# 1.3: Rates of Change and Behavior of Graphs

- Page ID
- 15568

Skills to Develop

- Find the average rate of change of a function.
- Use a graph to determine where a function is increasing, decreasing, or constant.
- Use a graph to locate local maxima and local minima.
- Use a graph to locate the absolute maximum and absolute minimum.

Gasoline costs have experienced some wild fluctuations over the last several decades. Table \(\PageIndex{1}\) lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year.

\(x\) | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 |
---|---|---|---|---|---|---|---|---|

\(C(x)\) | 2.31 | 2.62 | 2.84 | 3.30 | 2.41 | 2.84 | 3.58 | 3.68 |

If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these.

## Finding the Average Rate of Change of a Function

The price change per year is a **rate of change** because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table \(\PageIndex{1}\) did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the **average rate of change** over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value.

\[\begin{align*} \text{Average rate of change}&=\dfrac{\text{Change in output}}{\text{Change in input}} \\[5pt] &=\dfrac{\Delta y}{\Delta x}\\[5pt] &=\dfrac{y_2-y_1}{x_2-x_1}\\[5pt] &=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\end{align*} \label{1.3.1}\]

The Greek letter \(\Delta\) (delta) signifies the change in a quantity; we read the ratio as “delta-\(y\) over delta-\(x\)” or “the change in \(y\) divided by the change in \(x\).” Occasionally we write \(\Delta f\) instead of \(\Delta y\), which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function.

In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was

\[\dfrac{\Delta y}{\Delta x}=\dfrac{$1.37}{7 \text{years}}\approx \text{0.196 dollars per year.} \label{1.3.2} \nonumber\]

On average, the price of gas increased by about 19.6¢ each year. Other examples of rates of change include:

- A population of rats increasing by 40 rats per week
- A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes)
- A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon)
- The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage
- The amount of money in a college account decreasing by $4,000 per quarter

Definition: Rate of Change

A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.”

The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.

\[\dfrac{\Delta y}{\Delta x}=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\nonumber\]

Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values \(x_1\) and \(x_2\)

- Calculate the difference \(y_2−y_1=\Delta y\).
- Calculate the difference \(x_2−x_1=\Delta x\).
- Find the ratio \(\dfrac{\Delta y}{\Delta x}\).

Example \(\PageIndex{1}\): Computing an Average Rate of Change

Using the data in Table \(\PageIndex{1}\), find the average rate of change of the price of gasoline between 2007 and 2009.

**Solution**

In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is

\[\begin{align*} \dfrac{\Delta y}{\Delta x}&=\dfrac{y_2−y_1}{x_2−x_1} \\[5pt] &=\dfrac{$2.41−$2.84}{2009−2007} \\[5pt] &=\dfrac{−$0.43}{2 \text{ years}} \\[5pt] &=−$0.22 \text{ per year} \end{align*}\]

**Analysis**

Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases.

\(\PageIndex{1}\)

Using the data in Table \(\PageIndex{1}\), find the average rate of change between 2005 and 2010.

**Answer**-
\(\dfrac{$2.84−$2.315}{5 \text{ years}} =\dfrac{$0.535}{5 \text{ years}} =$0.106 \text{ per year}\)

Example \(\PageIndex{2}\): Computing Average Rate of Change from a Graph

Given the function \(g(t)\) shown in Figure \(\PageIndex{1}\), find the average rate of change on the interval \([−1,2]\).

*Figure \(\PageIndex{1}\): **Graph of a parabola.*

**Solution**

At \(t=−1\), Figure \(\PageIndex{2}\) shows \(g(−1)=4\). At \(t=2\),the graph shows \(g(2)=1\).

