Apply Newton's method to \( f(x)=x^{\frac{1}{3}} \) with \( x_{0}=1 \) and calculate \( x_{1}, x_{2}, x_{3} \), and \( x_{4} \). Find a formula for \( \left|x_{n}\right| \). What happens to \( \left|x_{n}\right| \) as \( n \rightarrow \infty \) ? Draw a picture that shows what is going on. \( x_{2}=\square \) (Simplify your answer.)

To apply Newton's method to the function \( f(x) = x^{\frac{1}{3}} \), we first need to compute its derivative: \[ f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3 \sqrt[3]{x^2}}. \] Newton's method uses the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. \] Starting with \( x_0 = 1 \): 1. **Calculate \( x_1 \)**: \[ f(x_0) = f(1) = 1^{\frac{1}{3}} = 1, \] \[ f'(x_0) = f'(1) = \frac{1}{3 \sqrt[3]{1^2}} = \frac{1}{3}. \] Now apply Newton's method: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{1}{\frac{1}{3}} = 1 - 3 = -2. \] 2. **Calculate \( x_2 \)**: \[ f(x_1) = f(-2) = (-2)^{\frac{1}{3}} = -\sqrt[3]{2}, \] \[ f'(-2) = f'(-2) = \frac{1}{3 \sqrt[3]{(-2)^2}} = \frac{1}{3 \sqrt[3]{4}}. \] Now apply Newton's method: \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -2 - \frac{-\sqrt[3]{2}}{\frac{1}{3 \sqrt[3]{4}}} = -2 + 3 \sqrt[3]{2} \cdot \sqrt[3]{4} = -2 + 3 \cdot \sqrt[3]{8} = -2 + 6 = 4. \] 3. **Calculate \( x_3 \)**: \[ f(x_2) = f(4) = 4^{\frac{1}{3}} = \sqrt[3]{4}, \] \[ f'(4) = f'(4) = \frac{1}{3 \sqrt[3]{16}}. \] Now apply Newton's method: \[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 4 - \frac{\sqrt[3]{4}}{\frac{1}{3 \sqrt[3]{16}}} = 4 - 3 \cdot \sqrt[3]{4} \cdot \sqrt[3]{16} = 4 - 3 \cdot \sqrt[3]{64} = 4 - 3 \cdot 4 = 4 - 12 = -8. \] 4. **Calculate \( x_4 \)**: \[ f(x_3) = f(-8) = (-8)^{\frac{1}{3}} = -2, \] \[ f'(-8) = f'(-8) = \frac{1}{3 \sqrt[3]{64}} = \frac{1}{12}. \] Now apply Newton's method: \[ x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} = -8 - \frac{-2}{\frac{1}{12}} = -8 + 24 = 16. \] Now we have the following values: - \( x_1 = -2 \) - \( x_2 = 4 \) - \( x_3 = -8 \) - \( x_4 = 16 \) Next, we can observe a pattern in the absolute values: \[ |x_n| = 2^n. \] As \( n \rightarrow \infty \), \( |x_n| \) grows without bound, i.e., \( |x_n| \rightarrow \infty \). ### Summary: - \( x_2 = 4 \) - The formula for \( |x_n| \) is \( |x_n| = 2^n \). - As \( n \rightarrow \infty \), \( |x_n| \rightarrow \infty \). ### Visualization: To visualize this, you can plot the points \( (n, |x_n|) \) for \( n = 0, 1, 2, 3, 4 \) and observe the exponential growth of \( |x_n| \). The points would be: - \( (0, 1) \) - \( (1, 2) \) - \( (2, 4) \) - \( (3, 8) \) - \( (4, 16) \) This would show a steep upward curve, indicating the rapid growth of \( |x_n| \).