9.2: Exponential Growth
This activity is intended to supplement Algebra I, Chapter 8, Lesson 5.
ID: 9467
Time required: 45 minutes
Activity Overview
The main objective of the activity is to find an approximation for the value of the mathematical constant e and to apply it to exponential growth and decay problems. To accomplish this, students are asked to search for the base,
Topic: Exponential & Logarithmic Functions

Graph exponential functions of the form
f(x)=bx . 
Evaluate the exponential function
f(x)=bx for any value ofx .  Calculate the doubling time or halflife in a problem involving exponential growth or decay.
Teacher Preparation and Notes
 Students encounter the exponential constant e at various levels in their mathematics schooling. It may happen well before they reach calculus, and it is often used without an appreciation for where it originates (or why it is important). A good time to use this activity is when students first encounter e, but it is also appropriate for Precalculus and Calculus students when they are studying derivatives and instantaneous rate of change.

Prerequisites for the students are: familiarity with graphing and tracing functions on the calculator; an understanding of functions and function notation (both “
y= ” and “f(x)= ”); and an intuitive understanding of rate of change.
Associated Materials
 Student Worksheet: Exponential Growth, http://www.ck12.org/flexr/chapter/9618, scroll down to the second activity.
Problem 1
The activity begins with an investigation of how the value of
The last question posed in this problem asks students to explain why the value of
Problem 2
In this problem, students work specifically with the graph of
Problem 3
Students are again asked to observe the changing values of the slope of the tangent line and the value of the function—and how they are related.
Students will discover that there is exactly one value of
Applications
Students are given a series of application problems to apply the knowledge about what they have learned by doing completing this activity.
Student Solutions
Problem 1
 Answers may vary. Possible observations: graph gets “steeper” as
b increases and “flatter” asb decreases; always passes through the point(0, 1) ; increasing whenb>1 and decreasing when0<b<1 
b=1  Answers may vary. Possible explanation: Even roots of negative numbers are not real numbers. Consider, for example,
(−1)0.5=−1−−−√ , which is not a real number.
Problem 2

x ,f(x) , and slope will vary  the slope is less than
f(x)  Answers may vary. Possible observations: slope is always positive; as
x increases, the slope increases; curve never reaches thex− axis



slope of tangent at 

















Problem 3
 Answers will vary. Possible answer:
3  Answers will vary. Possible answer:
2 
b≈2.718  Answers will vary.

\begin{align*}\frac{\text{slope of tangent at}\ x}{f(x)}\end{align*} 

\begin{align*}2\end{align*} 
\begin{align*}0.693\end{align*} 
\begin{align*}3\end{align*} 
\begin{align*}1.099\end{align*} 
\begin{align*}0.5\end{align*} 
\begin{align*}0.693\end{align*} 
\begin{align*}0.25\end{align*} 
\begin{align*}1.381\end{align*} 
\begin{align*}b\end{align*} 
\begin{align*}x\end{align*} 
\begin{align*}f(x)\end{align*} 
slope of tangent at \begin{align*}x\end{align*} 
\begin{align*}\frac{\text{slope of tangent at}\ x}{f(x)}\end{align*} 

\begin{align*}2.2\end{align*} 
\begin{align*}1\end{align*} 
\begin{align*}0.788\end{align*} 
\begin{align*}0.788\end{align*} 

\begin{align*}2.4\end{align*} 
\begin{align*}1\end{align*} 
\begin{align*}2.4\end{align*} 
\begin{align*}2.101\end{align*} 
\begin{align*}0.875\end{align*} 
\begin{align*}2.6\end{align*} 
\begin{align*}1\end{align*} 
\begin{align*}0.956\end{align*} 
\begin{align*}0.956\end{align*} 

\begin{align*}2.8\end{align*} 
\begin{align*}2\end{align*} 
\begin{align*}7.84\end{align*} 
\begin{align*}8.072\end{align*} 
\begin{align*}1.030\end{align*} 
\begin{align*}2.7\end{align*} 
\begin{align*}2.7\end{align*} 
\begin{align*}0.993\end{align*} 
\begin{align*}0.993\end{align*} 

\begin{align*}2.75\end{align*} 
\begin{align*}2.75\end{align*} 
\begin{align*}1.0116\end{align*} 
\begin{align*}1.0116\end{align*} 
Applications
 Modeling equation: \begin{align*}P=1,000e^{0.05t}\end{align*}
P=1,000e0.05t (where \begin{align*}P\end{align*}P is the value and \begin{align*}t\end{align*}t is the time in years); one year: \begin{align*}\$1,051.27\end{align*}$1,051.27 ; two years: \begin{align*}\$1,105.17\end{align*}$1,105.17 ; five years: \begin{align*}\$1,284.03\end{align*}$1,284.03  Modeling equation: \begin{align*}P=500e^{0.5 \cdot 24}\end{align*}
P=500e0.5⋅24 (where \begin{align*}P\end{align*}P is population); about \begin{align*}81,000,000\end{align*}81,000,000  Modeling equation: \begin{align*}P=1,000,000e^{0.15 \cdot 10}\end{align*}
P=1,000,000e−0.15⋅10 (where \begin{align*}P\end{align*}P is the volume); about \begin{align*}22.3\%\end{align*}22.3%  Modeling equation for growing snowball: \begin{align*}P=2e^{0.1t}\end{align*}
P=2e0.1t (where \begin{align*}P\end{align*}P is the weight and \begin{align*}t\end{align*}t is the time in seconds); \begin{align*}10 \ seconds\end{align*}10 seconds : \begin{align*}5.43 \ pounds\end{align*}5.43 pounds ; \begin{align*}20 \ seconds\end{align*}20 seconds : \begin{align*}14.78 \ pounds\end{align*}14.78 pounds ; \begin{align*}45 \ seconds\end{align*}45 seconds : \begin{align*}180.03 \ pounds\end{align*}180.03 pounds ; \begin{align*}1 \ minute\end{align*}1 minute : \begin{align*}806.86 \ pounds\end{align*}806.86 pounds
Possible limitations: the modeling equation might not be appropriate after too long a period of time, for example—the snowball may break apart if it gets too big, or it might reach the end of the hill.
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