*Figure \(\PageIndex{2}\): **Graph of a parabola with a line from points (-1, 4) and (2, 1) to show the changes for g(t) and t.*

The horizontal change \(\Delta t=3\) is shown by the red arrow, and the vertical change \(\Delta g(t)=−3\) is shown by the turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of

\[\dfrac{1−4}{2−(−1)}=\dfrac{−3}{3}=−1 \nonumber\]

**Analysis**

Note that the order we choose is very important. If, for example, we use \(\dfrac{y_2−y_1}{x_1−x_2}\), we will not get the correct answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as \((x_1,y_1)\) and \((x_2,y_2)\).

Example \(\PageIndex{3}\): Computing Average Rate of Change from a Table

After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values are shown in Table \(\PageIndex{2}\). Find her average speed over the first 6 hours.

t (hours) |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|---|

D(t)(miles) | 10 | 55 | 90 | 153 | 214 | 240 | 292 | 300 |

**Solution**

Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours, for an average speed of

\[\begin{align*}\dfrac{292−10}{6−0}&=\dfrac{282}{6}\\[5pt] &=47\end{align*}\]

The average speed is 47 miles per hour.

**Analysis**

Because the speed is not constant, the average speed depends on the interval chosen. For the interval \([2,3]\), the average speed is 63 miles per hour.

Example \(\PageIndex{4}\): Computing Average Rate of Change for a Function Expressed as a Formula

Compute the average rate of change of \(f(x)=x^2−\frac{1}{x}\) on the interval \([2, 4]\).

**Solution**

We can start by computing the function values at each endpoint of the interval.

\[\begin{align*}f(2)\quad&= \quad 2^2−\frac{1}{2}\quad\quad\quad f(4)&= \quad 4^2−\frac{1}{4} \\[5pt] &= \quad4−\frac{1}{2} &= \quad 16−\frac{1}{4} \\[5pt] &= \quad \frac{7}{2} &= \quad\frac{63}{4}\end{align*}\]

Now we compute the average rate of change.

\[\begin{align*}\text{Average rate of change} &=\dfrac{f(4)−f(2)}{4−2} \\[5pt] &=\dfrac{\frac{63}{4}-\frac{7}{2}}{4-2} \\[5pt] &=\dfrac{\frac{49}{4}}{2} \\[5pt] &= \dfrac{49}{8}\end{align*}\]

\(\PageIndex{2}\)

Find the average rate of change of \(f(x)=x−2\sqrt{x}\) on the interval \([1, 9]\).

**Answer**-
\(\frac{1}{2}\)

Example \(\PageIndex{5}\): Finding the Average Rate of Change of a Force

The electrostatic force \(F\) between two charged particles, measured in newtons, can be related to the distance between the particles \(d\), in centimeters, by the formula \(F(d)=\frac{2}{d^2}\). Find the average rate of change of force if the distance between the particles is increased from 2 cm to 6 cm.

**Solution**

We are computing the average rate of change of \(F(d)=\dfrac{2}{d^2}\) on the interval \([2,6]\).

\[\begin{align*}\text{Average rate of change }&=\dfrac{F(6)−F(2)}{6−2} \\[5pt] &=\dfrac{\frac{2}{6^2}-\frac{2}{2^2}}{6-2} & \text{Simplify} \\[5pt] &=\dfrac{\frac{2}{36}-\frac{2}{4}}{4} \\[5pt] &=\dfrac{-\frac{16}{36}}{4} & \text{Combine numerator terms} \\[5pt] &=−\dfrac{1}{9} & \text{Simplify}\end{align*}\]

The average rate of change is \(−\frac{1}{9}\) newton per centimeter.

Example \(\PageIndex{6}\): Finding an Average Rate of Change as an Expression

Find the average rate of change of \(g(t)=t^2+3t+1\) on the interval \([0, a]\). The answer will be an expression involving \(a\). This will be valuable for **calculus**.

**Solution**

We use the average rate of change formula.

