3.1: Exponential Growth and Decay
Motivating Example: Modeling Climate Change
Previously, we modeled the global temperature anomaly relative to time with a linear function. To do so, we assumed that the rate of increase in temperature was constant. As the data shows in the graph below (Figure 3-1) 1 , the actual temperature data does not increase at a constant rate. In fact, the global temperature anomalies have increased at an larger and larger rate over time . If we are going to accurately model the the global temperature anomaly over time, we are going to need to develop a new type of function to more accurately describe this type of increase. In this chapter, we will explore modeling relationships with exponential functions that will allow us to do this.
Focus: Understanding Exponential Growth and Decay
Can you fold a piece of paper in half 15 times? Try it. What do you notice?
It gets pretty hard to fold after about 6-7 folds. Why is this? There is a a relationship between the number of folds and the sheets of thickness of our paper is described in the table below.
| x = Number of folds | 0 | 1 | 2 | 3 | 4 |
| y = Sheets of thickness | 1 | 2 | 4 | 8 | 16 |
Do you notice a pattern? For each fold or increase in x of 1, the sheets of thickness, y, is doubled or multiplied by 2. If we fold the paper a fifth time (x= 5), then the sheets of thickness will double again to y = 2(16) = 32. To find the number of sheets of thickness with 15 folds, we could continue to multiply y by 2 for each increase in x by 1 until we get to x= 15 folds. However, we can calculate this faster if we notice that each output, y, can be calculated from the initial value y = 1 and the number of times we have doubled the sheets of thickness as shown in the table below.
| x = Number of folds | 0 | 1 | 2 | 3 | 4 | 15 | x |
| y = Sheets of thickness | \(1 \) | \(2=1(2)^{1} \) | \(4=1(2)^{2} \) | \(8=1(2)^{3} \) | \(16=1(2)^{4} \) | \(1(2)^{15} = 32,768 \) | \(1(2)^{x} \) |
We can calculate the sheets of thickness of the paper after 15 folds by multiplying 1 by 2 fifteen times or \[1(2)^{15} \nonumber\] to get a paper 32,768 sheets thick. To put this thickness into perspective, two hundred sheets of standard white printer paper is approximately 2 inches thick, or 100 sheets = 1 inch of thickness. The thickness after 15 folds is
\[1(2)^{15}=32768 \text { sheets } \cdot \frac{1 \text { in. }}{100 \text { sheets }} \cdot \frac{1 \text { ft .}}{12 \text { in. }} \approx 27.3 \text { ft. }\nonumber\]
This is the thickness of some of the giant sequoia redwood trees in California which no human can fold in half with their hands.
In general, if we fold a piece of paper x times, the sheets of thickness will double x times and can be described by the formula below.
In Summary:
The number of folds of a piece of paper x generates a thickness \(y=1(2)^{x} \). Notice the parameter in the formula a = 1 represents the initial or starting value. The parameter in the formula b = 2 represents the constant value we are multiplying each output y by for each increase in x by one. We call b = 2 the growth factor.
What is the pattern in the data below?
| x | 0 | 1 | 2 | 3 | 15 |
| y | 8 | 4 | 2 | 1 | ? |
For each increase in x by 1, or \(\Delta x = 1\), y is multiplied by \(\frac{1}{2}\) . To find the value of y when x =15, we could continue to multiply by \(\frac{1}{2}\) for each increase in x of 1 until we got to x= 15. However, we can calculate this faster if we notice that each output y, can be calculated from the initial value y = 8 and the number of times we multiply by \(\frac{1}{2}\) as show in the table below.
| x | 0 | 1 | 2 | 3 | 15 | x |
| y | \(8 \) | \(4=8 \cdot (\frac{1}{2})^{1} \) | \(2=8 \cdot (\frac{1}{2})^{2} \) | \(1=8 \cdot (\frac{1}{2})^{3} \) | \(8 \cdot (\frac{1}{2})^{15} \approx 0.000061035 \) | \(8 \cdot (\frac{1}{2})^{x}\ \) |
In Summary:
The relationship can be described by the formula \(y=8(\frac{1}{2})^{x} \). Notice the parameter in the formula a = 8 represents the initial value. The parameter in the formula the b = \(\frac{1}{2}\) represents the constant value we are multiplying each output by for each increase in x by one. We call the parameter b = \(\frac{1}{2}\) the growth factor.
