1.6 Measuring How Functions Change

iPod photo by mikefoster on Pixabay iPod photo by mikefoster on Pixabay

Introduction

When the stock market opened on Monday, October 22, 2001, stock in Enron (a large Fortune-500 energy company) was trading for about $20 per share.

At the same time stock in Apple (the electronics manufacturer) could also be purchased for $20.

Without additional information, it would be natural to assume that state of both companies was the same. In reality, however, they were headed in opposite directions.

Later that day, the SEC announced they were investigating Enron for accounting irregularities. Within weeks the company declared bankruptcy and several Enron executives were sent to prison for fraud and conspiracy.

Apple, on the other hand, had an important announcement of their own. The next day, on October 23, 2001, Apple CEO Steve Jobs unveiled the very first iPod. It was an iconic success that revolutionized the consumer electronics industry and helped establish Apple as one of the most innovative technology companies in the world.

As we've seen earlier, functions, just like stocks, can increase and decrease in value. Our goal in this section is to develop tools that allow us to measure, compare and categorize functions based on how they change.

Intervals

In section 1.3 we saw how to visually identify regions where a function is increasing, decreasing or constant. We also saw portions of graphs that were concave up or concave down.

Since functions can behave differently on different subsets of their domain, we will need to describe the specific intervals we are inspecting.

For instance, on this graph

the function is increasing on the interval $(a,b)$, constant on $(b,c)$, decreasing over the interval $(c,d)$ and increasing again on $(d,e)$. Generally a function is not increasing, decreasing or constant at the endpoints of an interval so we usually write those intervals with parenthesis $( \thinspace\thinspace)$, and not brackets $[ \thinspace\thinspace ]$ unless circumstances dictate otherwise.

Take special note that these intervals are only on the x-axis. That is because they are parts of the domain of the function. We will use y-values in a moment to measure how much the function has increased or decreased.

In addition to identifying a specific interval, we might also need to measure how long it is. The length of an interval from $a$ to $b$ is given by

[\Delta x = b-a]

The upper-case Greek leter delta ($\Delta$) is used throughout math and science to indicate change. If we mention the "change in the x-values" then that is the same as $\Delta x$ which is also the same thing as the length of the interval $[a, b]$.

QUICK CHECK

  1. Determine the specific intervals where this graph is increasing, decreasing and/or constant.

    Answer
    The graph is constant on the interval $(-\infty,-1)$, increasing on $(-1,0)$, and decreasing on $(0,\infty)$.

  2. What is the length of the interval $[-2, 5]$?

    Answer
    [ \Delta x = 5-(-2) = 7]

  3. What is the change in the x-values from $x=4$ to $x=10$?

    Answer
    [\begin{align} \Delta x &= 10-4 \newline &= 6 \end{align}]

Total Change

The amount that a function has increased or decreased on an interval is called the total change of the function. The symbol $\Delta y$ means the change in the y-values and will be used to represent total change.

To calculate the total change of a function on some interval $[a, b]$ we find the difference between the starting value $f(a)$ and the ending value $f(b)$.

[\Delta y = f(b)-f(a)]

Notice that the total change does not care about what happens in the middle of the interval, it only uses the values of the function at the ends of the interval.

Both $\Delta y$ and $\Delta x$ are illustrated in the diagram below.

QUICK CHECK

Use the graph of $f(x)$ to answer the questions below.

  1. Find $\Delta y$ on the interval $[-1, 3]$.

    Answer
    [\begin{align} f(b)-f(a)&=f(3)-f(-1)\newline &=-4-0 \newline &=-4 \end{align}]

  2. Calculate $\Delta y$ on the interval $[3, 6]$.

    Answer
    [\begin{align} f(b)-f(a)&=f(6)-f(3) \newline &= 4-(-4) \newline &= 8 \end{align}]

Percent Change

If you won a million dollars you might feel like that's a really big change. But if someone told you the government was going to increase national spending on education by a million dollars you'd probably feel like that is a very small change, even though the amount is exactly the same. That is one of the limitations of total change; it doesn't relate the change to the starting value. To do that we need a percent change formula.

