$\displaystyle f(x) = \frac{1}{2} x^8 + \frac{8}{5} x ^5+ 3 x +10$ near the point $x = 3.$
Use $x_1 = 3$ as the initial approximation. Find the next two approximations, $x_2$ and $x_3$, to four decimal places each.
And I get both of them wrong, why?
I used formula that is derived from linearization:
$x_1 = x_0 - \frac {f(x_0)}{f'(x_0)} $
For the $x_2$ this is how I solve:
$x_2$ = $3 - \frac {f(3)}{f'(3)}$ = $3 - \frac {\frac{36883}{10}}{9399}$ = 2.6076
f(3) = $\frac{36883}{10}$
f'(3) = 9399
but it's not correct, why?