Increasing and Decreasing Functions

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. Figure 1 shows examples of increasing and decreasing intervals on a function.

Figure 1. The function [latex]f\left(x\right)={x}^{3}-12x[/latex] is increasing on [latex]\left(-\infty \text{,}-\text{2}\right){{\cup }^{\text{ }}}^{\text{ }}\left(2,\infty \right)[/latex] and is decreasing on [latex]\left(-2\text{,}2\right)[/latex].

This video further explains how to find where a function is increasing or decreasing.

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 called a 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 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 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 in Figure 2, the local maximum is 16, and it occurs at [latex]x=-2[/latex]. The local minimum is [latex]-16[/latex] and it occurs at [latex]x=2[/latex].

Figure 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. Figure 3 illustrates these ideas for a local maximum.

Figure 3. Definition of a local maximum.

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

Example 1: Finding Increasing and Decreasing Intervals on a Graph

Given the function [latex]p\left(t\right)[/latex] in the graph below, identify the intervals on which the function appears to be increasing.

Figure 4

Show 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 [latex]t=1[/latex] to [latex]t=3[/latex] and from [latex]t=4[/latex] on.

In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval [latex]\left(4,\infty \right)[/latex].

Analysis of the Solution

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 [latex]t=1[/latex] , [latex]t=3[/latex] , and [latex]t=4[/latex] . These points are the local extrema (two minima and a maximum).

Example 2: Finding Local Extrema from a Graph

Graph the function [latex]f\left(x\right)=\frac{2}{x}+\frac{x}{3}[/latex]. Then use the graph to estimate the local extrema of the function and to determine the intervals on which the function is increasing.

Show Solution

Using technology, we find that the graph of the function looks like that in Figure 5. It appears there is a low point, or local minimum, between [latex]x=2[/latex] and [latex]x=3[/latex], and a mirror-image high point, or local maximum, somewhere between [latex]x=-3[/latex] and [latex]x=-2[/latex].

Figure 5

Analysis of the Solution

Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 6 provides screen images from two different technologies, showing the estimate for the local maximum and minimum.

Figure 6

Based on these estimates, the function is increasing on the interval [latex]\left(-\infty ,-{2.449}\right)[/latex]
and [latex]\left(2.449\text{,}\infty \right)[/latex]. Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at [latex]\pm \sqrt{6}[/latex], but determining this requires calculus.)

Example 3: Finding Local Maxima and Minima from a Graph

For the function [latex]f[/latex] whose graph is shown in Figure 7, find all local maxima and minima.

Figure 7

Show Solution

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

The graph attains a local minimum at [latex]\text{ }x=-1\text{ }[/latex] because it is the lowest point in an open interval around [latex]x=-1[/latex]. The local minimum is the y-coordinate at [latex]x=-1[/latex], which is [latex]-2[/latex].

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 [latex]y\text{-}[/latex] 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 Figure 10.

Figure 10

Not every function has an absolute maximum or minimum value. The toolkit function [latex]f\left(x\right)={x}^{3}[/latex] is one such function.

A General Note: Absolute Maxima and Minima

The absolute maximum of [latex]f[/latex] at [latex]x=c[/latex] is [latex]f\left(c\right)[/latex] where [latex]f\left(c\right)\ge f\left(x\right)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex].

The absolute minimum of [latex]f[/latex] at [latex]x=d[/latex] is [latex]f\left(d\right)[/latex] where [latex]f\left(d\right)\le f\left(x\right)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex].

Example 4: Finding Absolute Maxima and Minima from a Graph

For the function [latex]f[/latex] shown in Figure 11, find all absolute maxima and minima.

Figure 11

Show Solution

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

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

Gasoline costs have experienced some wild fluctuations over the last several decades. The table below^[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.

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 the table above 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.

The Greek letter [latex]\Delta[/latex] (delta) signifies the change in a quantity; we read the ratio as “delta-y over delta-x” or “the change in [latex]y[/latex] divided by the change in [latex]x[/latex].” Occasionally we write [latex]\Delta f[/latex] instead of [latex]\Delta y[/latex], 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

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

Example 5: Computing an Average Rate of Change

Using the data in the table below, find the average rate of change of the price of gasoline between 2007 and 2009.

[latex]y[/latex]	2005	2006	2007	2008	2009	2010	2011	2012
[latex]C\left(y\right)[/latex]	2.31	2.62	2.84	3.30	2.41	2.84	3.58	3.68

Show Solution

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

[latex]\begin{align}\frac{\Delta y}{\Delta x}&=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\[1mm] &=\frac{2.41-2.84}{2009 - 2007} \\[1mm] &=\frac{-0.43}{2\text{ years}} \\[1mm] &={-0.22}\text{ per year}\end{align}[/latex]

Analysis of the Solution

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.

