Answer:
[tex]x_{3}=2.35[/tex]
Explanation:
Given [tex] x^2-2x-1=0,x_1=2[/tex]
From Newton's method
[tex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}[/tex]
[tex]f(x)=x^2-2x-1[/tex]
[tex]f'(x)=2x-2[/tex]
Now
[tex]x_{2}=x_1-\dfrac{f(x_1)}{f'(x_1)}[/tex]
[tex]f(x)=x^2-2x-1[/tex]
[tex]f(2)=2^2-2\times 2-1[/tex]
f(x)=-1
[tex]f'(2)=2x-2[/tex]
[tex]f'(1)=2\times 2-2[/tex]
f'(1)=2
[tex]x_{2}=2+\dfrac{1}{2}[/tex]
[tex]x_{2}=2.5[/tex]
[tex]x_{3}=x_2-\dfrac{f(x_2)}{f'(x_2)}[/tex]
[tex]f(2.5)=2.5^2-2\times 2.5-1[/tex]
f(2.5)=0.45
[tex]f'(2.5)=2\times 2.5-2[/tex]
[tex]f'(2.5)=3[/tex]
[tex]x_{3}=2.5-\dfrac{0.45}{3}[/tex]
So
[tex]x_{3}=2.35[/tex]
