Answer
$x$ $=$ $3.732051$ $or$ $0.267949$
Work Step by Step
$x$ $=$ $x^{2}$$-$$3x$ $+$$1$
$x_{n+1}$ $=$ $x_{n}$ $-$ $\frac{f(x)}{f^{'}(x)}$
$f(x)$ $=$ $x$ $-$ $x^{2}$ $+$ $3x$ $-$ $1$
${f^{'}(x)}$ $=$ $-2x$ $+$ $4$
$Let$ $x_{1}$ $=$ $3$
$x_{2}$ $=$ $3$ $-$ $\frac{f(1)}{f'(1)}$ $=$ $4$
$x_{3}$ $=$ $4$ $-$ $\frac{f(2)}{f'(2)}$ $=$ $3.75$
$x_{4}$ $=$ $3.75$ $-$ $\frac{f(3)}{f'(3)}$ $=$ $3.73214$
$x_{5}$ $=$ $3.73214$ $-$ $\frac{f(4)}{f'(4)}$ $=$ $3.7320508$
$x_{6}$ $=$ $3.7320508$ $-$ $\frac{f(5)}{f'(5)}$ $=$ $3.7320508$ $=$ $x_{5}$
$Let$ $x_{1}$ $=$ $0.3$
$x_{2}$ $=$ $0.3$ $-$ $\frac{f(1)}{f'(1)}$ $=$ $0.267647$
$x_{3}$ $=$ $0.267647$ $-$ $\frac{f(2)}{f'(2)}$ $=$ $0.26794917$
$x_{4}$ $=$ $0.26794917$ $-$ $\frac{f(3)}{f'(3)}$ $=$ $0.26794919$
$x_{5}$ $=$ $0.26794919$ $-$ $\frac{f(4)}{f'(4)}$ $=$ $0.26794919$ $=$ $x_{4}$
$ Keep$ $six$ $decimal$ $places$
$So,$ $x$ $=$ $3.732051$ $or$ $0.267949$