Chapter 3 - The Derivative In Graphing And Applications - 3.7 Newton's Method - Exercises Set 3.7 - Page 250: 6

$$x \approx 0.6823278038$$

$$\eqalign{ & {x^3} + x - 1 = 0 \cr & {\text{Let }}f\left( x \right) = {x^3} + x - 1,{\text{ so }}f'\left( x \right) = 2x + 1 \cr & {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ we have}} \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr & {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{x_n^3 + {x_n} - 1}}{{3x_n^2 + 1}} \cr & \cr & {\text{Evaluating }}f\left( 0 \right){\text{ and }}f\left( 1 \right) \cr & f\left( 0 \right) = {\left( 1 \right)^3} + \left( 0 \right) - 1 = - 1 \cr & f\left( 1 \right) = {\left( 1 \right)^3} + \left( 1 \right) - 1 = 1 \cr & {\text{Then}}{\text{, Thus this solution lies between 0 and 1}} \cr & {\text{We will use }}{x_1}{\text{ = 0 as our first approximation}} \cr & {x_1} = 0 \cr & {x_{1 + 1}} = {x_2} = 0 - \frac{{{{\left( 0 \right)}^3} + \left( 0 \right) - 1}}{{3{{\left( 0 \right)}^2} + 1}} = 1 \cr & {x_{2 + 1}} = {x_3} = 1 - \frac{{{{\left( 1 \right)}^3} + \left( 1 \right) - 1}}{{3{{\left( 1 \right)}^2} + 1}} = 0.75 \cr & {x_{3 + 1}} = {x_4} = 0.75 - \frac{{{{\left( {0.75} \right)}^3} + \left( {0.75} \right) - 1}}{{3{{\left( {0.75} \right)}^2} + 1}} \approx 0.6860465116 \cr & {x_{4 + 1}} = \cr & {x_5} = 0.6860465116 - \frac{{{{\left( {0.6860465116} \right)}^3} + \left( {0.6860465116} \right) - 1}}{{3{{\left( {0.6860465116} \right)}^2} + 1}} \approx 0.6823395826 \cr & {x_{5 + 1}} = \cr & {x_6} = 0.6823395826 - \frac{{{{\left( {0.6823395826} \right)}^3} + \left( {0.6823395826} \right) - 1}}{{3{{\left( {0.6823395826} \right)}^2} + 1}} \approx 0.6823278039 \cr & {x_{6 + 1}} = \cr & {x_7} = 0.6823278039 - \frac{{{{\left( {0.6823278039} \right)}^3} + \left( {0.6823278039} \right) - 1}}{{3{{\left( {0.6823278039} \right)}^2} + 1}} \approx 0.6823278038 \cr & {x_{7 + 1}} = \cr & {x_8} = 0.6823278038 - \frac{{{{\left( {0.6823278038} \right)}^3} + \left( {0.6823278038} \right) - 1}}{{3{{\left( {0.6823278038} \right)}^2} + 1}} \approx 0.6823278038 \cr & {\text{We obtain that }}{x_7} \approx {x_8} \cr & {\text{Thus}}{\text{, the real solution is:}} \cr & x \approx 0.6823278038 \cr} $$