To determine whether the function satisfies the hypotheses of the Mean Value Theorem (MVT) and find the values of $c$, follow these steps:

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Step 1: Recall the Mean Value Theorem

The MVT states that if a function $f(x)$ satisfies the following two conditions on a closed interval $[a, b]$:

1. Continuity: $f(x)$ is continuous on $[a, b]$,
2. Differentiability: $f(x)$ is differentiable on $(a, b)$,

then there exists at least one point $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}.$

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Step 2: Check Continuity and Differentiability of $f(x)$

Given $f(x) = x^3 - 3x + 4$:

• Polynomials are continuous and differentiable everywhere on $\mathbb{R}$. Therefore:
  • $f(x)$ is continuous on $[-2, 2]$.
  • $f(x)$ is differentiable on $(-2, 2)$.

Thus, $f(x)$ satisfies the hypotheses of the MVT.

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Step 3: Find $f'(x)$

Compute the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(x^3 - 3x + 4) = 3x^2 - 3.$

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Step 4: Compute $f(b) - f(a)$

Substitute $a = -2$ and $b = 2$: $f(2) = 2^3 - 3(2) + 4 = 8 - 6 + 4 = 6,$ $f(-2) = (-2)^3 - 3(-2) + 4 = -8 + 6 + 4 = 2.$ $\frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{6 - 2}{4} = \frac{4}{4} = 1.$

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Step 5: Solve $f'(c) = 1$

Set $f'(c) = 3c^2 - 3$ equal to $1$: $3c^2 - 3 = 1.$ Simplify: $3c^2 = 4, \quad c^2 = \frac{4}{3}, \quad c = \pm \frac{2}{\sqrt{3}}.$ Simplify further: $c = \pm \frac{2\sqrt{3}}{3}.$

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Final Answer

The function satisfies the hypotheses of the Mean Value Theorem, and the values of $c$ that satisfy the conclusion are: $c = \pm \frac{2\sqrt{3}}{3}.$

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Do you have questions or want clarification on any part of this?

Here are five related questions to deepen understanding:

1. Why is continuity necessary for the Mean Value Theorem?
2. What happens if a function is not differentiable on the interval?
3. How do polynomials generally satisfy the hypotheses of MVT?
4. What is the geometric interpretation of the Mean Value Theorem?
5. How does the result change if the interval is different (e.g., $[-1, 1]$)?

Tip: Always verify both continuity and differentiability when checking MVT, as violating either condition invalidates the theorem.