Question

Use Newton's method to find the second and third approximation of a root of x3+x+2=0 starting with x1=−1 as the initial approximation. The second approximation is x2 = -18/13 The third approximation is x3 =

Answer 1

Answer:

The third approximation is $x_3=-1.08259$

Step-by-step explanation:

We are given that  

$f(x)=x^3+x+2=0$

$x_1=-2$

We have to find the second and third approximation of a root of given equation by using Newton's method.

We know that Newton's method , if nth approximation is given $x_n$ and $f'(x_n)\neq 0$ then, the next approximation is given by  

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

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

Substitute $x_1=-2$

$f(x_1)=f(-2)=(-2)^3+(-2)+2=-8$

$f'(x_1)=f'(-2)=3(-2)^2+1=13$

Substitute the value n=1 then, we get  

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}$

Substitute the values then , we get  the second approximation

$x_2=-2-\frac{-8}{13}=-\frac{18}{13}$

For n=2

$f(x_2)=f(-\frac{18}{13})=(-\frac{18}{13})^3-\frac{18}{13}+2=-2.03914$

$f'(x_2)=f'(-\frac{18}{13})=3(-\frac{18}{13})^2+1=6.75148$

$x_3=x_2-\frac{f(x_2)}{f'(x_2)}$

$x_3=-\frac{18}{13}-\frac{-2.03914}{6.75148}=-1.08259$

Hence, the third approximation is $x_3=-1.08259$