Chapter 8.3, Problem 119E

a)

To determine

Sketch the graph and identify the asymptote function of the graph and determine its relationship to the sum.

Answer to Problem 119E

The sum is equal to the asymptote of the given function

Explanation of Solution

Given:

  $f(x)=6\left[\frac{1-{(0.5)}^x}{1-0.5}\right]$

The graph of the function $f(x)=6\left[\frac{1-{(0.5)}^x}{1-0.5}\right]$ is,

  

From the figure, it is seen that the horizontal asymptote of the function is,

  $y=12$

The infinite sum of the sequence ${\displaystyle \sum }_{n=0}^{\infty }6{\left(\frac{1}{2}\right)}^n$ is,

The common ratio of this geometric series is,

  $\frac{1}{2}$ The first term is $6.$

Therefore, the sum is,

  $\begin{array}{cc}{S}_x& =\frac{a_1}{1-r}\\ & =\frac{6}{1-\frac{1}{2}}\\ & =12\end{array}$

The sum is equal to the asymptote of the given function.

b)

To determine

Sketch the graph and identify the asymptote function of the graph and determine its relationship to the sum.

Answer to Problem 119E

The sum of the series is $S_n={\displaystyle \sum }_{n=0}^{2}2{\left(\frac{4}{5}\right)}^n$

Explanation of Solution

Given:

  $f(x)=2\left[\frac{1-{(0.8)}^x}{1-(0.8)}\right]$

Consider the function,

  $f(x)=2\left[\frac{1-{(0.8)}^x}{1-(0.8)}\right]$

Enter the equation in T1-83

  

Click on WINDOW button and set the minimum and maximum values of labels (axis)

  

Click on the GRAPH button then below screen will be obtained,

  

Hence the Horizontal asymptote is $y=10$ .

The series is in the form of sum of geometric finite series that is,

  $S_m=a\left[\frac{1-r^{\ast }}{1-r}\right]$

Implies that $a=2,r=0.8$

Then the $n^{\text{th}}$ term is,

  $\begin{array}{cc}{a}_x& =a{r}^x\\ & =2{(0.8)}^x\\ & =2{\left(\frac{8}{10}\right)}^x\\ & =2{\left(\frac{4}{5}\right)}^a\end{array}$

Hence the sum of the series is $S_n={\displaystyle \sum }_{n=0}^{2}2{\left(\frac{4}{5}\right)}^n$