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<h4>Activity (20 minutes)</h4>
<p>Earlier in the course, students looked at constant rates of change in linear functions. In this activity, they begin
to explore average rates of change in an exponential growth function. In particular, the focus for this activity is on
how exponential functions have different rates of change for different input intervals, which was not the case for
linear functions.</p>
<p>In the next activity, they investigate average rates of growth in a decay function. In both cases, they revisit
contexts from earlier lessons.</p>
<h4> Student Activity</h4>
<p>Here is a table and a graph that show the number of coffee shops worldwide that a company had in its first 10 years,
between 1987 and 1997. The growth in the number of shops was roughly exponential.</p>
<p><img alt="Graph of a roughly exponential function, origin O. Years since 1987 and number of coffee shops." src="https://k12.openstax.org/contents/raise/resources/bdfc755f5f75d82877c744868cc66af19e924b2e"></p>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">Year</th>
<th scope="col">Number Of Shops</th>
</tr>
</thead>
<tbody>
<tr>
<td>
1987
</td>
<td>
17
</td>
</tr>
<tr>
<td>
1988
</td>
<td>
33
</td>
</tr>
<tr>
<td>
1989
</td>
<td>
55
</td>
</tr>
<tr>
<td>
1990
</td>
<td>
84
</td>
</tr>
<tr>
<td>
1991
</td>
<td>
116
</td>
</tr>
<tr>
<td>
1992
</td>
<td>
165
</td>
</tr>
<tr>
<td>
1993
</td>
<td>
272
</td>
</tr>
<tr>
<td>
1994
</td>
<td>
425
</td>
</tr>
<tr>
<td>
1995
</td>
<td>
677
</td>
</tr>
<tr>
<td>
1996
</td>
<td>
1,015
</td>
</tr>
<tr>
<td>
1997
</td>
<td>
1,412
</td>
</tr>
</tbody>
</table>
<br>
<ol class="os-raise-noindent">
<li> Find the average rate of change for each period of time. Show your reasoning.
<ol class="os-raise-noindent" type="a">
<li> 1987 to 1990 </li>
<li> 1987 to 1993 </li>
<li> 1987 to 1997 </li>
</ol>
<br>
<strong>Answer: </strong>
<ol class="os-raise-noindent" type="a">
<li>\(22\frac13\) stores per year since \(\frac{(84−17)}{(3−0)}=22 \frac13\). </li>
<li> 42.5 stores per year since \(\frac{(272−17)}{(6−0)}=42.5\).</li>
<li>139.5 stores per year since \(\frac{(1412−17)}{(10−0)}=139.5\). </li>
</ol>
</li>
</ol>
<ol class="os-raise-noindent" start="2">
<li> Make some observations about the rates of change you calculated. What do these average rates tell us about how
the company was growing during this time period? </li>
<br>
<strong>Answer: </strong> For example: The average rate of change of the first 6 years is almost double that of the
first 3 years. The average rate of change of the 10-year period is almost 7 times greater than that of the first 3
years and more than 3 times greater than the first 6 years. The company was growing at an increasing rate over the
10-year time period.<br>
<br>
<li> Use the graph to support your answers to these questions. How well do the average rates of change describe the
growth of the company in:
<ol class="os-raise-noindent" type="a">
<li> the first 3 years? </li>
<li> the first 6 years? </li>
<li> the entire 10 years? </li>
</ol>
</li>
<p>
<strong>Answer: </strong> Compare your answer:
</p>
<ol class="os-raise-noindent" type="a">
<li> An average rate of change of \(22\frac{1}{3}\) is a fairly accurate prediction of growth over the first 3
years, but it is a poor prediction of growth much beyond 1992. </li>
<li>An average rate of change of 42.5 is an okay prediction of growth over the first 6 years, but it overestimates
early on and greatly underestimates the growth in years after 1994. </li>
<li> An average rate of change of 139.5 does not accurately predict the actual growth of the company between 1987
and 1997. </li>
</ol>
</ol>
<ol class="os-raise-noindent" start="4">
<li> Let \(f\) be the function so that \(f(t)\) represents the number of stores \(t\) years since 1987. The value
of \(f(20)\) is 15,011. Find \(\frac{f(20)−f(10)}{20−10}\) and say what it tells us about the change in
the number of stores.</li>
<br>
<strong>Answer: </strong> For example: The average rate of growth from 1997 to 2007 is 1,359.9 stores per year. This
is 60 times greater than the average rate of change over the first 3 years and almost 10 times greater than the
average rate of change over the first 10 years of the time period. The number of stores continued to grow at an
increasing rate from 1997 to 2007.
