Let's break down the solution into **two steps** as requested, and provide explanations at each step. --- ### **Step 1: Compute the Central Difference Approximations** The central difference formula for the first derivative is: \[ N_1(h) = \frac{f(x_0 + h) - f(x_0 - h)}{2h} \] We will use \( h = 0.4 \), \( h/2 = 0.2 \), and \( h/4 = 0.1 \): #### **Calculate \( f(x) \) at needed points:** Given \( f(x) = x^3 \cos x \), \( x_0 = 2.3 \): - \( f(2.3) = (2.3)^3 \cos(2.3) \) - \( f(2.3 + 0.4) = (2.7)^3 \cos(2.7) \) - \( f(2.3 - 0.4) = (1.9)^3 \cos(1.9) \) - \( f(2.3 + 0.2) = (2.5)^3 \cos(2.5) \) - \( f(2.3 - 0.2) = (2.1)^3 \cos(2.1) \) - \( f(2.3 + 0.1) = (2.4)^3 \cos(2.4) \) - \( f(2.3 - 0.1) = (2.2)^3 \cos(2.2) \) **Evaluate:** - \( f(1.9) \approx 6.859 \times (-0.323) = -2.216 \) - \( f(2.1) \approx 9.261 \times (-0.504) = -4.669 \) - \( f(2.2) \approx 10.648 \times (-0.588) = -6.263 \) - \( f(2.3) \approx 12.167 \times (-0.666) = -8.104 \) - \( f(2.4) \approx 13.824 \times (-0.737) = -10.189 \) - \( f(2.5) \approx 15.625 \times (-0.801) = -12.515 \) - \( f(2.7) \approx 19.683 \times (-0.904) = -17.793 \) Now, plug these into the central difference formula: \[ N_1(h) = \frac{f(2.7) - f(1.9)}{2 \times 0.4} = \frac{-17.793 - (-2.216)}{0.8} = \frac{-15.577}{0.8} = -19.471 \] \[ N_1(h/2) = \frac{f(2.5) - f(2.1)}{2 \times 0.2} = \frac{-12.515 - (-4.669)}{0.4} = \frac{-7.846}{0.4} = -19.615 \] \[ N_1(h/4) = \frac{f(2.4) - f(2.2)}{2 \times 0.1} = \frac{-10.189 - (-6.263)}{0.2} = \frac{-3.926}{0.2} = -19.630 \] --- **Explanation:** In this step, we compute the first-level central difference approximations for three step sizes: \( h \), \( h/2 \), and \( h/4 \). --- ### **Step 2: Apply Richardson Extrapolation** Richardson extrapolation improves the accuracy using: \[ N_2(h) = \frac{4N_1(h/2) - N_1(h)}{3} \] \[ N_3(h) = \frac{16N_2(h/2) - N_2(h)}{15} \] First, compute \( N_2(h) \) and \( N_2(h/2) \): \[ N_2(h) = \frac{4 \times (-19.615) - (-19.471)}{3} = \frac{-78.46 + 19.471}{3} = \frac{-58.989}{3} = -19.663 \] \[ N_2(h/2) = \frac{4 \times (-19.630) - (-19.615)}{3} = \frac{-78.52 + 19.615}{3} = \frac{-58.905}{3} = -19.635 \] Now, compute \( N_3(h) \): \[ N_3(h) = \frac{16 \times (-19.635) - (-19.663)}{15} = \frac{-314.16 + 19.663}{15} = \frac{-294.497}{15} = -19.633 \] --- **Explanation:** In this step, we use Richardson extrapolation to improve our approximation accuracy, step by step, to reach \( N_3(h) \). --- ## **Final Answer** \[ \boxed{N_3(h) \approx -19.633} \] This is the extrapolated approximation to \( f'(2.3) \) for \( f(x) = x^3 \cos x \) using the given values.