Chapter 3 - Applications of Differentiation - 3.8 Exercises - Page 229: 10

$4.828$

$$\eqalign{ & f\left( x \right) = x - 2\sqrt {x + 1} \cr & {\text{Differentiating}} \cr & f'\left( x \right) = 1 - 2\left( {\frac{1}{{2\sqrt {x + 1} }}} \right) \cr & f'\left( x \right) = 1 - \frac{1}{{\sqrt {x + 1} }} \cr & {\text{Using the Newton's Method}} \cr & {\text{The iterative formula is}} \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f\left( {{x_n}} \right){\text{ into }}\left( {\bf{1}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{{x_n} - 2\sqrt {{x_n} + 1} }}{{1 - \frac{1}{{\sqrt {{x_n} + 1} }}}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} = 5 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_2} = 5 - \frac{{5 - 2\sqrt {5 + 1} }}{{1 - \frac{1}{{\sqrt {5 + 1} }}}} \approx 4.82929 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 4.82843 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_3} - {x_2}} \right| \approx 0.00086 < 0.001 \cr & {\text{We can estimate the zero of }}f\left( x \right){\text{ to be }} \cr & x = 4.82843 \cr & {\text{taking three decimal places}} \cr & x \approx 4.828 \cr & \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx 4.828427125 \cr} $$