## Calculus with Applications (10th Edition)

$$3.87$$
\eqalign{ & {x^4} + {x^3} - 14{x^2} - 15x - 15 = 0;\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {3,4} \right] \cr & {\text{L et }}f\left( x \right) = {x^4} + {x^3} - 14{x^2} - 15x - 15.{\text{ So that}}{\text{,}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{x^4} + {x^3} - 14{x^2} - 15x - 15} \right) \to f'\left( x \right) = 4{x^3} + 3{x^2} - 28x - 15 \cr & \cr & {\text{Interval }}\left[ {3,4} \right]{\text{ then }}a = 3{\text{ and }}b = 4.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 3 \right) = {\left( 3 \right)^4} + {\left( 3 \right)^3} - 14{\left( 3 \right)^2} - 15\left( 3 \right) - 15 = - 78 < 0 \cr & f\left( 4 \right) = {\left( 4 \right)^4} + {\left( 4 \right)^3} - 14{\left( 4 \right)^2} - 15\left( 4 \right) - 15 = 21 > 0 \cr & f\left( 3 \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( {3,4} \right) \cr & \cr & {\text{As an initial guess }}{c_1} = 3{\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)}} = 3 - \frac{{{{\left( 3 \right)}^4} + {{\left( 3 \right)}^3} - 14{{\left( 3 \right)}^2} - 15\left( 3 \right) - 15}}{{4{{\left( 3 \right)}^3} + 3{{\left( 3 \right)}^2} - 28\left( 3 \right) - 15}} \approx 5.166 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 5.166 - \frac{{{{\left( {5.166} \right)}^4} + {{\left( {5.166} \right)}^3} - 14{{\left( {5.166} \right)}^2} - 15\left( {5.166} \right) - 15}}{{4{{\left( {5.166} \right)}^3} + 3{{\left( {5.166} \right)}^2} - 28\left( {5.166} \right) - 15}} \approx 4.352 \cr & {\text{In the same way}} \cr & {c_4} = 4.352 - \frac{{{{\left( {4.352} \right)}^4} + {{\left( {4.352} \right)}^3} - 14{{\left( {4.352} \right)}^2} - 15\left( {4.352} \right) - 15}}{{4{{\left( {4.352} \right)}^3} + 3{{\left( {4.352} \right)}^2} - 28\left( {4.352} \right) - 15}} \approx 3.968 \cr & {c_5} = 3.968 - \frac{{{{\left( {3.968} \right)}^4} + {{\left( {3.968} \right)}^3} - 14{{\left( {3.968} \right)}^2} - 15\left( {3.968} \right) - 15}}{{4{{\left( {3.968} \right)}^3} + 3{{\left( {3.968} \right)}^2} - 28\left( {3.968} \right) - 15}} \approx 3.877 \cr & {c_6} = 3.877 - \frac{{{{\left( {3.877} \right)}^4} + {{\left( {3.877} \right)}^3} - 14{{\left( {3.877} \right)}^2} - 15\left( {3.877} \right) - 15}}{{4{{\left( {3.877} \right)}^3} + 3{{\left( {3.877} \right)}^2} - 28\left( {3.877} \right) - 15}} \approx 3.872 \cr & {c_7} = 3.872 - \frac{{{{\left( {3.872} \right)}^4} + {{\left( {3.872} \right)}^3} - 14{{\left( {3.872} \right)}^2} - 15\left( {3.872} \right) - 15}}{{4{{\left( {3.872} \right)}^3} + 3{{\left( {3.872} \right)}^2} - 28\left( {3.872} \right) - 15}} \approx 3.872 \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 3.87 \cr}