1. **State the problem:** We want to use Newton's Method to approximate the zero(s) of the function $$f(x) = x - 2\sqrt{x} + 1$$ until two successive approximations differ by less than 0.001. 2. **Recall Newton's Method formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ We need the derivative of $$f(x)$$. 3. **Find the derivative:** $$f(x) = x - 2x^{1/2} + 1$$ Using power rule: $$f'(x) = 1 - 2 \cdot \frac{1}{2} x^{-1/2} = 1 - \frac{1}{\sqrt{x}}$$ 4. **Choose an initial guess:** Since $$f(1) = 1 - 2(1) + 1 = 0$$, 1 is a root, but let's try Newton's Method starting at $$x_0 = 0.5$$ to approximate the root. 5. **Apply Newton's Method iterations:** Calculate $$x_1$$: $$f(0.5) = 0.5 - 2\sqrt{0.5} + 1 = 0.5 - 2(0.7071) + 1 = 0.5 - 1.4142 + 1 = 0.0858$$ $$f'(0.5) = 1 - \frac{1}{0.7071} = 1 - 1.4142 = -0.4142$$ $$x_1 = 0.5 - \frac{0.0858}{-0.4142} = 0.5 + 0.2072 = 0.7072$$ Calculate $$x_2$$: $$f(0.7072) = 0.7072 - 2\sqrt{0.7072} + 1 \approx 0.7072 - 2(0.841) + 1 = 0.7072 - 1.682 + 1 = 0.0252$$ $$f'(0.7072) = 1 - \frac{1}{0.841} = 1 - 1.188 = -0.188$$ $$x_2 = 0.7072 - \frac{0.0252}{-0.188} = 0.7072 + 0.134 = 0.8412$$ Calculate $$x_3$$: $$f(0.8412) = 0.8412 - 2\sqrt{0.8412} + 1 \approx 0.8412 - 2(0.917) + 1 = 0.8412 - 1.834 + 1 = 0.0072$$ $$f'(0.8412) = 1 - \frac{1}{0.917} = 1 - 1.090 = -0.090$$ $$x_3 = 0.8412 - \frac{0.0072}{-0.090} = 0.8412 + 0.08 = 0.9212$$ Calculate $$x_4$$: $$f(0.9212) = 0.9212 - 2\sqrt{0.9212} + 1 \approx 0.9212 - 2(0.96) + 1 = 0.9212 - 1.92 + 1 = 0.0012$$ $$f'(0.9212) = 1 - \frac{1}{0.96} = 1 - 1.042 = -0.042$$ $$x_4 = 0.9212 - \frac{0.0012}{-0.042} = 0.9212 + 0.0286 = 0.9498$$ Calculate $$x_5$$: $$f(0.9498) = 0.9498 - 2\sqrt{0.9498} + 1 \approx 0.9498 - 2(0.9746) + 1 = 0.9498 - 1.9492 + 1 = 0.0006$$ $$f'(0.9498) = 1 - \frac{1}{0.9746} = 1 - 1.0259 = -0.0259$$ $$x_5 = 0.9498 - \frac{0.0006}{-0.0259} = 0.9498 + 0.0232 = 0.9730$$ Difference between $$x_5$$ and $$x_4$$ is $$|0.9730 - 0.9498| = 0.0232 > 0.001$$, continue. Calculate $$x_6$$: $$f(0.9730) = 0.9730 - 2\sqrt{0.9730} + 1 \approx 0.9730 - 2(0.9864) + 1 = 0.9730 - 1.9728 + 1 = 0.0002$$ $$f'(0.9730) = 1 - \frac{1}{0.9864} = 1 - 1.0137 = -0.0137$$ $$x_6 = 0.9730 - \frac{0.0002}{-0.0137} = 0.9730 + 0.0146 = 0.9876$$ Difference $$|0.9876 - 0.9730| = 0.0146 > 0.001$$, continue. Calculate $$x_7$$: $$f(0.9876) = 0.9876 - 2\sqrt{0.9876} + 1 \approx 0.9876 - 2(0.9938) + 1 = 0.9876 - 1.9876 + 1 = 0$$ $$f'(0.9876) = 1 - \frac{1}{0.9938} = 1 - 1.0062 = -0.0062$$ $$x_7 = 0.9876 - \frac{0}{-0.0062} = 0.9876$$ Difference $$|0.9876 - 0.9876| = 0 < 0.001$$, stop iterations. 6. **Conclusion:** The approximate zero is $$x \approx 0.9876$$. 7. **Graphing utility check:** Graphing $$f(x) = x - 2\sqrt{x} + 1$$ shows a root near $$x=1$$, confirming our approximation. **Final answer:** The zero of the function is approximately $$x = 0.988$$ (rounded to three decimal places).