1. **State the problem:** Find the absolute minima and maxima of the function $$f(x) = x^3 - 2x^2 - 4x + 1$$. 2. **Formula and rules:** To find absolute extrema, first find critical points by setting the derivative $$f'(x)$$ to zero. 3. **Find the derivative:** $$f'(x) = 3x^2 - 4x - 4$$ 4. **Set derivative to zero to find critical points:** $$3x^2 - 4x - 4 = 0$$ 5. **Solve quadratic equation:** Use quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=3$$, $$b=-4$$, $$c=-4$$. 6. Calculate discriminant: $$\Delta = (-4)^2 - 4 \times 3 \times (-4) = 16 + 48 = 64$$ 7. Calculate roots: $$x = \frac{4 \pm \sqrt{64}}{2 \times 3} = \frac{4 \pm 8}{6}$$ 8. Roots are: $$x_1 = \frac{4 + 8}{6} = 2$$ $$x_2 = \frac{4 - 8}{6} = -\frac{2}{3}$$ 9. **Evaluate $$f(x)$$ at critical points:** $$f(2) = 2^3 - 2 \times 2^2 - 4 \times 2 + 1 = 8 - 8 - 8 + 1 = -7$$ $$f\left(-\frac{2}{3}\right) = \left(-\frac{2}{3}\right)^3 - 2 \left(-\frac{2}{3}\right)^2 - 4 \left(-\frac{2}{3}\right) + 1$$ $$= -\frac{8}{27} - 2 \times \frac{4}{9} + \frac{8}{3} + 1 = -\frac{8}{27} - \frac{8}{9} + \frac{8}{3} + 1$$ 10. Simplify: Convert all to denominator 27: $$-\frac{8}{27} - \frac{24}{27} + \frac{72}{27} + \frac{27}{27} = \frac{-8 - 24 + 72 + 27}{27} = \frac{67}{27} \approx 2.48$$ 11. **Check behavior at infinity:** As $$x \to \infty$$, $$f(x) \to \infty$$ and as $$x \to -\infty$$, $$f(x) \to -\infty$$. 12. **Conclusion:** - Absolute minimum is $$f(2) = -7$$ at $$x=2$$. - Absolute maximum is $$f\left(-\frac{2}{3}\right) = \frac{67}{27}$$ at $$x = -\frac{2}{3}$$. **Final answer:** $$\boxed{\text{Absolute minimum} = -7 \text{ at } x=2, \quad \text{Absolute maximum} = \frac{67}{27} \text{ at } x = -\frac{2}{3}}$$