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.

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.