APC The Average Rate of Change of a Function

Section 1.3 The Average Rate of Change of a Function

Motivating Questions

What do we mean by the average rate of change of a function on an interval?
What does the average rate of change of a function measure? How do we interpret its meaning in context?
How is the average rate of change of a function connected to a line that passes through two points on the curve?

Given a function that models a certain phenomenon, it’s natural to ask such questions as “how is the function changing on a given interval” or “on which interval is the function changing more rapidly?” The concept of average rate of change enables us to make these questions more mathematically precise. Initially, we will focus on the average rate of change of an object moving along a straight-line path.

For a function \(s\) that tells the location of a moving object along a straight path at time \(t\text{,}\) we define the average rate of change of \(s\) on the interval \([a,b]\) to be the quantity

\begin{equation*} AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}\text{.} \end{equation*}

Note particularly that the average rate of change of \(s\) on \([a,b]\) is measuring the change in position divided by the change in time.

Preview Activity 1.3.1.

Let the height function for a ball tossed vertically be given by \(s(t) = 64 - 16(t-1)^2\text{,}\) where \(t\) is measured in seconds and \(s\) is measured in feet above the ground.

Compute the value of \(AV_{[1.5,2.5]}\text{.}\)
What are the units on the quantity \(AV_{[1.5,2.5]}\text{?}\) What is the meaning of this number in the context of the rising/falling ball?
In Desmos, plot the function \(s(t) = 64 - 16(t-1)^2\) along with the points \((1.5,s(1.5))\) and \((2.5, s(2.5))\text{.}\) Make a copy of your plot on the axes in Figure 1.3.1, labeling key points as well as the scale on your axes. What is the domain of the model? The range? Why?

Figure 1.3.1. Axes for plotting the position function.
Work by hand to find the equation of the line through the points \((1.5,s(1.5))\) and \((2.5, s(2.5))\text{.}\) Write the line in the form \(y = mt + b\) and plot the line in Desmos, as well as on the axes above.
What is a geometric interpretation of the value \(AV_{[1.5,2.5]}\) in light of your work in the preceding questions?
How do your answers in the preceding questions change if we instead consider the interval \([0.25, 0.75]\text{?}\) \([0.5, 1.5]\text{?}\) \([1,3]\text{?}\)

Subsection 1.3.1 Defining and interpreting the average rate of change of a function

In the context of a function that measures height or position of a moving object at a given time, the meaning of the average rate of change of the function on a given interval is the average velocity of the moving object because it is the ratio of change in position to change in time. For example, in Preview Activity 1.3.1, the units on \(AV_{[1.5,2.5]} = -32\) are “feet per second” since the units on the numerator are “feet” and on the denominator “seconds”. Morever, \(-32\) is numerically the same value as the slope of the line that connects the two corresponding points on the graph of the position function, as seen in Figure 1.3.2. The fact that the average rate of change is negative in this example indicates that the ball is falling.

Figure 1.3.2. The average rate of change of \(s\) on \([1.5,2.5]\) for the function in Preview Activity 1.3.1.

Figure 1.3.3. The average rate of change of an abstract function \(f\) on the interval \([a,b]\text{.}\)

While the average rate of change of a position function tells us the moving object’s average velocity, in other contexts, the average rate of change of a function can be similarly defined and has a related interpretation. We make the following formal definition.

Definition 1.3.4.

For a function \(f\) defined on an interval \([a,b]\text{,}\) the average rate of change of \(f\) on \([a,b]\) is the quantity

\begin{equation*} AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}\text{.} \end{equation*}

In every situation, the units on the average rate of change help us interpret its meaning, and those units are always “units of output per unit of input.” Moreover, the average rate of change of \(f\) on \([a,b]\) always corresponds to the slope of the line between the points \((a,f(a))\) and \((b,f(b))\text{,}\) as seen in Figure 1.3.3.

Activity 1.3.2.

According to the US census, the populations of Kent and Ottawa Counties in Michigan where GVSU is located

Grand Rapids is in Kent, Allendale in Ottawa.

from 1960 to 2010 measured in \(10\)-year intervals are given in the following tables.

Table 1.3.5. Kent County population data.

1960	1970	1980	1990	2000	2010
363,187	411,044	444,506	500,631	574,336	602,622

Table 1.3.6. Ottawa county population data.

