Answer:
The third approximation is [tex]x_3=-1.08259[/tex]
Step-by-step explanation:
We are given that
[tex]f(x)=x^3+x+2=0[/tex]
[tex]x_1=-2[/tex]
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 [tex]x_n[/tex] and [tex]f'(x_n)\neq 0[/tex] then, the next approximation is given by
[tex]x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}[/tex]
[tex]f'(x)=3x^2+1[/tex]
Substitute [tex]x_1=-2[/tex]
[tex]f(x_1)=f(-2)=(-2)^3+(-2)+2=-8[/tex]
[tex]f'(x_1)=f'(-2)=3(-2)^2+1=13[/tex]
Substitute the value n=1 then, we get
[tex]x_2=x_1-\frac{f(x_1)}{f'(x_1)}[/tex]
Substitute the values then , we get the second approximation
[tex]x_2=-2-\frac{-8}{13}=-\frac{18}{13}[/tex]
For n=2
[tex]f(x_2)=f(-\frac{18}{13})=(-\frac{18}{13})^3-\frac{18}{13}+2=-2.03914[/tex]
[tex]f'(x_2)=f'(-\frac{18}{13})=3(-\frac{18}{13})^2+1=6.75148[/tex]
[tex]x_3=x_2-\frac{f(x_2)}{f'(x_2)}[/tex]
[tex]x_3=-\frac{18}{13}-\frac{-2.03914}{6.75148}=-1.08259[/tex]
Hence, the third approximation is [tex]x_3=-1.08259[/tex]
