1. **State the problem:** We are given values of a function $f(x)$ at points $x = 0.12, 0.16, 0.20, 0.24, 0.28, 0.32$ and corresponding function values $f(x) = 0.6144, 0.6256, 0.6400, 0.6576, 0.6784, 0.7024$. We want to find $f(0.14)$ using Newton-Gregory finite difference interpolation. 2. **Formula and explanation:** The Newton-Gregory forward interpolation formula is: $$ P(x) = f(x_0) + s\Delta f(x_0) + \frac{s(s-1)}{2!}\Delta^2 f(x_0) + \frac{s(s-1)(s-2)}{3!}\Delta^3 f(x_0) + \cdots $$ where $s = \frac{x - x_0}{h}$, $h$ is the uniform spacing between $x$ values, and $\Delta$ denotes forward differences. 3. **Calculate $h$ and $s$:** $$ h = 0.16 - 0.12 = 0.04 $$ $$ s = \frac{0.14 - 0.12}{0.04} = \frac{0.02}{0.04} = 0.5 $$ 4. **Construct the forward difference table:** \begin{align*} \text{At } x: & 0.12 & 0.16 & 0.20 & 0.24 & 0.28 & 0.32 \\ f(x): & 0.6144 & 0.6256 & 0.6400 & 0.6576 & 0.6784 & 0.7024 \\ \Delta f: & 0.0112 & 0.0144 & 0.0176 & 0.0208 & 0.0240 \\ \Delta^2 f: & 0.0032 & 0.0032 & 0.0032 & 0.0032 \\ \Delta^3 f: & 0 & 0 & 0 \\ \Delta^4 f: & 0 & 0 \\ \Delta^5 f: & 0 \end{align*} 5. **Apply the interpolation formula:** $$ P(0.14) = f(0.12) + s\Delta f(0.12) + \frac{s(s-1)}{2!}\Delta^2 f(0.12) $$ Since higher differences are zero, terms beyond second difference vanish. Substitute values: $$ P(0.14) = 0.6144 + 0.5 \times 0.0112 + \frac{0.5(0.5-1)}{2} \times 0.0032 $$ Calculate step by step: $$ 0.5 \times 0.0112 = 0.0056 $$ $$ \frac{0.5 \times (-0.5)}{2} = \frac{-0.25}{2} = -0.125 $$ $$ -0.125 \times 0.0032 = -0.0004 $$ So, $$ P(0.14) = 0.6144 + 0.0056 - 0.0004 = 0.6196 $$ 6. **Final answer:** $$ f(0.14) \approx 0.6196 $$