Write the formula for Newton's method and use the given initial approximation to compute the approximations \( x_{1} \) and \( x_{2} \). \( f(x)=x^{2}+2 x-15, x_{0}=4 \) B. \( x_{n+1}=x_{n}+\frac{x_{n}^{2}+2 x_{n}-15}{2 x_{n}+2} \) C. \( x_{n+1}=x_{n}-\frac{2 x_{n}+2}{x_{n}^{2}+2 x_{n}-15} \) D. \( x_{n+1}=x_{n}-\frac{x_{n}^{2}+2 x_{n}-15}{2 x_{n}+2} \) Use the given initial approximation to compute the approximations \( x_{1} \) and \( x_{2} \). \( x_{1}=\square \) (Do not round until the final answer. Then round to three decimal places as needed.) \( x_{2}=\square \) (Do not round until the final answer. Then round to three decimal places as needed.) (D)

To compute the approximations \( x_{1} \) and \( x_{2} \) using Newton's method with the given function \( f(x) = x^2 + 2x - 15 \) and initial approximation \( x_{0} = 4 \), we will use formula (D): \[ x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \] First, we need to find \( f(x) \) and its derivative \( f'(x) \): 1. **Evaluate \( f(x_0) \)**: \[ f(4) = 4^2 + 2(4) - 15 = 16 + 8 - 15 = 9 \] 2. **Calculate \( f'(x) \)**: \[ f'(x) = 2x + 2 \] Now, evaluate \( f'(x_0) \): \[ f'(4) = 2(4) + 2 = 8 + 2 = 10 \] 3. **Calculate \( x_{1} \)**: \[ x_{1} = x_{0} - \frac{f(4)}{f'(4)} = 4 - \frac{9}{10} = 4 - 0.9 = 3.1 \] 4. **Now compute \( x_{2} \)** by repeating the process using \( x_{1} = 3.1 \): - **Evaluate \( f(x_1) \)**: \[ f(3.1) = (3.1)^2 + 2(3.1) - 15 = 9.61 + 6.2 - 15 = 0.81 \] - **Calculate \( f'(3.1) \)**: \[ f'(3.1) = 2(3.1) + 2 = 6.2 + 2 = 8.2 \] - **Calculate \( x_{2} \)**: \[ x_{2} = x_{1} - \frac{f(3.1)}{f'(3.1)} = 3.1 - \frac{0.81}{8.2} \approx 3.1 - 0.0988 \approx 3.0012 \] Finally, rounding these results to three decimal places: \( x_{1} = 3.100 \) \( x_{2} = 3.001 \)