We take f ( x ) = x 4 − 3 {\displaystyle f(x)=x^{4}-3} and x 0 = 1 {\displaystyle x_{0}=1} . (We choose this x 0 {\displaystyle x_{0}} because f ( x 0 ) {\displaystyle f(x_{0})} and f ′ ( x 0 ) {\displaystyle f'(x_{0})} are easy to evaluate, and clearly 1 < 3 4 ≪ 2 {\displaystyle 1<{\sqrt[{4}]{3}}\ll 2} .)
Then f ′ ( x ) = 4 x 3 {\displaystyle f'(x)=4x^{3}} .
We use Newton's method:
x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) = 1 − 1 − 3 4 = 3 2 . {\displaystyle x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}=1-{\frac {1-3}{4}}={\frac {3}{2}}.}
The final answer is
x 2 = x 1 − f ( x 1 ) f ′ ( x 1 ) = 3 2 − ( 3 2 ) 4 − 3 4 ( 3 2 ) 3 . {\displaystyle x_{2}=x_{1}-{\frac {f(x_{1})}{f'(x_{1})}}={\frac {3}{2}}-{\frac {\left({\frac {3}{2}}\right)^{4}-3}{4\left({\frac {3}{2}}\right)^{3}}}.}
![{\displaystyle 1<{\sqrt[{4}]{3}}\ll 2}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/15ee1165d8098d784d4c72834436c299a351238b)
