To verify Rolle’s Theorem for the function $f(x) = e^{-x} \sin x$ on the interval $[0, \pi]$, we need to check the following conditions of Rolle's Theorem:

Conditions of Rolle's Theorem:

1. Continuity on the closed interval $[a, b]$: The function must be continuous on the entire interval.
2. Differentiability on the open interval $(a, b)$: The function must be differentiable on the open interval.
3. Equal function values at the endpoints of the interval: $f(a) = f(b)$.

Step 1: Check the continuity on $[0, \pi]$

We check if the function $f(x) = e^{-x} \sin x$ is continuous on $[0, \pi]$.

• The exponential function $e^{-x}$ is continuous for all real values of $x$.
• The sine function $\sin x$ is also continuous for all real values of $x$.
• The product of two continuous functions is continuous. Thus, $f(x) = e^{-x} \sin x$ is continuous on the closed interval $[0, \pi]$.

Step 2: Check the differentiability on $(0, \pi)$

We check if $f(x) = e^{-x} \sin x$ is differentiable on the open interval $(0, \pi)$.

• Both $e^{-x}$ and $\sin x$ are differentiable on $(0, \pi)$, and the product of two differentiable functions is differentiable. Thus, $f(x) = e^{-x} \sin x$ is differentiable on $(0, \pi)$.

Step 3: Check if $f(0) = f(\pi)$

We now check if $f(x)$ takes the same value at the endpoints of the interval, i.e., whether $f(0) = f(\pi)$.

• $f(0) = e^{0} \sin(0) = 1 \cdot 0 = 0$
• $f(\pi) = e^{-\pi} \sin(\pi) = e^{-\pi} \cdot 0 = 0$

Since $f(0) = f(\pi) = 0$, this condition is satisfied.

Step 4: Find $c$ such that $f'(c) = 0$

Now that all conditions are satisfied, we can apply Rolle's Theorem. We need to find $c \in (0, \pi)$ such that $f'(c) = 0$.

Differentiate $f(x) = e^{-x} \sin x$

To differentiate $f(x) = e^{-x} \sin x$, we apply the product rule: $f'(x) = \frac{d}{dx} \left( e^{-x} \right) \sin x + e^{-x} \frac{d}{dx} \left( \sin x \right)$ $f'(x) = -e^{-x} \sin x + e^{-x} \cos x$ $f'(x) = e^{-x} (-\sin x + \cos x)$

Now, set $f'(x) = 0$ to find critical points: $e^{-x} (-\sin x + \cos x) = 0$ Since $e^{-x} \neq 0$ for any real $x$, we have: $-\sin x + \cos x = 0$ $\sin x = \cos x$ x = \frac{\pi}{4} \quad \text{(in the interval (0, \pi))}

Thus, $c = \frac{\pi}{4}$ satisfies $f'(c) = 0$.

Conclusion:

Rolle’s Theorem is verified for the function $f(x) = e^{-x} \sin x$ on the interval $[0, \pi]$, and the value of $c$ where $f'(c) = 0$ is $c = \frac{\pi}{4}$.

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Would you like further explanation or details on any part of the solution? Here are some related questions to explore:

1. How does Rolle's Theorem relate to the Mean Value Theorem?
2. What are the steps involved in applying the product rule to differentiate functions like $f(x) = e^{-x} \sin x$?
3. Can you verify Rolle's Theorem for different functions, such as $f(x) = \cos x$ on $[0, \pi]$?
4. How do you check the continuity of a function involving exponential and trigonometric terms?
5. What is the geometric interpretation of Rolle's Theorem?

Tip: Rolle's Theorem guarantees a point where the tangent line is horizontal, but only if the function satisfies the three conditions (continuity, differentiability, and equal function values at the endpoints).