# Newton's Method

f(x) f'(x) Starting x value


$$n$$ $$x_n$$ $$f(x_n)$$ $$f'(x_n)$$ $$x_n - {f(x_n) \over f'(x_n)}$$

Note: Parentheses are sometimes removed in the function display. So $${x^2 - \left(x - 1\right)}$$ will show up as $${x^2 - {{x - 1}}},$$ but will still work as intended.