use newtons method with initial approximation ( x_1=-1 ) to find ( x_2 ), the second approximation to the…

use newtons method with initial approximation ( x_1=-1 ) to find ( x_2 ), the second approximation to the solution of the following equation.\n\n( x^{3}+x + 4 = 0 )\n( x_2=)

use newtons method with initial approximation ( x_1=-1 ) to find ( x_2 ), the second approximation to the solution of the following equation.\n\n( x^{3}+x + 4 = 0 )\n( x_2=)

Answer

Explanation:

Step1: Find the derivative of the function

Let (f(x)=x^{3}+x + 4). Then (f^{\prime}(x)=3x^{2}+1).

Step2: Calculate (f(x_1)) and (f^{\prime}(x_1))

When (x_1=-1), (f(-1)=(-1)^{3}+(-1)+4=-1 - 1+4 = 2), (f^{\prime}(-1)=3(-1)^{2}+1=3 + 1=4).

Step3: Apply Newton's method formula (x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})})

For (n = 1), (x_{2}=x_{1}-\frac{f(x_{1})}{f^{\prime}(x_{1})}). Substitute (x_1=-1), (f(x_1) = 2), (f^{\prime}(x_1)=4) into the formula: (x_{2}=-1-\frac{2}{4}=-1 - 0.5=-1.5).

Answer:

(-1.5)