\(\begin{align*}\text{Average rate of change} &=\dfrac{g(a)−g(0)}{a−0} & \text{Evaluate} \\[5pt] &=\dfrac{(a^2+3a+1)−(0^2+3(0)+1)}{a−0} & \text{Simplify} \\[5pt] &=\dfrac{a^2+3a+1−1}{a} & \text{Simplify and factor}\\[5pt] &= \dfrac{a(a+3)}{a} & \text{Divide by the common factor } a\\[5pt] &= a+3 \end{align*}\)

This result tells us the average rate of change (in terms of \(a\)) between \(t=0\) and any other point \(t=a\). For example, on the interval \([0,5]\), the average rate of change would be \(a+3 = 5+3=8\).\(\PageIndex{3}\)

Find the average rate of change of \(f(x)=x^2+2x−8\) on the interval \([5, a]\).

**Answer**-
\(a+7\)

## Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant

As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an interval where a function is increasing is positive, and the average rate of change of an interval where a function is decreasing is negative. Figure \(\PageIndex{3}\) shows examples of increasing and decreasing intervals on a function. This will be valuable for **calculus**.

**Figure \(\PageIndex{3}\): **The function \(f(x)=x^3−12x\) is increasing on \((−\infty, −2)\cup (2,\infty)\) and is decreasing on \((−2, 2)\).

While some functions are increasing (or decreasing) over their entire domain, many others are not. (**Note**: if you can think of an example of a function that is increasing on part of its domain and decreasing on another part of its domain, write it down and give it to your instructor for ** Extra Credit.**) If there is an input value at which a function changes from increasing to decreasing (as we go from left to right; that is, as the input variable increases), then the corresponding output value is called a

**local maximum**. If a function has more than one local maximum, we say it has local

**maxima**. Similarly, if there is an input value at which a function changes from decreasing to increasing as the input variable increases, then the corresponding output value is called a

**local minimum**. The plural form is “local

**minima**.” Together, local maxima and minima are called

**local extrema**, or local extreme values, of the function. (The singular form is “extremum.”) Often, the term "local" is replaced by the term "relative." In this text, we will use the term "local."

Clearly, a function is neither increasing nor decreasing on an interval where the outputs are constant. A function is also considered neither increasing nor decreasing at extrema. Note that because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain, we have to speak of local extrema. If the function **does** have a highest output over the entire domain, this output is called a **global maximum **(to be discussed later in this Section).

For the function whose graph is shown in Figure \(\PageIndex{4}\), the local maximum is 16, and it occurs at \(x=−2\). The local minimum is −16 and it occurs at \(x=2\).

**Figure \(\PageIndex{4}\):** Graph of a polynomial that shows the increasing and decreasing intervals and local maximum.

To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure \(\PageIndex{5}\) illustrates these ideas for a local maximum.

**Figure \(\PageIndex{5}\): Definition of a local maximum**

These observations lead us to a formal definition of local extrema. This will be valuable for **calculus**.

Increasing and Decreasing Functions, Local Minima and Local Maxima

- A function \(f\) is an
**increasing function**on an open interval \( (a,b) \) if \(f(x_2)>f(x_1)\) for every \(x_1\), \(x_2\) in \( (a,b) \) where \(x_2>x_1\). - A function \(f\) is a
**decreasing function**on an open interval \( (a,b) \) if \(f(x_2)<f(x_1)\) for every \(x_1\), \(x_2\) in \( (a,b) \) where \(x_2>x_1\).

A function \(f\) has a **local maximum** at a point \(c\) in an open interval \((a,b)\) if \(f(c)\) is greater than or equal to \(f(x)\) for every point \(x\) in \((a,b)\). Likewise, \(f\) has a **local minimum** at a point \(c\) in \((a,b)\) if \(f(c)\) is less than or equal to \(f(x)\) for every \(x\) in \((a,b)\).

Example \(\PageIndex{7}\) Finding Increasing and Decreasing Intervals on a Graph

Given the function \(p(t)\) in Figure \(\PageIndex{6}\), identify the intervals on which the function appears to be increasing.

**Figure \(\PageIndex{6}\): **Graph of a polynomial.

**Solution**

We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from \(t=1\) to \(t=3\) and from \(t=4\) on.

