1. **State the problem:** Find the absolute minimum and absolute maximum values of the function $$f(x) = x^3 - 2x^2 - 4x + 1$$ on the closed interval $$[-1, 3]$$. 2. **Formula and rules:** To find absolute extrema on a closed interval, evaluate the function at critical points inside the interval and at the endpoints. Critical points occur where the derivative $$f'(x)$$ is zero or undefined. 3. **Find the derivative:** $$f'(x) = 3x^2 - 4x - 4$$ 4. **Find critical points by solving:** $$3x^2 - 4x - 4 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}$$ So, $$x_1 = \frac{4 + 8}{6} = \frac{12}{6} = 2$$ $$x_2 = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}$$ Both critical points $$x = 2$$ and $$x = -\frac{2}{3}$$ lie within the interval $$[-1, 3]$$. 5. **Evaluate $$f(x)$$ at critical points and endpoints:** - At $$x = -1$$: $$f(-1) = (-1)^3 - 2(-1)^2 - 4(-1) + 1 = -1 - 2 + 4 + 1 = 2$$ - At $$x = -\frac{2}{3}$$: $$f\left(-\frac{2}{3}\right) = \left(-\frac{2}{3}\right)^3 - 2\left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) + 1 = -\frac{8}{27} - 2 \cdot \frac{4}{9} + \frac{8}{3} + 1$$ Simplify step-by-step: $$-\frac{8}{27} - \frac{8}{9} + \frac{8}{3} + 1 = -\frac{8}{27} - \frac{24}{27} + \frac{72}{27} + \frac{27}{27} = \frac{-8 - 24 + 72 + 27}{27} = \frac{67}{27} \approx 2.481$$ - At $$x = 2$$: $$f(2) = 2^3 - 2 \cdot 2^2 - 4 \cdot 2 + 1 = 8 - 8 - 8 + 1 = -7$$ - At $$x = 3$$: $$f(3) = 3^3 - 2 \cdot 3^2 - 4 \cdot 3 + 1 = 27 - 18 - 12 + 1 = -2$$ 6. **Determine absolute extrema:** - Values: $$f(-1) = 2$$, $$f(-\frac{2}{3}) \approx 2.481$$, $$f(2) = -7$$, $$f(3) = -2$$ - Absolute maximum is approximately $$2.481$$ at $$x = -\frac{2}{3}$$. - Absolute minimum is $$-7$$ at $$x = 2$$. **Final answer:** Absolute maximum: $$f\left(-\frac{2}{3}\right) = \frac{67}{27}$$ Absolute minimum: $$f(2) = -7$$