Chapter 3 - Applications of Differentiation - 3.8 Newton's Method - 3.8 Exercises - Page 276: 8

$x_{3}\approx-1.2917$

Let $f(x)=x^{7}+4$, $f'(x)=7x^{6}$ We have $x_{1}=-1$, apply Newton's Method: $x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{1})}= -1-\frac{(-1)^{7}+4}{7\times(-1)^{6}}=\frac{-10}{7}$ $x_{3}=x_{2}-\frac{f(x_{2})}{f'(x_{2})}=\frac{-10}{7}-\frac{(\frac{-10}{7})^{7}+4}{7\times(\frac{-10}{7})^{6}}\approx-1.2917$