To determine if Rolle's Theorem can be applied to the function $f(x) = 1 - 9 \sin(5x)$ on the interval $[\pi/5, 4\pi/5]$, we need to check the three conditions for Rolle's Theorem:

1. Continuity: The function $f(x)$ is continuous on the closed interval $[a, b]$.
2. Differentiability: The function $f(x)$ is differentiable on the open interval $(a, b)$.
3. Equal values at the endpoints: $f(a) = f(b)$, where $a = \pi/5$ and $b = 4\pi/5$.

Step 1: Check continuity and differentiability

The function $f(x) = 1 - 9 \sin(5x)$ is a composition of sine, which is continuous and differentiable everywhere. Thus, $f(x)$ is continuous on $[\pi/5, 4\pi/5]$ and differentiable on $(\pi/5, 4\pi/5]$.

Step 2: Check if $f(\pi/5) = f(4\pi/5)$

Evaluate $f(x)$ at $x = \pi/5$ and $x = 4\pi/5$:

$f(\pi/5) = 1 - 9 \sin\left(5 \cdot \frac{\pi}{5}\right) = 1 - 9 \sin(\pi) = 1 - 9 \cdot 0 = 1$ $f(4\pi/5) = 1 - 9 \sin\left(5 \cdot \frac{4\pi}{5}\right) = 1 - 9 \sin(4\pi) = 1 - 9 \cdot 0 = 1$

Since $f(\pi/5) = f(4\pi/5)$, the third condition is satisfied.

Step 3: Apply Rolle's Theorem

Since all three conditions of Rolle's Theorem are satisfied, there exists at least one $c \in (\pi/5, 4\pi/5)$ such that $f'(c) = 0$.

Now, we compute $f'(x)$:

$f'(x) = \frac{d}{dx} \left(1 - 9 \sin(5x)\right) = -9 \cdot 5 \cos(5x) = -45 \cos(5x)$

We need to solve $f'(c) = 0$:

$-45 \cos(5c) = 0$ $\cos(5c) = 0$

The cosine function is zero at $5c = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

Thus, $5c = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$ $c = \frac{\pi}{10}, \frac{3\pi}{10}, \dots$

Now, we check which of these values for $c$ lie in the interval $(\pi/5, 4\pi/5) = (\frac{\pi}{5}, \frac{4\pi}{5})$.

1. $c = \frac{\pi}{10}$ does not lie in $(\pi/5, 4\pi/5)$.
2. $c = \frac{3\pi}{10}$ lies in the interval $(\frac{\pi}{5}, \frac{4\pi}{5})$.
3. $c = \frac{5\pi}{10} = \frac{\pi}{2}$ lies in the interval $(\frac{\pi}{5}, \frac{4\pi}{5})$.
4. $c = \frac{7\pi}{10}$ lies in the interval $(\frac{\pi}{5}, \frac{4\pi}{5})$.

Thus, the values of $c$ are $\frac{3\pi}{10}, \frac{\pi}{2}, \frac{7\pi}{10}$.

Final Answer:

The values of $c$ are: $c = \frac{3\pi}{10}, \frac{\pi}{2}, \frac{7\pi}{10}$

Would you like more details or have any other questions?

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Here are five related questions:

1. How does Rolle's Theorem relate to the Mean Value Theorem?
2. What happens if the function is not differentiable at any point within the interval?
3. Can Rolle's Theorem be applied to functions with discontinuities?
4. How does changing the interval affect the number of values $c$?
5. What are the graphical interpretations of Rolle’s Theorem?

Tip: When using Rolle's Theorem, always check the endpoint values to ensure they are equal. This is often the step where the theorem may fail to apply.