## Calculus with Applications (10th Edition)

$$1.81$$
\eqalign{ & 2\ln x + x - 3 = 0;\,\,\,\,\,{\text{in the interval }}\left[ {1,4} \right] \cr & {\text{Let }}f\left( x \right) = 2\ln x + x - 3,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {2\ln x + x - 3} \right) \cr & f'\left( x \right) = \frac{2}{x} + 1 \cr & {\text{Interval }}\left[ {1,4} \right]{\text{ then }}a = 1{\text{ and }}b = 4.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 1 \right) = 2\ln \left( 1 \right) + \left( 1 \right) - 3 = - 2 < 0 \cr & f\left( 4 \right) = 2\ln \left( 4 \right) + \left( 4 \right) - 3 \approx 3.77 > 0 \cr & f\left( 1 \right){\text{ and }}f\left( 4 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr & {\text{There is a solution for the equation in the interval }}\left( {1,4} \right) \cr & {\text{As an initial guess }}{c_1} = 1.{\text{ A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr & {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{2\ln \left( 1 \right) + \left( 1 \right) - 3}}{{2/\left( 1 \right) + 1}} \approx 1.666 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 1.666 - \frac{{2\ln \left( {1.666} \right) + \left( {1.666} \right) - 3}}{{2/\left( {1.666} \right) + 1}} \approx 1.808 \cr & {\text{In the same way}} \cr & {c_4} = 1.808 - \frac{{2\ln \left( {1.808} \right) + \left( {1.808} \right) - 3}}{{2/\left( {1.808} \right) + 1}} \approx 1.811 \cr & {c_5} = 1.811 - \frac{{2\ln \left( {1.811} \right) + \left( {1.811} \right) - 3}}{{2/\left( {1.811} \right) + 1}} \approx 1.811 \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}}{\text{. A solution for the given interval is }}x \approx 1.81 \cr}