1.4: Behavior of Graphs of Functions
In this section, you will:
- 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.
- Determine whether a function is even, odd, or neither from its graph.
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.
| \( y \) | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 |
| \( C ( y ) \) | 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.
Average Rate of Change
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.\[ \text{Average rate of change} = \dfrac{\text{Change in output}}{\text{Change in input}} = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 − y_1}{x_2 − x_1} = \dfrac{f ( x_2 ) − f ( x_1 )}{x_2 − x_1}. \nonumber \]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\[ \dfrac{\Delta y}{\Delta x} = \dfrac{\$ 1.37}{7 \text{years}} \approx 0.196 \text{ dollars per year} .\nonumber \]On average, the price of gas increased by about 19.6 cents 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
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.\[ \dfrac{\Delta y}{\Delta x} = \dfrac{f ( x_2 ) − f ( x_1 )}{x_2 − x_1}. \nonumber \]
Using the data in Table \( \PageIndex{ 1 } \), find the average rate of change of the price of gasoline between 2007 and 2009.
- Solution
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In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is\[ \begin{array}{rclcl}
\dfrac{\Delta y}{\Delta x} & = & \dfrac{y_2 - y_1}{x_2 - x_1} & & \\[6pt]
& = & \dfrac{\$2.41 - \$2.84}{2009 - 2007} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{-\$0.43}{2 \, \text{years}} & \quad & \left( \text{simplifying} \right) \\[6pt]
& \approx & -\$0.22 \, \text{ per year} & & \\[6pt]
\end{array} \nonumber \]
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.
Using the data in Table \( \PageIndex{ 1 } \), find the average rate of change between 2005 and 2010.
Given the function \( g ( t ) \) shown in the following figure, find the average rate of change on the interval \( [ − 1 , 2 ] \).
- Solution
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The following figure shows \( g ( −1 ) = 4 \). At \( t = 2 \), the graph shows \( g ( 2 ) = 1 \).
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\[ \dfrac{1 − 4}{2 − ( − 1 )} = \dfrac{− 3}{3} = −1. \nonumber \]
Example \( \PageIndex{ 2 } \) illustrates the fact that the order we choose is very important. If, for example, we used \( \frac{y_2 − y_1}{x_1 − x_2} \), we would 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 \( ( x_2 , y_2 ) \).
After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values are shown in the following table. 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
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Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours, for an average speed of\[ \dfrac{292 -10}{6 - 0} = \dfrac{282}{6} = 47. \nonumber \]The average speed is 47 miles per hour.
In Example \( \PageIndex{ 3 } \), 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.
Compute the average rate of change of \( f ( x ) = x^2 − \frac{1}{x} \) on the interval \( [2, 4] \).
- Solution
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We can start by computing the function values at each endpoint of the interval.\[ \begin{array}{rclcrcl}
f(2) & = & 2^2 - \dfrac{1}{2} & \quad & f(4) & = & 4^2 - \dfrac{1}{4} \\[6pt]
& = & 4 - \dfrac{1}{2} & \quad & & = & 16 - \dfrac{1}{4} \\[6pt]
& = & \dfrac{7}{2} & \quad & & = & \dfrac{63}{4} \\[6pt]
\end{array} \nonumber \]Now we compute the average rate of change.\[ \begin{array}{rclcl}
\text{Average rate of change} & = & \dfrac{f(4) - f(2)}{4 - 2} & & \\[6pt]
& = & \dfrac{\frac{63}{4} - \frac{7}{2}}{4 - 2} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{4}{4} \cdot \left( \dfrac{\frac{63}{4} - \frac{7}{2}}{2} \right) & \quad & \left( \text{simplifying the compound fraction} \right) \\[6pt]
& = & \dfrac{63 - 14}{8} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{49}{8} & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]
Find the average rate of change of \( f ( x ) = x − 2 \sqrt{ x } \) on the interval \( [ 1 , 9 ] \).
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.
- Solution
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We are computing the average rate of change of \( \frac{2}{d^2} \) on the interval \( [ 2 , 6 ] \).\[ \begin{array}{rclcl}
\text{Average rate of change} & = & \dfrac{F(6) - F(2)}{6 - 2} & & \\[6pt]
& = & \dfrac{\frac{2}{6^2} - \frac{2}{2^2}}{4} & \quad & \left( \text{substituting and simplifying} \right) \\[6pt]
& = & \dfrac{\frac{2}{36} - \frac{2}{4}}{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{\frac{1}{18} - \frac{1}{2}}{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{18}{18} \cdot \left(\dfrac{\frac{1}{18} - \frac{1}{2}}{4}\right) & \quad & \left( \text{simplifying the compound fraction} \right) \\[6pt]
& = & \dfrac{1 - 9}{72} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & -\dfrac{8}{72} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & -\dfrac{1}{9} & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]The average rate of change is \( −\frac{1}{9}\) newton per centimeter.
Find the average rate of change of \( g ( t ) = t^2 + 3 t + 1 \) on the interval \( [ 0 , a ] \). The answer will be an expression involving \( a \).
- Solution
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We use the average rate of change formula.\[ \begin{array}{rclcl}
\text{Average rate of change} & = & \dfrac{g(a) - g(0)}{a - 0} & & \\[6pt]
& = & \dfrac{(a^2 + 3a + 1) - (0^2 + 3(0) + 1)}{a} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{a^2 + 3a + 1 - 1}{a} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{a^2 + 3a}{a} & \quad & \left( \text{combining like terms} \right) \\[6pt]
& = & \dfrac{a(a + 3)}{a} & \quad & \left( \text{factoring }a\text{ out of the numerator} \right) \\[6pt]
& = & \dfrac{\cancelto{1}{a}(a + 3)}{\cancelto{1}{a}} & \quad & \left( \text{canceling like factors} \right) \\[6pt]
& = & a + 3 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]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 \( 5 + 3 = 8 \).
