Chapter 3 - The Derivative In Graphing And Applications - 3.7 Newton's Method - Exercises Set 3.7 - Page 250: 7

$$x \approx 1.22443955$$

$$\eqalign{ & {x^5} + {x^4} - 5 = 0 \cr & {\text{Let }}f\left( x \right) = {x^5} + {x^4} - 5,{\text{ so }}f'\left( x \right) = 5{x^4} + 4{x^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 + x_n^4 - 5}}{{5x_n^4 + 4x_n^3}} \cr & \cr & {\text{Evaluating }}f\left( 1 \right){\text{ and }}f\left( 2 \right) \cr & f\left( 1 \right) = {\left( 1 \right)^5} + {\left( 1 \right)^4} - 5 = - 3 \cr & f\left( 2 \right) = {\left( 2 \right)^5} + {\left( 2 \right)^4} - 5 = 43 \cr & {\text{Then}}{\text{, Thus this solution lies between 1 and 2}} \cr & {\text{We will use }}{x_1}{\text{ = 1 as our first approximation}} \cr & {x_1} = 1 \cr & {x_{1 + 1}} = {x_2} = 1 - \frac{{{{\left( 1 \right)}^5} + {{\left( 1 \right)}^4} - 5}}{{5{{\left( 1 \right)}^4} + 4{{\left( 1 \right)}^3}}} \approx 1.333333333 \cr & {x_{2 + 1}} = \cr & {x_3} = 1.333333333 - \frac{{{{\left( {1.333333333} \right)}^5} + {{\left( {1.333333333} \right)}^4} - 5}}{{5{{\left( {1.333333333} \right)}^4} + 4{{\left( {1.333333333} \right)}^3}}} \approx 1.239420573 \cr & {x_{3 + 1}} = \cr & {x_4} = 1.239420573 - \frac{{{{\left( {1.239420573} \right)}^5} + {{\left( {1.239420573} \right)}^4} - 5}}{{5{{\left( {1.239420573} \right)}^4} + 4{{\left( {1.239420573} \right)}^3}}} \approx 1.224762689 \cr & {x_{4 + 1}} = \cr & {x_5} = 1.224762689 - \frac{{{{\left( {1.224762689} \right)}^5} + {{\left( {1.224762689} \right)}^4} - 5}}{{5{{\left( {1.224762689} \right)}^4} + 4{{\left( {1.224762689} \right)}^3}}} \approx 1.224439704 \cr & {x_{5 + 1}} = \cr & {x_6} = 1.224439704 - \frac{{{{\left( {1.224439704} \right)}^5} + {{\left( {1.224439704} \right)}^4} - 5}}{{5{{\left( {1.224439704} \right)}^4} + 4{{\left( {1.224439704} \right)}^3}}} \approx 1.22443955 \cr & {x_{6 + 1}} = \cr & {x_7} = 1.22443955 - \frac{{{{\left( {1.22443955} \right)}^5} + {{\left( {1.22443955} \right)}^4} - 5}}{{5{{\left( {1.22443955} \right)}^4} + 4{{\left( {1.22443955} \right)}^3}}} \approx 1.22443955 \cr & {\text{We obtain that }}{x_6} \approx {x_7} \cr & {\text{Thus}}{\text{, the real solution is:}} \cr & x \approx 1.22443955 \cr} $$