Let's break down the solution as requested: --- ## (a)(i) Find \( f'(x_0) \) at \( x_0 = -2.6 \) using the three-point method **Step 1: Identify the three points around \( x_0 = -2.6 \):** - The closest points are \( x_{i-1} = -3.0 \), \( x_i = -2.6 \), and \( x_{i+1} = -2.2 \). - The table values: - \( x_{i-1} = -3.0 \), \( f(x_{i-1}) = 9.3679 \) - \( x_{i} = -2.6 \), \( f(x_i) = 7.1804 \) - \( x_{i+1} = -2.2 \), \( f(x_{i+1}) = 5.3203 \) - The spacing \( h = x_{i} - x_{i-1} = -2.6 - (-3.0) = 0.4 \) **Step 2: Use the three-point central difference formula:** \[ f'(x_0) \approx \frac{f(x_{i+1}) - f(x_{i-1})}{2h} \] \[ f'(-2.6) \approx \frac{f(-2.2) - f(-3.0)}{2 \times 0.4} = \frac{5.3203 - 9.3679}{0.8} = \frac{-4.0476}{0.8} = -5.0595 \] **Explanation:** We used the central difference formula with equally spaced points to estimate the derivative at \( x_0 = -2.6 \). --- ## (a)(ii) Find \( f''(x_0) \) at \( x_0 = -2.6 \) using the five-point midpoint method **Step 1: Gather the five points around \( x_0 = -2.6 \):** - The points are \( x_{i-2} = -3.4 \), \( x_{i-1} = -3.0 \), \( x_i = -2.6 \), \( x_{i+1} = -2.2 \), \( x_{i+2} = -1.8 \) - The corresponding \( f(x) \) values: - \( f(-3.4) = 11.8820 \) - \( f(-3.0) = 9.3679 \) - \( f(-2.6) = 7.1804 \) - \( f(-2.2) = 5.3203 \) - \( f(-1.8) = 3.7888 \) - The spacing \( h = 0.4 \) **Step 2: Use the five-point midpoint method for the second derivative:** \[ f''(x_0) \approx \frac{-f(x_{i-2}) + 16f(x_{i-1}) - 30f(x_{i}) + 16f(x_{i+1}) - f(x_{i+2})}{12h^2} \] Substitute the values: \[ = \frac{-11.8820 + 16 \times 9.3679 - 30 \times 7.1804 + 16 \times 5.3203 - 3.7888}{12 \times (0.4)^2} \] Calculate numerator: \[ -11.8820 + 149.8864 - 215.412 + 85.1248 - 3.7888 = 4.9284 \] Denominator: \[ 12 \times 0.16 = 1.92 \] \[ f''(-2.6) \approx \frac{4.9284}{1.92} = 2.569 \] **Explanation:** We applied the five-point midpoint formula for the second derivative using the given values. --- ## (b) Find the actual error in (a)(i) if the data was generated from \( f(x) = e^{x/3} + x^2 \) **Step 1: Compute the exact derivative at \( x_0 = -2.6 \):** \[ f(x) = e^{x/3} + x^2 \] \[ f'(x) = \frac{1}{3}e^{x/3} + 2x \] At \( x = -2.6 \): \[ f'(-2.6) = \frac{1}{3}e^{-2.6/3} + 2 \times (-2.6) \] \[ = \frac{1}{3}e^{-0.8667} - 5.2 \] \[ e^{-0.8667} \approx 0.4200 \] \[ f'(-2.6) \approx \frac{1}{3} \times 0.4200 - 5.2 = 0.1400 - 5.2 = -5.060 \] **Step 2: Find the actual error:** \[ \text{Error} = |\text{Numerical value} - \text{Exact value}| \] \[ = |-5.0595 - (-5.060)| = |0.0005| \] **Explanation:** We found the true derivative from the formula and subtracted the numerical estimate to get the actual error. --- ## **Final Answers** **(a)(i):** \[ f'(-2.6) \approx -5.0595 \] **(a)(ii):** \[ f''(-2.6) \approx 2.569 \] **(b):** \[ \text{Actual error in (a)(i)} \approx 0.0005 \]