Chapter 12.7, Problem 19E

To determine

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

Answer to Problem 19E

2.99

Explanation of Solution

Given information:

  $e^{1.1}$

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:

  $\begin{array}{c}{e}^{1.1}\approx 1+1.1+\frac{{\left(1.1\right)}^2}{2!}+\frac{{\left(1.1\right)}^3}{3!}+\frac{{\left(1.1\right)}^4}{4!}\\ \approx 1+1.1+0.605+0.222+0.06\\ \approx 2.99\end{array}$

Thus, the value is 2.99.