1. **Problem Statement:** We are given the cubic function $$f(x) = \frac{1}{3}x^{3} - \frac{1}{2}x^{2} - 6x + 1$$ and asked to evaluate or approximate the value of $$f\left(\frac{2019}{2003}\right)$$ which is approximately $$10^{-3}$$. We also have a point $$x_0 = 1$$. 2. **Understanding the function:** This is a cubic polynomial function. To evaluate it at any point, substitute the value of $$x$$ into the formula and simplify. 3. **Substitution:** Substitute $$x = \frac{2019}{2003}$$ into the function: $$ f\left(\frac{2019}{2003}\right) = \frac{1}{3} \left(\frac{2019}{2003}\right)^3 - \frac{1}{2} \left(\frac{2019}{2003}\right)^2 - 6 \left(\frac{2019}{2003}\right) + 1 $$ 4. **Approximate powers:** Since $$\frac{2019}{2003} \approx 1.00799$$ (a value slightly greater than 1), we can approximate powers: - $$\left(1.00799\right)^3 \approx 1.024$$ - $$\left(1.00799\right)^2 \approx 1.016$$ 5. **Calculate each term:** - $$\frac{1}{3} \times 1.024 \approx 0.3413$$ - $$-\frac{1}{2} \times 1.016 \approx -0.508$$ - $$-6 \times 1.00799 \approx -6.048$$ - $$+1$$ remains as is. 6. **Sum all terms:** $$ 0.3413 - 0.508 - 6.048 + 1 = (0.3413 - 0.508) + (1 - 6.048) = -0.1667 - 5.048 = -5.2147 $$ 7. **Reconsider approximation:** The problem states $$f\left(\frac{2019}{2003}\right) \approx 10^{-3}$$ which is about 0.001, but our rough calculation gave about -5.21. This suggests the problem might be referring to the difference $$f\left(\frac{2019}{2003}\right) - f(1)$$ or a linear approximation near $$x_0=1$$. 8. **Calculate $$f(1)$$:** $$ f(1) = \frac{1}{3} (1)^3 - \frac{1}{2} (1)^2 - 6(1) + 1 = \frac{1}{3} - \frac{1}{2} - 6 + 1 = 0.3333 - 0.5 - 6 + 1 = -5.1667 $$ 9. **Difference:** $$ f\left(\frac{2019}{2003}\right) - f(1) \approx -5.2147 - (-5.1667) = -0.048$$ 10. **Linear approximation (using derivative):** - Derivative: $$ f'(x) = \frac{d}{dx} \left( \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x + 1 \right) = x^2 - x - 6 $$ - Evaluate at $$x_0=1$$: $$ f'(1) = 1^2 - 1 - 6 = 1 - 1 - 6 = -6 $$ - Approximate change: $$ \Delta x = \frac{2019}{2003} - 1 = \frac{2019 - 2003}{2003} = \frac{16}{2003} \approx 0.00799 $$ - Linear approximation: $$ f\left(\frac{2019}{2003}\right) \approx f(1) + f'(1) \times \Delta x = -5.1667 + (-6)(0.00799) = -5.1667 - 0.0479 = -5.2146 $$ This matches the direct substitution result. 11. **Conclusion:** The value $$f\left(\frac{2019}{2003}\right)$$ is approximately $$-5.2146$$, not $$10^{-3}$$. The given approximation $$10^{-3}$$ might refer to a different context or error term. --- **Final answer:** $$f\left(\frac{2019}{2003}\right) \approx -5.2146$$