Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 355: 28

$$\eqalign{ & 0.24852414 \cr & 4.05010984 \cr} $$

$$\eqalign{ & \ln \left( {{x^2} + 2} \right) = \frac{{3x}}{{\sqrt {{x^2} + 1} }} \cr & {\text{Subtract }}\frac{{3x}}{{\sqrt {{x^2} + 1} }}{\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 & \ln \left( {{x^2} + 2} \right) - \frac{{3x}}{{\sqrt {{x^2} + 1} }} = 0 \cr & {\text{Let }}f\left( x \right) = \ln \left( {{x^2} + 2} \right) - \frac{{3x}}{{\sqrt {{x^2} + 1} }} \cr & f\left( x \right) = 0 \cr & {\text{The graph of the equation is shown below}} \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {{x^2} + 2} \right) - \frac{{3x}}{{\sqrt {{x^2} + 1} }}} \right] \cr & f'\left( x \right) = \frac{{2x}}{{{x^2} + 2}} - \frac{{3\sqrt {{x^2} + 1} - 3x\left( {\frac{{2x}}{{2\sqrt {{x^2} + 1} }}} \right)}}{{{x^2} + 1}} \cr & f'\left( x \right) = \frac{{2x}}{{{x^2} + 2}} - \frac{{3\left( {{x^2} + 1} \right) - 3{x^2}}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr & f'\left( x \right) = \frac{{2x}}{{{x^2} + 2}} - \frac{3}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr & {\text{Use the procedure Newton's Method for Approximations}} \cr & {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 352}}} \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{{\ln \left( {x_n^2 + 2} \right) - \frac{{3{x_n}}}{{\sqrt {x_n^2 + 1} }}}}{{\frac{{2{x_n}}}{{x_n^2 + 2}} - \frac{3}{{{{\left( {x_n^2 + 1} \right)}^{3/2}}}}}} \cr & \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x \approx 0 \cr & {\text{If we choose }}{x_1} \approx 0{\text{ as the initial approximation}},{\text{then }} \cr & {\text{we obtain:}} \cr & {x_2} \approx 0.23104906 \cr & {x_3} \approx 0.24835770 \cr & {x_4} \approx 0.24852412 \cr & {x_5} \approx 0.24852414 \cr & {x_6} \approx 0.24852414 \cr & {x_6}{\text{ and }}{x_5}{\text{ are agree to eight decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation is }} \approx 0.24852414 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x \approx 4 \cr & {\text{If we choose }}{x_1} \approx 4{\text{ as the initial approximation}},{\text{then }} \cr & {\text{we obtain:}} \cr & {x_2} \approx 4.04993412 \cr & {x_3} \approx 4.05010983 \cr & {x_4} \approx 4.05010984 \cr & {x_5} \approx 4.05010984 \cr & {x_5}{\text{ and }}{x_4}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation is }} \approx 4.05010984 \cr} $$