In interval notation, we would say the function appears to be increasing on the interval \((1,3)\) and on the interval \((4,\infty)\).

**Analysis**

Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at \(t=1\), \(t=3\), and \(t=4\). At these points there are local extrema (two minima and a maximum).

Example \(\PageIndex{8}\): Finding Local Maxima and Minima from a Graph

For the function \(f\) whose graph is shown in Figure \(\PageIndex{7}\), find all local maxima and minima.

**Figure \(\PageIndex{7}\):** Graph of a polynomial.

**Solution**

Observe the graph of \(f\). The graph attains a local maximum at \(x=1\) because it is the highest point in an open interval around \(x=1\). The local maximum is the \(y\)-coordinate at \(x=1\), which is \(2\).

The graph attains a local minimum at \(x=−1\) because it is the lowest point in an open interval around \(x=−1\). The local minimum is the \(y\)-coordinate at \(x=−1\), which is \(−2\).

## Analyzing the Toolkit Functions for Increasing or Decreasing Intervals

We will now return to our toolkit functions and discuss their graphical behavior in Figure \(\PageIndex{8}\). (Note: there is an error in interval notation in this chart. If you can find it, write up your correction and give it to your instructor for **Extra Credit**.)

**Figure \(\PageIndex{8}\)**

## Use A Graph to Locate the Global Maximum and Global Minimum

There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally) and locating the highest and lowest points on the graph for the entire domain. The \(y\)-coordinates (output) at the highest and lowest points are called the **global maximum** and **global minimum**, respectively. (The term "absolute" is also used in place of the term "global.") To locate global maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function (Figure \(\PageIndex{9}\)).

**Figure \(\PageIndex{9}\)**: Graph of a segment of a parabola with a global minimum at (0, -2) and global maximum at (2, 2).

Not every function has a global maximum or minimum value. The toolkit function \(f(x)=x^3\) is one such function.

Definition: Global Maxima and Minima

- The
**global maximum**of \(f\) at \(x=c\) is \(f(c)\) where \(f(c)≥f(x)\) for all \(x\) in the domain of \(f\). - The
**global minimum**of \(f\) at \(x=d\) is \(f(d)\) where \(f(d)≤f(x)\) for all \(x\) in the domain of \(f\).

Example \(\PageIndex{9}\): Finding Absolute Maxima and Minima from a Graph

For the function \(f\) shown in Figure \(\PageIndex{10}\), find all global maxima and minima.

*Figure \(\PageIndex{14}\): **Graph of a polynomial defined on the interval \([-2.5, 3]\).*

**Solution**

Observe the graph of \(f\). The graph attains a global maximum at two locations, \(x=−2\) and \(x=2\), because at these locations, the graph attains its highest point on the domain of the function. The global maximum is the \(y\)-coordinate at \(x=−2\) and \(x=2\), which is 16.

The graph attains a global minimum at \(x=3,\) because it is the lowest point within the domain of the function's graph. The global minimum is the \(y\)-coordinate at \(x=3,\) which is \(-10\).

**Note**: In addition to the global minimum at \(x=3\), there are two local minima. If you see them, write down the \(x\)-value at which each one occurs, as well as the \(y\)-value, and give it to your instructor for *Extra Credit.*

## Key Equations

- Average rate of change: \(\dfrac{\Delta y}{\Delta x}=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\)

## Key Concepts

- A rate of change relates a change in an output quantity to a change in an input quantity.
- Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example 1.3.2.
- Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Examples 1.3.1 and 1.3.3.
- An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Examples 1.3.4 and 1.3.5.
- The average rate of change can sometimes be determined as an expression. See Example 1.3.6.
- A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See Example 1.3.7.
- A local (or relative) maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values.
- A local (or relative) minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values.
- Minima and maxima are also called extrema.
- We can find local extrema from a graph. See Example 1.3.8.
- The highest and lowest points on a graph indicate the global (or absolute) maxima and minima. See Example 1.3.9.

## Contributors

Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.