1. **Problem:** Verify Rolle's Theorem for the function $f(x) = e^{-x} \sin x$ on the interval $[0, \pi]$. 2. **Rolle's Theorem states:** If a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. 3. **Check conditions:** - $f(x) = e^{-x} \sin x$ is continuous and differentiable everywhere. - Evaluate $f(0) = e^{0} \sin 0 = 0$. - Evaluate $f(\pi) = e^{-\pi} \sin \pi = 0$. 4. **Since $f(0) = f(\pi)$, Rolle's Theorem applies.** 5. **Find $f'(x)$:** $$ f'(x) = \frac{d}{dx} (e^{-x} \sin x) = e^{-x} \cos x - e^{-x} \sin x = e^{-x} (\cos x - \sin x) $$ 6. **Set $f'(c) = 0$:** $$ e^{-c} (\cos c - \sin c) = 0 \implies \cos c - \sin c = 0 \implies \cos c = \sin c $$ 7. **Solve for $c$:** $$ \tan c = 1 \implies c = \frac{\pi}{4} \quad \text{(since } c \in (0, \pi)\text{)} $$ 8. **Conclusion:** There exists $c = \frac{\pi}{4} \in (0, \pi)$ such that $f'(c) = 0$, verifying Rolle's Theorem. Final answer: $c = \frac{\pi}{4}$.