1. **Problem:** Verify that the function $f(x) = \sin(9\pi x)$ satisfies the three hypotheses of Rolle's Theorem on the interval $\left[-\frac{2}{9}, \frac{2}{9}\right]$ and find all numbers $c$ that satisfy the conclusion of Rolle's Theorem. 2. **Rolle's Theorem states:** If a function $f$ is continuous on a closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. 3. **Check hypotheses:** - $f(x) = \sin(9\pi x)$ is continuous and differentiable everywhere because sine is continuous and differentiable. - Evaluate $f$ at the endpoints: $$f\left(-\frac{2}{9}\right) = \sin\left(9\pi \cdot -\frac{2}{9}\right) = \sin(-2\pi) = 0$$ $$f\left(\frac{2}{9}\right) = \sin\left(9\pi \cdot \frac{2}{9}\right) = \sin(2\pi) = 0$$ - Since $f\left(-\frac{2}{9}\right) = f\left(\frac{2}{9}\right)$, the third hypothesis holds. 4. **Find $c$ such that $f'(c) = 0$:** - Compute derivative: $$f'(x) = 9\pi \cos(9\pi x)$$ - Set derivative equal to zero: $$9\pi \cos(9\pi c) = 0 \implies \cos(9\pi c) = 0$$ - Solve for $c$: $$9\pi c = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ $$c = \frac{\frac{\pi}{2} + k\pi}{9\pi} = \frac{1}{18} + \frac{k}{9}$$ 5. **Find all $c$ in $\left(-\frac{2}{9}, \frac{2}{9}\right)$:** - Convert interval to decimals for clarity: $$-\frac{2}{9} \approx -0.222, \quad \frac{2}{9} \approx 0.222$$ - Test integer values of $k$: - For $k = -1$: $c = \frac{1}{18} - \frac{1}{9} = \frac{1}{18} - \frac{2}{18} = -\frac{1}{18} \approx -0.0556$ (inside interval) - For $k = 0$: $c = \frac{1}{18} = 0.0556$ (inside interval) - For $k = 1$: $c = \frac{1}{18} + \frac{1}{9} = \frac{1}{18} + \frac{2}{18} = \frac{3}{18} = \frac{1}{6} \approx 0.1667$ (inside interval) - For $k = -2$: $c = \frac{1}{18} - \frac{2}{9} = \frac{1}{18} - \frac{4}{18} = -\frac{3}{18} = -\frac{1}{6} \approx -0.1667$ (inside interval) - For $k = 2$: $c = \frac{1}{18} + \frac{2}{9} = \frac{1}{18} + \frac{4}{18} = \frac{5}{18} \approx 0.2778$ (outside interval) - For $k = -3$: $c = \frac{1}{18} - \frac{3}{9} = \frac{1}{18} - \frac{6}{18} = -\frac{5}{18} \approx -0.2778$ (outside interval) 6. **Conclusion:** The values of $c$ in $\left(-\frac{2}{9}, \frac{2}{9}\right)$ satisfying Rolle's Theorem are: $$c = -\frac{1}{6}, -\frac{1}{18}, \frac{1}{18}, \frac{1}{6}$$ **Final answer:** $$\boxed{c = -\frac{1}{6}, -\frac{1}{18}, \frac{1}{18}, \frac{1}{6}}$$