Let's solve the problem step by step as requested. --- ### Step 1: Compute \( f'(x_0) \) at \( x_0 = -2.6 \) using the three-point method The **three-point central difference formula** for the first derivative at \( x_0 \) (where \( x_0 \) is not an endpoint) is: \[ f'(x_0) \approx \frac{f(x_1) - f(x_{-1})}{2h} \] where \( x_{-1} = x_0 - h \), \( x_1 = x_0 + h \). From the table: - \( x_0 = -2.6 \) - \( x_{-1} = -3.0 \), \( f(x_{-1}) = 9.3679 \) - \( x_1 = -2.2 \), \( f(x_1) = 5.3203 \) - Spacing \( h = 0.4 \) (since \( -2.2 - (-2.6) = 0.4 \)) **Calculation:** \[ f'(x_0) \approx \frac{f(-2.2) - f(-3.0)}{2 \times 0.4} = \frac{5.3203 - 9.3679}{0.8} = \frac{-4.0476}{0.8} = -5.0595 \] **Explanation:** We used the central difference approximation and the points immediately before and after \( x_0 \). The method gives a good estimate of the derivative at interior points. --- ### Step 2: Compute \( f''(x_0) \) at \( x_0 = -2.6 \) using the five-point midpoint method The **five-point midpoint formula** for the second derivative at \( x_0 \) is: \[ f''(x_0) \approx \frac{-f(x_{-2}) + 16f(x_{-1}) - 30f(x_0) + 16f(x_1) - f(x_2)}{12h^2} \] where: - \( x_{-2} = x_0 - 2h = -3.4 \), \( f(x_{-2}) = 11.8820 \) - \( x_{-1} = x_0 - h = -3.0 \), \( f(x_{-1}) = 9.3679 \) - \( x_0 = -2.6 \), \( f(x_0) = 7.1804 \) - \( x_1 = x_0 + h = -2.2 \), \( f(x_1) = 5.3203 \) - \( x_2 = x_0 + 2h = -1.8 \), \( f(x_2) = 3.7888 \) - \( h = 0.4 \) **Calculation:** \[ f''(x_0) \approx \frac{-11.8820 + 16 \times 9.3679 - 30 \times 7.1804 + 16 \times 5.3203 - 3.7888}{12 \times (0.4)^2} \] Calculate numerator: \[ -11.8820 + 149.8864 - 215.412 + 85.1248 - 3.7888 = 3.9284 \] Calculate denominator: \[ 12 \times 0.16 = 1.92 \] \[ f''(x_0) \approx \frac{3.9284}{1.92} = 2.047 \] **Explanation:** The five-point midpoint formula provides a more accurate estimate of the second derivative by using more data points around \( x_0 \). --- ### **Final Answers** **(a)(i) \( f'(x_0) \) at \( x_0 = -2.6 \) using the three-point method:** \[ \boxed{-5.0595} \] **(a)(ii) \( f''(x_0) \) at \( x_0 = -2.6 \) using the five-point midpoint method:** \[ \boxed{2.047} \] --- If you need the answer for part (b), let me know!