1. **Problem Statement:** We are given data points for $x$ and $f(x)$ and asked to find $f(0.58)$ using Newton's interpolation method of order 3. 2. **Newton's Interpolation Formula:** For order 3, the interpolation polynomial is: $$P_3(x) = f[x_0] + f[x_0,x_1](x-x_0) + f[x_0,x_1,x_2](x-x_0)(x-x_1) + f[x_0,x_1,x_2,x_3](x-x_0)(x-x_1)(x-x_2)$$ where $f[x_i]$ are function values and $f[x_i,x_j,...]$ are divided differences. 3. **Select Points:** Choose four points near $x=0.58$ for order 3 interpolation. The closest points are $x_1=0.3$, $x_2=0.5$, $x_3=0.7$, and $x_4=0.9$ with corresponding $f(x)$ values. 4. **Calculate Divided Differences:** - $f[x_1] = 0.067$ - $f[x_2] = 0.148$ - $f[x_3] = 0.248$ - $f[x_4] = 0.370$ First order: $$f[x_1,x_2] = \frac{0.148 - 0.067}{0.5 - 0.3} = 0.405$$ $$f[x_2,x_3] = \frac{0.248 - 0.148}{0.7 - 0.5} = 0.5$$ $$f[x_3,x_4] = \frac{0.370 - 0.248}{0.9 - 0.7} = 0.61$$ Second order: $$f[x_1,x_2,x_3] = \frac{0.5 - 0.405}{0.7 - 0.3} = 0.2375$$ $$f[x_2,x_3,x_4] = \frac{0.61 - 0.5}{0.9 - 0.5} = 0.275$$ Third order: $$f[x_1,x_2,x_3,x_4] = \frac{0.275 - 0.2375}{0.9 - 0.3} = 0.0625$$ 5. **Construct Polynomial:** $$P_3(x) = 0.067 + 0.405(x-0.3) + 0.2375(x-0.3)(x-0.5) + 0.0625(x-0.3)(x-0.5)(x-0.7)$$ 6. **Evaluate at $x=0.58$:** Calculate each term: - $x-0.3 = 0.28$ - $x-0.5 = 0.08$ - $x-0.7 = -0.12$ So, $$P_3(0.58) = 0.067 + 0.405(0.28) + 0.2375(0.28)(0.08) + 0.0625(0.28)(0.08)(-0.12)$$ Calculate stepwise: - $0.405 \times 0.28 = 0.1134$ - $0.2375 \times 0.28 \times 0.08 = 0.00532$ - $0.0625 \times 0.28 \times 0.08 \times (-0.12) = -0.000168$ Sum all: $$0.067 + 0.1134 + 0.00532 - 0.000168 = 0.185552$$ **Final answer:** $$f(0.58) \approx 0.186$$