Answer
$$0.44$$
Work Step by Step
$$\eqalign{
& {e^x} + x - 2 = 0;\,\,\,\,{\text{in the interval }}\left[ {0,3} \right] \cr
& {\text{Let }}f\left( x \right) = {e^x} + x - 2,{\text{ so that}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{e^x} + x - 2} \right) \cr
& f'\left( x \right) = {e^x} + 1 \cr
& {\text{Interval }}\left[ {0,3} \right]{\text{ then }}a = 0{\text{ and }}b = 3.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr
& f\left( 0 \right) = {e^0} + 0 - 2 = - 1 < 0 \cr
& f'\left( 3 \right) = {e^3} + 3 - 2 \approx 21 > 0 \cr
& f\left( 0 \right){\text{ and }}f\left( 3 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr
& {\text{There is a solution for the equation in the interval }}\left( {0,3} \right) \cr
& {\text{As an initial guess }}{c_1} = 1.{\text{ A better guess}}{\text{, }}{c_2},{\text{ can be found as follows}}:{\text{ }} \cr
& {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{{e^1} + 1 - 2}}{{{e^1} + 1}} = 0.5378 \cr
& {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr
& {c_3} = 0.5378 - \frac{{{e^{0.5378}} + 0.5378 - 2}}{{{e^{0.5378}} + 1}} = 0.4456 \cr
& {\text{In the same way}} \cr
& {c_4} = 0.4456 - \frac{{{e^{0.4456}} + 0.4456 - 2}}{{{e^{0.4456}} + 1}} = 0.4428 \cr
& \cr
& {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr
& {\text{Then}}{\text{. A solution for the given interval is }}x \approx 0.44 \cr} $$