1. **State the problem:** We are given the polynomial function $$g(x) = -x^3 - 2x^2 + 2x + 6$$ which is differentiable on $$(-\infty, \infty)$$. We need to find the value(s) $$c$$ in the interval $$[-3,1]$$ guaranteed by the Mean Value Theorem (MVT). 2. **Recall the Mean Value Theorem:** If a function $$g$$ 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 slope of the secant line:** Given $$a = -3$$ and $$b = 1$$, $$$g(-3) = 9, \quad g(1) = 5,$$$ so $$$\frac{g(1) - g(-3)}{1 - (-3)} = \frac{5 - 9}{1 + 3} = \frac{-4}{4} = -1.$$$ 4. **Find the derivative of $$g(x)$$:** $$$g'(x) = \frac{d}{dx}(-x^3 - 2x^2 + 2x + 6) = -3x^2 - 4x + 2.$$$ 5. **Set the derivative equal to the slope from step 3:** $$$-3x^2 - 4x + 2 = -1.$$$ 6. **Solve for $$x$$:** $$$-3x^2 - 4x + 2 = -1 \implies -3x^2 - 4x + 3 = 0.$$$ Multiply both sides by $$-1$$ to simplify: $$$3x^2 + 4x - 3 = 0.$$$ 7. **Use the quadratic formula:** $$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} = \frac{-4 \pm \sqrt{16 + 36}}{6} = \frac{-4 \pm \sqrt{52}}{6} = \frac{-4 \pm 2\sqrt{13}}{6} = \frac{-2 \pm \sqrt{13}}{3}.$$$ 8. **Calculate approximate values:** $$$x_1 = \frac{-2 + \sqrt{13}}{3} \approx 0.535,$$$ $$$x_2 = \frac{-2 - \sqrt{13}}{3} \approx -1.868.$$$ 9. **Check if values lie in the interval $$[-3,1]$$:** Both $$0.535$$ and $$-1.868$$ are within $$[-3,1]$$. 10. **Sum of the values:** $$$0.535 + (-1.868) = -1.333.$$$ **Final answer:** The values guaranteed by the Mean Value Theorem are approximately $$x = 0.535$$ and $$x = -1.868$$, and their sum is $$-1.333$$.