Skip to main content

Section 1.3 Average Rate of Change

Objectives: Student Learning Outcomes

After completing this lesson you will be able to:

  1. Perform calculations using function notation including Average Rate of Change.

  2. Define an interval and understand the use of inequality notation to denote an interval.

In this activity we introduce the Greek letter \(\Delta\) that represents change, and we develop the concept of average rate of change.

The prerequisites for this lesson are knowing how to calculate the slope of a line and understanding the meaning of slope.

Subsection 1.3.1 Intuitive Knowledge

Let's start by realizing you already understand the concept of average rate of change, but maybe you have never heard the phrase before.

It turns out, average rate of change is just another way to talk about slope. As the name implies, it measures how quickly things change, on average.

Common examples of average rate of change:

  • Traveling from Portland, OR to Seattle, WA (about \(200\) miles) in \(4\) hours is an average speed of 50 mph.

  • Pressure at the surface of the ocean is about 14.6 (pounds per square inch). If you dive under water \(20\) feet the pressure is about 23.6 . That's an average rate of pressure increase of about 0.45 pounds per square inch, per foot underwater.

  • Earning \(\$108\) for \(6\) hours of work is an average hourly pay rate of \(18\) dollars per hour.

In each example there is a function that relates an input with an output. When the input changes by some amount, the output will also change (though usually by some other amount).

The change in the output divided by the change in the input is called an average rate of change of the function.

Be careful to remember that we are only talking about an average. For instance, no one can drive from Portland to Seattle at a constant speed. There is always stop and go traffic, variations in speed limits, etc.

The interactive graph below is that of a function \(D = f(t)\) where D is the number of miles a train is from downtown at \(t\) hours after 8 am.

In math we use the Greek letter \(\Delta\) to mean “change in”. Change between two values is typically measured by subtracting an old value from a new value.

In our train story where \(D = f(t)\text{,}\) the expression \(\Delta f\) means “change in miles from downtown”. As the train travels away from downtown, the distance from downtown increases. New distance is greater than old distance, so traveling away from downtown is a positive change in distance.

However, the distance from downtown decreases as the train travels towards downtown. New distance is less than old distance, therefore traveling toward downtown is a negative change in distance.

Subsection 1.3.2 Computing Average Rate of Change

As we move from one point to another point on a function, there are two changes that take place: The change in the input is called \(\Delta\text{Input}\text{,}\) and the change in the output is called \(\Delta\text{Output}\text{.}\)

If we divide these two changes we get an average rate of change.

An average rate of change \(=\frac{\Delta Output}{\Delta Input}\text{.}\)

The function below shows a Demand Curve, which shows the relationship between the price at which an item is sold, and the number of items people will buy at that price. When the price is low, people will buy more of them — when the price is high, people won't buy as many.

Figure 1.3.5. Demand Curve

The labeled slope triangle indicates that when the price of the item increases from \(\$2\) to \(\$5\text{,}\) the number sold will decrease from \(2.5\) thousand to \(1.5\) thousand. The average rate of change of thousands sold with respect to price in dollars is:

\begin{equation*} \frac{-1000}{3} \approx -333.33 \frac{\text{sold}}{\text{\$}}\text{.} \end{equation*}

Does this look familiar? Think “rise over run”.

Once again, average rate of change is just slope. You should be very familiar with the concept of slope, especially as it relates to lines and the famous equation

\begin{equation*} y = mx + b \end{equation*}

where \(m\) represents the slope of the line. Now we are going to use our newly learned function notation to write slope in a different way.

Remember, the notation \(f(3)\) represents an output value; it's a number. It's the output when the input is \(3\text{.}\)

Next is the easy part: making the actual calculations. We need to find the actual value of our two outputs and we need to find the difference between them. In math “difference” essentially means to subtract.

As demonstrated in the previous exercise, the average rate of change of a function depends on what interval (section) of the input axis we use.

As a practical example, consider the average daily temperature in Portland as a function of the month. Temperatures tend to climb as winter transitions into summer (a positive average rate of change), while temperatures tend to drop as summer transitions back into winter (a negative rate of change).

Let \(T = f(m)\) be the function that relates the average high temperatures, \(T\) degrees Fahrenheit, in Portland as a function of \(m\text{,}\) the number of months after December 31 in any given year.

Earlier you found the outputs by estimating the results from a graph. Next you will find the outputs by calculating them directly using a formula.

One question that comes up a lot is, “Does it matter what order we evaluate \(\Delta f\) and \(\Delta x\) ?”

To answer this question recall earlier you used \(f(x) = (x-2)(x+3)\) to calculate \(f(0)\) and \(f(-2)\text{.}\) Use these values to calculate the average rate of change in two different directions.

The only thing that matters is that the ordered pairs are kept together. In the previous exercise the ordered pairs were \((0, f(0))\) and \((-2,f(-2))\text{.}\) If you choose to start with \(x = 0\) in the denominator then you must start with \(f(0)\) in the numerator.

On the other hand, if you choose to start with \(x = -2\) in the denominator then you must start with \(f(-2)\) in the numerator. We say that the ordered pairs must be vertically aligned.

Let's finish this activity by going back to the train example where the distance (miles) of the train from downtown is a function of the time (hours) after 8 a.m. In other words \(D = f(t)\text{.}\)