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Rates of Change and Behavior of Graphs

Find the average rate of change of a function.

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 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.

\[\begin{array}{lll}\text{Average rate of change} & = & \frac{\text{Change in output}}{\text{Change in input}} \\ & = & \frac{\Delta y}{\Delta x} \\ & = & \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\ & = & \frac{f({x}_{2})-f({x}_{1})}{{x}_{2}-{x}_{1}}\end{array}\]

The Greek letter $\text{\Delta }$ (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 $\text{\Delta }f$ instead of $\text{\Delta }y,$ 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

\[\frac{\text{\Delta }y}{\text{\Delta }x}=\frac{\text{\$}1.37}{\text{7 years}}\approx 0.196\ \text{dollars per year}\]

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

Condensed — the full section is in OpenStax College Algebra 2e.

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 increasing function is positive, and the average rate of change of a decreasing function is negative. shows examples of increasing and decreasing intervals on a function.

While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is the location of a local maximum. The function value at that point is the local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is the location of a local minimum. The function value at that point is the 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 it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain.

For the function whose graph is shown in , the local maximum is 16, and it occurs at $x=-2.$ The local minimum is $-16$ and it occurs at $x=2.$

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. illustrates these ideas for a local maximum.

These observations lead us to a formal definition of local extrema.

Condensed — the full section is in OpenStax College Algebra 2e.

Use A Graph to Locate the Absolute Maximum and Absolute 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 $y\text{-}$ coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively.

To locate absolute 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. See .

Not every function has an absolute maximum or minimum value. The toolkit function $f(x)={x}^{3}$ is one such function.

Example

Try it.

For the function $f$ shown in , find all absolute maxima and minima.

Solution

Observe the graph of $f.$ The graph attains an absolute maximum in two locations, $x=-2$ and $x=2,$ because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at $x=-2$ and $x=2,$ which is $16.$

The graph attains an absolute minimum at $x=3,$ because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at $x=3,$ which is $-10.$

Key Concepts

A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See .
Identifying points that mark the interval on a graph can be used to find the average rate of change. See .
Comparing pairs of input and output values in a table can also be used to find the average rate of change. See .
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 and .
The average rate of change can sometimes be determined as an expression. See .
A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See .
A local 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 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 and .
The highest and lowest points on a graph indicate the maxima and minima. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the slope of the line shown.

Bonisa impendulo

Locate two points on the graph whose coordinates are integers.
$(0,5)=({x}_{1},{y}_{1})$
$(3,3)=({x}_{2},{y}_{2})$
It may help to visualize this change as $m=\frac{\text{rise}}{\text{run}}$ . Count the rise between the points: down 2 units. Then count the run, or horizontal change: to the right 3 units. Note, since the line goes down, the slope is negative. Use the slope formula: $m=\frac{\text{rise}}{\text{run}}=\frac{({y}_{2}-{y}_{1})}{({x}_{2}-{x}_{1})}=\frac{3-5}{3-0}\frac{-2}{3}$
Complete this table’s y-values, and then graph the line. $y=\frac{2}{3}x+1$

$x$ $y$
-6
-3
0
3
6

$x$ $y$
-6
-3
0
3
6
What is the slope of this line? ________ What is the y-intercept of this line? ( ________ , ________ )
The following table shows the number of Associates degrees awarded (in thousands) in the US for several years.
Year Number of Associate’s Degrees Earned (in thousands)
2000 569
2001 579
2005 668
2010 719
2014 1,003
Find the following average rates of change, being careful to attach units to your answers.
ⓐ Between 2005 and 2000. Average rate of change= $\frac{f({x}_{2})-f({x}_{1})}{{x}_{2}-{x}_{1}}$
ⓑ Between 2001 and 2010. Average rate of change= $\frac{f({x}_{2})-f({x}_{1})}{{x}_{2}-{x}_{1}}$
ⓒ Between 2014 and 2010. Average rate of change= $\frac{f({x}_{2})-f({x}_{1})}{{x}_{2}-{x}_{1}}$
ⓐ Complete the following table of values for $f(x)={x}^{2}+2x-8$

x f(x)
–2
2
4

ⓑ Use the table above to find the average rate of change between x=–2 and x=2.
ⓒ Use the table above to find the average rate of change between x=2 and x=4
ⓓ Sketch the graph of f(x) below and graph the lines with slopes equal to the average rates of change found in parts ⓑ and ⓒ.
1. ⓐ Complete the following table of values for $g(x)=\frac{1}{x+3}$
  
  x g(x)
  –1
  0
  2
2. ⓑ Use the table above to find the average rate of change between x=–1 and x=0.
3. ⓒ Use the table above to find the average rate of change between x=–1 and x=2.
4. ⓓ Sketch the graph of g(x) below and graph the lines with slopes equal to the average rates of change found in parts ⓑ and ⓒ.
Using the data in , find the average rate of change of the price of gasoline between 2007 and 2009.

