Chapter 3 - The Derivative In Graphing And Applications - 3.7 Newton's Method - Exercises Set 3.7 - Page 250: 8

$$x \approx - 1.495771348$$

$$\eqalign{ & {x^5} - 3x + 3 = 0 \cr & {\text{Let }}f\left( x \right) = {x^5} - 3x + 3,{\text{ so }}f'\left( x \right) = 5{x^4} - 3 \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^5 - 3{x_n} + 3}}{{5x_n^4 - 3}} \cr & \cr & {\text{Evaluating }}f\left( { - 2} \right){\text{ and }}f\left( { - 1} \right) \cr & f\left( { - 1} \right) = {\left( { - 1} \right)^5} - 3\left( { - 1} \right) + 3 = 5 \cr & f\left( { - 2} \right) = {\left( { - 2} \right)^5} - 3\left( { - 2} \right) + 3 = - 23 \cr & {\text{Then}}{\text{, Thus this solution lies between }} - 2{\text{ and }} - 1 \cr & {\text{We will use }}{x_1}{\text{ = }} - 1{\text{ as our first approximation}} \cr & {x_1} = - 1 \cr & {x_{1 + 1}} = {x_2} = - 1 - \frac{{{{\left( { - 1} \right)}^5} - 3\left( { - 1} \right) + 3}}{{5{{\left( { - 1} \right)}^4} - 3}} = - 3.5 \cr & {x_{2 + 1}} = {x_3} = - 3.5 - \frac{{{{\left( { - 3.5} \right)}^5} - 3\left( { - 3.5} \right) + 3}}{{5{{\left( { - 3.5} \right)}^4} - 3}} \approx - 2.815254663 \cr & {x_{3 + 1}} = \cr & {x_4} = - 2.815254663 - \frac{{{{\left( { - 2.815254663} \right)}^5} - 3\left( { - 2.815254663} \right) + 3}}{{5{{\left( { - 2.815254663} \right)}^4} - 3}} \approx - 2.283567368 \cr & {x_{4 + 1}} = \cr & {x_5} = - 2.283567368 - \frac{{{{\left( { - 2.283567368} \right)}^5} - 3\left( { - 2.283567368} \right) + 3}}{{5{{\left( { - 2.283567368} \right)}^4} - 3}} \approx - 1.890634572 \cr & {x_{5 + 1}} = \cr & {x_6} = - 1.890634572 - \frac{{{{\left( { - 1.890634572} \right)}^5} - 3\left( { - 1.890634572} \right) + 3}}{{5{{\left( { - 1.890634572} \right)}^4} - 3}} \approx - 1.636306556 \cr & {x_{6 + 1}} = \cr & {x_7} = - 1.636306556 - \frac{{{{\left( { - 1.636306556} \right)}^5} - 3\left( { - 1.636306556} \right) + 3}}{{5{{\left( { - 1.636306556} \right)}^4} - 3}} \approx - 1.519949004 \cr & {x_{7 + 1}} = \cr & {x_8} = - 1.519949004 - \frac{{{{\left( { - 1.519949004} \right)}^5} - 3\left( { - 1.519949004} \right) + 3}}{{5{{\left( { - 1.519949004} \right)}^4} - 3}} \approx - 1.496624276 \cr & {x_{8 + 1}} = \cr & {x_9} = - 1.496624276 - \frac{{{{\left( { - 1.496624276} \right)}^5} - 3\left( { - 1.496624276} \right) + 3}}{{5{{\left( { - 1.496624276} \right)}^4} - 3}} \approx - 1.495772451 \cr & {x_{9 + 1}} = \cr & {x_{10}} = - 1.495772451 - \frac{{{{\left( { - 1.495772451} \right)}^5} - 3\left( { - 1.495772451} \right) + 3}}{{5{{\left( { - 1.495772451} \right)}^4} - 3}} \approx - 1.495771348 \cr & {x_{10 + 1}} = \cr & {x_{11}} = - 1.495771348 - \frac{{{{\left( { - 1.495771348} \right)}^5} - 3\left( { - 1.495771348} \right) + 3}}{{5{{\left( { - 1.495771348} \right)}^4} - 3}} \approx - 1.495771348 \cr & \cr & {\text{We obtain that }}{x_{10}} \approx {x_1} \cr & {\text{Thus}}{\text{, the real solution is:}} \cr & x \approx - 1.495771348 \cr} $$