To verify Rolle's theorem for the function \( f(x) = e^{-x} \sin x \) on the interval \([0, \pi]\), we will follow the steps outlined below:
### Step 1: Check the conditions of Rolle's theorem
Rolle's theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).
### Step 2: Evaluate \( f(0) \) and \( f(\pi) \)
First, we need to calculate \( f(0) \) and \( f(\pi) \):
\[
f(0) = e^{-0} \sin(0) = 1 \cdot 0 = 0
\]
\[
f(\pi) = e^{-\pi} \sin(\pi) = e^{-\pi} \cdot 0 = 0
\]
Since \( f(0) = f(\pi) = 0 \), the first condition \( f(0) = f(\pi) \) is satisfied.
### Step 3: Check continuity and differentiability
The function \( f(x) = e^{-x} \sin x \) is the product of two functions, \( e^{-x} \) and \( \sin x \). Both of these functions are continuous and differentiable everywhere. Therefore, \( f(x) \) is continuous on \([0, \pi]\) and differentiable on \((0, \pi)\).
### Step 4: Find the derivative \( f'(x) \)
Now, we will find the derivative \( f'(x) \) using the product rule:
\[
f'(x) = \frac{d}{dx}(e^{-x}) \sin x + e^{-x} \frac{d}{dx}(\sin x)
\]
Calculating the derivatives:
\[
\frac{d}{dx}(e^{-x}) = -e^{-x}
\]
\[
\frac{d}{dx}(\sin x) = \cos x
\]
Thus, we have:
\[
f'(x) = -e^{-x} \sin x + e^{-x} \cos x
\]
\[
= e^{-x} (\cos x - \sin x)
\]
### Step 5: Set \( f'(c) = 0 \)
To find \( c \) such that \( f'(c) = 0 \):
\[
e^{-c} (\cos c - \sin c) = 0
\]
Since \( e^{-c} \) is never zero, we can set:
\[
\cos c - \sin c = 0
\]
This simplifies to:
\[
\cos c = \sin c
\]
Dividing both sides by \( \cos c \) (assuming \( \cos c \neq 0 \)) gives:
\[
\tan c = 1
\]
### Step 6: Solve for \( c \)
The equation \( \tan c = 1 \) holds true when:
\[
c = \frac{\pi}{4}
\]
### Conclusion
Since \( c = \frac{\pi}{4} \) lies in the interval \((0, \pi)\), we have verified that there exists at least one \( c \) in \((0, \pi)\) such that \( f'(c) = 0 \). Therefore, by the conditions of Rolle's theorem, we conclude that Rolle's theorem is satisfied for the function \( f(x) = e^{-x} \sin x \) on the interval \([0, \pi]\).
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