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The equation f(x) is given as x2 – 4 = 0. Considering the
initial approximation at x=6 then the value of next
approximation correct upto 2 decimal


Sagot :

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]