Bonisa impendulo

In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is
\[\begin{array}{lll}\frac{\Delta y}{\Delta x} & = & \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\ & = & \frac{\$2.41-\$2.84}{2009-2007} \\ & = & \frac{-\$0.43}{2\ \text{years}} \\ & = & -\$0.22\ \text{per year}\end{array}\]
Using the data in , find the average rate of change between 2005 and 2010.

Bonisa impendulo

$\frac{\$2.84-\$2.31}{5\ \text{years}}=\frac{\$0.53}{5\ \text{years}}=\$0.106$ per year.
Given the function $g(t)$ shown in , find the average rate of change on the interval $[-1,2].$

Bonisa impendulo

At $t=-1,$ shows $g(-1)=4.$ At $t=2,$ the graph shows $g(2)=1.$
The horizontal change $\text{\Delta }t=3$ is shown by the red arrow, and the vertical change $\text{\Delta }g(t)=-3$ is shown by the turquoise arrow. The average rate of change is shown by the slope of the orange line segment. The output changes by –3 while the input changes by 3, giving an average rate of change of
\[\frac{1-4}{2-(-1)}=\frac{-3}{3}=-1\]
After picking up a friend who lives 10 miles away and leaving on a trip, Anna records her distance from home over time. The values are shown in . 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

Bonisa impendulo

Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours.
\[\begin{array}{lll}\frac{292-10}{6-0} & = & \frac{282}{6} \\ & = & 47\end{array}\]
The average speed is 47 miles per hour.
Compute the average rate of change of $f(x)={x}^{2}-\frac{1}{x}$ on the interval $\text{[2,}\ \text{4].}$

Bonisa impendulo

We can start by computing the function values at each endpoint of the interval.
\[\begin{array}{llllll}f(2) & = & {2}^{2}-\frac{1}{2} & \ f(4) & = & {4}^{2}-\frac{1}{4} \\ & = & 4-\frac{1}{2} & & = & 16-\frac{1}{4} \\ & = & \frac{7}{2} & & = & \frac{63}{4}\end{array}\]
Now we compute the average rate of change.
\[\begin{array}{lll}\text{Average rate of change} & = & \frac{f(4)-f(2)}{4-2} \\ & = & \frac{\frac{63}{4}-\frac{7}{2}}{4-2} \\ & = & \frac{\frac{49}{4}}{2} \\ & = & \frac{49}{8}\end{array}\]
Find the average rate of change of $f(x)=x-2\sqrt{x}$ on the interval $[1,\ 9].$

Bonisa impendulo

$\frac{1}{2}$
The electrostatic force $F,$ measured in newtons, between two charged particles 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.

Bonisa impendulo

We are computing the average rate of change of $F(d)=\frac{2}{{d}^{2}}$ on the interval $[2,6].$
\[\begin{array}{llll}\text{Average rate of change} & = & \ \frac{F(6)-F(2)}{6-2} & \\ & = & \frac{\frac{2}{{6}^{2}}-\frac{2}{{2}^{2}}}{6-2} & \text{Simplify}. \\ & = & \frac{\frac{2}{36}-\frac{2}{4}}{4} & \\ & = & \frac{-\frac{16}{36}}{4} & \text{Combine numerator terms}. \\ & = & -\frac{1}{9} & \text{Simplify}\end{array}\]
The average rate of change is $-\frac{1}{9}$ newton per centimeter.
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 $a$ in simplest form.

Bonisa impendulo

We use the average rate of change formula.
\[\begin{array}{llll}\text{Average rate of change} & = & \frac{g(a)-g(0)}{a-0} & \text{Evaluate}. \\ & = & \frac{({a}^{2}+3a+1)-({0}^{2}+3(0)+1)}{a-0} & \text{Simplify}. \\ & = & \frac{{a}^{2}+3a+1-1}{a} & \text{Simplify and factor}. \\ & = & \frac{a(a+3)}{a} & \text{Divide by the common factor }a. \\ & = & a+3 & \end{array}\]
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 $5+3=8.$
Find the average rate of change of $f(x)={x}^{2}+2x-8$ on the interval $[5,a]$ in simplest forms in terms of $a.$

Bonisa impendulo

$a+7$
Given the function $p(t)$ in , identify the intervals on which the function appears to be increasing.

