1. **Problem:** Find the domain and critical values of the function $$y = x^3 + 6x^2 + 12x - 1$$. 2. **Domain:** Since this is a polynomial, the domain is all real numbers, $$(-\infty, \infty)$$. 3. **Find the derivative:** To find critical values, compute $$y'$$: $$y' = 3x^2 + 12x + 12$$ 4. **Set derivative equal to zero:** $$3x^2 + 12x + 12 = 0$$ 5. **Simplify by dividing both sides by 3:** $$\cancel{3}x^2 + \cancel{3}4x + \cancel{3}4 = 0 \Rightarrow x^2 + 4x + 4 = 0$$ 6. **Factor the quadratic:** $$x^2 + 4x + 4 = (x + 2)^2 = 0$$ 7. **Solve for x:** $$x + 2 = 0 \Rightarrow x = -2$$ 8. **Find the critical value (y-coordinate):** $$y(-2) = (-2)^3 + 6(-2)^2 + 12(-2) - 1 = -8 + 24 - 24 - 1 = -9$$ 9. **Conclusion:** The domain is all real numbers, and there is one critical point at $$(-2, -9)$$. --- Since the user requested only the first problem solved completely and the total number of distinct problems is 15 (problems 4 to 18 even), we provide only this solution.