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(A) Given the function $f(x) = e^{-2x} \sin 2x$. First,we find the derivative $f'(x)$ using the product rule: $f'(x) = \frac{d}{dx}(e^{-2x}) \cdot \si

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$\pi/8$

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(A) Given the function $f(x) = e^{-2x} \sin 2x$. First,we find the derivative $f'(x)$ using the product rule: $f'(x) = \frac{d}{dx}(e^{-2x}) \cdot \sin 2x + e^{-2x} \cdot \frac{d}{dx}(\sin 2x)$ $f'(x) = -2e^{-2x} \sin 2x + e^{-2x} \cdot 2 \cos 2x$ $f'(x) = 2e^{-2x} (\cos 2x - \sin 2x)$. According to Rolle's theorem,there exists a $c \in (0, \pi/2)$ such that $f'(c) = 0$. Setting $f'(c) = 0$: $2e^{-2c} (\cos 2c - \sin 2c) = 0$. Since $2e^{-2c} 
eq 0$ for any real $c$,we must have: $\cos 2c - \sin 2c = 0$ $\cos 2c = \sin 2c$ $	an 2c = 1$. Since $c \in (0, \pi/2)$,then $2c \in (0, \pi)$. The value of $2c$ for which $	an 2c = 1$ is $2c = \pi/4$. Therefore,$c = \pi/8$.

Consider the function $f(x) = e^{-2x} \sin 2x$ over the interval $(0, \pi/2)$. $A$ real number $c \in (0, \pi/2)$,as guaranteed by Rolle's theorem,such that $f'(c) = 0$ is

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