Answer
$$ 3.3166$$
Work Step by Step
Given $$\sqrt{11} $$
Let $$f(x) =x^2-11 $$
and suppose that $x_0=3$, then
Since
$$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
Then
\begin{align*}
x_{n+1}&= x_n -\frac{x_n^{2}-11}{2x_n}\\
&=\frac{x_n^{2}+11}{2x_n}
\end{align*}
Hence
\begin{aligned}
x_{1}&=\frac{x_0^{2}+11}{2x_0}= \frac{(3)^{2}+11}{2(3)} \approx 3.3333333 \\
x_{2}&=\frac{x_1^{2}+11}{2x_1}= \frac{( 3.3333333)^{2}+11}{2( 3.3333333)} \approx 3.31666\\
x_{3}&=\frac{x_2^{2}+11}{2x_2} = \frac{(3.31666)^{2}+11}{2(3.31666)} \approx3.31662479
\end{aligned}
Since $x_2=x_3 $, then $\sqrt{11}\approx 3.3166$
By calculator, we get
$$\sqrt{11}= 3.31662479$$