The answer in Example \( 6 \) is a repeated concept from the previous section - it is an alternative form of the difference quotient!
Find the average rate of change of \( f ( x ) = x^2 + 2 x − 8 \) on the interval \( [ 5 , a ] \).
Focusing on Calculus - Increasing, Decreasing, or Constant Graphs of 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 \( \PageIndex{ 1 } \) shows examples of increasing and decreasing intervals on a function.
Figure \( \PageIndex{ 1 } \): The function \( f ( x ) = x^3 − 12 x \) 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. 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 below, 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. The following figure illustrates these ideas for a local maximum.
These observations lead us to a formal definition of local extrema.
A function \( f \) is an increasing function on an open interval if \( f ( b ) > f ( a ) \) for every \( a \) and \( b \) in the interval where \( b > a \).
A function \( f \) is a decreasing function on an open interval if \( f ( b ) < f ( a ) \) for every \( a \) and \( b \) in the interval where \( b > a \).
A function \( f \) has a local maximum at a point \( b \) in an open interval \( ( a , c ) \) if \( f ( b ) \) is greater than or equal to \( f ( x ) \) for every point \( x \) where ( \( x \) does not equal \( b \) ) in the interval. Likewise, \( f \) has a local minimum at a point \( b \) in \( ( a , c ) \) if \( f ( b ) \) is less than or equal to \( f ( x ) \) for every \( x \) where ( \( x \) does not equal \( b \) ) in the interval.
Given the function \( p ( t ) \) in the following figure, identify the intervals on which the function appears to be increasing.
- Solution
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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 \( (1,3) \cup ( 4 , \infty ) \).
Notice in Example \( \PageIndex{ 7 } \) 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 \( t = 4 \). These points are the local extrema (two minima and a maximum).
Use graphing technology to 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.
- Solution
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Using technology, we find that the graph of the function looks like that in the figure below. 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 \( x = −2 \).
Most graphing calculators and graphing utilities can estimate the location of maxima and minima. The following figure provides screen images from two different technologies, showing the estimate for the local maximum and minimum.
Based on these estimates, the function is increasing on the interval \( ( − \infty , −2.449) \) and \( ( 2.449 , \infty ) \). 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 \( \pm \sqrt{ 6 } \), but determining this requires Calculus.)
Graph the function \( f ( x ) = x^3 − 6 x^2 − 15 x + 20 \) to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.
For the function \( f \) whose graph is shown below, find all local maxima and minima.
- Solution
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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 \( −2 \).
Revisiting the Toolkit Functions
We will now return to our toolkit functions and discuss their graphical behavior.
Focusing on Calculus - Absolute Extrema of Graphs of Functions
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 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.
Not every function has an absolute maximum or minimum value. The toolkit function \( f ( x ) = x^3 \) is one such function.
The absolute maximum of \( f \) at \( x = c \) is \( f ( c ) \) where \( f ( c ) \geq f ( x ) \) for all \( x \) in the domain of \( f \).
The absolute minimum of \( f \) at \( x = d \) is \( f ( d ) \) where \( f ( d ) \leq f ( x ) \) for all \( x \) in the domain of \( f \).
For the function \( f \) shown below, find all absolute maxima and minima.
- Solution
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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 \( −10 \).
Even and Odd Functions
Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions \( f ( x ) = x^2 \) or \( f ( x ) = | x | \) 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 \( f ( x ) = x^3 \) or \( f ( x ) = \frac{1}{x} \) were reflected over both axes, the result would be the original graph, as shown in Figre \( \PageIndex{ 2 } \).
Figure \( \PageIndex{ 2(a) } \): The cubic toolkit function.
Figure \( \PageIndex{ 2(b) } \): Horizontal reflection of the cubic toolkit function.
Figure \( \PageIndex{ 2(c) } \): 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 .
A function can be neither even nor odd if it does not exhibit either symmetry. For example, \( f ( x ) = 2^x \) is neither even nor odd. Also, the only function that is both even and odd is the constant function \( f ( x ) = 0 \).
A function is called an even function if for every input \( x \)\[ f ( -x ) = f ( x ). \nonumber \]The graph of an even function is symmetric about the \( y \)- axis.
A function is called an odd function if for every input \( x \)\[ f ( -x ) = − f ( x ). \nonumber \]The graph of an odd function is symmetric about the origin.
Is the function \( f ( x ) = x^3 + 2 x \) even, odd, or neither?
- Solution
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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.\[ \begin{array}{rclcl}
f(-x) & = & (-x)^3 + 2(-x) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & -x^3 - 2x & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & -(x^3 + 2x) & \quad & \left( \text{factoring out }-1 \right) \\[6pt]
& = & -f(x) & & \\[6pt]
\end{array} \nonumber \]Because \( f ( − x ) = -f ( x ) \), this is an odd function.
The graph of the function from Example \( \PageIndex{ 11 } \) is shown below. Notice that the graph is symmetric about the origin. For every point \( ( x , y ) \) on the graph, the corresponding point \( ( − x , − y ) \) is also on the graph. For example, \( (1, 3) \) is on the graph of \( f \), and the corresponding point \( ( −1 , −3 ) \) is also on the graph.
Is the function \( f ( s ) = s^4 + 3 s^2 + 7 \) even, odd, or neither?
Access this online resource for additional instruction and practice with rates of change.