Answer
$$0.96078944$$
Work Step by Step
$$\eqalign{
& {e^{ - 0.04}} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 1}} \cr
& f\left( x \right) = {e^{ - 2x}}{\text{ we found that}} \cr
& {P_4}\left( x \right) = 1 - 2x + 2{x^2} - \frac{4}{3}{x^3} + \frac{2}{3}{x^4} \cr
& {\text{To approximate }}{e^{ - 0.04}},{\text{ we must evaluate }}{e^{ - 2\left( {0.02} \right)}},{\text{ then taking }}x = 0.02 \cr
& {P_4}\left( {0.02} \right) = 1 - 2\left( {0.02} \right) + 2{\left( {0.02} \right)^2} - \frac{4}{3}{\left( {0.02} \right)^3} + \frac{2}{3}{\left( {0.02} \right)^4} \cr
& {P_4}\left( {0.02} \right) = 1 - 0.04 + 0.0008 - 0.000010666 + 0.000000106 \cr
& {P_4}\left( {0.02} \right) = 0.96078944 \cr
& {\text{Thus}}{\text{, }}{e^{ - 0.04}} \approx 0.96078944 \cr} $$