1960	1970	1980	1990	2000	2010
98,719	128,181	157,174	187,768	238,313	263,801

Let \(K(Y)\) represent the population of Kent County in year \(Y\) and \(W(Y)\) the population of Ottawa County in year \(Y\text{.}\)

Compute \(AV_{[1990,2010]}\) for both \(K\) and \(W\text{.}\)
What are the units on each of the quantities you computed in (a.)?
Write a careful sentence that explains the meaning of the average rate of change of the Ottawa county population on the time interval \([1990,2010]\text{.}\) Your sentence should begin something like “In an average year between 1990 and 2010, the population of Ottawa County was \(\ldots\)”
Which county had a greater average rate of change during the time interval \([2000,2010]\text{?}\) Were there any intervals in which one of the counties had a negative average rate of change?
Using the given data, what do you predict will be the population of Ottawa County in 2018? Why?

The average rate of change of a function on an interval gives us an excellent way to describe how the function behaves, on average. For instance, if we compute \(AV_{[1970,2000]}\) for Kent County, we find that

\begin{equation*} AV_{[1970,2000]} = \frac{574,336 - 411,044}{30} = 5443.07\text{,} \end{equation*}

which tells us that in an average year from 1970 to 2000, the population of Kent County increased by about \(5443\) people. Said differently, we could also say that from 1970 to 2000, Kent County was growing at an average rate of \(5443\) people per year. These ideas also afford the opportunity to make comparisons over time. Since

\begin{equation*} AV_{[1990,2000]} = \frac{574,336 - 500,631}{10} = 7370.5\text{,} \end{equation*}

we can not only say that Kent county’s population increased by about \(7370\) in an average year between 1990 and 2000, but also that the population was growing faster from 1990 to 2000 than it did from 1970 to 2000.

Finally, we can even use the average rate of change of a function to predict future behavior. Since the population was changing on average by \(7370.5\) people per year from 1990 to 2000, we can estimate that the population in 2002 is

\begin{equation*} K(2002) \approx K(2000) + 2 \cdot 7370.5 = 574,336 + 14,741 = 589,077\text{.} \end{equation*}

Subsection 1.3.2 How average rate of change indicates function trends

We have already seen that it is natural to use words such as “increasing” and “decreasing” to describe a function’s behavior. For instance, for the tennis ball whose height is modeled by \(s(t) = 64 - 16(t-1)^2\text{,}\) we computed that \(AV_{[1.5,2.5]} = -32\text{,}\) which indicates that on the interval \([1.5,2.5]\text{,}\) the tennis ball’s height is decreasing at an average rate of \(32\) feet per second. Similarly, for the population of Kent County, since \(AV_{[1990,2000]} = 7370.5\text{,}\) we know that on the interval \([1990,2000]\) the population is increasing at an average rate of \(7370.5\) people per year.

We make the following formal definitions to clarify what it means to say that a function is increasing or decreasing.

Definition 1.3.7.

Let \(f\) be a function defined on an interval \((a,b)\) (that is, on the set of all \(x\) for which \(a \lt x \lt b\)). We say that \(f\) is increasing on \((a,b)\) provided that the function is always rising as we move from left to right. That is, for any \(x\) and \(y\) in \((a,b)\text{,}\) if \(x \lt y\text{,}\) then \(f(x) \lt f(y)\text{.}\)

Similarly, we say that \(f\) is decreasing on \((a,b)\) provided that the function is always falling as we move from left to right. That is, for any \(x\) and \(y\) in \((a,b)\text{,}\) if \(x \lt y\text{,}\) then \(f(x) \gt f(y)\text{.}\)

If we compute the average rate of change of a function on an interval, we can decide if the function is increasing or decreasing on average on the interval, but it takes more work

Calculus offers one way to justify that a function is always increasing or always decreasing on an interval.

to decide if the function is increasing or decreasing always on the interval.

Activity 1.3.3.

Let’s consider two different functions and see how different computations of their average rate of change tells us about their respective behavior. Plots of \(q\) and \(h\) are shown in Figures 1.3.8 and 1.3.9.

