Answer
$$4.73$$
Work Step by Step
$$\eqalign{
& {x^3} - 8{x^2} + 18x - 12 = 0;\,\,\,\,\,\,\,{\text{in the interval }}\left[ {4,5} \right] \cr
& {\text{Let }}f\left( x \right) = {x^3} - 8{x^2} + 18x - 12.{\text{ So that}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{x^3} - 8{x^2} + 18x - 12} \right) \to f'\left( x \right) = 3{x^2} - 16x + 18 \cr
& \cr
& {\text{Interval }}\left[ {4,5} \right]{\text{ then }}a = 4{\text{ and }}b = 5.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr
& f\left( 4 \right) = {\left( 4 \right)^3} - 8{\left( 4 \right)^2} + 18\left( 4 \right) - 12 = - 4 < 0 \cr
& f\left( 5 \right) = {\left( 5 \right)^3} - 8{\left( 5 \right)^2} + 18\left( 5 \right) - 12 = 3 > 0 \cr
& f\left( 4 \right){\text{ and }}f\left( 5 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr
& {\text{There is a solution for the equation in the interval }}\left( {4,5} \right) \cr
& \cr
& {\text{As an initial guess }}{c_1} = 4.{\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
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 4 - \frac{{{{\left( 4 \right)}^3} - 8{{\left( 4 \right)}^2} + 18\left( 4 \right) - 12}}{{3{{\left( 4 \right)}^2} - 16\left( 4 \right) + 18}} = 6 \cr
& {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr
& {c_3} = 6 - \frac{{{{\left( 6 \right)}^3} - 8{{\left( 6 \right)}^2} + 18\left( 6 \right) - 12}}{{3{{\left( 6 \right)}^2} - 16\left( 6 \right) + 18}} \approx 5.2 \cr
& {\text{In the same way}} \cr
& {c_4} = 5.2 - \frac{{{{\left( {5.2} \right)}^3} - 8{{\left( {5.2} \right)}^2} + 18\left( {5.2} \right) - 12}}{{3{{\left( {5.2} \right)}^2} - 16\left( {5.2} \right) + 18}} \approx 4.83 \cr
& {c_5} = 4.83 - \frac{{{{\left( {4.83} \right)}^3} - 8{{\left( {4.83} \right)}^2} + 18\left( {4.83} \right) - 12}}{{3{{\left( {4.83} \right)}^2} - 16\left( {4.83} \right) + 18}} \approx 4.737 \cr
& {c_6} = 4.737 - \frac{{{{\left( {4.737} \right)}^3} - 8{{\left( {4.737} \right)}^2} + 18\left( {4.737} \right) - 12}}{{3{{\left( {4.737} \right)}^2} - 16\left( {4.737} \right) + 18}} \approx 4.732 \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 4.73 \cr} $$