</ol>
<h4>Anticipated Misconceptions</h4>
<p>If students struggle with a rate of change that is not constant like the slope they saw in linear relationships, ask
if they can draw a straight line connecting all the points on the graph in question 3. Spend time on the graph in the
synthesis so the lines can help them see why the rate of change varies with an exponential function.</p>
<h4> Activity Synthesis</h4>
<p>For the third question, make sure students see that when they calculated the average rate of change for each of the
three time periods, they were in effect finding the slope of the line that goes through two points that represent the
starting year and the ending year. Display a graph like the one shown here to help illustrate this point:</p>
<p><img alt="11 points on coordinate plane. Lines connect first point to fourth, seventh, and eleventh points. Horizontal axis from 0 to 10, time in years. Vertical axis from 0 to 1500, number of coffee shops." src="https://k12.openstax.org/contents/raise/resources/69e0f30cf89cfc35f420c63da1e7be40107ad268"></p>
<p>The line that connects the points for 1987 and 1990 fits the data for that period fairly well, so the slope of that
line (the average rate of change between those two points) describes the growth in those three years fairly
accurately. In contrast, the line that connects the points for 1987 and 1997 does not at all fit the data, so the
slope of that line does not paint an accurate picture of how the company was growing that decade.</p>
<p>Here are some questions for discussion:</p>
<ol class="os-raise-noindent">
<li>“Would the average rate of change between 1994 and 1997 accurately depict how the company was growing in the
last three years in the data?” (Yes, because the average rate of change from 1994 to 1997 is 329 shops per
year since \(\frac{1412−425}{10−7}=329\). The actual number of shops increased by 252, 338, and then 397
from 1994 to 1997, respectively.) </li>
<li>“Is there a single period of time where the average rate of change would accurately summarize how the
company was growing from 1987 to 1997?” (No, since the number of shops is increasing at an increasing rate, an
average of two values is not enough to summarize the growth of the company.) </li>
</ol>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for English Language Learners</p>
<p class="os-raise-extrasupport-name">MLR 8 Discussion Supports: Speaking, Representing</p>
</div>
<div class="os-raise-extrasupport-body">
<p> Use this routine to amplify mathematical uses of language to communicate how the rate of change differs during each
specified time period. Encourage students to demonstrate mathematical thinking and problem solving by referencing
two points on the graph in their explanation. Press for further detail by asking, “How can you use the graph
to determine which time period is represented best by the rate of change?” This will help students understand
how the rate of change differs in exponential functions through a whole-class discussion.</p>
<p class="os-raise-text-italicize">Design Principle(s): Support sense-making</p>
</div>
</div>
<br>
<h4> Video: Analyzing Average Rate of Change of an Exponential Function</h4>
<p>Watch the following video to learn more about the average rate of change of exponential functions:</p>
<div class="os-raise-d-flex-nowrap os-raise-justify-content-center">
<div class="os-raise-video-container"><video controls="true" crossorigin="anonymous">
<source src="https://k12.openstax.org/contents/raise/resources/f012f2931d956078be6f22f90b24f63f50921333">
<track default="true" kind="captions" label="On" src="https://k12.openstax.org/contents/raise/resources/f92dfb251ff3725fb06898b80dae14782cba7df4" srclang="en_us">
https://k12.openstax.org/contents/raise/resources/f012f2931d956078be6f22f90b24f63f50921333
</video></div>
</div>
<br>
<br>
<h3>5.10.2: Self Check</h3>
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their understanding of the
concepts explored in the activity.</em></p>
<p class="os-raise-text-bold">QUESTION: </p>
<p>Suppose a population of cells starts at 500 and triples every day. The number of cells each day can be
calculated as follows:</p>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">Number Of Days</th>
<th scope="col">Number Of Cells</th>
</tr>
</thead>
<tbody>
<tr>
<td>
0
</td>
<td>
500
</td>
</tr>
<tr>
<td>
1
</td>
<td>
1,500
</td>
</tr>
<tr>
<td>
2
</td>
<td>
4,500
</td>
</tr>
<tr>
<td>
3
</td>
<td>
13,500
</td>
</tr>
</tbody>
</table>
<br>
<p>Find the average rate of change from day 0 to day 2.</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td>
1,000
</td>
<td>
Incorrect. Let’s try again a different way: This is the average rate of change from day 0 to day 1. The
answer is 2,000.
