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

*f*(*x*) = *x*3 − 3*x* +3,   [−2, 2] If it satisfies the hypotheses, find all numbers *c* that satisfy the conclusion of the Mean Value Theorem.

To determine if the function satisfies the hypotheses of the **Mean Value Theorem (MVT)** on the interval \([-2, 2]\), we proceed as follows:

### **Step 1: Verify the hypotheses of the MVT**
The Mean Value Theorem requires that:
1. \( f(x) \) is **continuous** on \([a, b]\),
2. \( f(x) \) is **differentiable** on \((a, b)\).

#### 1. Continuity:
The function \( f(x) = x^3 - 3x + 3 \) is a polynomial. Polynomials are continuous everywhere. Therefore, \( f(x) \) is continuous on \([-2, 2]\).

#### 2. Differentiability:
The function \( f(x) = x^3 - 3x + 3 \) is a polynomial. Polynomials are differentiable everywhere. Therefore, \( f(x) \) is differentiable on \((-2, 2)\).

Since both conditions are satisfied, the function \( f(x) \) satisfies the hypotheses of the Mean Value Theorem on \([-2, 2]\).

---

### **Step 2: Apply the Mean Value Theorem**
The MVT guarantees that there exists at least one \( c \in (-2, 2) \) such that:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}.
\]
Here, \( a = -2 \) and \( b = 2 \). Compute the values of \( f(a) \) and \( f(b) \):
\[
f(-2) = (-2)^3 - 3(-2) + 3 = -8 + 6 + 3 = 1,
\]
\[
f(2) = (2)^3 - 3(2) + 3 = 8 - 6 + 3 = 5.
\]
The slope of the secant line is:
\[
\frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{5 - 1}{4} = 1.
\]

Thus, we need to find \( c \in (-2, 2) \) such that:
\[
f'(c) = 1.
\]

---

### **Step 3: Find \( c \) such that \( f'(c) = 1 \)**
First, compute \( f'(x) \):
\[
f'(x) = 3x^2 - 3.
\]
Set \( f'(x) = 1 \) to solve for \( c \):
\[
3x^2 - 3 = 1.
\]
Simplify:
\[
3x^2 = 4 \quad \implies \quad x^2 = \frac{4}{3}.
\]
Take the square root:
\[
x = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}.
\]

---

### **Step 4: Verify \( c \) is in the interval**
The solutions are:
\[
c = \frac{2\sqrt{3}}{3}, \quad c = -\frac{2\sqrt{3}}{3}.
\]
Both values are in the interval \((-2, 2)\).

---

### **Final Answer**
The function satisfies the hypotheses of the Mean Value Theorem on \([-2, 2]\). The values of \( c \) that satisfy the conclusion of the theorem are:
\[
c = \pm \frac{2\sqrt{3}}{3}.
\]

---

Would you like more details or clarification on any step?  

Here are 5 related questions for practice:
1. How do you prove a function is continuous and differentiable?
2. Can the Mean Value Theorem apply to functions that are not differentiable everywhere?
3. What does the Mean Value Theorem geometrically represent?
4. How would you apply the Mean Value Theorem to a trigonometric function?
5. What changes in the process if the interval is not symmetric about the origin?

**Tip:** For polynomials, continuity and differentiability are always guaranteed! Focus on finding the slope and derivative conditions for MVT.

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

Explore whether the polynomial function f(x) = x^3 − 3x + 3 satisfies the Mean Value Theorem on the interval [−2, 2]. Includes a detailed step-by-step solution with slope calculation and critical points. Perfect for calculus students studying the application of the MVT.

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