Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 9

$$ - 0.58$$

$$\eqalign{ & 4{x^{1/3}} - 2{x^2} + 4 = 0{\text{ in the interval }}\left[ { - 3,0} \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[ { - 3,0} \right]{\text{ then }}a = - 3{\text{ and }}b = 0.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f'\left( { - 3} \right) = 4{\left( { - 3} \right)^{1/3}} - 2{\left( { - 3} \right)^2} + 4 \approx - 19 < 0 \cr & f\left( 0 \right) = 4{\left( 0 \right)^{1/3}} - 2{\left( 0 \right)^2} + 4 = 4 > 0 \cr & f\left( { - 3} \right){\text{ and }}f\left( 0 \right){\text{ have opposite signs}}{\text{, then there is a solution for the equation }} \cr & {\text{in the interval }}\left( { - 3,0} \right). \cr & {\text{As an initial guess }}{c_1} = - 3,{\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 & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = - 3 - \frac{{4{{\left( { - 3} \right)}^{1/3}} - 2{{\left( { - 3} \right)}^2} + 4}}{{\frac{4}{3}{{\left( { - 3} \right)}^{ - 2/3}} - 4\left( { - 3} \right)}} = - 1.4361 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = - 1.4361 - \frac{{4{{\left( { - 1.4361} \right)}^{1/3}} - 2{{\left( { - 1.4361} \right)}^2} + 4}}{{\frac{4}{3}{{\left( { - 1.4361} \right)}^{ - 2/3}} - 4\left( { - 1.4361} \right)}} = - 0.7532 \cr & {\text{In the same way}} \cr & {c_4} = - 0.7532 - \frac{{4{{\left( { - 0.7532} \right)}^{1/3}} - 2{{\left( { - 0.7532} \right)}^2} + 4}}{{\frac{4}{3}{{\left( { - 0.7532} \right)}^{ - 2/3}} - 4\left( { - 0.7532} \right)}} = - 0.5857 \cr & {c_5} = - 0.5857 - \frac{{4{{\left( { - 0.5857} \right)}^{1/3}} - 2{{\left( { - 0.5857} \right)}^2} + 4}}{{\frac{4}{3}{{\left( { - 0.5857} \right)}^{ - 2/3}} - 4\left( { - 0.5857} \right)}} = - 0.5779 \cr & {c_6} = - 0.5779 - \frac{{4{{\left( { - 0.5779} \right)}^{1/3}} - 2{{\left( { - 0.5779} \right)}^2} + 4}}{{\frac{4}{3}{{\left( { - 0.5779} \right)}^{ - 2/3}} - 4\left( { - 0.5779} \right)}} = - 0.5779 \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 - 0.58 \cr} $$