Example 1: Use Newton's method for approximate a root of f(x) =$x^{3}$ -3x + 1 in the interval [1,2]. $x_{0}$= 2 , then find $x_{1},x_{2}$
Solution : f(x) =$x^{3}$ -3x + 1
f(2) = $2^{3}$ - 3(2) + 1 = 3
f '(x) = $3x^{2} -3 \Rightarrow f '(2) = 3(2)^{2} -3 $ = 9
f(x) =$x^{3}$ -3x + 1 to find $x_{1},x_{2},x_{3}$ we will use the formula of Newton's method.
$x_{n+1} = x_{n} - \frac{f(x_{n})}{f '(x_{n})}$
Plug in n= 0
$x_{1} = x_{0} - \frac{f(x_{0})}{f '(x_{0})}$
= 2 - $\frac{3}{9}$
$x_{1} \approx $ 1.6666667
For $x_{2}$
f(x) =$x^{3}$ -3x + 1
f '(x) = $3x^{2}$ -3
f(1.66667) = $(1.6666667)^{3}$ -3(1.6666667) + 1 = 0.629629
f '(1.6666667) = $3(1.6666667)^{2}$ -3 = 5.3336667
$x_{n+1} = x_{n} - \frac{f(x_{n})}{f '(x_{n})}$
Plug in x= 1
$x_{2} = x_{1} - \frac{f(x_{1})}{f '(x_{1})}$
= 1.6666667 - $\frac{0.629629}{5.3336667}$
$x_{2}$ = 1.5486111
Example 2: Using Newton's method find the value of $\sqrt{5}$
Solution : x = $\sqrt{5}$
$x^{2}$ = 5
f(x) = $x^{2}$ - 5 = 0
x = $\pm \sqrt{5}$
$\sqrt{4} < \sqrt{5} < \sqrt{9}$
2 < $\sqrt{5}$ <3
Let $x_{0}$ = 2
f($x_{0}$) = 4 - 5 = -1
f(2) = -1
f(x) = $x^{2}$ - 5
f '(x) = 2x
f '(2) = 4
Using Newton's formula
$x_{n+1} = x_{n} - \frac{f(x_{n})}{f '(x_{n})}$
Plug in x = 0
$x_{1} = x_{0} - \frac{f(x_{0})}{f '(x_{0})}$
$x_{1} = 2 - \frac{-1}{4}$
$x_{1} = 2 + \frac{1}{4}$
$x_{1} \approx $ 2.25