The percent change of a function $f$ over an interval $[a, b]$ is the ratio of the total change to the starting value. We calculate percent change as follows:

[\text{percent change} = \frac{f(b)-f(a)}{f(a)}]

With this formula the percent change is given in decimal form. In other words, a value such as $0.12$ should be interpreted as $12\%$. Also note that the percent change cannot be computed if $f(a)=0$.

QUICK CHECK

  1. Suppose the number of scholarships given by a university increased from 125 to 180. What is the percent change?

    Answer
    [ \begin{align}\frac{f(b)-f(a)}{f(a)} \thinspace &= \thinspace \frac{180-125}{125} \newline \thinspace &= \thinspace \frac{55}{125} \newline \thinspace &= \thinspace 0.44\end{align}] which we interpret as a $44\text{%}$ increase.

  2. The cost per person of sharing a \$30 pizza between $x$ people is given by the function $f(x) = \frac{30}{x}$. Find the percent change between sharing it with 3 people instead of 2.

    Answer
    [ \begin{align}\frac{f(b)-f(a)}{f(a)} &= \frac{f(3)-f(2)}{f(2)} \newline &= \frac{\frac{30}{3}-\frac{30}{2}}{\frac{30}{2}} \newline &= \frac{10-15}{15} \newline &= \frac{-5}{15} \newline &\approx -0.333 \end{align}] which we interpret as roughly a $33.3\text{%}$ decrease.

Average Rate of Change

While percent change helps you see the relative increase or decrease in values, it does not indicate how quickly the change is happening. To see how rapidly function values change we need to average the total change across the length of the interval.

You are probably already familiar with the basic process of finding averages. For instance, if you drove a distance of 100 miles in 2 hours then your average speed would be $\frac{100 \text{ miles}}{2 \text{ hours}} = 50 \text{ miles per hour}$.

Your actual speed may have gone up and down during the trip but, on average, your speed was 50 miles per hour.

To find the average change of a function we follow the same process and divide the total change of the function $\Delta y$ by the corresponding change $\Delta x$ in the input values to get $\frac{\Delta y}{\Delta x}$.

[ \frac{\Delta y}{\Delta x} = \frac{f(b)-f(a)}{b-a}]

Since $\frac{\Delta y}{\Delta x}$ is a ratio, it is often called the average rate of change of a function. The resulting quantities describe per unit changes. That is to say, by averaging we can tell what will happen to $f(x)$ every time $x$ increases by $1$.

To see how this works let's find the average rate of change of $f(x) = \sqrt{x}$ on the interval $[9, 100]$. Using the formula above we see that

[ \begin{align} \frac{\Delta y}{\Delta x} &= \frac{\sqrt{100} - \sqrt{9}} {100 - 9} \newline &= \frac{10-3}{91} \newline &= \frac{7}{91} \newline &\approx 0.077 \end{align}]

This means that from $x=9$ to $x=100$ the function $f(x)=\sqrt x$ increases, on average, by $0.077$ units every time $x$ increases by $1$ unit.

It's important to point out that unless the function is linear, the average rate of change of a function will be different over different intervals. The value we get depends on the specific interval $[a, b]$ chosen.

QUICK CHECK

Calculate the average rate of change of $f(x)=\sqrt x$ over the interval $[0,16]$.
Answer
[\begin{align}\frac{\Delta y}{\Delta x} &= \frac{\sqrt{16}-\sqrt{0}}{16-0} \newline &= \frac{4-0}{16} \newline &= \frac{1}{4} \end{align}]

The average rate of change formula might look familiar. That's because it is the same thing as the slope of the line connecting the points $(a, f(a))$ and $(b, f(b))$. This line is called a secant line.

When a function is defined by a graph, the average rate of change can always be found by calculating the slope of the secant line.

QUICK CHECK

  1. Use the graph above to find $\frac{\Delta y}{\Delta x}$ on the interval $[-1, 1]$.

    Answer
    [\begin{align} \frac{\Delta y}{\Delta x} &= \frac{f(1)-f(-1)}{1-(-1)} \newline &= \frac{2-0}{2} \newline &= 1\end{align}]

  2. From the graph, what is $\frac{\Delta y}{\Delta x}$ on the interval $[1, 6]$?

    Answer
    [\begin{align}\frac{\Delta y}{\Delta x} &= \frac{f(6)-f(1)}{6-1} \newline &= \frac{4-2}{6-1} \newline &= \frac{2}{5}\end{align}]

Concavity and Rates that Change

We've seen before that functions can curve up or curve down allowing the function to change a different speeds. In other words, concavity is an indication that the rates of change themselves can change.

