Chapter 12.7, Problem 22E

To determine

To find: the value to the nearest hundredth by using the first five terms of the exponential series.

Answer to Problem 22E

1.73

Explanation of Solution

Given information:

  $e^{0.55}$

Calculation:

Use $e^x\approx 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}$ to evaluate the given value as follows:

  $\begin{array}{c}{e}^{0.55}\approx 1+0.55+\frac{{\left(0.55\right)}^2}{2!}+\frac{{\left(0.55\right)}^3}{3!}+\frac{{\left(0.55\right)}^4}{4!}\\ \approx 1+0.55+0.151+0.028+0.004\\ \approx 1.73\end{array}$

Thus, the value is 1.73.