1. **Problem (三):** Find values of $a$ and $b$ such that $(2,3)$ is an inflection point of $y = ax^3 + bx^2$. 2. **Recall:** An inflection point occurs where the second derivative $y''$ changes sign, so $y''(x) = 0$ at that point. 3. **Calculate derivatives:** $$y = ax^3 + bx^2$$ $$y' = 3ax^2 + 2bx$$ $$y'' = 6ax + 2b$$ 4. **Set $y''(2) = 0$ for inflection point:** $$6a(2) + 2b = 0 \Rightarrow 12a + 2b = 0 \Rightarrow 6a + b = 0$$ 5. **Use point $(2,3)$ on the curve:** $$3 = a(2)^3 + b(2)^2 = 8a + 4b$$ 6. **Solve system:** From step 4: $b = -6a$ Substitute into step 5: $$3 = 8a + 4(-6a) = 8a - 24a = -16a \Rightarrow a = -\frac{3}{16}$$ Then $$b = -6 \times -\frac{3}{16} = \frac{18}{16} = \frac{9}{8}$$ --- 7. **Problem (四):** Use Newton's Algorithm to find the third approximation of root of $f(x) = x^3 + 5x - 2$ starting with $x_1 = \frac{1}{2}$. 8. **Newton's formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ 9. **Calculate derivative:** $$f'(x) = 3x^2 + 5$$ 10. **Calculate $x_2$:** $$f(0.5) = (0.5)^3 + 5(0.5) - 2 = 0.125 + 2.5 - 2 = 0.625$$ $$f'(0.5) = 3(0.5)^2 + 5 = 3(0.25) + 5 = 0.75 + 5 = 5.75$$ $$x_2 = 0.5 - \frac{0.625}{5.75} = 0.5 - 0.1087 = 0.3913$$ 11. **Calculate $x_3$:** $$f(0.3913) = (0.3913)^3 + 5(0.3913) - 2 \approx 0.0599 + 1.9565 - 2 = 0.0164$$ $$f'(0.3913) = 3(0.3913)^2 + 5 = 3(0.1531) + 5 = 0.4593 + 5 = 5.4593$$ $$x_3 = 0.3913 - \frac{0.0164}{5.4593} = 0.3913 - 0.0030 = 0.3883$$ 12. **Calculate $x_4$ (third approximation):** $$f(0.3883) = (0.3883)^3 + 5(0.3883) - 2 \approx 0.0585 + 1.9415 - 2 = 0.0000$$ $$f'(0.3883) = 3(0.3883)^2 + 5 = 3(0.1508) + 5 = 0.4524 + 5 = 5.4524$$ $$x_4 = 0.3883 - \frac{0.0000}{5.4524} = 0.3883$$ **Final answers:** - (三) $a = -\frac{3}{16}$, $b = \frac{9}{8}$ - (四) Third approximation $x_4 \approx 0.3883$