</td>
</tr>
<tr>
<td>
4,500
</td>
<td>
Incorrect. Let’s try again a different way: This is the population at day 2, but it is not the rate of
change from day 0 to day 2. The answer is 2,000.
</td>
</tr>
<tr>
<td>
2,000
</td>
<td>
That’s correct! Check yourself: Find the difference between 4,500 and 500 and then divide by 2 days.
</td>
</tr>
<tr>
<td>
–2,000
</td>
<td>
Incorrect. Let’s try again a different way: Make sure to align the corresponding values in the formula
and watch the signs. The cell population is growing, not decaying. The answer is 2,000.
</td>
</tr>
</tbody>
</table>
<br>
<h3> 5.10.2: Additional Resources</h3>
<p class="os-raise-text-bold"><em>The following content is available to students who would like more support based on their experience with
the self check. Students will not automatically have access to this content, so you may wish to share it with
those who could benefit from it. </em></p>
<h4> Average Rate of Change from a Table</h4>
<p>Let’s look at an exponential function we studied earlier. Let \(A(t)\) be the function that models the area
\(A\), in square yards, of algae covering a pond \(t\) weeks after beginning treatment to control the algae bloom.
Here is a table showing approximately how many square yards of algae remain during the first 5 weeks of treatment.</p>
<table class="os-raise-skinnytable">
<thead>
<tr>
<th scope="col">\(t\) </th>
<th scope="col">\(A(t)\) </th>
</tr>
</thead>
<tbody>
<tr>
<td>
0
</td>
<td>
240
</td>
</tr>
<tr>
<td>
1
</td>
<td>
80
</td>
</tr>
<tr>
<td>
2
</td>
<td>
27
</td>
</tr>
<tr>
<td>
3
</td>
<td>
9
</td>
</tr>
<tr>
<td>
4
</td>
<td>
3
</td>
</tr>
</tbody>
</table>
<br>
<p>The average rate of change of \(A\) from the start of treatment to week 2 is about –107 square yards per week
since \(\frac{A(2)-A(0)}{2-0} \approx -107\). The average rate of change of \(A\) from week 2 to week 4, however, is
only about –12 square yards per week since \(\frac{A(4)-A(2)}{4-2} \approx -12\) .</p>
<h4> Try It: Average Rate of Change from a Table</h4>
<p>Using the same table from above, what is the average rate of change of \(A\) from week 1 to week 3?</p>
<table class="os-raise-skinnytable">
<thead>
<tr>
<th scope="col">\(t\) </th>
<th scope="col">\(A(t)\) </th>
</tr>
</thead>
<tbody>
<tr>
<td>
0
</td>
<td>
240
</td>
</tr>
<tr>
<td>
1
</td>
<td>
80
</td>
</tr>
<tr>
<td>
2
</td>
<td>
27
</td>
</tr>
<tr>
<td>
3
</td>
<td>
9
</td>
</tr>
<tr>
<td>
4
</td>
<td>
3
</td>
</tr>
</tbody>
</table>
<br>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<h5> Solution</h5>
<p>Here is how to use the table to find the average rate of change:</p>
<table class="os-raise-horizontaltable">
<thead></thead>
<tbody>
<tr>
<th scope="row"> Find the points on the table when \(t=1\) and \(t=3\).</th>
<td>
\((1, 80)\), \((3,9)\)
</td>
</tr>
<tr>
<th scope="row">Substitute into the slope formula to find the rate of change.</th>
<td>
\(\frac{A(3)-A(1)}{3-1}=\frac{9-80}{2}\)
</td>
</tr>
<tr>
<th scope="row"> Simplify to find the rate of change.</th>
<td>
\(\frac{-71}{2}=-35.5\) square yards per week
</td>
</tr>
</tbody>
</table>