Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 35

$$x \approx 0.179295$$

$$\eqalign{ & f\left( x \right) = {e^{ - x}} - \frac{{x + 4}}{5} \cr & {\text{Let }}f\left( x \right) = 0 \cr & {e^{ - x}} - \frac{{x + 4}}{5} = 0 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - x}} - \frac{{x + 4}}{5}} \right] \cr & f'\left( x \right) = - {e^{ - x}} - \frac{1}{5} \cr & {\text{Use the procedure Newton's Method for Approximations}} \cr & {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 311}}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr & {\text{form}} \cr & {x_{n + 1}} = {x_n} - \frac{{{e^{ - {x_n}}} - \frac{{{x_n} + 4}}{5}}}{{ - {e^{ - {x_n}}} - \frac{1}{5}}} \cr & {x_{n + 1}} = {x_n} - \frac{{5{e^{ - {x_n}}} - \left( {{x_n} + 4} \right)}}{{ - 5{e^{ - {x_n}}} - 1}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = 0 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 0 \cr & {x_1} = 0 - \frac{{5{e^{ - 0}} - \left( {0 + 4} \right)}}{{ - 5{e^{ - 0}} - 1}} = \frac{1}{6} \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 0.179231 \cr & {x_3} \approx 0.179295 \cr & {x_4} \approx 0.179295 \cr & \cr & {\text{The approximation of the root is:}} \cr & x \approx 0.179295 \cr} $$