Let's solve this step-by-step.

Step 1: Evaluate $f(-8)$ and $f(8)$

The given function is:

$f(x) = 4 - x^{2/3}$

Now, substitute $x = -8$ and $x = 8$ to find $f(-8)$ and $f(8)$.

For $f(-8)$:

$f(-8) = 4 - (-8)^{2/3}$

The cube root of $-8$ is $-2$, and squaring $-2$ gives $4$. So,

$f(-8) = 4 - 4 = 0$

For $f(8)$:

$f(8) = 4 - (8)^{2/3}$

The cube root of $8$ is $2$, and squaring $2$ gives $4$. So,

$f(8) = 4 - 4 = 0$

Thus, $f(-8) = 0$ and $f(8) = 0$.

Step 2: Find $f'(x)$

Now, we differentiate the function $f(x) = 4 - x^{2/3}$.

Using the power rule for differentiation:

$f'(x) = \frac{d}{dx}\left(4 - x^{2/3}\right) = 0 - \frac{2}{3}x^{-1/3} = -\frac{2}{3}x^{-1/3}$

So,

$f'(x) = -\frac{2}{3} \cdot \frac{1}{\sqrt[3]{x}}$

Step 3: Find $c$ such that $f'(c) = 0$

Set $f'(x) = 0$:

$-\frac{2}{3} \cdot \frac{1}{\sqrt[3]{x}} = 0$

For this equation to hold, there is no value of $x$ that makes $-\frac{2}{3} \cdot \frac{1}{\sqrt[3]{x}} = 0$. This implies that there is no $c$ in the interval $(-8, 8)$ such that $f'(c) = 0$.

Therefore, the answer for $c$ is:

$c = \text{DNE (Does Not Exist)}$

Step 4: Rolle's Theorem

Rolle's Theorem states that if a function $f$ is continuous on the closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

In this case:

• $f(-8) = f(8) = 0$, so the first condition is satisfied.
• However, the derivative $f'(x)$ involves $x^{-1/3}$, which is undefined at $x = 0$. This means the function is not differentiable at $x = 0$.

Thus, this does not contradict Rolle's Theorem, since the function is not differentiable on the entire interval $(-8, 8)$.

The correct conclusion is: $\boxed{\text{This does not contradict Rolle's Theorem, since } f'(0) \text{ does not exist, and so } f \text{ is not differentiable on } (-8, 8).}$

---

Do you need more details on any part of this explanation?

Here are some related questions you could explore:

1. How is the cube root of a negative number defined in general?
2. What are the key conditions of Rolle's Theorem?
3. What happens if a function is not differentiable at a single point in the interval?
4. Can a function still be continuous even if it's not differentiable everywhere?
5. How do we find critical points where the derivative doesn't exist?

Tip: Always check if the function is differentiable across the entire interval when applying Rolle's Theorem!