Question

In Exercises 87 and 88 , use a graphing utility to graph the function. Identify the horizontal asymptote of the graph and determine its relationship to the sum of the series. $$ \frac{\text { Function }}{f(x)=3\left[\frac{1-(0.5)^{x}}{1-0.5}\right]} \frac{\text { Series }}{\sum_{n=0}^{\infty} 3\left(\frac{1}{2}\right)^{n}} $$

Short answer

The horizontal asymptote of the graph is \(y = 6\), which indicates that as x approaches infinity, the function will get indefinitely close to this line but never reach it or cross it. The sum of the given infinite geometric series is also \(6\). Hence, the horizontal asymptote of the graph of the function and the sum of the series are equal.

Step-by-step solution

Simplify the Function

Simplify the given function \(f(x) = 3\[\frac{1-(0.5)^{x}}{1-0.5}\]\) to \(f(x) = 6(1-(0.5)^x)\). This simplification is done to get a clear view of the function which will help in drawing the graph.

Graph the Function

Graph the simplified function \(f(x) = 6(1-(0.5)^x)\) using a graphing utility. Observe its behavior at extremes, this gives us an idea about the presence of any horizontal asymptotes.

Identify the Horizontal Asymptote

The horizontal asymptote of the function is the value which the function approaches but never quite reaches as x goes to infinity or to negative infinity. From the graph of the function, the horizontal asymptote can be identified as \(y = 6\).

Sum of the Geometric Series

The given series \(\sum_{n=0}^{\infty} 3\(\frac{1}{2}\)^{n}\) is a geometric series with first term, \(a = 3\) and common ratio \(r = 0.5\). The sum of an infinite geometric series can be computed using the formula \(S = \frac{a}{1-r}\). Computing using the values, we have \(S = \frac{3}{1 - 0.5} = 6\).

Relate the Asymptote to the Sum of the Series

The horizontal asymptote of the graph, \(y = 6\) corresponds to the sum of the infinite geometric series which is also 6.