1. Problem 17: Given $y = 2x^3 - 3x^2 - 36x + 5$, find the range of $x$ for which $\frac{dy}{dx} < 0$. 2. Differentiate $y$ with respect to $x$: $$\frac{dy}{dx} = 6x^2 - 6x - 36$$ 3. Set the inequality: $$6x^2 - 6x - 36 < 0$$ 4. Divide both sides by 6: $$x^2 - x - 6 < 0$$ 5. Factorize the quadratic: $$ (x - 3)(x + 2) < 0 $$ 6. The product of two factors is less than zero when one is positive and the other is negative. This happens between the roots: $$ -2 < x < 3 $$ --- 7. Problem 18: Given $y = 4x^3 + 3x^2 - 6x - 9$, find the range of $x$ for which $\frac{dy}{dx} \geq 0$. 8. Differentiate $y$: $$\frac{dy}{dx} = 12x^2 + 6x - 6$$ 9. Set the inequality: $$12x^2 + 6x - 6 \geq 0$$ 10. Divide by 6: $$2x^2 + x - 1 \geq 0$$ 11. Factorize or use quadratic formula for $2x^2 + x - 1 = 0$: $$x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ 12. Roots are: $$x = \frac{2}{4} = 0.5, \quad x = \frac{-4}{4} = -1$$ 13. Since the parabola opens upwards ($a=2>0$), $2x^2 + x -1 \geq 0$ outside the roots: $$x \leq -1 \quad \text{or} \quad x \geq 0.5$$ --- 14. Problem 19: Given $y = 3x^3 + 6x^2 + 4x - 5$, show the gradient is never negative. 15. Differentiate $y$: $$\frac{dy}{dx} = 9x^2 + 12x + 4$$ 16. Check if $9x^2 + 12x + 4 < 0$ has any solution. 17. Calculate discriminant: $$\Delta = 12^2 - 4 \times 9 \times 4 = 144 - 144 = 0$$ 18. Since $\Delta = 0$, the quadratic touches the x-axis at one point: $$x = \frac{-12}{2 \times 9} = -\frac{2}{3}$$ 19. Because $a=9 > 0$, the parabola opens upwards and the minimum value is zero at $x = -\frac{2}{3}$. 20. Therefore, $\frac{dy}{dx} \geq 0$ for all $x$, so the gradient is never negative. Final answers: - For problem 17: $-2 < x < 3$ - For problem 18: $x \leq -1$ or $x \geq 0.5$ - For problem 19: Gradient $\geq 0$ for all $x$ (never negative).