1. **State the problem:** We are given the function $f(x) = x^3 + 3x^2 - 4x$ with roots at $x = -4, 0, 1$. We need to find the maximum and minimum points of this function and sketch its graph. 2. **Find the derivative:** To find the critical points (maxima and minima), compute the first derivative: $$f'(x) = \frac{d}{dx}(x^3 + 3x^2 - 4x) = 3x^2 + 6x - 4$$ 3. **Set the derivative equal to zero:** Solve for $x$ where $f'(x) = 0$: $$3x^2 + 6x - 4 = 0$$ Divide both sides by 3: $$x^2 + 2x - \frac{4}{3} = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=2$, $c=-\frac{4}{3}$. Calculate the discriminant: $$\Delta = 2^2 - 4 \times 1 \times \left(-\frac{4}{3}\right) = 4 + \frac{16}{3} = \frac{28}{3}$$ So, $$x = \frac{-2 \pm \sqrt{\frac{28}{3}}}{2} = \frac{-2 \pm \frac{2\sqrt{21}}{3}}{2} = -1 \pm \frac{\sqrt{21}}{3}$$ 5. **Find the critical points:** $$x_1 = -1 + \frac{\sqrt{21}}{3} \approx 0.527$$ $$x_2 = -1 - \frac{\sqrt{21}}{3} \approx -2.527$$ 6. **Determine if these points are maxima or minima:** Compute the second derivative: $$f''(x) = \frac{d}{dx}(3x^2 + 6x - 4) = 6x + 6$$ Evaluate at $x_1$: $$f''(0.527) = 6(0.527) + 6 = 3.162 + 6 = 9.162 > 0$$ Since $f''(x_1) > 0$, $x_1$ is a local minimum. Evaluate at $x_2$: $$f''(-2.527) = 6(-2.527) + 6 = -15.162 + 6 = -9.162 < 0$$ Since $f''(x_2) < 0$, $x_2$ is a local maximum. 7. **Find the function values at critical points:** $$f(0.527) = (0.527)^3 + 3(0.527)^2 - 4(0.527) \approx 0.146 + 0.833 - 2.108 = -1.129$$ $$f(-2.527) = (-2.527)^3 + 3(-2.527)^2 - 4(-2.527) \approx -16.139 + 19.164 + 10.108 = 13.133$$ 8. **Summary:** - Local maximum at $\left(-2.527, 13.133\right)$ - Local minimum at $\left(0.527, -1.129\right)$ 9. **Graph sketch notes:** - Roots at $x = -4, 0, 1$ - Increasing and decreasing behavior changes at critical points - The graph rises to a maximum near $x = -2.527$, falls to a minimum near $x = 0.527$, then rises again Final answer: Maximum point: $\left(-1 - \frac{\sqrt{21}}{3}, f\left(-1 - \frac{\sqrt{21}}{3}\right)\right)$ Minimum point: $\left(-1 + \frac{\sqrt{21}}{3}, f\left(-1 + \frac{\sqrt{21}}{3}\right)\right)$