To prove Rolle's Theorem using the Lagrange Mean Value Theorem (LMVT), we first need to recall the statements of both theorems.

Rolle's Theorem:

Let $f(x)$ be a function that satisfies the following conditions:

1. $f(x)$ is continuous on the closed interval $[a, b]$,
2. $f(x)$ is differentiable on the open interval $(a, b)$,
3. $f(a) = f(b)$.

Then, there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

Lagrange Mean Value Theorem:

Let $f(x)$ be a function that satisfies the following conditions:

1. $f(x)$ is continuous on the closed interval $[a, b]$,
2. $f(x)$ is differentiable on the open interval $(a, b)$.

Then, there exists at least one $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}.$

Proof of Rolle's Theorem from LMVT:

We assume the conditions of Rolle's Theorem:

• $f(x)$ is continuous on $[a, b]$,
• $f(x)$ is differentiable on $(a, b)$,
• $f(a) = f(b)$.

Now, we apply the Lagrange Mean Value Theorem to $f(x)$, which is continuous on $[a, b]$ and differentiable on $(a, b)$. Therefore, by LMVT, there exists some point $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}.$

Since $f(a) = f(b)$, the right-hand side of this equation becomes: $f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{0}{b - a} = 0.$

Thus, $f'(c) = 0$ for some $c \in (a, b)$, which proves Rolle's Theorem.

Conclusion:

Rolle's Theorem is a special case of the Lagrange Mean Value Theorem when $f(a) = f(b)$, resulting in $f'(c) = 0$ for some $c \in (a, b)$.

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Would you like more details or have any questions?

Here are 5 questions that build on this explanation:

1. How can we apply Rolle’s Theorem to polynomial functions?
2. What happens if $f(a) \neq f(b)$? Does Rolle’s Theorem still apply?
3. Can you give an example of a function that satisfies the conditions of Rolle's Theorem?
4. How does Rolle's Theorem help in finding stationary points of a function?
5. What is the geometrical interpretation of Rolle's Theorem?

Tip: Whenever you use Rolle's Theorem, check the boundary conditions carefully to ensure that $f(a) = f(b)$.