1. **State the problem:** Use Newton's method to estimate the real root of the equation $$7x^3 + x - 5 = 0$$ accurate to five decimal places. 2. **Recall Newton's method formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ where $f(x) = 7x^3 + x - 5$ and $f'(x)$ is its derivative. 3. **Find the derivative:** $$f'(x) = 21x^2 + 1$$ 4. **Choose an initial guess:** By inspection, try $x_0 = 1$ because $7(1)^3 + 1 - 5 = 3$ (positive) and $x=0$ gives $-5$ (negative), so root is near 1. 5. **Apply Newton's method iterations:** - Iteration 1: $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{7(1)^3 + 1 - 5}{21(1)^2 + 1} = 1 - \frac{3}{22} = 1 - \frac{\cancel{3}}{\cancel{22}} = 1 - 0.13636 = 0.86364$$ - Iteration 2: Calculate $f(0.86364)$ and $f'(0.86364)$: $$f(0.86364) = 7(0.86364)^3 + 0.86364 - 5 = 7(0.644) + 0.86364 - 5 = 4.508 + 0.86364 - 5 = 0.37164$$ $$f'(0.86364) = 21(0.86364)^2 + 1 = 21(0.746) + 1 = 15.666 + 1 = 16.666$$ Then: $$x_2 = 0.86364 - \frac{0.37164}{16.666} = 0.86364 - 0.0223 = 0.84134$$ - Iteration 3: Calculate $f(0.84134)$ and $f'(0.84134)$: $$f(0.84134) = 7(0.84134)^3 + 0.84134 - 5 = 7(0.596) + 0.84134 - 5 = 4.172 + 0.84134 - 5 = 0.01334$$ $$f'(0.84134) = 21(0.84134)^2 + 1 = 21(0.708) + 1 = 14.868 + 1 = 15.868$$ Then: $$x_3 = 0.84134 - \frac{0.01334}{15.868} = 0.84134 - 0.00084 = 0.84050$$ - Iteration 4: Calculate $f(0.84050)$ and $f'(0.84050)$: $$f(0.84050) = 7(0.84050)^3 + 0.84050 - 5 = 7(0.594) + 0.84050 - 5 = 4.158 + 0.84050 - 5 = 0.00013$$ $$f'(0.84050) = 21(0.84050)^2 + 1 = 21(0.706) + 1 = 14.826 + 1 = 15.826$$ Then: $$x_4 = 0.84050 - \frac{0.00013}{15.826} = 0.84050 - 0.000008 = 0.84049$$ 6. **Conclusion:** The root accurate to five decimal places is approximately $$\boxed{0.84049}$$