1. **Problem Statement:** Find the first derivative $f'(x)$ and second derivative $f''(x)$ at $x=1$ using the forward difference operator from the given table: $$\begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ f(x) & 3 & 4 & 19 & 84 & 259 & 628 \\\end{array}$$ 2. **Formulas:** - First derivative approximation: $$f'(x) = \frac{1}{h} \left[ \Delta - \frac{\Delta^2}{2} + \frac{\Delta^3}{3} - \frac{\Delta^4}{4} + \cdots \right] f(x)$$ - Second derivative approximation: $$f''(x) = \frac{1}{h^2} \left[ \Delta^2 - 2 \frac{\Delta^3}{2} + \frac{11}{12} \Delta^4 - \cdots \right] f(x)$$ where $h$ is the step size (distance between $x$ values), and $\Delta$ is the forward difference operator. 3. **Step size:** Given $x$ values are equally spaced by 1, so $h=1$. 4. **Calculate forward differences at $x=1$:** - $\Delta f(1) = f(2) - f(1) = 19 - 4 = 15$ - $\Delta^2 f(1) = \Delta f(2) - \Delta f(1) = (f(3) - f(2)) - (f(2) - f(1)) = (84 - 19) - 15 = 65 - 15 = 50$ - $\Delta^3 f(1) = \Delta^2 f(2) - \Delta^2 f(1) = ((f(4) - f(3)) - (f(3) - f(2))) - 50 = ((259 - 84) - (84 - 19)) - 50 = (175 - 65) - 50 = 110 - 50 = 60$ - $\Delta^4 f(1) = \Delta^3 f(2) - \Delta^3 f(1) = (((f(5) - f(4)) - (f(4) - f(3))) - ((f(4) - f(3)) - (f(3) - f(2)))) - 60 = ((628 - 259) - (259 - 84)) - 60 = (369 - 175) - 60 = 194 - 60 = 134$ 5. **Calculate $f'(1)$:** $$f'(1) = \frac{1}{1} \left[ 15 - \frac{50}{2} + \frac{60}{3} - \frac{134}{4} \right] = 15 - 25 + 20 - 33.5 = -23.5$$ 6. **Calculate $f''(1)$:** $$f''(1) = \frac{1}{1^2} \left[ 50 - 2 \times \frac{60}{2} + \frac{11}{12} \times 134 \right] = 50 - 60 + 122.8333 = 112.8333$$ 7. **Final answers:** $$f'(1) = -23.5$$ $$f''(1) \approx 112.83$$