Answer
$${e^{0.06}} \approx 1.06186654$$
Work Step by Step
$$\eqalign{
& {e^{0.06}} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 2}} \cr
& f\left( x \right) = {e^{3x}}{\text{ we found that}} \cr
& {P_4}\left( x \right) = 1 + 3x + \frac{9}{2}{x^2} + \frac{9}{2}{x^3} + \frac{{27}}{8}{x^4} \cr
& {\text{To approximate }}{e^{0.06}},{\text{ we must evaluate }}{e^{3\left( {0.02} \right)}},{\text{ then taking }}x = 0.02 \cr
& {P_4}\left( {0.02} \right) = 1 + 3\left( {0.02} \right) + \frac{9}{2}{\left( {0.02} \right)^2} + \frac{9}{2}{\left( {0.02} \right)^3} + \frac{{27}}{8}{\left( {0.02} \right)^4} \cr
& {P_4}\left( {0.02} \right) = 1 + 0.06 + 0.0018 + 0.000036 + 0.00000054 \cr
& {P_4}\left( {0.02} \right) = 1.06186654 \cr
& {\text{Thus}}{\text{, }}{e^{0.06}} \approx 1.06186654 \cr} $$
