## Calculus with Applications (10th Edition)

$$3.58$$
\eqalign{ & 2{x^2} - 8x + 3 = 0;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {3,4} \right] \cr & {\text{let }}f\left( x \right) = 2{x^2} - 8x + 3 = 0,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {2{x^2} - 8x + 3} \right) \cr & f'\left( x \right) = 4x - 8 \cr & {\text{interval }}\left[ {3,4} \right]{\text{ then }}a = 3{\text{ and }}b = 4 \cr & {\text{check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 3 \right) = 2{\left( 3 \right)^2} - 8\left( 3 \right) + 3 = - 3 < 0 \cr & f'\left( 4 \right) = 2{\left( 4 \right)^2} - 8\left( 4 \right) + 3 = 3 > 0 \cr & f\left( 3 \right){\text{ and }}f\left( 4 \right){\text{ have opposite signs}}{\text{, then}} \cr & {\text{there is a solution for the equaion in the interval }}\left( {3,4} \right) \cr & {\text{As an initial guess }}{c_1} = 3 \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 & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 3 - \frac{{2{{\left( 3 \right)}^2} - 8\left( 3 \right) + 3}}{{4\left( 3 \right) - 8}} = 3.75 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 3.75 - \frac{{2{{\left( {3.75} \right)}^2} - 8\left( {3.75} \right) + 3}}{{4\left( {3.75} \right) - 8}} = 3.589 \cr & {\text{In the same way}} \cr & {c_4} = 3.58 - \frac{{5{{\left( {3.58} \right)}^2} - 3\left( {3.58} \right) - 3}}{{10\left( {3.58} \right) - 3}} = 3.581 \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}} \cr & {\text{A solution for the given interval is }}x \approx 3.58 \cr}