Let's revisit the problem and ensure we clearly follow the steps to apply Rolle's Theorem correctly to the function $f(x) = 1 + \frac{(x-5)^8}{5}$ on the interval $[3, 6]$.

Rolle's Theorem Recap:

Rolle's Theorem states that if a function $f(x)$ satisfies the following three conditions on a closed interval $[a, b]$:

1. Continuity on $[a, b]$,
2. Differentiability on $(a, b)$,
3. $f(a) = f(b)$ (the function has equal values at the endpoints), then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

Step 1: Continuity and Differentiability

The given function is: $f(x) = 1 + \frac{(x-5)^8}{5}$ Since this is a polynomial function, it is both continuous and differentiable everywhere, including on the interval $[3, 6]$. Thus, the first two conditions are satisfied.

Step 2: Check if $f(3) = f(6)$

We need to check the values of $f(x)$ at $x = 3$ and $x = 6$.

Compute $f(3)$:

$f(3) = 1 + \frac{(3-5)^8}{5} = 1 + \frac{(-2)^8}{5} = 1 + \frac{256}{5} = 1 + 51.2 = 52.2$

Compute $f(6)$:

$f(6) = 1 + \frac{(6-5)^8}{5} = 1 + \frac{1^8}{5} = 1 + \frac{1}{5} = 1 + 0.2 = 1.2$

Since $f(3) \neq f(6)$ (i.e., $52.2 \neq 1.2$), the third condition for Rolle's Theorem is not satisfied.

Conclusion:

Rolle's Theorem cannot be applied because the values of the function at the endpoints of the interval are not equal.

Would you like further clarification or have any other questions?

Here are 5 related questions you may want to explore:

1. What happens if $f(a) \neq f(b)$ in the context of Rolle's Theorem?
2. Can you apply Rolle's Theorem to trigonometric functions like sine or cosine?
3. How is Rolle's Theorem a special case of the Mean Value Theorem?
4. What are the conditions for the Mean Value Theorem, and how do they differ from Rolle's Theorem?
5. What is the significance of finding a point where $f'(c) = 0$ in Rolle's Theorem?

Tip: If $f(a) \neq f(b)$, check if the Mean Value Theorem could still apply to analyze the function's behavior.