1. The first problem involves the function $f(x) = 2x^3 - 3x^2 - 36x$ and inequalities involving $x$. 2. We need to analyze the inequalities given: $x < 2$ or $x > 3$, $x < 2$ or $x > -3$, $x < -2$ or $x > 3$, $x < -2$ or $x > -3$, $x > -2$ or $x > 3$. 3. These inequalities describe ranges of $x$ values. The "or" means the solution set is the union of the two intervals. 4. For example, $x < 2$ or $x > 3$ means all $x$ less than 2 plus all $x$ greater than 3. 5. The question likely asks which inequality correctly describes the domain or solution related to $f(x)$, but since no explicit question is stated, we focus on the first problem only. 6. The second problem about $\sin \alpha = 30^\circ$ and $\cos \alpha = 30^\circ$ is incorrect because $\sin 30^\circ = 0.5$ and $\cos 30^\circ = \sqrt{3}/2$, so this is inconsistent. We ignore this as per instructions. Since the first problem is about the function and inequalities, let's analyze the function's critical points to understand the behavior and possible solution intervals. 7. Find the derivative: $$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x) = 6x^2 - 6x - 36$$ 8. Set derivative to zero to find critical points: $$6x^2 - 6x - 36 = 0 \implies x^2 - x - 6 = 0$$ 9. Solve quadratic: $$x = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}$$ 10. So critical points are: $$x = 3 \quad \text{or} \quad x = -2$$ 11. These points divide the number line into intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. 12. Test the sign of $f'(x)$ in each interval to determine increasing/decreasing behavior: - For $x < -2$, pick $x = -3$: $f'(-3) = 6(9) - 6(-3) - 36 = 54 + 18 - 36 = 36 > 0$ (increasing) - For $-2 < x < 3$, pick $x=0$: $f'(0) = 0 - 0 - 36 = -36 < 0$ (decreasing) - For $x > 3$, pick $x=4$: $f'(4) = 6(16) - 6(4) - 36 = 96 - 24 - 36 = 36 > 0$ (increasing) 13. So $f(x)$ increases on $(-\infty, -2)$, decreases on $(-2, 3)$, and increases on $(3, \infty)$. 14. The function has local maximum at $x = -2$ and local minimum at $x = 3$. 15. The inequalities given correspond to these critical points: - $x < -2$ or $x > 3$ matches the intervals where $f(x)$ is increasing. 16. Therefore, the correct inequality describing where $f(x)$ is increasing is: $$x < -2 \text{ or } x > 3$$ Final answer: $$\boxed{x < -2 \text{ or } x > 3}$$