Question

Use Newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 + x + 6 = 0. (Round your answer to four decimal places.)

Answer 1

Answer:

The value of $x_2=-1.6923$.

Step-by-step explanation:

Consider the provided information.

The provided formula is $f(x)=x^3+x+6$

Substitute $x_1=-2$ in above equation.

$f(x_1)=(-2)^3+(-2)+6$

$f(x_1)=-8-2+6$

$f(x_1)=-4$

Differentiate the provided function and calculate the value of $f'(x_1)$

$f'(x)=3x^2+1$

$f'(x)=3(-2)^2+1$

$f'(x)=13$

The Newton iteration formula: $x_2=x_1-\frac{f(x_1)}{f'(x_1)}$

Substitute the respective values in the above formula.

$x_2=-2-\frac{(-4)}{13}$

$x_2=-2+0.3077$

$x_2=-1.6923$

Hence, the value of $x_2=-1.6923$.