Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 88

$x \approx 1.146{\text{ and }}x \approx 7.854$

$$\eqalign{ & f\left( x \right) = 3\sqrt {x - 1} - x \cr & {\text{Differentiating}} \cr & f'\left( x \right) = 3\left( {\frac{1}{{2\sqrt {x - 1} }}} \right) - 1 \cr & f'\left( x \right) = \frac{3}{{2\sqrt {x - 1} }} - 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{{3\sqrt {{x_n} - 1} - {x_n}}}{{\frac{3}{{2\sqrt {{x_n} - 1} }} - 1}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} \approx - 1.000,{\text{ the derivative is not defined}} \cr & {\text{at }}x = - 1,{\text{ let }}{x_1} \approx 1.01 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx - 1.010 \cr & {x_2} \approx 1.061 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 1.124 \cr & {x_4} \approx 1.145 \cr & {x_5} \approx 1.146 \cr & {x_6} \approx 1.146 \cr & {\text{The successive approximations }}{x_6}{\text{ and }}{x_5}{\text{ differ by}} \cr & \left| {{x_6} - {x_5}} \right| < 0.001 \cr & {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr & x \approx 1.146 \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx 1.145898034 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}{x_1} \approx 8 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 8.000 \cr & {x_2} \approx 7.855 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 7.854 \cr & {x_4} \approx 7.854 \cr & {\text{The successive approximations }}{x_4}{\text{ and }}{x_3}{\text{ differ by}} \cr & \left| {{x_4} - {x_3}} \right| < 0.001 \cr & {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr & x \approx 7.854 \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx 7.854101966 \cr & \cr & {\text{The solutions are}} \cr & x \approx 1.146{\text{ and }}x \approx 7.854 \cr} $$