To find the value(s) of $c$ that satisfy Rolle's Theorem for $f(x) = \sin(2x)$ on the interval $[0, 2\pi]$, we proceed as follows:

Steps:

1. Check the conditions of Rolle's Theorem:

   • The function $f(x)$ is continuous on $[0, 2\pi]$ and differentiable on $(0, 2\pi]$.
   • $f(0) = \sin(2 \cdot 0) = 0$ and $f(2\pi) = \sin(2 \cdot 2\pi) = 0$, so $f(0) = f(2\pi)$.

   Since $f(x)$ satisfies all conditions of Rolle's Theorem, there exists at least one $c \in (0, 2\pi)$ such that $f'(c) = 0$.

2. Find the derivative $f'(x)$: $f'(x) = \frac{d}{dx} \sin(2x) = 2\cos(2x).$

3. Set $f'(c) = 0$ and solve: $2\cos(2c) = 0 \quad \Rightarrow \quad \cos(2c) = 0.$

   The cosine function is zero at $2c = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$, which simplifies to: $2c = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}.$

   Dividing by 2: $c = \frac{\pi}{4} + \frac{n\pi}{2}, \quad n \in \mathbb{Z}.$

4. Find $c$ within the interval $(0, 2\pi)$: Substitute $n = 0, 1, 2, 3, \dots$ and keep $c$ in $(0, 2\pi)$:

   • $n = 0$: $c = \frac{\pi}{4}$.
   • $n = 1$: $c = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$.
   • $n = 2$: $c = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$.
   • $n = 3$: $c = \frac{\pi}{4} + \frac{3\pi}{2} = \frac{7\pi}{4}$.
   • $n = 4$: $c = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4}$, which is outside $(0, 2\pi)$.

Thus, the values of $c$ are: $c = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.$

Final Answer:

The values of $c$ that satisfy Rolle's Theorem are: $c = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.$

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Related Questions:

1. How does differentiability ensure the existence of a critical point in Rolle's Theorem?
2. What is the geometrical interpretation of Rolle's Theorem for $\sin(2x)$?
3. What happens if $f(a) \neq f(b)$ in Rolle’s Theorem?
4. How does $\cos(2x) = 0$ relate to the critical points of $\sin(2x)$?
5. Can Rolle's Theorem be applied to non-polynomial functions, and why?

Tip:

To apply Rolle's Theorem, always verify the conditions (continuity, differentiability, and equal endpoints) before solving.