Answer
$${\text{ }}{e^{ - 0.07}} \approx 0.932393834$$
Work Step by Step
$$\eqalign{
& {e^{ - 0.07}} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 4}} \cr
& f\left( x \right) = {e^{ - x}}{\text{ we found that}} \cr
& {P_4}\left( x \right) = 1 - x + \frac{1}{2}{x^2} - \frac{1}{6}{x^3} + \frac{1}{{24}}{x^4} \cr
& {\text{To approximate }}{e^{ - 0.07}},{\text{ we must evaluate }}{e^{ - \left( {0.07} \right)}},{\text{ then taking }}x = 0.07 \cr
& {P_4}\left( {0.07} \right) = 1 - \left( {0.07} \right) + \frac{1}{2}{\left( {0.07} \right)^2} - \frac{1}{6}{\left( {0.07} \right)^3} + \frac{1}{{24}}{\left( {0.07} \right)^4} \cr
& {P_4}\left( {0.07} \right) = 1 - 0.07 + 0.00245 - 0.000057166 + 0.000001 \cr
& {P_4}\left( {0.07} \right) = 0.932393834 \cr
& {\text{Thus}}{\text{, }}{e^{ - 0.07}} \approx 0.932393834 \cr} $$