Generalizing further, we arrive at the standard form of the formula for an exponential function.
Key Points:
1) Exponential growth or decay by a function is characterized by each increase in the input x by one,( \(\Delta x = 1\)) causes the output y to be multiplied by a constant value b. The constant value b is called the growth factor .
2) A function that exhibits exponential growth or decay is called an exponential function . The formula for an exponential function can be written in the form
\[f(x)=ab^{x} \]
where \(a\) is the initial or starting value of the function and \(b\) is the growth factor. We limit \(b\) to positive values or \(b>0\).
Notice the variable input x of this function is in the exponent and is why we call these types of functions exponential functions.
3) An exponential function \(f(x)=ab^{x} \) increases when the growth factor b > 1 and decreases when 0<b<1.
The population of a county over time and the number of people the county can feed with its agricultural output over time are listed in the table below.
| t = Decades from 1960 | P = Population of the County | F = Food Supply of the County (in people fed) |
| 0 | 1,000 | 5,000 |
| 1 | 3,000 | 8,000 |
| 2 | 6,000 | 11,000 |
| 3 | 18,000 | 14,000 |
| 4 | 54,000 | 17,000 |
a) Which quantity is experiencing exponential growth over time? Which quantity is experiencing linear growth over time?
b) Find a formula for the quantity that grows exponentially over time. Then find a formula for the quantity that grows linearly growing over time.
Solution
a) The population of the county is growing exponentially since the population is multiplied by a factor of 3 (the growth factor, b = 3) for each \(\Delta t = 1\). The food supply of the county is growing linearly since the food supply increases by 3000, \( \Delta F = 3000\), for each \( \Delta t = 1\).
b) The population of the county has an initial value a = 1000 and the growth factor b = 3. The exponential formula for the population of the county can be written as \[P=1000(3)^{t} .\nonumber\] The food supply of the county has a initial value of a = 5000 and a rate of change \(\frac{\Delta F}{\Delta t} = 3000 \). The linear formula for the food supply can be written \[F=1000+3000t . \nonumber\]
Now you try the following example. Determine which functions are exponential. Then write a formula of the form \(f(x)=ab^{x} \) for each exponential function.
| x | f(x) | g(x) | h(x) |
| 0 | 3 | 54 | 12 |
| 1 | 12 | 18 | 10 |
| 2 | 48 | 6 | 8 |
| 3 | 192 | 2 | 6 |
| 4 | 768 | \(\frac{2}{3}\) | 4 |
Solution
The functions (\f \) and (\g \) are exponential with a growth factor b = 4 and b = \(\frac{1}{3}\), respectively. The function (\h \) is not exponential, since it has a constant rate of change \(\frac{\Delta y}{\Delta x} = -2 \) and not a constant growth factor.
The formulas for the exponential functions are \(f(x)=3(4)^{x} \) and \(g(x)=54(\frac{1}{3})^{x} \).
Focus: Graphing Exponential Functions
Now let's explore the graphs of exponential functions to help us further model relationships with exponential functions.
| x | 0 | 1 | 2 | 3 | 4 |
| \(y = f(x) = 1(2)^{x} \) | 1 | 2 | 4 | 8 | 16 |
Notice f is increasing for all values of x since we are multiplying by a growth factor b = 2 for each increase in x by 1. Notice the right end behavior of the graph: as x gets larger and larger approaching infinity , y gets larger and larger approaching infinity ( x \(\rightarrow \infty \) , y=f(x) \(\rightarrow \infty \) ). Also, for negative values of x we have.
| x | -1 | -2 | -3 |
| \(y = 1(2)^{x} \) | \(1(2)^{-1} = \frac{1}{2^{1}} =\frac{1}{2}\) | \(1(2)^{-2} = \frac{1}{2^{2}} = \frac{1}{4}\) | \(1(2)^{-3} = \frac{1}{2^{3}} = \frac{1}{8}\) |
Notice the left end behavior of the graph: as x becomes larger and larger magnitude negative number approaching negative infinity , y gets smaller and smaller in size approaching zero ( x \(\rightarrow - \infty \) , y=f(x) \(\rightarrow 0 \) ).
