Question

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

Answer 1

Answer:

The correct answer is "$\frac{10}{3}$".

Explanation:

The given function is:

$f(x)=x^2-4=0$

and,

$x_0=6$

By differentiating with the help of Newton's Raphson method, we get

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

then,

⇒ $x_1=x_0-\frac{f(x_0)}{f'(x_0)}$

        $=6-\frac{f(6)}{f'(6)}$

        $=6-\frac{(6)^2-4}{2\times 6}$

        $=6-\frac{8}{3}$

        $=\frac{10}{3}$