## Calculus with Applications (10th Edition)

$$0.81$$
\eqalign{ & 4{x^3} - 5{x^2} - 6x + 6 = 0;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {1,2} \right] \cr & {\text{let }}f\left( x \right) = 4{x^3} - 5{x^2} - 6x + 6,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {4{x^3} - 5{x^2} - 6x + 6} \right) \cr & f'\left( x \right) = 12{x^2} - 10x - 6 \cr & {\text{interval }}\left[ {1,2} \right]{\text{ then }}a = 1{\text{ and }}b = 2 \cr & {\text{check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 1 \right) = 4{\left( 1 \right)^3} - 5{\left( 1 \right)^2} - 6\left( 1 \right) + 6 = - 1 < 0 \cr & f'\left( 2 \right) = 4{\left( 2 \right)^3} - 5{\left( 2 \right)^2} - 6\left( 2 \right) + 6 = 6 > 0 \cr & f\left( 1 \right){\text{ and }}f\left( 2 \right){\text{ have opposite signs}}{\text{, then}} \cr & {\text{there is a solution for the equaion in the interval }}\left( {1,2} \right) \cr & {\text{As an initial guess }}{c_1} = 1 \cr & {\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 & {\text{then }} \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{4{{\left( 1 \right)}^3} - 5{{\left( 1 \right)}^2} - 6\left( 1 \right) + 6}}{{12{{\left( 1 \right)}^2} - 10\left( 1 \right) - 6}} = 0.75 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 0.75 - \frac{{4{{\left( {0.75} \right)}^3} - 5{{\left( {0.75} \right)}^2} - 6\left( {0.75} \right) + 6}}{{12{{\left( {0.75} \right)}^2} - 10\left( {0.75} \right) - 6}} = 0.8055 \cr & {\text{In the same way}} \cr & {c_4} = 0.8055 - \frac{{4{{\left( {0.8055} \right)}^3} - 5{{\left( {0.8055} \right)}^2} - 6\left( {0.8055} \right) + 6}}{{12{{\left( {0.8055} \right)}^2} - 10\left( {0.8055} \right) - 6}} = 0.807 \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}} \cr & {\text{A solution for the given interval is }}x \approx 0.81 \cr}