1. **Problem Statement:** Use Newton's Backward Interpolation formula to find $f(45095)$ given the data points: $(45000, 9.648583)$, $(45020, 9.648696)$, $(45040, 9.648810)$, $(45060, 9.648923)$, $(45080, 9.649036)$, $(45100, 9.649150)$. 2. **Formula:** Newton's Backward Interpolation formula is: $$ P_n(x) = f(x_n) + p\Delta f(x_n) + \frac{p(p+1)}{2!}\Delta^2 f(x_n) + \frac{p(p+1)(p+2)}{3!}\Delta^3 f(x_n) + \cdots $$ where $p = \frac{x - x_n}{h}$, $h$ is the equal spacing between $x$ values, and $\Delta$ denotes backward differences. 3. **Step 1: Calculate $h$:** $$h = 45020 - 45000 = 20$$ 4. **Step 2: Calculate $p$ for $x=45095$:** $$p = \frac{45095 - 45100}{20} = \frac{-5}{20} = -0.25$$ 5. **Step 3: Construct backward difference table:** \begin{align*} \text{Given } f(x): &\\ f_0 = 9.648583 & (45000)\\ f_1 = 9.648696 & (45020)\\ f_2 = 9.648810 & (45040)\\ f_3 = 9.648923 & (45060)\\ f_4 = 9.649036 & (45080)\\ f_5 = 9.649150 & (45100) \end{align*} Calculate backward differences $\Delta f$: \begin{align*} \Delta f_5 &= f_5 - f_4 = 9.649150 - 9.649036 = 0.000114 \\ \Delta f_4 &= f_4 - f_3 = 9.649036 - 9.648923 = 0.000113 \\ \Delta f_3 &= f_3 - f_2 = 9.648923 - 9.648810 = 0.000113 \\ \Delta f_2 &= f_2 - f_1 = 9.648810 - 9.648696 = 0.000114 \\ \Delta f_1 &= f_1 - f_0 = 9.648696 - 9.648583 = 0.000113 \end{align*} Calculate second backward differences $\Delta^2 f$: \begin{align*} \Delta^2 f_5 &= \Delta f_5 - \Delta f_4 = 0.000114 - 0.000113 = 0.000001 \\ \Delta^2 f_4 &= \Delta f_4 - \Delta f_3 = 0.000113 - 0.000113 = 0 \\ \Delta^2 f_3 &= \Delta f_3 - \Delta f_2 = 0.000113 - 0.000114 = -0.000001 \\ \Delta^2 f_2 &= \Delta f_2 - \Delta f_1 = 0.000114 - 0.000113 = 0.000001 \end{align*} Calculate third backward differences $\Delta^3 f$: \begin{align*} \Delta^3 f_5 &= \Delta^2 f_5 - \Delta^2 f_4 = 0.000001 - 0 = 0.000001 \\ \Delta^3 f_4 &= \Delta^2 f_4 - \Delta^2 f_3 = 0 - (-0.000001) = 0.000001 \\ \Delta^3 f_3 &= \Delta^2 f_3 - \Delta^2 f_2 = -0.000001 - 0.000001 = -0.000002 \end{align*} Calculate fourth backward differences $\Delta^4 f$: \begin{align*} \Delta^4 f_5 &= \Delta^3 f_5 - \Delta^3 f_4 = 0.000001 - 0.000001 = 0 \\ \Delta^4 f_4 &= \Delta^3 f_4 - \Delta^3 f_3 = 0.000001 - (-0.000002) = 0.000003 \end{align*} Calculate fifth backward difference $\Delta^5 f$: \begin{align*} \Delta^5 f_5 &= \Delta^4 f_5 - \Delta^4 f_4 = 0 - 0.000003 = -0.000003 \end{align*} 6. **Step 4: Apply Newton's Backward formula:** $$ P_5(45095) = f_5 + p\Delta f_5 + \frac{p(p+1)}{2!} \Delta^2 f_5 + \frac{p(p+1)(p+2)}{3!} \Delta^3 f_5 + \frac{p(p+1)(p+2)(p+3)}{4!} \Delta^4 f_5 + \frac{p(p+1)(p+2)(p+3)(p+4)}{5!} \Delta^5 f_5 $$ Calculate each term: - $f_5 = 9.649150$ - $p\Delta f_5 = -0.25 \times 0.000114 = -0.0000285$ - $\frac{p(p+1)}{2} \Delta^2 f_5 = \frac{-0.25 \times 0.75}{2} \times 0.000001 = -0.00000009375$ - $\frac{p(p+1)(p+2)}{6} \Delta^3 f_5 = \frac{-0.25 \times 0.75 \times 1.75}{6} \times 0.000001 = -0.0000000547$ - $\frac{p(p+1)(p+2)(p+3)}{24} \Delta^4 f_5 = \frac{-0.25 \times 0.75 \times 1.75 \times 2.75}{24} \times 0 = 0$ - $\frac{p(p+1)(p+2)(p+3)(p+4)}{120} \Delta^5 f_5 = \frac{-0.25 \times 0.75 \times 1.75 \times 2.75 \times 3.75}{120} \times (-0.000003) \approx 0.000000030$ Sum all terms: $$ 9.649150 - 0.0000285 - 0.00000009375 - 0.0000000547 + 0 + 0.000000030 \approx 9.6491214 $$ 7. **Final answer:** $$ f(45095) \approx 9.6491214 $$