1. **Problem Statement:** Verify that each function satisfies Rolle's theorem on the given interval and find all values $c$ where $f'(c) = 0$. 2. **Rolle's Theorem Conditions:** - The function $f$ must be continuous on the closed interval $[a,b]$. - The function $f$ must be differentiable on the open interval $(a,b)$. - The function values at the endpoints must be equal: $f(a) = f(b)$. If these hold, there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. --- ### (a) $f(x) = x^2 + 2x$ on $[-2,0]$ 3. **Check endpoint values:** $$f(-2) = (-2)^2 + 2(-2) = 4 - 4 = 0$$ $$f(0) = 0^2 + 2(0) = 0$$ So, $f(-2) = f(0) = 0$. 4. **Continuity and differentiability:** $f$ is a polynomial, so it is continuous and differentiable everywhere. 5. **Find $f'(x)$:** $$f'(x) = 2x + 2$$ 6. **Solve $f'(c) = 0$ for $c \in (-2,0)$:** $$2c + 2 = 0 \implies 2c = -2 \implies c = -1$$ Since $-1 \in (-2,0)$, this satisfies Rolle's theorem. --- ### (b) $f(x) = \frac{1}{3}x^3 - 4x$ on $[-2,2]$ 7. **Check endpoint values:** $$f(-2) = \frac{1}{3}(-2)^3 - 4(-2) = \frac{1}{3}(-8) + 8 = -\frac{8}{3} + 8 = \frac{16}{3}$$ $$f(2) = \frac{1}{3}(2)^3 - 4(2) = \frac{8}{3} - 8 = -\frac{16}{3}$$ Since $f(-2) \neq f(2)$, Rolle's theorem does not apply directly. 8. **However, if the problem expects to find $c$ where $f'(c) = 0$, we proceed:** 9. **Find $f'(x)$:** $$f'(x) = x^2 - 4$$ 10. **Solve $f'(c) = 0$:** $$c^2 - 4 = 0 \implies c^2 = 4 \implies c = \pm 2$$ Since $c = \pm 2$ are endpoints, no $c$ strictly inside $(-2,2)$ satisfies $f'(c) = 0$. **Summary:** - (a) Rolle's theorem applies, $c = -1$. - (b) Rolle's theorem does not apply because $f(-2) \neq f(2)$.