## Calculus with Applications (10th Edition)

$$2.33$$
\eqalign{ & 3{x^3} - 4{x^2} - 4x - 7 = 0;\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {2,3} \right] \cr & {\text{Let }}f\left( x \right) = 3{x^3} - 4{x^2} - 4x - 7.{\text{ So that}}{\text{,}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {3{x^3} - 4{x^2} - 4x - 7} \right) \to f'\left( x \right) = 9{x^2} - 8x - 4 \cr & \cr & {\text{Interval }}\left[ {2,3} \right]{\text{ then }}a = 2{\text{ and }}b = 3.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 2 \right) = 3{\left( 2 \right)^3} - 4{\left( 2 \right)^2} - 4\left( 2 \right) - 7 = - 7 < 0 \cr & f\left( 3 \right) = 3{\left( 3 \right)^3} - 4{\left( 3 \right)^2} - 4\left( 3 \right) - 7 = 26 > 0 \cr & f\left( 2 \right){\text{ and }}f\left( 3 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr & {\text{There is a solution for the equation in the interval }}\left( {2,3} \right) \cr & \cr & {\text{As an initial guess }}{c_1} = 2.{\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)}} = 2 - \frac{{3{{\left( 2 \right)}^3} - 4{{\left( 2 \right)}^2} - 4\left( 2 \right) - 7}}{{9{{\left( 2 \right)}^2} - 8\left( 2 \right) - 4}} \approx 2.437 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 2.437 - \frac{{3{{\left( {2.437} \right)}^3} - 4{{\left( {2.437} \right)}^2} - 4\left( {2.437} \right) - 7}}{{9{{\left( {2.437} \right)}^2} - 8\left( {2.437} \right) - 4}} \approx 2.339 \cr & {\text{In the same way}} \cr & {c_4} = 2.339 - \frac{{3{{\left( {2.339} \right)}^3} - 4{{\left( {2.339} \right)}^2} - 4\left( {2.339} \right) - 7}}{{9{{\left( {2.339} \right)}^2} - 8\left( {2.339} \right) - 4}} \approx 2.333 \cr & {c_5} = 2.333 - \frac{{3{{\left( {2.333} \right)}^3} - 4{{\left( {2.333} \right)}^2} - 4\left( {2.333} \right) - 7}}{{9{{\left( {2.333} \right)}^2} - 8\left( {2.333} \right) - 4}} \approx 2.333 \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 2.33 \cr}