Answer
$$0.47$$
Work Step by Step
$$\eqalign{
& {e^{2x}} + 3x - 4 = 0;\,\,\,\,\,\,{\text{in the interval }}\left[ {0,3} \right] \cr
& {\text{Let }}f\left( x \right) = {e^x} + x - 2,{\text{ so that}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{e^{2x}} + 3x - 4} \right) \cr
& f'\left( x \right) = 2{e^{2x}} + 3 \cr
& {\text{Interval }}\left[ {0,3} \right]{\text{ then }}a = 0{\text{ and }}b = 3.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr
& f\left( 0 \right) = {e^{2\left( 0 \right)}} + 3\left( 0 \right) - 4 = - 3 < 0 \cr
& f'\left( 3 \right) = {e^3} + 3 - 2 \approx 408 > 0 \cr
& f\left( 0 \right){\text{ and }}f\left( 3 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr
& {\text{There is a solution for the equation in the interval }}\left( {0,3} \right) \cr
& {\text{As an initial guess }}{c_1} = 0.{\text{ A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr
& {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 0 - \frac{{{e^{2\left( 0 \right)}} + 3\left( 0 \right) - 4}}{{2{e^{2\left( 0 \right)}} + 3}} = 0.6 \cr
& {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr
& {c_3} = 0.6 - \frac{{{e^{2\left( {0.6} \right)}} + 3\left( {0.6} \right) - 4}}{{2{e^{2\left( {0.6} \right)}} + 3}} = 0.4838 \cr
& {\text{In the same way}} \cr
& {c_4} = 0.4838 - \frac{{{e^{2\left( {0.4838} \right)}} + 3\left( {0.4838} \right) - 4}}{{2{e^{2\left( {0.4838} \right)}} + 3}} = 0.4737 \cr
& {c_5} = 0.4737 - \frac{{{e^{2\left( {0.4737} \right)}} + 3\left( {0.4737} \right) - 4}}{{2{e^{2\left( {0.4737} \right)}} + 3}} = 0.4736 \cr
& \cr
& {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr
& {\text{Then}}{\text{. A solution for the given interval is }}x \approx 0.47 \cr} $$
