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. **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 continuity and differentiability:** The denominator factors as $x^2 + 2x - 3 = (x+3)(x-1)$. The function is undefined at $x = -3$ and $x = 1$. Since $-3$ lies in $[-6,2]$, $f$ is not continuous on $[-6,2]$. Rolle's Theorem does not apply. 4. **Evaluate $f(-6)$ and $f(2)$:** $$f(-6) = \frac{36 - 24 - 12}{36 - 12 - 3} = \frac{0}{21} = 0$$ $$f(2) = \frac{4 + 8 - 12}{4 + 4 - 3} = \frac{0}{5} = 0$$ 5. Although $f(-6) = f(2)$, the function is not continuous on $[-6,2]$ due to the vertical asymptote at $x = -3$. 6. **Conclusion:** Rolle's Theorem does not apply, so no $c$ exists in $(-6,2)$ such that $f'(c) = 0$. **Final answer:** (E) No such value exists.