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Continuity is an essential condition for the Mean Value Theorem (MVT) to apply to a function on a closed interval. A function is continuous on an interval if there are no breaks, jumps, or holes in its graph over that interval. For a function to be continuous on the interval \([-2, 2]\), it must be defined at every point in that interval.
\Polynomial functions, such as \(F(x) = \frac{x^3}{3}\), are continuous everywhere on the real number line. This means that they smoothly pass through all points without interruption.

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• No gaps or jumps exist.
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• Function values approach from both the left and right at any point.
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This is why the function \(F(x) = \frac{x^3}{3}\) is continuous on \([-2, 2]\). This fulfills the first requirement for the Mean Value Theorem.

Differentiability is the second requirement for applying the Mean Value Theorem. This condition states that a function must have a defined derivative at every point in an open interval \((-2, 2)\). A function is differentiable when it has a tangent line that approximates the function at any given point, meaning the function's graph does not have any sharp turns or corners.
Since polynomial functions like \(F(x) = \frac{x^3}{3}\) are smooth, they are differentiable at all points on the real number line. This includes the interval \((-2, 2)\), where the Mean Value Theorem is checked. Thus, the second condition is satisfied, ensuring that the polynomial has a derivative at every point within this interval.

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• The derivative provides information about the function's slope.
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• No abrupt changes in direction occur.
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This ensures that \(F(x) = \frac{x^3}{3}\) meets the differentiability condition of the Mean Value Theorem.

A polynomial function consists of terms that are powers of the variable, with coefficients. A basic characteristic of polynomials is their smooth and continuous curve, without breaks or sharp turns. For example, \(F(x) = \frac{x^3}{3}\) is a simple cubic polynomial.
Polynomials are among the most straightforward functions to work with because:

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• Their behavior is predictable.
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• The degree of the polynomial indicates the number of roots or intersections with the x-axis.
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• Higher powers result in more curves and turns in the graph.
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In our scenario, \(F(x) = \frac{x^3}{3}\) has a cubic term, which means its graph crosses the origin. The function is continuous and differentiable across anywhere on the real number line, making it an ideal candidate for applying calculus concepts like the Mean Value Theorem.

To apply the Mean Value Theorem, it's crucial to calculate the derivative of the function accurately. The derivative represents the function's rate of change or slope at any given point. For the polynomial \(F(x) = \frac{x^3}{3}\), its derivative is found using basic differentiation rules.
The differentiation of \(F(x)\) gives:\[\mathbf{F'(x) = \frac{d}{dx} \left( \frac{x^3}{3} \right) = x^2}\]This derivative, \(x^2\), provides crucial information for finding the specific value of \(c\) that satisfies the Mean Value Theorem.

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• The derivative must be set equal to the difference quotient calculated between the interval's endpoints.
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• Solving for \(c\) ensures the tangent line's slope equals the secant line's slope over the interval.
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Thus, solving \(c^2 = \frac{4}{3}\) yields the possible values of \(c\) within the interval \((-2, 2)\). These solutions align with the theorem, confirming the continuity and differentiability conditions.