Consider the function \(q(x) = 4-(x-2)^2\text{.}\) Compute \(AV_{[0,1]}\text{,}\) \(AV_{[1,2]}\text{,}\) \(AV_{[2,3]}\text{,}\) and \(AV_{[3,4]}\text{.}\) What do your last two computations tell you about the behavior of the function \(q\) on \([2,4]\text{?}\)
Consider the function \(h(t) = 3 - 2(0.5)^t\text{.}\) Compute \(AV_{[-1,1]}\text{,}\) \(AV_{[1,3]}\text{,}\) and \(AV_{[3,5]}\text{.}\) What do your computations tell you about the behavior of the function \(h\) on \([-1,5]\text{?}\)
On the graphs in Figures 1.3.8 and 1.3.9, plot the line segments whose respective slopes are the average rates of change you computed in (a) and (b).

Figure 1.3.8. Plot of \(q\) from part (a).

Figure 1.3.9. Plot of \(h\) from part (b).
True or false: Since \(AV_{[0,3]} = 1\text{,}\) the function \(q\) is increasing on the interval \((0,3)\text{.}\) Justify your decision.
Give an example of a function that has the same average rate of change no matter what interval you choose. You can provide your example through a table, a graph, or a formula; regardless of your choice, write a sentence to explain.

It is helpful be able to connect information about a function’s average rate of change and its graph. For instance, if we have determined that \(AV_{[-3,2]} = 1.75\) for some function \(f\text{,}\) this tells us that, on average, the function rises between the points \(x = -3\) and \(x = 2\) and does so at an average rate of \(1.75\) vertical units for every horizontal unit. Moreover, we can even determine that the difference between \(f(2)\) and \(f(-3)\) is

\begin{equation*} f(2)-f(-3) = 1.75 \cdot 5 = 8.75 \end{equation*}

since \(\frac{f(2)-f(-3)}{2-(-3)} = 1.75\text{.}\)

Activity 1.3.4.

Sketch at least two different possible graphs that satisfy the criteria for the function stated in each part. Make your graphs as significantly different as you can. If it is impossible for a graph to satisfy the criteria, explain why.

\(f\) is a function defined on \([-1,7]\) such that \(f(1) = 4\) and \(AV_{[1,3]} = -2\text{.}\)
\(g\) is a function defined on \([-1,7]\) such that \(g(4) = 3\text{,}\) \(AV_{[0,4]} = 0.5\text{,}\) and \(g\) is not always increasing on \((0,4)\text{.}\)
\(h\) is a function defined on \([-1,7]\) such that \(h(2) = 5\text{,}\) \(h(4) = 3\) and \(AV_{[2,4]} = -2\text{.}\)

Subsection 1.3.3 Summary

For a function \(f\) defined on an interval \([a,b]\text{,}\) the average rate of change of \(f\) on \([a,b]\) is the quantity

\begin{equation*} AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}\text{.} \end{equation*}
The value of \(AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}\) tells us how much the function rises or falls, on average, for each additional unit we move to the right on the graph. For instance, if \(AV_{[3,7]} = 0.75\text{,}\) this means that for additional \(1\)-unit increase in the value of \(x\) on the interval \([3,7]\text{,}\) the function increases, on average, by \(0.75\) units. In applied settings, the units of \(AV_{[a,b]}\) are “units of output per unit of input”.
The value of \(AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}\) is also the slope of the line that passes through the points \((a,f(a))\) and \((b,f(b))\) on the graph of \(f\text{,}\) as shown in Figure 1.3.3.

Exercises 1.3.4 Exercises

1.

Let \(P_1\) and \(P_2\) be the populations (in hundreds) of Town 1 and Town 2, respectively. The table below shows data for these two populations for five different years.

Year	1980	1983	1987	1993	1999
\(P_1\)	49	53	57	61	65
\(P_2\)	79	72	65	58	51

Find the average rate of change of each population over each of the time intervals below.

(a) From 1980 to 1987, the average rate of change of the population of Town 1 was hundred people per year, and the average rate of change of the population of Town 2 was hundred people per year.

(b) From 1987 to 1999, the average rate of change of the population of Town 1 was hundred people per year, and the average rate of change of the population of Town 2 was hundred people per year.

(c) From 1980 to 1999, the average rate of change of the population of Town 1 was hundred people per year, and the average rate of change of the population of Town 2 was hundred people per year.

2.

(a) What is the average rate of change of \(g(x) = -6 - 5 x\) between the points \((-4, 14)\) and \((5,-31)\text{?}\)

answer =

(b) The function \(g\) is

increasing
decreasing

on the interval \(-4 \leq x \leq 5\text{.}\)

3.

