https://file.aiquickdraw.com/mathbot/file/172962621221946wla9xe.jpg

Let's break down and solve the problem in the image.

We are asked to use the **Mean Value Theorem (MVT)** for the function:

\[
f(x) = x^3 - 3x + 5
\]
on the interval \([-2, 2]\).

### Step 1: Apply the Mean Value Theorem formula

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

\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]

In this case:
- \(a = -2\)
- \(b = 2\)

We need to calculate \(f(-2)\) and \(f(2)\).

### Step 2: Calculate \(f(a)\) and \(f(b)\)

For \(f(-2)\):
\[
f(-2) = (-2)^3 - 3(-2) + 5 = -8 + 6 + 5 = 3
\]

For \(f(2)\):
\[
f(2) = (2)^3 - 3(2) + 5 = 8 - 6 + 5 = 7
\]

### Step 3: Compute the slope \(\frac{f(b) - f(a)}{b - a}\)

\[
\frac{f(2) - f(-2)}{2 - (-2)} = \frac{7 - 3}{2 + 2} = \frac{4}{4} = 1
\]

So, we are looking for \(c\) such that:

\[
f'(c) = 1
\]

### Step 4: Find \(f'(x)\)

First, calculate the derivative of \(f(x)\):

\[
f'(x) = \frac{d}{dx}(x^3 - 3x + 5) = 3x^2 - 3
\]

### Step 5: Solve for \(c\)

We now set \(f'(c) = 1\) and solve for \(c\):

\[
3c^2 - 3 = 1
\]
\[
3c^2 = 4
\]
\[
c^2 = \frac{4}{3}
\]
\[
c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}
\]

Thus, the possible values of \(c\) are:

\[
c = \frac{2\sqrt{3}}{3}, \, -\frac{2\sqrt{3}}{3}
\]

### Step 6: Check if \(c\) is in the interval

The values \(c = \pm \frac{2\sqrt{3}}{3}\) are approximately \( \pm 1.15\), which both lie within the interval \((-2, 2)\).

### Final Answer:

The possible values for \(c\) are:

\[
c = \frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}
\]

Let me know if you would like any more details or clarifications.

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Here are five related questions for further exploration:
1. How does the Mean Value Theorem help in analyzing the behavior of a function?
2. What are the conditions necessary for applying the Mean Value Theorem?
3. Can there be multiple values of \(c\) that satisfy the Mean Value Theorem?
4. How does the derivative \(f'(x)\) inform the slope of the tangent line to the curve at a point?
5. How would the Mean Value Theorem change if the function had a discontinuity?

**Tip:** When solving MVT problems, always ensure that the function meets the continuity and differentiability conditions before applying the theorem.

Find all possible value(s) for c given f(x) = x^3 - 3x + 5, -2 ≤ x ≤ 2 using the Mean Value Theorem.

This solution demonstrates how to apply the Mean Value Theorem to find the values of c for the function f(x) = x^3 - 3x + 5 in the interval -2 ≤ x ≤ 2. The problem covers key calculus concepts, including differentiation and polynomial analysis.

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