Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 19

$${x_1} \approx 0.062997,{\text{ }}{x_2} \approx 2.230120$$

$$\eqalign{ & y = 4\sqrt x ,{\text{ }}y = {x^2} + 1 \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & 4\sqrt x = {x^2} + 1 \cr & {\text{Subtract }}{x^2} + 1{\text{ from both sides of equation to write the }} \cr & {\text{functions in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr & f\left( x \right) = 0 \cr & 4\sqrt x - {x^2} - 1 = 0 \cr & {\text{Let }}f\left( x \right) = 4\sqrt x - {x^2} - 1,{\text{ differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {4\sqrt x - {x^2} - 1} \right] \cr & f'\left( x \right) = 4\left( {\frac{1}{{2\sqrt x }}} \right) - 2x \cr & f'\left( x \right) = \frac{2}{{\sqrt x }} - 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{{4\sqrt {{x_n}} - x_n^2 - 1}}{{\frac{2}{{\sqrt {{x_n}} }} - 2{x_n}}} \cr & {\text{From the graph we can see that the first 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 0.058378 \cr & n = 1,{\text{ }}{x_1} = 0.058378 \cr & {x_2} \approx \left( {0.058378} \right) - \frac{{\frac{1}{{\left( {0.058378} \right)}} - 4 + {{\left( {0.058378} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.058378} \right)}^2}}} + 2\left( {0.058378} \right)}} \cr & {x_2} \approx 0.062905 \cr & n = 2,{\text{ }}{x_2} = 0.062905 \cr & {x_3} \approx \left( {0.062905} \right) - \frac{{\frac{1}{{\left( {0.062905} \right)}} - 4 + {{\left( {0.062905} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.062905} \right)}^2}}} + 2\left( {0.062905} \right)}} \cr & {x_3} \approx 0.062997 \cr & n = 3,{\text{ }}{x_1} = 0.062997 \cr & {x_4} \approx \left( {0.062997} \right) - \frac{{\frac{1}{{\left( {0.062997} \right)}} - 4 + {{\left( {0.062997} \right)}^2}}}{{ - \frac{1}{{{{\left( {0.062997} \right)}^2}}} + 2\left( {0.062997} \right)}} \cr & {x_4} \approx 0.062997 \cr & \cr & {\text{From the graph we can see that the second 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 2.254024 \cr & n = 1,{\text{ }}{x_1} = 2.254024 \cr & {x_2} \approx \left( {2.254024} \right) - \frac{{\frac{1}{{\left( {2.254024} \right)}} - 4 + {{\left( {2.254024} \right)}^2}}}{{ - \frac{1}{{{{\left( {2.254024} \right)}^2}}} + 2\left( {2.254024} \right)}} \cr & {x_2} \approx 2.230326 \cr & n = 2,{\text{ }}{x_2} = 2.230326 \cr & {x_3} \approx \left( {2.230326} \right) - \frac{{\frac{1}{{\left( {2.230326} \right)}} - 4 + {{\left( {2.230326} \right)}^2}}}{{ - \frac{1}{{{{\left( {2.230326} \right)}^2}}} + 2\left( {2.230326} \right)}} \cr & {x_3} \approx 2.230120 \cr & n = 3,{\text{ }}{x_1} = 2.230120 \cr & {x_4} \approx \left( {2.230120} \right) - \frac{{\frac{1}{{\left( {2.230120} \right)}} - 4 + {{\left( {2.230120} \right)}^2}}}{{ - \frac{1}{{{{\left( {2.230120} \right)}^2}}} + 2\left( {2.230120} \right)}} \cr & {x_4} \approx 2.230120 \cr & \cr & {\text{The approximation of the solutions are:}} \cr & {x_1} \approx 0.062997,{\text{ }}{x_2} \approx 2.230120 \cr} $$