1. **State the problem:** We are given values of a function $f(x)$ at points $x = 3, 3.2, 3.4, 3.6, 3.8, 4.0$ and need to find the first derivative $f'(3)$ and the second derivative $f''(3)$ using Newton's forward difference formula. 2. **Recall Newton's forward difference formula for derivatives:** For equally spaced points with spacing $h$, the first derivative at $x_0$ is approximated by $$f'(x_0) \approx \frac{1}{h} \left(\Delta f_0 - \frac{1}{2} \Delta^2 f_0 + \frac{1}{3} \Delta^3 f_0 - \cdots \right)$$ and the second derivative by $$f''(x_0) \approx \frac{1}{h^2} \left(\Delta^2 f_0 - \Delta^3 f_0 + \frac{11}{12} \Delta^4 f_0 - \cdots \right)$$ where $\Delta$ denotes the forward difference operator. 3. **Calculate the forward differences:** Given $h = 0.2$ (since $3.2 - 3 = 0.2$), the function values are: $f_0 = f(3) = -14$ $f_1 = f(3.2) = -10.032$ $f_2 = f(3.4) = -5.296$ $f_3 = f(3.6) = -0.256$ $f_4 = f(3.8) = 6.672$ $f_5 = f(4.0) = 14$ Calculate first forward differences: $$\Delta f_0 = f_1 - f_0 = -10.032 - (-14) = 3.968$$ $$\Delta f_1 = f_2 - f_1 = -5.296 - (-10.032) = 4.736$$ $$\Delta f_2 = f_3 - f_2 = -0.256 - (-5.296) = 5.04$$ $$\Delta f_3 = f_4 - f_3 = 6.672 - (-0.256) = 6.928$$ $$\Delta f_4 = f_5 - f_4 = 14 - 6.672 = 7.328$$ Calculate second forward differences: $$\Delta^2 f_0 = \Delta f_1 - \Delta f_0 = 4.736 - 3.968 = 0.768$$ $$\Delta^2 f_1 = \Delta f_2 - \Delta f_1 = 5.04 - 4.736 = 0.304$$ $$\Delta^2 f_2 = \Delta f_3 - \Delta f_2 = 6.928 - 5.04 = 1.888$$ $$\Delta^2 f_3 = \Delta f_4 - \Delta f_3 = 7.328 - 6.928 = 0.4$$ Calculate third forward differences: $$\Delta^3 f_0 = \Delta^2 f_1 - \Delta^2 f_0 = 0.304 - 0.768 = -0.464$$ $$\Delta^3 f_1 = \Delta^2 f_2 - \Delta^2 f_1 = 1.888 - 0.304 = 1.584$$ $$\Delta^3 f_2 = \Delta^2 f_3 - \Delta^2 f_2 = 0.4 - 1.888 = -1.488$$ Calculate fourth forward differences: $$\Delta^4 f_0 = \Delta^3 f_1 - \Delta^3 f_0 = 1.584 - (-0.464) = 2.048$$ $$\Delta^4 f_1 = \Delta^3 f_2 - \Delta^3 f_1 = -1.488 - 1.584 = -3.072$$ 4. **Apply the formulas:** First derivative at $x=3$: $$f'(3) \approx \frac{1}{0.2} \left(3.968 - \frac{1}{2}(-0.464) + \frac{1}{3}(2.048)\right)$$ Calculate inside the parentheses: $$3.968 + 0.232 + 0.6827 = 4.8827$$ So, $$f'(3) \approx \frac{4.8827}{0.2} = 24.4135$$ Second derivative at $x=3$: $$f''(3) \approx \frac{1}{0.2^2} \left(0.768 - (-0.464) + \frac{11}{12} (2.048)\right)$$ Calculate inside the parentheses: $$0.768 + 0.464 + 1.8787 = 3.1107$$ Since $0.2^2 = 0.04$, $$f''(3) \approx \frac{3.1107}{0.04} = 77.7675$$ 5. **Final answers:** $$\boxed{f'(3) \approx 24.41}$$ $$\boxed{f''(3) \approx 77.77}$$