## Calculus with Applications (10th Edition)

$$6.8895$$
\eqalign{ & {e^{1.93}} \cr & {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found }} \cr & {\text{in the exercise 17 above }}\left( {page{\text{ 662}}} \right){\text{. }}f\left( x \right) = {e^{2 - x}}{\text{ we found that}} \cr & {P_4}\left( x \right) = {e^2} - {e^2}x + \frac{{{e^2}}}{2}{x^2} - \frac{{{e^2}}}{6}{x^3} + \frac{{{e^2}}}{{24}}{x^4} \cr & {\text{To approximate }}{e^{1.93}},{\text{ we must evaluate }}{e^{2 - 0.07}},{\text{ }} \cr & {\text{then taking }}x = 0.07 \cr & {P_4}\left( {0.07} \right) = {e^2} - {e^2}\left( {0.07} \right) + \frac{{{e^2}}}{2}{\left( {0.07} \right)^2} - \frac{{{e^2}}}{6}{\left( {0.07} \right)^3} + \frac{{{e^2}}}{{24}}{\left( {0.07} \right)^4} \cr & {P_4}\left( {0.07} \right) = 7.38905 - 0.517233 + 0.0181031 - 0.00042 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 0.000007392 \cr & {P_4}\left( {0.07} \right) = 6.8895 \cr & {\text{Thus}}{\text{, }}{e^{1.93}} \approx 6.8895 \cr}