#### Answer

$$
f(x)=\frac{2}{x}-x^{2}+1
$$
The third approximation to the root of the given equation is
$\approx 1.5215$

#### Work Step by Step

$$
f(x)=\frac{2}{x}-x^{2}+1
$$
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
$$
\begin{aligned}
x_{3}&=x_{2}-\frac{2 / x_{2}-x_{2}^{2}+1}{-2 / x_{2}^{2}-2 x_{2}}\\
&=\frac{14}{9}-\frac{2 /(14 / 9)-(14 / 9)^{2}+1}{-2(14 / 9)^{2}-2(14 / 9)} \\
&\approx 1.5215
\end{aligned}
$$