To verify Rolle's Theorem for the function $F(x) = (x - a)^m (x - b)^n$ on the interval $[a, b]$, we need to check the following three conditions:

1. Continuity on the closed interval $[a, b]$.
2. Differentiability on the open interval $(a, b)$.
3. $F(a) = F(b) = 0$.

Let's go through these steps:

1. Continuity on $[a, b]$

The function $F(x) = (x - a)^m (x - b)^n$ is a product of polynomials, and polynomials are continuous everywhere on $\mathbb{R}$. Thus, $F(x)$ is continuous on the closed interval $[a, b]$.

2. Differentiability on $(a, b)$

Polynomials are also differentiable everywhere on $\mathbb{R}$, so $F(x)$ is differentiable on the open interval $(a, b)$.

3. $F(a) = F(b) = 0$

• At $x = a$, the function becomes: $F(a) = (a - a)^m (a - b)^n = 0 \cdot (a - b)^n = 0$
• At $x = b$, the function becomes: $F(b) = (b - a)^m (b - b)^n = (b - a)^m \cdot 0 = 0$

Thus, $F(a) = F(b) = 0$.

Conclusion

Since all three conditions of Rolle's Theorem are satisfied, there must be at least one point $c \in (a, b)$ where $F'(c) = 0$.

Now, let's compute the derivative $F'(x)$ and find where $F'(x) = 0$.

Derivative of $F(x)$:

Using the product rule for differentiation, we have: $F(x) = (x - a)^m (x - b)^n$ $F'(x) = \frac{d}{dx}\left((x - a)^m\right) \cdot (x - b)^n + (x - a)^m \cdot \frac{d}{dx}\left((x - b)^n\right)$ The derivatives of each term are: $\frac{d}{dx}\left((x - a)^m\right) = m(x - a)^{m-1}$ $\frac{d}{dx}\left((x - b)^n\right) = n(x - b)^{n-1}$ Thus, the derivative becomes: $F'(x) = m(x - a)^{m-1}(x - b)^n + n(x - a)^m(x - b)^{n-1}$

To find the critical points:

We need to solve $F'(x) = 0$, which leads to: $m(x - a)^{m-1}(x - b)^n + n(x - a)^m(x - b)^{n-1} = 0$ Factorizing the common terms: $(x - a)^{m-1}(x - b)^{n-1} \left[m(x - b) + n(x - a)\right] = 0$ This gives two possible cases:

1. $(x - a)^{m-1} = 0$ or $(x - b)^{n-1} = 0$ → $x = a$ or $x = b$, but these are the endpoints, and we are looking for a critical point in the open interval $(a, b)$.
2. Solve $m(x - b) + n(x - a) = 0$: $mx - mb + nx - na = 0$ $(m + n)x = ma + nb$ $x = \frac{ma + nb}{m + n}$

Thus, the point $c = \frac{ma + nb}{m + n}$ is in the open interval $(a, b)$, provided $m, n \in \mathbb{N}$, which satisfies Rolle's Theorem.

Conclusion:

Rolle's Theorem is verified, and the critical point $c = \frac{ma + nb}{m + n}$ exists in $(a, b)$.

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Do you want more details on this solution or have any questions? Here are some related questions to explore further:

1. What happens if $m = 1$ and $n = 1$? How does the critical point change?
2. How does the form of the derivative change if $m$ and $n$ are different natural numbers?
3. Can you apply Rolle’s theorem if $m$ or $n$ are not natural numbers?
4. What is the geometric interpretation of Rolle’s theorem for this function?
5. How does the critical point behave when $a = b$?

Tip: Rolle's Theorem guarantees a critical point only if the function satisfies all three conditions, especially at the boundaries.