To determine if we can apply Rolle's Theorem to the function $f(x) = 1 + \frac{(1-x)^4}{3}$ on the interval $[0, 2]$, we need to verify the three conditions of Rolle's Theorem:

1. The function is continuous on the closed interval $[0, 2]$.
2. The function is differentiable on the open interval $(0, 2)$.
3. The function satisfies $f(a) = f(b)$, where $a = 0$ and $b = 2$.

Let's go through these steps:

Step 1: Continuity of $f(x)$

Since $f(x) = 1 + \frac{(1-x)^4}{3}$, this is a polynomial function and polynomials are continuous everywhere. Therefore, $f(x)$ is continuous on $[0, 2]$.

Step 2: Differentiability of $f(x)$

Polynomials are also differentiable everywhere, so $f(x)$ is differentiable on the open interval $(0, 2)$.

Step 3: Checking if $f(0) = f(2)$

Now we compute $f(0)$ and $f(2)$.

• $f(0) = 1 + \frac{(1-0)^4}{3} = 1 + \frac{1^4}{3} = 1 + \frac{1}{3} = \frac{4}{3}$
• $f(2) = 1 + \frac{(1-2)^4}{3} = 1 + \frac{(-1)^4}{3} = 1 + \frac{1}{3} = \frac{4}{3}$

Since $f(0) = f(2)$, the third condition is satisfied.

Conclusion:

All the conditions of Rolle's Theorem are satisfied. Therefore, we can apply Rolle's Theorem to the function $f(x)$ on the interval $[0, 2]$.

Rolle's Theorem guarantees that there exists at least one $c \in (0, 2)$ such that $f'(c) = 0$.

Would you like me to calculate the value of $c$ where $f'(c) = 0$?

Related Questions:

1. Can you find the derivative of $f(x) = 1 + \frac{(1-x)^4}{3}$?
2. What is the geometric interpretation of Rolle's Theorem?
3. Can you generalize Rolle's Theorem for different types of functions?
4. How does Rolle's Theorem relate to the Mean Value Theorem?
5. What is an example of a function where Rolle's Theorem does not apply?

Tip: When using Rolle's Theorem, ensure the function is differentiable at every point in the interval; any discontinuity or sharp corner can prevent its application.