## Calculus 8th Edition

$$f(x)=\frac{2}{x}-x^{2}+1$$ The third approximation to the root of the given equation is $\approx 1.5215$
$$f(x)=\frac{2}{x}-x^{2}+1$$ We apply Newton’s method with $$f(x)=\frac{2}{x}-x^{2}+1$$ and $$f^{\prime}(x)=-\frac{2}{x^{2}}-2 x$$ $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}=x_{n}-\frac{\frac{2}{x_{n}}-x_{n}^{2}+1 }{-\frac{2}{x_{n}^{2}}-2 x_{n}}$$ So $$x_{n+1}=x_{n}-\frac{2 / x_{n}-x_{n}^{2}+1}{-2 / x_{n}^{2}-2 x_{n}}$$ Now, with $n= 1 , x_{1}=2$ we have \begin{aligned} x_{2} &=x_{1}-\frac{2 / x_{1}-x_{1}^{2}+1}{-2 / x_{1}^{2}-2 x_{1}}\\ &=2-\frac{1-4+1}{-1 / 2-4} \\ &=2-\frac{-2}{-9 / 2} \\ &=\frac{14}{9} \end{aligned} with $n= 2 , x_{2}=\frac{14}{9}$ we have \begin{aligned} x_{3}&=x_{2}-\frac{2 / x_{2}-x_{2}^{2}+1}{-2 / x_{2}^{2}-2 x_{2}}\\ &=\frac{14}{9}-\frac{2 /(14 / 9)-(14 / 9)^{2}+1}{-2(14 / 9)^{2}-2(14 / 9)} \\ &\approx 1.5215 \end{aligned}