -8 + 30 + 2 = 24 \)
3. \( f(0) = 0^3 - 15(0) + 2 = 2 \)
4. \( f(2) = (2)^3 - 15(2) + 2 = 8 - 30 + 2 = -20 \)
5. \( f(4) = (4)^3 - 15(4) + 2 = 64 - 60 + 2 = 6 \)

Sign changes occur between:
- \( f(-4) = -2 \) and \( f(-2) = 24 \)
- \( f(0) = 2 \) and \( f(2) = -20 \)
- \( f(2) = -20 \) and \( f(4) = 6 \)

Thus, by the Intermediate Value Theorem, there are at least three roots in the interval \([-4, 4]\) where the function changes sign.

### Part 2: Identifying where the solutions exist
The second part of the question asks us to select the correct condition for where the solutions to the equation exist, based on the properties of the function \( f(x) \).

Since \( f(x) \) is continuous and we know solutions exist where \( f(x) \) changes sign (i.e., crossing the x-axis), we are looking for intervals where the function transitions from positive to negative or vice versa.

The correct answer is likely **Option A**, which states:
> Solutions exist between values of \( x_1 \) and \( x_2 \) if \( f(x_1) \leq 0 \) and \( f(x_2) \geq 0 \), or if \( f(x_1) \geq 0 \) and \( f(x_2) \leq 0 \).

This reflects the behavior described by the Intermediate Value Theorem: solutions exist where the function crosses the x-axis, requiring sign changes between \( x_1 \) and \( x_2 \).

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Would you like more details on any of these steps? Here are five related questions that may help deepen your understanding:

1. How does the Intermediate Value Theorem guarantee the existence of roots?
2. Why do polynomials like \( f(x) = x^3 - 15x + 2 \) have continuous behavior?
3. How can we find the exact solutions (roots) for the equation \( x^3 - 15x + 2 = 0 \)?
4. How does sign change between intervals correspond to finding roots of a function?
5. What methods can be used to approximate the roots of this cubic equation?

**Tip:** When analyzing polynomial functions, always consider their continuity and behavior at the boundary of the interval for finding roots.

Show that the equation x^3 - 15x + 2 = 0 has three solutions in the interval [-4, 4]. Let f(x) be equal to the left side of the equation. The function f is a polynomial, which is everywhere-continuous. Where do the solutions to the equation exist?

This problem explores the cubic equation x^3 - 15x + 2 = 0, demonstrating the existence of three solutions in the interval [-4, 4]. Using the Intermediate Value Theorem, it examines key polynomial behavior and where the solutions exist.

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