Question

Use Newton's method with initial approximation x1 = −1 to find x2, the second approximation to the root of the equation x3 + x + 8 = 0. (Round your answer to four decimal places.) x2 =

Answer 1

Answer:

The second approximation to the root of the equation $x^{3}+x+8 = 0$ is -1.5000.

Step-by-step explanation:

The Newton's method is a numerical method by approximation that help find roots of a equation of the form $f(x) = 0$ with the help of the equation itself and its first derivative. The Newton's formula is:

$x_{i+1} = x_{i} - \frac{f(x_{i})}{f'(x_{i})}$

Where:

$x_{i}$ - i-th approximation, dimensionless.

$x_{i+1}$ - (i+1)-th approximation, dimensionless.

$f(x_{i})$ - Function evaluated at the i-th approximation, dimensionless.

$f'(x_{i})$ - First derivative of the function evaluated at the i-th approximation, dimensionless.

The function and its first derivative are $f(x) = x^{3}+x+8$ and $f'(x) = 3\cdot x^{2}+1$, respectively. Now, the Newton's formula is expanded:

$x_{i+1} = x_{i}-\frac{x_{i}^{3}+x_{i}+8}{3\cdot x_{i}^{2}+1}$

If $x_{1} = -1$, the value of $x_{2}$ is:

$x_{2} = -1 - \frac{(-1)^{3}+(-1)+8}{3\cdot (-1)^{2}+1}$

$x_{2} = -1.5000$

The second approximation to the root of the equation $x^{3}+x+8 = 0$ is -1.5000.