1. **Problem:** Find the value of $c$ that satisfies Rolle's Theorem for $$f(x) = \frac{x^2 + 4x - 12}{x^2 + 2x - 3}$$ on the interval $[-6, 2]$. 2. **Rolle's Theorem states:** 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 continuity and differentiability:** The denominator $x^2 + 2x - 3 = (x+3)(x-1)$ is zero at $x = -3$ and $x = 1$. Since $-3$ and $1$ lie inside the interval $[-6,2]$, $f$ is not continuous on $[-6,2]$ (discontinuities at $x=-3$ and $x=1$). 4. **Evaluate $f(-6)$ and $f(2)$:** $$f(-6) = \frac{36 - 24 - 12}{36 - 12 - 3} = \frac{0}{21} = 0$$ $$f(2) = \frac{4 + 8 - 12}{4 + 4 - 3} = \frac{0}{5} = 0$$ So, $f(-6) = f(2) = 0$. 5. **Since $f$ is not continuous on $[-6,2]$ due to vertical asymptotes at $x=-3$ and $x=1$, Rolle's Theorem does not apply.** 6. **Conclusion:** No value $c$ in $(-6,2)$ satisfies Rolle's Theorem for this function. **Final answer:** (E) No such value exists.