Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 326: 88

$x \approx 2.208$

$$\eqalign{ & y = \ln x,{\text{ }}y = 3 - x \cr & {\text{Find the intersection points let }}y = y \cr & \ln x = 3 - x \cr & \ln x - 3 + x = 0 \cr & {\text{Let }}f\left( x \right) = \ln x - 3 + x \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{1}{x} + 1 \cr & {\text{Using the Newton's Method}} \cr & {\text{The iterative formula is}} \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f\left( {{x_n}} \right){\text{ into }}\left( {\bf{1}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{\ln {x_n} - 3 + {x_n}}}{{\frac{1}{{{x_n}}} + 1}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} \approx 2.2 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 2.2 \cr & {x_2} = 0.6 - \frac{{\ln \left( {0.6} \right) - 3 + \left( {0.6} \right)}}{{\frac{1}{{0.6}} + 1}} \cr & {x_2} \approx 2.208 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 2.208 \cr & {\text{The successive approximations }}{x_3}{\text{ and }}{x_2}{\text{ differ by}} \cr & \left| {{x_3} - {x_2}} \right| \approx 0 < 0.001 \cr & {\text{We can estimate the intersect point is}} \cr & x \approx 2.208 \cr & \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx 2.207940032 \cr} $$