For instance, an increasing function could grow at a steady rate, at a rate that quickens, or at a rate that slows. All three options are shown below.

Functions can also decrease at constant, increasing, and decreasing rates.

While working with these changes in the rate is a topic best studied in calculus, it's not too difficult to recognize situations where rates change.

QUICK CHECK

For each of the scenarios below, decide if the rates of change would be increasing, decreasing or constant.

  1. The temperature of a turkey as it roasts in an oven.

    Answer
    The temperature would increase at a decreasing rate.

  2. The world record in the 100 meter dash.

    Answer
    This is decreasing at a decreasing rate.

  3. The altitude of a space ship as it is launched from Earth.

    Answer
    Altitude would increase at an increasing rate.

  4. The height of a ball dropped off a cliff.

    Answer
    Decreases at an increasing rate.

Difference Quotients

The formulas we have used so far show what happens on a specific interval. We'll now build a mechanism that has the potential to describe the behavior of a function on every interval all at once.

The tool that does this is called the difference quotient. The difference quotient $D(x)$ is a function that lets us calculate the average rate of change on a generic interval from $x$ to $x+h$, where $h$ is any positive number. This is the same thing as the average rate of change, but with a change in notation.

To see how the difference quotient is derived from the average rate of change formula, we substitute $\color{red}{x}$ in place of $\color{red}{a}$, and $\color{blue}{x+h}$ in place of $\color{blue}{b}$ and simplify.

[ \begin{align} \frac{f(\color{blue}{b})-f(\color{red}{a})}{\color{blue}{b}-\color{red}{a}} &= \frac{f(\color{blue}{x+h})-f(\color{red}{x})}{\color{blue}{(x+h)}-\color{red}{x}} \newline \newline &= \frac{f(x+h)-f(x)}{h} \end{align} ]

The behavior of $D(x)$ is connected to how the function $f(x)$ changes in two important ways. First, whether the function is strictly increasing, decreasing or constant on an interval impacts the sign of $D(x)$.

  • If $f(x)$ is increasing on an interval, then the difference quotient is positive on that same interval.
  • If $f(x)$ is decreasing on an interval, then the difference quotient will be negative on that interval.
  • If $f(x)$ is constant over an interval, then the difference quotient is $0$ over that interval.

Secondly, the concavity of the function tells us if $D(x)$ itself is increasing, decreasing or constant. That is to say,

  • If $f(x)$ is concave up on an interval, then $D(x)$ increases over that interval.
  • If $f(x)$ is concave down on an interval, then $D(x)$ decreases on that interval.
  • On any interval where $f(x)$ is a straight line, $D(x)$ will be constant.

Let's examine a few simple examples. The function $f(x)=3x+2$ is an increasing linear function, so we expect the difference quotient to be a positive constant value, and it is.

It is interesting to note that the value of the difference quotient is exactly the same as the value of the slope of the line.

For our final example, we'll look at the square function $f(x)=x^{2}$. We know that this function is concave up over its entire domain of all real numbers, so we expcect its difference quotient will always be increasing.

Notice that no matter what value $h$ happens to be, $D(x)$ is always increasing since it is a linear function with a slope of $2$.

Looking Ahead

Exploring how functions change is a fascinating topic and is foundational to calculus. Not only will calculus teach you more about efficiently computing average rates of change, you will also learn to use the difference quotient to calculate instantaneous rates change, leading to the concept of a derivative.

But for now, it is enough to know that functions with similar behaviors have similar difference quotients, allowing us to group them together into different families that we will encounter in the next chapters.

Creative Commons LicenseThis work copyrighted by Nolan Mitchell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Source files are available at https://github.com/duragauge/duragauge.github.io.

You can also download a PDF copy of the entire book here. The PDF was last updated on January 21, 2024, and may not reflect changes to the online content made after that date.