Chapter 4 - Applications of the Derivative - Review Exercises - Page 331: 36

$${x_1} = - 0.816162,{\text{ }}{x_2} \approx 1.079172$$

$$\eqalign{ & f\left( x \right) = {e^{ - 2x}} + 2{e^x} - 6 \cr & {\text{Let }}f\left( x \right) = 0 \cr & {e^{ - 2x}} + 2{e^x} - 6 = 0 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - 2x}} + 2{e^x} - 6} \right] \cr & f'\left( x \right) = - 2{e^{ - 2x}} + 2{e^x} \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^{ - 2{x_n}}} + 2{e^{{x_n}}} - 6}}{{ - 2{e^{ - 2{x_n}}} + 2{e^{{x_n}}}}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = - 1 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = - 1 \cr & {x_{n + 1}} = \left( { - 1} \right) - \frac{{{e^{ - 2\left( { - 1} \right)}} + 2{e^{\left( { - 1} \right)}} - 6}}{{ - 2{e^{ - 2\left( { - 1} \right)}} + 2{e^{\left( { - 1} \right)}}}} \approx - 0.845685 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx - 0.817331 \cr & {x_3} \approx - 0.816164 \cr & {x_4} \approx - 0.816162 \cr & {x_5} \approx - 0.816162 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 1 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 1 \cr & {x_{n + 1}} = \left( 1 \right) - \frac{{{e^{ - 2\left( 1 \right)}} + 2{e^{\left( 1 \right)}} - 6}}{{ - 2{e^{ - 2\left( 1 \right)}} + 2{e^{\left( 1 \right)}}}} \approx 1.082871 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 1.079179 \cr & {x_3} \approx 1.079172 \cr & {x_4} \approx 1.079172 \cr & \cr & {\text{The approximation of the roots are:}} \cr & {x_1} = - 0.816162,{\text{ }}{x_2} \approx 1.079172 \cr} $$