$($Final$)$ Newton's Method for approximating zeros of a differentiable function $f(x)$ works as follows.

Guess a zero $x_0.$

If $x_n$ is the last approximation, then

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

is the next $($likely better$)$ approximation.

Under certain mild conditions, the approximations get better and better. $($Note: There are examples were Newton's Method fails badly.$)$

Example. Use two iterates of Newton's method to approximate $\sqrt[\phantom{2}]{66}$ using

$f(x)=x^2-66.$

Solution. Since $f^{\text{'}}(x)=2x$ the formula for the next iterate is

$x_{n+1}=x_n-\frac{x_n^2-66}{2x_n}$

My first guess is $x_0=8($since $8^2=64$ is near $66).$ The first iterate is

$x_1=8-\frac{8^2-66}{2*8}=8\mathrm{.}125$

The next iterate is

$x_2=8\mathrm{.}125-\frac{{8\mathrm{.}125}^2-66}{2*8\mathrm{.}125}=8\mathrm{.}124038462.$

which is already very close to the actual

$\sqrt[\phantom{2}]{66}$~~$8\mathrm{.}124038405.$

The error here is only $0\mathrm{.}000000057$ which is crazy small. This is a powerful trick! Newton's method converges rapidly when it works! Your turn: Use two iterates of Newton's Method to approximate $\sqrt[\phantom{2}]{47}.$