Chapter 3 - Applications of Differentiation - 3.8 Exercises - Page 229: 9

$ 1.250{\text{ and }} 5.000$

$$\eqalign{ & f\left( x \right) = 5\sqrt {x - 1} - 2x \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{5}{{2\sqrt {x - 1} }} - 2 \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{{5\sqrt {{x_n} - 1} - 2{x_n}}}{{\frac{5}{{2\sqrt {{x_n} - 1} }} - 2}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} = 1,{\text{ but the derivative is not defined}} \cr & {\text{at }}x = 1,{\text{ then set }}{x_1} = 1.1 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_2} = 1 - \frac{{5\sqrt {1.1 - 1} - {{\left( {1.1} \right)}_n}}}{{\frac{5}{{2\sqrt {1.1 - 1} }} - 2}} \approx 1.20479 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 1.26647 \cr & {x_4} \approx 1.24998 \cr & {x_5} \approx 1.25000 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_4} - {x_3}} \right| \approx 0.00002 < 0.001 \cr & {\text{We can estimate the zero of }}f\left( x \right){\text{ to be }} \cr & x = 1.250 \cr & {\text{Using a graphing utility we obtain}} \cr & x = 1.250000000 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}{x_1} = 5 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_2} = 1 - \frac{{5\sqrt {5 - 1} - {{\left( 5 \right)}_n}}}{{\frac{5}{{2\sqrt {5 - 1} }} - 2}} = 5.000 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_2} - {x_1}} \right| = 0{\text{ }}\left( {{\text{exact solution}}} \right) \cr & x = 5.000 \cr & {\text{Using a graphing utility we obtain}} \cr & x = 5.0000000 \cr & {\text{The solutions are:}} \cr & x = 1.250{\text{ and }}x = 5.000 \cr} $$