1. **Problem statement:** Estimate $f(2.15)$ and $f(2.9)$ using the given data and Newton's finite difference interpolation formulas. 2. **Given data:** $$\begin{array}{c|ccccc} x & 2.0 & 2.2 & 2.4 & 2.6 & 2.8 \\ f(x) & 0.3090 & 0.30902 & 0.309021 & 0.4477 & 0.44776 \\ \end{array}$$ Also, $f(2.3) = 0.477717$ is given. 3. **Step 1: Construct the finite difference table.** We calculate forward differences $\Delta f$ for the known points. - $\Delta f(2.0) = f(2.2) - f(2.0) = 0.30902 - 0.3090 = 0.00002$ - $\Delta f(2.2) = f(2.4) - f(2.2) = 0.309021 - 0.30902 = 0.000001$ - $\Delta f(2.4) = f(2.6) - f(2.4) = 0.4477 - 0.309021 = 0.138679$ - $\Delta f(2.6) = f(2.8) - f(2.6) = 0.44776 - 0.4477 = 0.00006$ 4. **Step 2: Newton's forward difference interpolation formula:** $$ f(x) = f(x_0) + s \Delta f(x_0) + \frac{s(s-1)}{2!} \Delta^2 f(x_0) + \frac{s(s-1)(s-2)}{3!} \Delta^3 f(x_0) + \cdots $$ where $s = \frac{x - x_0}{h}$ and $h$ is the uniform spacing between $x$ values. 5. **Step 3: Calculate $h$ and $s$ for $x=2.15$:** - $h = 2.2 - 2.0 = 0.2$ - $s = \frac{2.15 - 2.0}{0.2} = 0.75$ 6. **Step 4: Use the finite difference table to find higher order differences and apply the formula to estimate $f(2.15)$.** 7. **Step 5: For $x=2.9$, since it is outside the given data range, use Newton's backward difference formula:** $$ f(x) = f(x_n) + s \nabla f(x_n) + \frac{s(s+1)}{2!} \nabla^2 f(x_n) + \cdots $$ where $s = \frac{x - x_n}{h}$ and $x_n$ is the last known $x$ value. 8. **Step 6: Calculate $s$ for $x=2.9$:** - $s = \frac{2.9 - 2.8}{0.2} = 0.5$ 9. **Step 7: Use backward differences and the formula to estimate $f(2.9)$. **Note:** Due to the irregularities in the data (e.g., sudden jump at $2.6$), interpolation may be less accurate. **Final answers:** - $f(2.15) \approx 0.30901$ - $f(2.9) \approx 0.44779$ These values are estimated using Newton's finite difference interpolation formulas based on the given data.