The following video provides another example of how to find the average rate of change between two points from a table of values.

Try It

Using the data in the table below, find the average rate of change between 2005 and 2010.

[latex]y[/latex]	2005	2006	2007	2008	2009	2010	2011	2012
[latex]C\left(y\right)[/latex]	2.31	2.62	2.84	3.30	2.41	2.84	3.58	3.68

Show Solution

[latex]\dfrac{$2.84-$2.31}{5\text{ years}}=\dfrac{$0.53}{5\text{ years}}=$0.106[/latex] per year.

Example 6: Computing Average Rate of Change from a Graph

Given the function [latex]g\left(t\right)[/latex] shown in Figure 5, find the average rate of change on the interval [latex]\left[-1,2\right][/latex].

Figure 5

Show Solution

Figure 6

At [latex]t=-1[/latex], the graph shows [latex]g\left(-1\right)=4[/latex]. At [latex]t=2[/latex], the graph shows [latex]g\left(2\right)=1[/latex].

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

[latex]\frac{1 - 4}{2-\left(-1\right)}=\frac{-3}{3}=-1[/latex]

Analysis of the Solution

Note that the order we choose is very important. If, for example, we use [latex]\frac{{y}_{2}-{y}_{1}}{{x}_{1}-{x}_{2}}[/latex], 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 [latex]\left({x}_{1},{y}_{1}\right)[/latex] and [latex]\left({x}_{2},{y}_{2}\right)[/latex].

Example 7: 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 the table below. 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	282	300

Show Solution

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

[latex]\frac{292 - 10}{6 - 0} =\frac{282}{6} =47[/latex]

The average speed is 47 miles per hour.

Analysis of the Solution

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 8: Computing Average Rate of Change for a Function Expressed as a Formula

Compute the average rate of change of [latex]f\left(x\right)={x}^{2}-\frac{1}{x}[/latex] on the interval [latex]\text{[2,}\text{4].}[/latex]

Show Solution

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

[latex]\begin{align}f\left(2\right)&={2}^{2}-\frac{1}{2} &&& f\left(4\right)&={4}^{2}-\frac{1}{4} \\ &=4-\frac{1}{2} &&&& =16-{1}{4} \\ &=\frac{7}{2} &&&& =\frac{63}{4} \end{align}[/latex]

Now we compute the average rate of change.

[latex]\begin{align}\text{Average rate of change}&=\frac{f\left(4\right)-f\left(2\right)}{4 - 2} \\[1mm] &=\frac{\frac{63}{4}-\frac{7}{2}}{4 - 2} \\[1mm] &=\frac{\frac{49}{4}}{2} \\[1mm] &=\frac{49}{8} \end{align}[/latex]

The following video provides another example of finding the average rate of change of a function given a formula and an interval.

Try It

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

Show Solution

[latex]\frac{1}{2}[/latex]

Try It

Example 9: Finding the Average Rate of Change of a Force

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

Show Solution

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

[latex]\begin{align}\text{Average rate of change }&=\frac{F\left(6\right)-F\left(2\right)}{6 - 2} \\[1mm] &=\frac{\frac{2}{{6}^{2}}-\frac{2}{{2}^{2}}}{6 - 2} && \text{Simplify}. \\[1mm] &=\frac{\frac{2}{36}-\frac{2}{4}}{4} \\[1mm] &=\frac{-\frac{16}{36}}{4} &&\text{Combine numerator terms}. \\[1mm] &=-\frac{1}{9}&&\text{Simplify}\end{align}[/latex]

The average rate of change is [latex]-\frac{1}{9}[/latex] newton per centimeter.

Example 10: Finding an Average Rate of Change as an Expression

Find the average rate of change of [latex]g\left(t\right)={t}^{2}+3t+1[/latex] on the interval [latex]\left[0,a\right][/latex]. The answer will be an expression involving [latex]a[/latex].

Show Solution

We use the average rate of change formula.

