1. **State the problem:** Find the value(s) $c$ in the interval $[-3,1]$ guaranteed by the Mean Value Theorem (MVT) for the function $$g(x) = -x^3 - 2x^2 + 2x + 6.$$ The function is differentiable on $(-\infty, \infty)$, so MVT applies. 2. **Recall the Mean Value Theorem:** If $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c \in (a,b)$ such that $$g'(c) = \frac{g(b) - g(a)}{b - a}.$$ 3. **Calculate the average rate of change on $[-3,1]$:** Given $g(-3) = 9$ and $g(1) = 5$, $$\frac{g(1) - g(-3)}{1 - (-3)} = \frac{5 - 9}{1 + 3} = \frac{-4}{4} = -1.$$ 4. **Find $g'(x)$:** $$g'(x) = \frac{d}{dx}(-x^3 - 2x^2 + 2x + 6) = -3x^2 - 4x + 2.$$ 5. **Set $g'(x)$ equal to the average rate of change and solve:** $$-3x^2 - 4x + 2 = -1$$ $$-3x^2 - 4x + 3 = 0$$ Multiply both sides by $-1$ for clarity: $$3x^2 + 4x - 3 = 0.$$ 6. **Solve the quadratic equation:** Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=3$, $b=4$, $c=-3$: $$x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} = \frac{-4 \pm \sqrt{16 + 36}}{6} = \frac{-4 \pm \sqrt{52}}{6} = \frac{-4 \pm 2\sqrt{13}}{6} = \frac{-2 \pm \sqrt{13}}{3}.$$ 7. **Evaluate the roots numerically:** $$x_1 = \frac{-2 + \sqrt{13}}{3} \approx 0.535, \quad x_2 = \frac{-2 - \sqrt{13}}{3} \approx -1.868.$$ 8. **Check if roots lie in the interval $[-3,1]$:** Both $0.535$ and $-1.868$ are within $[-3,1]$, so both values satisfy the MVT. 9. **Final answer:** The values guaranteed by the Mean Value Theorem are approximately $$x = 0.535 \text{ and } x = -1.868.$$ If asked for the sum of these values: $$0.535 + (-1.868) = -1.333.$$