1. **State the problem:** We need to show that the function $f(x) = x^3 - 3x^2 + 2x$ satisfies Rolle's Theorem on the interval $[0, 2]$ and then find the point $c$ in $(0, 2)$ such that $f'(c) = 0$. 2. **Recall Rolle's Theorem:** 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 the conditions:** - $f(x)$ is a polynomial, so it is continuous and differentiable everywhere, including $[0, 2]$ and $(0, 2)$. - Calculate $f(0)$ and $f(2)$: $$f(0) = 0^3 - 3\cdot0^2 + 2\cdot0 = 0$$ $$f(2) = 2^3 - 3\cdot2^2 + 2\cdot2 = 8 - 12 + 4 = 0$$ - Since $f(0) = f(2) = 0$, the conditions of Rolle's Theorem are satisfied. 4. **Find $c$ such that $f'(c) = 0$:** - Compute the derivative: $$f'(x) = 3x^2 - 6x + 2$$ - Set $f'(x) = 0$: $$3x^2 - 6x + 2 = 0$$ - Divide both sides by 3: $$x^2 - 2x + \frac{2}{3} = 0$$ - Use the quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \frac{2}{3}}}{2} = \frac{2 \pm \sqrt{4 - \frac{8}{3}}}{2} = \frac{2 \pm \sqrt{\frac{12}{3} - \frac{8}{3}}}{2} = \frac{2 \pm \sqrt{\frac{4}{3}}}{2}$$ - Simplify the square root: $$\sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$ - So, $$x = \frac{2 \pm \frac{2\sqrt{3}}{3}}{2} = 1 \pm \frac{\sqrt{3}}{3}$$ 5. **Check which solutions lie in $(0, 2)$:** - $1 - \frac{\sqrt{3}}{3} \approx 1 - 0.577 = 0.423$ (in $(0, 2)$) - $1 + \frac{\sqrt{3}}{3} \approx 1 + 0.577 = 1.577$ (in $(0, 2)$) 6. **Conclusion:** There are two points $c$ in $(0, 2)$ where $f'(c) = 0$, namely: $$c = 1 - \frac{\sqrt{3}}{3} \quad \text{and} \quad c = 1 + \frac{\sqrt{3}}{3}$$ Thus, the function satisfies Rolle's Theorem on $[0, 2]$ and the points $c$ where the derivative is zero are as above.