Answer
$${e^{0.06}} \approx 2.72372203$$
Work Step by Step
$$\eqalign{
& {e^{1.02}} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 3}} \cr
& f\left( x \right) = {e^{x + 1}}{\text{ we found that}} \cr
& {P_4}\left( x \right) = e + ex + \frac{e}{2}{x^2} + \frac{e}{6}{x^3} + \frac{e}{{24}}{x^4} \cr
& {\text{To approximate }}{e^{1.02}},{\text{ we must evaluate }}{e^{1 + 0.02}},{\text{ then taking }}x = 0.02 \cr
& {P_4}\left( {0.02} \right) = e + e\left( {0.02} \right) + \frac{e}{2}{\left( {0.02} \right)^2} + \frac{e}{6}{\left( {0.02} \right)^3} + \frac{e}{{24}}{\left( {0.02} \right)^4} \cr
& {P_4}\left( {0.02} \right) = 2.718281828 + 0.00543656 + 0.000003624 + 0.000000018 \cr
& {\text{Thus}}{\text{, }}{e^{0.06}} \approx 2.72372203 \cr} $$