Find the average rate of change of \(f(x) = 3 x^2 +7\) between each of the pairs of points below.

(a) Between \((3, 34)\) and \((5, 82)\)

answer =

(b) Between \((c, k)\) and \((q, t)\)

answer =

(c) Between \((x,\ f(x))\) and \((x+h,\ f(x+h))\)

answer =

4.

In 2005, you have 45 CDs in your collection. In 2008, you have 130 CDs. In 2012, you have 50 CDs. What is the average rate of change in the size of your CD collection between:

(a) 2005 and 2008?

(b) 2008 and 2012?

5.

Based on the graphs of \(f(x)\) and \(g(x)\) below, answer the following questions. You should not approximate any of your answers.

a) What is the average rate of change of \(f(x)\) over the interval \(\ 2.2 \leq x \leq 6.1 \ \) ?

b) What is the average rate of change of \(g(x)\) over the interval \(\ 2.2 \leq x \leq 6.1 \ \) ?

6.

The graph below shows the distance traveled, \(D\) (in miles) as a function of time, \(t\) (in hours).

(Click on the graph to get a larger version.)

a) For each of the intervals, find the values of \(\Delta D\) and \(\Delta t\) between the indicated start and end times. Enter your answers in their respective columns in the table below.

Time Interval	\(\Delta D\)	\(\Delta t\)
t = 1.5 to t = 4.5
t = 2 to t = 4.5
t = 1 to t = 3

b) Based on your results from (a) it follows that the average rate of change of \(D\) is constant, it does not depend over which interval of time you choose. What is the constant rate of change of \(D\text{?}\)

\(\frac{ \Delta D }{ \Delta t} =\)

c) Which of the statements below CORRECTLY explains the significance of your answer to part (b)? Select ALL that apply (more than one may apply).

It is the average velocity of the car over the first two hours.
It is the total distance the car travels in five hours.
It is how far the car will travel in a half-hour.
It represents the car’s velocity.
It is the acceleration of the car over the five hour time interval.
It is the slope of the line.
None of the above

7.

Let \(f(x) = 36 - x^2\text{.}\)

Compute each of the following expressions and interpret each as an average rate of change.
1. \(\displaystyle{\frac{f(4) - f(0)}{4 - 0} = {}}\)
2. \(\displaystyle{\frac{f(6) - f(4)}{6 - 4 } = {}}\)
3. \(\displaystyle{\frac{f(6) - f(0)}{6 - 0} = {}}\)
Based on the graph sketched below, match each of your answers in (i)–(iii) with one of the lines labeled A–F. Type the corresponding letter of the line segment next to the appropriate formula. Clearly not all letters will be used.

\(\displaystyle{\dfrac{f(4) - f(0)}{4 - 0}}\)
- A
- B
- C
- D
- E
- F
\(\displaystyle{\dfrac{f(6) - f(4)}{6 - 4}}\)
- A
- B
- C
- D
- E
- F
\(\displaystyle{\dfrac{f(6) - f(0)}{6 - 0}}\)
- A
- B
- C
- D
- E
- F

8.

The table below gives the average temperature, \(T\text{,}\) at a depth \(d\text{,}\) in a borehole in Belleterre, Quebec.

\(d \mbox{, depth} \ (m)\)	\(T \mbox{, temp} \ \left( C^{\circ} \right)\)
25	5.50
50	5.20
75	5.10
100	5.10
125	5.30
150	5.50
175	5.75
200	6.00
225	6.25
250	6.50
275	6.75
300	7.00

Evaluate \(\Delta T / \Delta d\) on the following intervals

a) \(125 \leq d \leq 275\) \(\Delta T / \Delta d\) =

b) \(25 \leq d \leq 125\) \(\Delta T / \Delta d\) =

c) \(75 \leq d \leq 200\) \(\Delta T / \Delta d\) =

d) Which of the statements below correctly explains the significance of your answer to part (c)? Select all that apply (more than one may apply).

On average, the temperature is changing at a rate of 0.0072 degrees Celsius per minute over the interval \(75 \leq d \leq 200\) .
0.0072 is the slope of the graph of at \(d = 75\text{.}\)
The temperature changes by a total of 0.0072 degrees Celsius when moving from a depth 75 meters to 200 meters.
Over the interval from 75 meters to 200 meters, the temperature changes on average at a rate of 0.0072 degrees Celsius per meter.
The temperature is changing at a rate of 0.0072 degrees Celsius per minute when the depth is 75 meters.
None of the above

9.