Graphing the function, we have
Features of the graph:
- The y-intercept is at y = 1. Equivalently, the y-intercept is the point (0,1).
- The functions is increasing at an increasing rate.
- The right end behavior of the graph: as x \(\rightarrow \infty \) , y=f(x) \(\rightarrow \infty \).
- The left end behavior of the graph: as x \(\rightarrow -\infty \) , y=f(x) \(\rightarrow 0 \).
Graphs of increasing exponential functions have the same general shape as this graph.
| x | 0 | 1 | 2 | 3 |
| \(y = 8(\frac{1}{2})^{x} \) | 8 | 4 | 2 | 1 |
Notice f is decreasing for all values of x since we are multiplying by the growth factor b = \(\frac{1}{2}) for each increase in x by 1. Also, for negative values of x we have.
| x | -1 | -2 | -3 |
| \(y = 8 (\frac{1}{2})^{x} \) | \(8 (\frac{1}{2})^{-1} = 8(2)^{1} =16\) | \(8 (\frac{1}{2})^{-2} = 8(2)^{2} =32\) | \(8 (\frac{1}{2})^{-3} = 8(2)^{1} =64 \) |
Notice the left end behavior of the graph: as x \(\rightarrow - \infty \) , y=f(x) \(\rightarrow \infty \).
Graphing the function, we have
Features of the graph:
- The y-intercept is at y = 8. Equivalently, the y-intercept is the point (0,8).
- The functions is decreasing. The rate of decrease is getting smaller in size.
- The right end behavior of the graph: as x \(\rightarrow \infty \) , y=f(x) \(\rightarrow 0 \).
- The left end behavior of the graph: as x \(\rightarrow -\infty \) , y=f(x) \(\rightarrow \infty \) .
Graphs of decreasing exponential functions have the same general shape as this graph.
Key Points:
The graph of the exponential function \(f(x)=ab^{x} \) has a y-intercept at y = a. The function is increasing when the growth factor b >1 and decreasing when 0<b<1.
The population of a county has the formula \(P=1000(3)^{t} \), where t is the decades from 1960. This is an increasing exponential function since our growth factor b = 3 > 1 with y-intercept at y = 1000. The food supply of the county has a formula \(F=1000+3000t \). This is an increasing linear function with slope m = 3000 and y-intercept at y = 1000.
Graphing the both functions, we have
Notice, that in the long run the exponential function increases at a faster rate than the linear function. In general, exponential growth will outpace linear growth in the long run. In the case of our example, this illustrates that an exponentially growing population will outpace its food supply if it grows linearly.
Example 3.1.4
Sketch a graph of \(f(x)=4\left(\dfrac{1}{3} \right)^{x}\).
Solution
This graph will have a y-intercept at y=4, and pass through the point \(\left(1,\dfrac{4}{3} \right)\). Since \(b < 1\), the function is decreasing.
Student Example 3.1.2
Graph the following functions on the same axis:
\(f(x)=(2)^{x}\) ; \(g(x)=2(2)^{x}\); \(h(x)=2(1/2)^{x}\).
- Answer
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Student Example 3.1.3
Match each equation with its graph.
\[{f(x)=2(1.3)^{x} }\nonumber\]
\[{g(x)=2(1.8)^{x} }\nonumber \]
\[{h(x)=4(1.3)^{x} }\nonumber \]
\[{k(x)=4(0.7)^{x} }\nonumber\]
Solution
The graph of \(k(x)\) is the easiest to identify, since it is the only equation with a growth factor between zero and one, which will produce a decreasing graph. The graph of \(h(x)\) can be identified as the only growing exponential function with a y-intercept at y = 4 or (0,4). The graphs of \(f(x)\) and \(g(x)\) both have a y-intercept intercept at y = 2 or (0,2), but since \(g(x)\) has a larger growth factor, we can identify it as the graph increasing faster.
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1- the data was reported by berkeleyearth at https://berkeleyearth.org on February 13, 2023.
Important Topics of this Section
- Exponential Growth and Decay
- Growth factor
- The formula for an exponential function
- Graphs of exponential functions