Let's solve the problem step by step as requested, including explanations and calculations for **forward difference**, **central difference**, and **Richardson's extrapolation**. Given: \[ f(x) = 6x^3 + 2.5x^2 + 1.2x + 3.5 \] Find \( f'(3.0) \). --- ## **Step 1: Analytical Derivative (For Error Calculation)** First, let's get the exact derivative for error calculation. \[ f'(x) = \frac{d}{dx}[6x^3 + 2.5x^2 + 1.2x + 3.5] = 18x^2 + 5x + 1.2 \] At \( x = 3.0 \): \[ f'(3.0) = 18 \times 9 + 5 \times 3 + 1.2 = 162 + 15 + 1.2 = \boxed{178.2} \] --- ## **Step 2: Forward Difference Method** ### **Choose \( \Delta x = 0.2 \)** **Formula:** \[ f'(x) \approx \frac{f(x+\Delta x) - f(x)}{\Delta x} \] Calculate \( f(3.0) \) and \( f(3.2) \): - \( f(3.0) = 6 \times 27 + 2.5 \times 9 + 1.2 \times 3 + 3.5 = 162 + 22.5 + 3.6 + 3.5 = 191.6 \) - \( f(3.2) = 6 \times (3.2)^3 + 2.5 \times (3.2)^2 + 1.2 \times 3.2 + 3.5 \) Calculate \( (3.2)^3 = 32.768 \), \( (3.2)^2 = 10.24 \): \[ f(3.2) = 6 \times 32.768 + 2.5 \times 10.24 + 1.2 \times 3.2 + 3.5 \] \[ = 196.608 + 25.6 + 3.84 + 3.5 = 229.548 \] So, \[ f'(3.0) \approx \frac{229.548 - 191.6}{0.2} = \frac{37.948}{0.2} = \boxed{189.74} \] --- ### **Now Reduce \( \Delta x \) to \( 0.2/3 = 0.0667 \)** \[ \Delta x_2 = 0.0667 \] \[ f(3.0667) = 6 \times (3.0667)^3 + 2.5 \times (3.0667)^2 + 1.2 \times 3.0667 + 3.5 \] Calculate: - \( (3.0667)^3 \approx 28.849 \) - \( (3.0667)^2 \approx 9.403 \) \[ f(3.0667) = 6 \times 28.849 + 2.5 \times 9.403 + 1.2 \times 3.0667 + 3.5 = 173.094 + 23.507 + 3.680 + 3.5 = 203.781 \] \[ f'(3.0) \approx \frac{203.781 - 191.6}{0.0667} = \frac{12.181}{0.0667} = \boxed{182.69} \] --- ### **Step 3: Richardson's Extrapolation** Richardson's formula for forward difference (order \( O(h) \)): \[ f'(x) \approx \frac{(3F_2 - F_1)}{2} \] Where: - \( F_1 \) = derivative with \( h = 0.2 \) (\( = 189.74 \)) - \( F_2 \) = derivative with \( h = 0.0667 \) (\( = 182.69 \)) \[ F_{refined} = \frac{(3 \times 182.69 - 189.74)}{2} = \frac{548.07 - 189.74}{2} = \frac{358.29}{2} = \boxed{179.15} \] --- ### **Percentage Error** \[ \text{Percentage Error} = \left| \frac{179.15 - 178.2}{178.2} \right| \times 100\% = \left| \frac{0.95}{178.2} \right| \times 100\% \approx \boxed{0.53\%} \] --- ## **Step 4: Central Difference Method** **Central Difference Formula:** \[ f'(x) \approx \frac{f(x+\Delta x) - f(x-\Delta x)}{2\Delta x} \] ### **For \( \Delta x = 0.2 \):** - \( f(2.8) = 6 \times (2.8)^3 + 2.5 \times (2.8)^2 + 1.2 \times 2.8 + 3.5 \) - \( (2.8)^3 = 21.952 \) - \( (2.8)^2 = 7.84 \) \[ f(2.8) = 6 \times 21.952 + 2.5 \times 7.84 + 1.2 \times 2.8 + 3.5 = 131.712 + 19.6 + 3.36 + 3.5 = 158.172 \] \[ f'(3.0) \approx \frac{229.548 - 158.172}{0.4} = \frac{71.376}{0.4} = \boxed{178.44} \] --- ### **For \( \Delta x = 0.0667 \):** - \( f(2.9333) = 6 \times (2.9333)^3 + 2.5 \times (2.9333)^2 + 1.2 \times 2.9333 + 3.5 \) - \( (2.9333)^3 \approx 25.234 \) - \( (2.9333)^2 \approx 8.603 \) \[ f(2.9333) = 6 \times 25.234 + 2.5 \times 8.603 + 1.2 \times 2.9333 + 3.5 = 151.404 + 21.507 + 3.52 + 3.5 = 179.931 \] \[ f'(3.0) \approx \frac{203.781 - 179.931}{2 \times 0.0667} = \frac{23.85}{0.1334} = \boxed{178.77} \] --- ### **Richardson's Extrapolation for Central Difference** Richardson's formula for central difference (order \( O(h^2) \)): \[ F_{refined} = \frac{4F_2 - F_1}{3} \] where - \( F_1 = 178.44 \) - \( F_2 = 178.77 \) \[ F_{refined} = \frac{4 \times 178.77 - 178.44}{3} = \frac{715.08 - 178.44}{3} = \frac{536.64}{3} = \boxed{178.88} \] --- ### **Percentage Error (Central Difference)** \[ \text{Percentage Error} = \left| \frac{178.88 - 178.2}{178.2} \right| \times 100\% = \left| \frac{0.68}{178.2} \right| \times 100\% \approx \boxed{0.38\%} \] --- ## **Summary Table** | Method | \( f'(3.0) \) | % Error | |-------------------------|---------------|---------| | Analytical | 178.20 | 0 | | Forward (\( h=0.2 \)) | 189.74 | 6.45% | | Forward (\( h=0.0667 \))| 182.69 | 2.52% | | Forward (Richardson) | 179.15 | 0.53% | | Central (\( h=0.2 \)) | 178.44 | 0.13% | | Central (\( h=0.0667 \))| 178.77 | 0.32% | | Central (Richardson) | 178.88 | 0.38% | --- ## **Final Answers** - **Forward Difference Richardson result:** \( \boxed{179.15} \), **% error:** \( \boxed{0.53\%} \) - **Central Difference Richardson result:** \( \boxed{178.88} \), **% error:** \( \boxed{0.38\%} \) **Central difference with Richardson's extrapolation gives a more accurate result.**

The derivatives are approximated by finite difference methods using function values at nearby points. Richardson's extrapolation improves accuracy by combining estimates with different step sizes. Central difference generally provides higher accuracy than forward difference due to its symmetric nature.