A cold can of soda is removed from a refrigerator. Its temperature \(F\) in degrees Fahrenheit is measured at \(5\)-minute intervals, as recorded in the following table.

Table 1.3.10. Data for the soda’s temperature as a function of time.

\(t\) (minutes)	\(0\)	\(5\)	\(10\)	\(15\)	\(20\)	\(25\)	\(30\)	\(35\)
\(F\) (Fahrenheit temp)	\(37.00\)	\(44.74\)	\(50.77\)	\(55.47\)	\(59.12\)	\(61.97\)	\(64.19\)	\(65.92\)

Determine \(AV_{[0,5]}\text{,}\) \(AV_{[5,10]}\text{,}\) and \(AV_{[10,15]}\text{,}\) including appropriate units. Choose one of these quantities and write a careful sentence to explain its meaning. Your sentence might look something like “On the interval \(\ldots\text{,}\) the temperature of the soda is \(\ldots\) on average by \(\ldots\) for each \(1\)-unit increase in \(\ldots\)”.
On which interval is there more total change in the soda’s temperature: \([10,20]\) or \([25,35]\text{?}\)
What can you observe about when the soda’s temperature appears to be changing most rapidly?
Estimate the soda’s temperature when \(t = 37\) minutes. Write at least one sentence to explain your thinking.

10.

The position of a car driving along a straight road at time \(t\) in minutes is given by the function \(y = s(t)\) that is pictured in Figure 1.3.11. The car’s position function has units measured in thousands of feet. For instance, the point \((2,4)\) on the graph indicates that after 2 minutes, the car has traveled 4000 feet.

Figure 1.3.11. The graph of \(y = s(t)\text{,}\) the position of the car (measured in thousands of feet from its starting location) at time \(t\) in minutes.

In everyday language, describe the behavior of the car over the provided time interval. In particular, carefully discuss what is happening on each of the time intervals \([0,1]\text{,}\) \([1,2]\text{,}\) \([2,3]\text{,}\) \([3,4]\text{,}\) and \([4,5]\text{,}\) plus provide commentary overall on what the car is doing on the interval \([0,12]\text{.}\)
Compute the average rate of change of \(s\) on the intervals \([3,4]\text{,}\) \([4,6]\text{,}\) and \([5,8]\text{.}\) Label your results using the notation “\(AV_{[a,b]}\)” appropriately, and include units on each quantity.
On the graph of \(s\text{,}\) sketch the three lines whose slope corresponds to the values of \(AV_{[3,4]}\text{,}\) \(AV_{[4,6]}\text{,}\) and \(AV_{[5,8]}\) that you computed in (b).
Is there a time interval on which the car’s average velocity is \(5000\) feet per minute? Why or why not?
Is there ever a time interval when the car is going in reverse? Why or why not?

11.

Consider an inverted conical tank (point down) whose top has a radius of \(3\) feet and that is \(2\) feet deep. The tank is initially empty and then is filled at a constant rate of \(0.75\) cubic feet per minute. Let \(V=f(t)\) denote the volume of water (in cubic feet) at time \(t\) in minutes, and let \(h= g(t)\) denote the depth of the water (in feet) at time \(t\text{.}\) It turns out that the formula for the function \(g\) is \(g(t) = \left( \frac{t}{\pi} \right)^{1/3}\text{.}\)

In everyday language, describe how you expect the height function \(h = g(t)\) to behave as time increases.
For the height function \(h = g(t) = \left( \frac{t}{\pi} \right)^{1/3}\text{,}\) compute \(AV_{[0,2]}\text{,}\) \(AV_{[2,4]}\text{,}\) and \(AV_{[4,6]}\text{.}\) Include units on your results.
Again working with the height function, can you determine an interval \([a,b]\) on which \(AV_{[a,b]} = 2\) feet per minute? If yes, state the interval; if not, explain why there is no such interval.
Now consider the volume function, \(V = f(t)\text{.}\) Even though we don’t have a formula for \(f\text{,}\) is it possible to determine the average rate of change of the volume function on the intervals \([0,2]\text{,}\) \([2,4]\text{,}\) and \([4,6]\text{?}\) Why or why not?

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