Answer:
The correct answer is "[tex]\frac{10}{3}[/tex]".
Explanation:
The given function is:
[tex]f(x)=x^2-4=0[/tex]
and,
[tex]x_0=6[/tex]
By differentiating with the help of Newton's Raphson method, we get
⇒ [tex]x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}[/tex]
then,
⇒ [tex]x_1=x_0-\frac{f(x_0)}{f'(x_0)}[/tex]
[tex]=6-\frac{f(6)}{f'(6)}[/tex]
[tex]=6-\frac{(6)^2-4}{2\times 6}[/tex]
[tex]=6-\frac{8}{3}[/tex]
[tex]=\frac{10}{3}[/tex]
