The Mean Value Theorem (MVT) states that if a function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one number $c \in (a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}.$

Step 1: Check the hypotheses of the Mean Value Theorem.

The function given is $f(x) = x^3 - 3x + 6$, which is a polynomial.

• Polynomials are continuous and differentiable everywhere. Thus, $f(x)$ is continuous on the closed interval $[-2, 2]$ and differentiable on the open interval $(-2, 2)$.
• Therefore, the function satisfies the hypotheses of the Mean Value Theorem.

Step 2: Apply the Mean Value Theorem.

Now, we need to find all values of $c$ such that:

$f'(c) = \frac{f(2) - f(-2)}{2 - (-2)}.$

Step 2.1: Calculate $f(2)$ and $f(-2)$.

• $f(2) = 2^3 - 3(2) + 6 = 8 - 6 + 6 = 8$.
• $f(-2) = (-2)^3 - 3(-2) + 6 = -8 + 6 + 6 = 4$.

Now compute the difference:

$\frac{f(2) - f(-2)}{2 - (-2)} = \frac{8 - 4}{4} = \frac{4}{4} = 1.$

Step 2.2: Differentiate $f(x)$ to find $f'(x)$.

$f'(x) = \frac{d}{dx}(x^3 - 3x + 6) = 3x^2 - 3.$

We need to solve for $c$ such that $f'(c) = 1$: $3c^2 - 3 = 1.$

Solve for $c$: $3c^2 = 4 \quad \Rightarrow \quad c^2 = \frac{4}{3} \quad \Rightarrow \quad c = \pm \frac{2}{\sqrt{3}}.$

Rationalizing the denominator: $c = \pm \frac{2\sqrt{3}}{3}.$

Final Answer:

The values of $c$ that satisfy the conclusion of the Mean Value Theorem are:

$c = \frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}.$

Would you like more details or have any questions?

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Here are 5 related questions for further exploration:

1. What are the geometric implications of the Mean Value Theorem?
2. How does Rolle's Theorem relate to the Mean Value Theorem?
3. Can you apply the Mean Value Theorem to non-polynomial functions?
4. What happens if a function is not differentiable at some point on the interval? Does it still satisfy the MVT?
5. How do higher-degree polynomials impact the number of solutions to the MVT?

Tip: Always ensure that both continuity and differentiability hold on the respective intervals before applying the Mean Value Theorem.