1. **Problem Statement:** We are given values of $y$ at points $x = 20, 25, 30, 35, 40, 45$ and want to find $f(22)$ using the Gauss forward interpolation formula. 2. **Given Data:** $$\begin{array}{c|cccccc} x & 20 & 25 & 30 & 35 & 40 & 45 \\ y & 354 & 332 & 291 & 260 & 231 & 204 \\\end{array}$$ 3. **Step Size:** The difference between consecutive $x$ values is $h = 25 - 20 = 5$. 4. **Locate $x_0$:** Since $22$ is close to $20$, we take $x_0 = 20$ and $f(x_0) = 354$. 5. **Calculate forward differences:** $$\begin{array}{c|cccccc} y & 354 & 332 & 291 & 260 & 231 & 204 \\ \Delta y & -22 & -41 & -31 & -29 & -27 \\ \Delta^2 y & -19 & 10 & 2 & 2 \\ \Delta^3 y & 29 & -8 & 0 \\ \Delta^4 y & -37 & 8 \\ \Delta^5 y & 45 \\ \end{array}$$ 6. **Calculate $p$:** $$p = \frac{x - x_0}{h} = \frac{22 - 20}{5} = 0.4$$ 7. **Gauss forward interpolation formula:** $$f(x) = f(x_0) + p\Delta y_0 + \frac{p(p-1)}{2!}\Delta^2 y_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 y_0 + \frac{p(p-1)(p-2)(p-3)}{4!}\Delta^4 y_0 + \cdots$$ 8. **Substitute values:** $$f(22) = 354 + 0.4(-22) + \frac{0.4(0.4-1)}{2}(-19) + \frac{0.4(0.4-1)(0.4-2)}{6}(29) + \frac{0.4(0.4-1)(0.4-2)(0.4-3)}{24}(-37)$$ 9. **Calculate each term:** - $0.4(-22) = -8.8$ - $\frac{0.4 \times (-0.6)}{2} \times (-19) = \frac{-0.24}{2} \times (-19) = -0.12 \times (-19) = 2.28$ - $\frac{0.4 \times (-0.6) \times (-1.6)}{6} \times 29 = \frac{0.384}{6} \times 29 = 0.064 \times 29 = 1.856$ - $\frac{0.4 \times (-0.6) \times (-1.6) \times (-2.6)}{24} \times (-37) = \frac{-0.9984}{24} \times (-37) = -0.0416 \times (-37) = 1.5392$ 10. **Sum all terms:** $$f(22) = 354 - 8.8 + 2.28 + 1.856 + 1.5392 = 350.8752$$ **Final answer:** $$\boxed{f(22) \approx 350.88}$$