​[latex]\begin{align}\text{Average rate of change}&=\frac{g\left(a\right)-g\left(0\right)}{a - 0}&&\text{Evaluate} \\[1mm] &​=\frac{\left({a}^{2}+3a+1\right)-\left({0}^{2}+3\left(0\right)+1\right)}{a - 0}&&\text{Simplify}​ \\[1mm] &=\frac{{a}^{2}+3a+1 - 1}{a}&&\text{Simplify and factor}​ \\[1mm] &=\frac{a\left(a+3\right)}{a}&&\text{Divide by the common factor }a​ \\[1mm] &=a+3 \end{align}[/latex]

This result tells us the average rate of change in terms of [latex]a[/latex] between [latex]t=0[/latex] and any other point [latex]t=a[/latex]. For example, on the interval [latex]\left[0,5\right][/latex], the average rate of change would be [latex]5+3=8[/latex].

Try It

Find the average rate of change of [latex]f\left(x\right)={x}^{2}+2x - 8[/latex]. on the interval [latex]\left[5,a\right][/latex].

Show Solution

[latex]a+7[/latex]

Determining Even and Odd Functions

Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions [latex]f\left(x\right)={x}^{2}[/latex] or [latex]f\left(x\right)=|x|[/latex] will result in the original graph. We say that these types of graphs are symmetric about the y-axis. Functions whose graphs are symmetric about the y-axis are called even functions.

If the graphs of [latex]f\left(x\right)={x}^{3}[/latex] or [latex]f\left(x\right)=\frac{1}{x}[/latex] were reflected over both axes, the result would be the original graph.

Figure 7. (a) The cubic toolkit function (b) Horizontal reflection of the cubic toolkit function (c) Horizontal and vertical reflections reproduce the original cubic function.

We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function.

We use the expressions, “odd” and “even” because of polynomials. A polynomial function with only odd degree terms (odd powers of x) will be an odd function. A polynomial function with only even degree terms (even powers of x) will be an even function.

Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, [latex]f\left(x\right)={2}^{x}[/latex] is neither even nor odd. Also, the only function that is both even and odd is the constant function [latex]f\left(x\right)=0[/latex].

Example 11: Determining whether a Function Is Even, Odd, or Neither

Is the function [latex]f\left(x\right)={x}^{3}+2x[/latex] even, odd, or neither?

Show Solution

Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Let’s begin with the rule for even functions.

[latex]f\left(-x\right)={\left(-x\right)}^{3}+2\left(-x\right)=-{x}^{3}-2x[/latex]

This does not return us to the original function, so this function is not even. We can now test the rule for odd functions.

[latex]-f\left(-x\right)=-\left(-{x}^{3}-2x\right)={x}^{3}+2x[/latex]

Because [latex]-f\left(-x\right)=f\left(x\right)[/latex], this is an odd function.

Analysis of the Solution

Consider the graph of [latex]f[/latex]. Notice that the graph is symmetric about the origin. For every point [latex]\left(x,y\right)[/latex] on the graph, the corresponding point [latex]\left(-x,-y\right)[/latex] is also on the graph. For example, (1, 3) is on the graph of [latex]f[/latex], and the corresponding point [latex]\left(-1,-3\right)[/latex] is also on the graph.

Figure 8

Section 2.3 Homework Exercises

1. Can the average rate of change of a function be constant?

2. If a function [latex]f[/latex] is increasing on [latex]\left(a,b\right)[/latex] and decreasing on [latex]\left(b,c\right)[/latex], then what can be said about the local extremum of [latex]f[/latex] on [latex]\left(a,c\right)?[/latex]

3. How are the absolute maximum and minimum similar to and different from the local extrema?

4. How does the graph of the absolute value function compare to the graph of the quadratic function, [latex]y={x}^{2}[/latex], in terms of increasing and decreasing intervals?

5. What is the meaning if the average rate of change is zero?

For exercises 6–10, find the average rate of change of each function on the interval specified for real numbers [latex]b[/latex] or [latex]h[/latex].

6. [latex]f\left(x\right)=4{x}^{2}-7[/latex] on [latex]\left[1,\text{ }b\right][/latex]

7. [latex]g\left(x\right)=2{x}^{2}-9[/latex] on [latex]\left[4,\text{ }b\right][/latex]

8. [latex]p\left(x\right)=3x+4[/latex] on [latex]\left[2,\text{ }2+h\right][/latex]

9. [latex]g\left(x\right)=3{x}^{2}-2[/latex] on [latex]\left[x,x+h\right][/latex]

For exercises 10–12, consider the graph of [latex]f[/latex].

10. Estimate the average rate of change from [latex]x=1[/latex] to [latex]x=4[/latex].

11. Estimate the average rate of change from [latex]x=2[/latex] to [latex]x=5[/latex].

12. Estimate the average rate of change from [latex]x=5[/latex] to [latex]x=7[/latex].

For the following exercises, determine whether the function is odd, even, or neither.