Bonisa impendulo

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 the interval $(4,\infty ).$
Graph the function $f(x)=\frac{2}{x}+\frac{x}{3}.$ Then use the graph to estimate the local extrema of the function and to determine the intervals on which the function is increasing.

Bonisa impendulo

Using technology, we find that the graph of the function looks like that in . It appears there is a low point, or local minimum, between $x=2$ and $x=3,$ and a mirror-image high point, or local maximum, somewhere between $x=-3$ and $x=-2.$
Graph the function $f(x)={x}^{3}-6{x}^{2}-15x+20$ to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.

Bonisa impendulo

The local maximum appears to occur at $(-1,28),$ and the local minimum occurs at $(5,-80).$ The function is increasing on $(-\infty ,-1)\cup (5,\infty )$ and decreasing on $(-1,5).$
For the function $f$ whose graph is shown in , find all local maxima and minima.

Bonisa impendulo

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 $-2.$
For the function $f$ shown in , find all absolute maxima and minima.

Bonisa impendulo

Observe the graph of $f.$ The graph attains an absolute maximum in two locations, $x=-2$ and $x=2,$ because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at $x=-2$ and $x=2,$ which is $16.$
The graph attains an absolute minimum at $x=3,$ because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at $x=3,$ which is $-10.$
Can the average rate of change of a function be constant?

Bonisa impendulo

Yes, the average rate of change of all linear functions is constant.
If a function $f$ is increasing on $(a,b)$ and decreasing on $(b,c),$ then what can be said about the local extremum of $f$ on $(a,c)?$
How are the absolute maximum and minimum similar to and different from the local extrema?

Bonisa impendulo

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.
How does the graph of the absolute value function compare to the graph of the quadratic function, $y={x}^{2},$ in terms of increasing and decreasing intervals?
$f(x)=4{x}^{2}-7$ on $[1,\ b]$

Bonisa impendulo

$4(b+1)$
$g(x)=2{x}^{2}-9$ on $[4,\ b]$
$p(x)=3x+4$ on $[2,\ 2+h]$

Bonisa impendulo

3
$k(x)=4x-2$ on $[3,\ 3+h]$
$f(x)=2{x}^{2}+1$ on $[x,x+h]$

Bonisa impendulo

$4x+2h$
$g(x)=3{x}^{2}-2$ on $[x,x+h]$
$a(t)=\frac{1}{t+4}$ on $[9,9+h]$

Bonisa impendulo

$\frac{-1}{13(13+h)}$
$b(x)=\frac{1}{x+3}$ on $[1,1+h]$
$j(x)=3{x}^{3}$ on $[1,1+h]$

Bonisa impendulo

$3{h}^{2}+9h+9$
$r(t)=4{t}^{3}$ on $[2,2+h]$
Find $\frac{f(x+h)-f(x)}{h}$ given $f(x)=2{x}^{2}-3x$ on $[x,x+h]$

Bonisa impendulo

$4x+2h-3$
Estimate the average rate of change from $x=1$ to $x=4.$
Estimate the average rate of change from $x=2$ to $x=5.$

Bonisa impendulo

$\frac{4}{3}$
Estimate the intervals where the function is increasing or decreasing.
Estimate the point(s) at which the graph of $f$ has a local maximum or a local minimum.

Bonisa impendulo

local maximum: $(-3,\ 60),$ local minimum: $(3,\ -60)$
If the complete graph of the function is shown, estimate the intervals where the function is increasing or decreasing.
If the complete graph of the function is shown, estimate the absolute maximum and absolute minimum.

Bonisa impendulo

absolute maximum at approximately $(7,\ 150),$ absolute minimum at approximately $(-7.5,\ -220)$

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Rates of Change and Behavior of Graphs

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.
Find the slope of a line (IA 3.2.1)

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Zama wena

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Year	Number of Associate’s Degrees Earned (in thousands)
2000	569
2001	579
2005	668
2010	719
2014	1,003

x	f(x)
–2
2
4

x	g(x)
–1
0
2

t (hours)	0	1	2	3	4	5	6	7
D(t) (miles)	10	55	90	153	214	240	292	300