1. **State the problem:** Find the local maximum and minimum points of the function $$f(x) = x^3 + 3x^2 - 4x$$ given its roots at $$x = -4, 0, 1$$. 2. **Find the derivative:** To find critical points, compute $$f'(x)$$: $$f'(x) = 3x^2 + 6x - 4$$ 3. **Set derivative to zero:** Solve $$3x^2 + 6x - 4 = 0$$ for $$x$$: Divide both sides by 3: $$x^2 + 2x - \frac{4}{3} = 0$$ 4. **Use quadratic formula:** $$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-\frac{4}{3})}}{2} = \frac{-2 \pm \sqrt{4 + \frac{16}{3}}}{2} = \frac{-2 \pm \sqrt{\frac{28}{3}}}{2}$$ Simplify the square root: $$\sqrt{\frac{28}{3}} = \frac{2\sqrt{21}}{3}$$ So critical points are: $$x = \frac{-2 \pm \frac{2\sqrt{21}}{3}}{2} = -1 \pm \frac{\sqrt{21}}{3}$$ 5. **Calculate approximate values:** $$\sqrt{21} \approx 4.58$$ So, $$x_1 = -1 + \frac{4.58}{3} \approx -1 + 1.53 = 0.53$$ $$x_2 = -1 - 1.53 = -2.53$$ 6. **Find function values at critical points:** $$f(0.53) = (0.53)^3 + 3(0.53)^2 - 4(0.53) \approx 0.15 + 0.84 - 2.12 = -1.13$$ $$f(-2.53) = (-2.53)^3 + 3(-2.53)^2 - 4(-2.53) \approx -16.17 + 19.20 + 10.12 = 13.15$$ 7. **Determine maxima and minima:** Use the second derivative: $$f''(x) = 6x + 6$$ At $$x = 0.53$$: $$f''(0.53) = 6(0.53) + 6 = 3.18 + 6 = 9.18 > 0$$ so local minimum. At $$x = -2.53$$: $$f''(-2.53) = 6(-2.53) + 6 = -15.18 + 6 = -9.18 < 0$$ so local maximum. **Final answer:** - Local maximum at $$\left(-2.53, 13.15\right)$$ - Local minimum at $$\left(0.53, -1.13\right)$$ The graph is a cubic curve crossing the x-axis at $$-4, 0, 1$$ with a peak near $$x = -2.53$$ and a valley near $$x = 0.53$$.