For exercises 22–25, use the graph of each function to estimate the intervals on which the function is increasing or decreasing.

22.

23.

24.

25.

For exercises 26–27, consider the graph shown below.

26. Estimate the intervals where the function is increasing or decreasing.

27. Estimate the point(s) at which the graph of [latex]f[/latex] has a local maximum or a local minimum.

For exercises 28–29, consider the graph below.

28. If the complete graph of the function is shown, estimate the intervals where the function is increasing or decreasing.

29. If the complete graph of the function is shown, estimate the absolute maximum and absolute minimum.

30. The table below gives the annual sales (in millions of dollars) of a product from 1998 to 2006. What was the average rate of change of annual sales (a) between 2001 and 2002, and (b) between 2001 and 2004?

31. The table below gives the population of a town (in thousands) from 2000 to 2008. What was the average rate of change of population (a) between 2002 and 2004, and (b) between 2002 and 2006?

For the following exercises, find the average rate of change of each function on the interval specified.

32. [latex]f\left(x\right)={x}^{2}[/latex] on [latex]\left[1,\text{ }5\right][/latex]

33. [latex]h\left(x\right)=5 - 2{x}^{2}[/latex] on [latex]\left[-2,\text{4}\right][/latex]

34. [latex]q\left(x\right)={x}^{3}[/latex] on [latex]\left[-4,\text{2}\right][/latex]

35. [latex]g\left(x\right)=3{x}^{3}-1[/latex] on [latex]\left[-3,\text{3}\right][/latex]

36. [latex]y=\frac{1}{x}[/latex] on [latex]\left[1,\text{ 3}\right][/latex]

37. [latex]p\left(t\right)=\frac{\left({t}^{2}-4\right)\left(t+1\right)}{{t}^{2}+3}[/latex] on [latex]\left[-3,\text{1}\right][/latex]

38. [latex]k\left(t\right)=6{t}^{2}+\frac{4}{{t}^{3}}[/latex] on [latex]\left[-1,3\right][/latex]

For the following exercises, use a graphing utility to estimate the local extrema of each function and to estimate the intervals on which the function is increasing and decreasing.

39. [latex]f\left(x\right)={x}^{4}-4{x}^{3}+5[/latex]

40. [latex]h\left(x\right)={x}^{5}+5{x}^{4}+10{x}^{3}+10{x}^{2}-1[/latex]

41. [latex]g\left(t\right)=t\sqrt{t+3}[/latex]

42. [latex]k\left(t\right)=3{t}^{\frac{2}{3}}-t[/latex]

43. [latex]m\left(x\right)={x}^{4}+2{x}^{3}-12{x}^{2}-10x+4[/latex]

44. [latex]n\left(x\right)={x}^{4}-8{x}^{3}+18{x}^{2}-6x+2[/latex]

45. The graph of the function [latex]f[/latex] is shown below.

Based on the calculator screenshot, the point [latex]\left(1.333,\text{ }5.185\right)[/latex] is which of the following?
A) a relative (local) maximum of the function
B) the vertex of the function
C) the absolute maximum of the function
D) a zero of the function.

46. Let [latex]f\left(x\right)=\frac{1}{x}[/latex]. Find a number [latex]c[/latex] such that the average rate of change of the function [latex]f[/latex] on the interval [latex]\left(1,c\right)[/latex] is [latex]-\frac{1}{4}[/latex].

47. Let [latex]f\left(x\right)=\frac{1}{x}[/latex] . Find the number [latex]b[/latex] such that the average rate of change of [latex]f[/latex] on the interval [latex]\left(2,b\right)[/latex] is [latex]-\frac{1}{10}[/latex].

48. At the start of a trip, the odometer on a car read 21,395. At the end of the trip, 13.5 hours later, the odometer read 22,125. Assume the scale on the odometer is in miles. What is the average speed the car traveled during this trip?

49. A driver of a car stopped at a gas station to fill up his gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was full and he noticed that he had pumped 10.7 gallons. What is the average rate of flow of the gasoline into the gas tank?

50. Near the surface of the moon, the distance that an object falls is a function of time. It is given by [latex]d\left(t\right)=2.6667{t}^{2}[/latex], where [latex]t[/latex] is in seconds and [latex]d\left(t\right)[/latex] is in feet. If an object is dropped from a certain height, find the average velocity of the object from [latex]t=1[/latex] to [latex]t=2[/latex].

51. The graph below illustrates the decay of a radioactive substance over [latex]t[/latex] days.

Use the graph to estimate the average decay rate from [latex]t=5[/latex] to [latex]t=15[/latex].