Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?

*f*(*x*) = *x*3 − 3*x* + 6,   [−2, 2]

Yes, it does not matter if *f* is continuous or differentiable; every function satisfies the Mean Value Theorem.Yes, *f* is continuous on [−2, 2] and differentiable on (−2, 2) since polynomials are continuous and differentiable on .     No, *f* is not continuous on [−2, 2].No, *f* is continuous on [−2, 2] but not differentiable on (−2, 2).There is not enough information to verify if this function satisfies the Mean Value Theorem.

If it satisfies the hypotheses, find all numbers *c* that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).

*c* =

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.

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = x^3 − 3x + 6, [−2, 2]. If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem.

Explore whether the function f(x) = x^3 − 3x + 6 satisfies the hypotheses of the Mean Value Theorem on the interval [−2, 2]. This problem involves checking continuity and differentiability, calculating the derivative, and solving for values of c that satisfy the theorem. Suitable for advanced high school or undergraduate calculus students.

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