Answer
$${x_1} \approx 1 - 2.114908,{\text{ }}{x_2} \approx 0.254102,{\text{ }}{x_3} \approx 1.860806$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{x},{\text{ }}y = 4 - {x^2} \cr
& {\text{The graphs of the functions are shown below}} \cr
& \cr
& {\text{Find the intersection points}}{\text{, let }}y = y \cr
& \frac{1}{x} = 4 - {x^2} \cr
& {\text{Subtract }}4 - {x^2}{\text{ from both sides of equation to write the functions}} \cr
& {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr
& f\left( x \right) = 0 \cr
& \frac{1}{x} - 4 + {x^2} = 0 \cr
& {\text{Let }}f\left( x \right) = \frac{1}{x} - 4 - {x^2},{\text{ differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{x} - 4 + {x^2}} \right] \cr
& f'\left( x \right) = - \frac{1}{{{x^2}}} + 2x \cr
& {\text{Use the procedure Newton's Method for Approximations}} \cr
& {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 311}}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr
& {\text{form}} \cr
& {x_{n + 1}} = {x_n} - \frac{{\frac{1}{{{x_n}}} - 4 + x_n^2}}{{ - \frac{1}{{x_n^2}} + 2{x_n}}} \cr
& {\text{From the graph we can see that the first possible initial }} \cr
& {\text{approximation is }}x = - 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& n = 0,{\text{ }}{x_0} = - 2 \cr
& {x_{n + 1}} = \left( { - 2} \right) - \frac{{\frac{1}{{\left( { - 2} \right)}} - 4 + {{\left( { - 2} \right)}^2}}}{{ - \frac{1}{{{{\left( { - 2} \right)}^2}}} + 2\left( { - 2} \right)}} \cr
& n = 1,{\text{ }}{x_1} = - 2.117647 \cr
& {x_2} \approx \left( { - 2.117647} \right) - \frac{{\frac{1}{{\left( { - 2.117647} \right)}} - 4 + {{\left( { - 2.117647} \right)}^2}}}{{ - \frac{1}{{{{\left( { - 2.117647} \right)}^2}}} + 2\left( { - 2.117647} \right)}} \cr
& {x_2} \approx - 2.114909 \cr
& n = 2,{\text{ }}{x_2} = - 2.114909 \cr
& {x_3} \approx \left( { - 2.114909} \right) - \frac{{\frac{1}{{\left( { - 2.114909} \right)}} - 4 + {{\left( { - 2.114909} \right)}^2}}}{{ - \frac{1}{{{{\left( { - 2.114909} \right)}^2}}} + 2\left( { - 2.114909} \right)}} \cr
& {x_3} \approx - 2.114908 \cr
& \cr
& {\text{From the graph we can see that the second possible initial }} \cr
& {\text{approximation is }}x = 0.1 \cr
& {\text{Therefore}}{\text{,}} \cr
& n = 0,{\text{ }}{x_0} = 0.1 \cr
& {x_1} \approx \left( {0.1} \right) - \frac{{\frac{1}{{\left( {0.1} \right)}} - 4 + {{\left( {0.1} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.1} \right)}^2}}} + 2\left( {0.1} \right)}} \approx 0.160220 \cr
& n = 1,{\text{ }}{x_1} = 0.160220 \cr
& {x_2} \approx \left( {0.160220} \right) - \frac{{\frac{1}{{\left( {0.160220} \right)}} - 4 + {{\left( {0.160220} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.160220} \right)}^2}}} + 2\left( {0.160220} \right)}} \cr
& {x_2} \approx 0.218899 \cr
& n = 2,{\text{ }}{x_1} = 0.218899 \cr
& {x_3} \approx \left( {0.218899} \right) - \frac{{\frac{1}{{\left( {0.218899} \right)}} - 4 + {{\left( {0.218899} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.218899} \right)}^2}}} + 2\left( {0.218899} \right)}} \cr
& {x_3} \approx 0.249059 \cr
& n = 3,{\text{ }}{x_1} = 0.249059 \cr
& {x_4} \approx \left( {0.249059} \right) - \frac{{\frac{1}{{\left( {0.249059} \right)}} - 4 + {{\left( {0.249059} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.249059} \right)}^2}}} + 2\left( {0.249059} \right)}} \cr
& {x_4} \approx 0.253996 \cr
& n = 4,{\text{ }}{x_1} = 0.253996 \cr
& {x_5} \approx \left( {0.253996} \right) - \frac{{\frac{1}{{\left( {0.253996} \right)}} - 4 + {{\left( {0.253996} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.253996} \right)}^2}}} + 2\left( {0.253996} \right)}} \cr
& {x_5} \approx 0.254102 \cr
& {\text{From the graph we can see that the third possible initial }} \cr
& {\text{approximation is }}x = 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& n = 0,{\text{ }}{x_0} = 2 \cr
& {x_1} \approx \left( 2 \right) - \frac{{\frac{1}{{\left( 2 \right)}} - 4 + {{\left( 2 \right)}^2}}}{{ - \frac{1}{{{{\left( 2 \right)}^2}}} + 2\left( 2 \right)}} \approx 1.866666 \cr
& n = 1,{\text{ }}{x_1} = 1.866666 \cr
& {x_2} \approx \left( {1.866666} \right) - \frac{{\frac{1}{{\left( {1.866666} \right)}} - 4 + {{\left( {1.866666} \right)}^2}}}{{ - \frac{1}{{{{\left( {1.866666} \right)}^2}}} + 2\left( {1.866666} \right)}} \cr
& {x_2} \approx 1.860817 \cr
& n = 2,{\text{ }}{x_1} = 1.860817 \cr
& {x_3} \approx \left( {1.860817} \right) - \frac{{\frac{1}{{\left( {1.860817} \right)}} - 4 + {{\left( {1.860817} \right)}^2}}}{{ - \frac{1}{{{{\left( {1.860817} \right)}^2}}} + 2\left( {1.860817} \right)}} \cr
& {x_3} \approx 1.860805 \cr
& n = 3,{\text{ }}{x_1} = 1.860805 \cr
& {x_4} \approx \left( {1.860805} \right) - \frac{{\frac{1}{{\left( {1.860805} \right)}} - 4 + {{\left( {1.860805} \right)}^2}}}{{ - \frac{1}{{{{\left( {1.860805} \right)}^2}}} + 2\left( {1.860805} \right)}} \cr
& {x_4} \approx 1.860806 \cr
& \cr
& {\text{The approximation of the solutions are:}} \cr
& {x_1} \approx 1 - 2.114908,{\text{ }}{x_2} \approx 0.254102,{\text{ }}{x_3} \approx 1.860806 \cr} $$
