1. **State the problem:** Given data points $(1,1)$, $(2,5)$, $(7,5)$, and $(8,4)$, find the Newton's divided difference interpolation polynomial $f(x)$ and evaluate $f(6)$. 2. **Formula and rules:** Newton's divided difference polynomial is given by $$f(x) = f[x_0] + f[x_0,x_1](x-x_0) + f[x_0,x_1,x_2](x-x_0)(x-x_1) + \cdots$$ where $f[x_i]$ are function values and $f[x_i,x_j,...]$ are divided differences. 3. **Calculate divided differences:** - $f[x_0] = f(1) = 1$ - $f[x_1] = f(2) = 5$ - $f[x_2] = f(7) = 5$ - $f[x_3] = f(8) = 4$ First order differences: $$f[x_0,x_1] = \frac{f[x_1]-f[x_0]}{x_1 - x_0} = \frac{5-1}{2-1} = 4$$ $$f[x_1,x_2] = \frac{5-5}{7-2} = 0$$ $$f[x_2,x_3] = \frac{4-5}{8-7} = -1$$ Second order differences: $$f[x_0,x_1,x_2] = \frac{f[x_1,x_2] - f[x_0,x_1]}{x_2 - x_0} = \frac{0-4}{7-1} = \frac{-4}{6} = -\frac{2}{3}$$ $$f[x_1,x_2,x_3] = \frac{f[x_2,x_3] - f[x_1,x_2]}{x_3 - x_1} = \frac{-1-0}{8-2} = \frac{-1}{6}$$ Third order difference: $$f[x_0,x_1,x_2,x_3] = \frac{f[x_1,x_2,x_3] - f[x_0,x_1,x_2]}{x_3 - x_0} = \frac{-\frac{1}{6} - (-\frac{2}{3})}{8-1} = \frac{-\frac{1}{6} + \frac{4}{6}}{7} = \frac{\frac{3}{6}}{7} = \frac{1}{14}$$ 4. **Construct the polynomial:** $$f(x) = 1 + 4(x-1) - \frac{2}{3}(x-1)(x-2) + \frac{1}{14}(x-1)(x-2)(x-7)$$ 5. **Evaluate $f(6)$:** Calculate each term: - $4(6-1) = 4 \times 5 = 20$ - $-\frac{2}{3}(6-1)(6-2) = -\frac{2}{3} \times 5 \times 4 = -\frac{2}{3} \times 20 = -\frac{40}{3}$ - $\frac{1}{14}(6-1)(6-2)(6-7) = \frac{1}{14} \times 5 \times 4 \times (-1) = -\frac{20}{14} = -\frac{10}{7}$ Sum all: $$f(6) = 1 + 20 - \frac{40}{3} - \frac{10}{7} = 21 - \frac{40}{3} - \frac{10}{7}$$ Find common denominator $21$: $$21 = \frac{441}{21}, \quad \frac{40}{3} = \frac{280}{21}, \quad \frac{10}{7} = \frac{30}{21}$$ So $$f(6) = \frac{441}{21} - \frac{280}{21} - \frac{30}{21} = \frac{441 - 280 - 30}{21} = \frac{131}{21} \approx 6.238$$ **Final answers:** $$f(x) = 1 + 4(x-1) - \frac{2}{3}(x-1)(x-2) + \frac{1}{14}(x-1)(x-2)(x-7)$$ $$f(6) = \frac{131}{21} \approx 6.238$$