Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 10

$$2.14$$

$$\eqalign{ & 4{x^{1/3}} - 2{x^2} + 4 = 0;\,\,\,{\text{in the interval }}\left[ {0,3} \right] \cr & {\text{let }}f\left( x \right) = 4{x^{1/3}} - 2{x^2} + 4,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {4{x^{1/3}} - 2{x^2} + 4} \right) \cr & f'\left( x \right) = \frac{4}{3}{x^{ - 2/3}} - 4x \cr & {\text{Interval }}\left[ {0,3} \right]{\text{ then }}a = 0{\text{ and }}b = 3.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 0 \right) = 4{\left( 0 \right)^{1/3}} - 2{\left( 0 \right)^2} + 4 = 4 > 0 \cr & f'\left( 3 \right) = 4{\left( 3 \right)^{1/3}} - 2{\left( 3 \right)^2} + 4 \approx - 8 < 0 \cr & f\left( 0 \right){\text{ and }}f\left( 0 \right){\text{ have opposite signs}}{\text{, then}} \cr & \cr & {\text{There is a solution for the equation in the interval }}\left( {0,3} \right){\text{ As an initial guess }}{c_1} = 1.{\text{ }} \cr & {\text{We cannot use as initial value 0 because }}f'\left( 0 \right) = 0.{\text{ Then}} \cr & {\text{A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr & {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr & {\text{then}} \cr & {\text{ }} \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{4{{\left( 1 \right)}^{1/3}} - 2{{\left( 1 \right)}^2} + 4}}{{\frac{4}{3}{{\left( 1 \right)}^{ - 2/3}} - 4\left( 1 \right)}} = 3.25 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 3.25 - \frac{{4{{\left( {3.25} \right)}^{1/3}} - 2{{\left( {3.25} \right)}^2} + 4}}{{\frac{4}{3}{{\left( {3.25} \right)}^{ - 2/3}} - 4\left( {3.25} \right)}} = 2.3462 \cr & {\text{In the same way}} \cr & {c_4} = 2.3462 - \frac{{4{{\left( {2.3462} \right)}^{1/3}} - 2{{\left( {2.3462} \right)}^2} + 4}}{{\frac{4}{3}{{\left( {2.3462} \right)}^{ - 2/3}} - 4\left( {2.3462} \right)}} = 2.1498 \cr & {c_5} = 2.1498 - \frac{{4{{\left( {2.1498} \right)}^{1/3}} - 2{{\left( {2.1498} \right)}^2} + 4}}{{\frac{4}{3}{{\left( {2.1498} \right)}^{ - 2/3}} - 4\left( {2.1498} \right)}} = 2.1394 \cr & {c_6} = 2.1394 - \frac{{4{{\left( {2.1394} \right)}^{1/3}} - 2{{\left( {2.1394} \right)}^2} + 4}}{{\frac{4}{3}{{\left( {2.1394} \right)}^{ - 2/3}} - 4\left( {2.1394} \right)}} = 2.1394 \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}}{\text{. A solution for the given interval is }}x \approx 2.14 \cr} $$