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.
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 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.
If we consider the two points [latex]\left({x}_{1}, {y}_{1}\right)[/latex] and [latex]\left({x}_{2}, {y}_{2}\right)[/latex] on the graph of a function [latex]f,[/latex] we can talk about the average rate of change on the interval of input values [latex]\left[{x}_{1}, {x}_{2}\right].[/latex]
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.
You may also see the interval given as [latex]\left[a,b\right].[/latex] This means that our points would be [latex]\left(a, f\left(a\right)\right)[/latex] and [latex]\left(b, f\left(b\right)\right).[/latex] This would lead to the alternate form for the average rate of change shown below.
[latex]\begin{align*}\text{ Average rate of change }=\frac{f\left(b\right)-f\left(a\right)}{b-a}\end{align*}[/latex]
You should be comfortable working with any of these presentations of the material.
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:
[latex]\\[/latex]
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 output values [latex]f\left({x}_{1}\right)[/latex] and [latex]f\left({x}_{2}\right)[/latex] divided by the change in the input values [latex]{x}_{1}[/latex] and [latex]{x}_{2}[/latex].
[latex]\frac{\Delta y}{\Delta x}=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}[/latex]
Given the value of a function at different points, calculate the average rate of change of a function for the interval between two input values [latex]{x}_{1}[/latex] and [latex]{x}_{2}.[/latex]
Using the data in Table 1 , find the average rate of change of the price of gasoline between 2007 and 2009.
In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is
Using the data in Table 1 , find the average rate of change between 2005 and 2010.
[latex]\frac{$2.84-$2.31}{5\text{ years}}=\frac{$0.53}{5\text{ years}}=$0.106[/latex] per year.
Given the function [latex]g\left(t\right)[/latex] shown in Figure 1 , find the average rate of change on the interval [latex]\left[-1,2\right].[/latex]
At [latex]t=-1,[/latex] Figure 2 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
After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values are shown in Table 2 . Find her average speed over the first 6 hours.
(hours) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
( ) (miles) | 10 | 55 | 90 | 153 | 214 | 240 | 292 | 300 |
Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours, for an average speed of
The average speed is 47 miles per hour.
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 [latex]f\left(x\right)={x}^{2}-\frac{1}{x}[/latex] on the interval [latex]\text{[2,}\text{ }\text{4].}[/latex]
We can start by computing the function values at each endpoint of the interval.
Now we compute the average rate of change from [latex]x=2[/latex] to [latex]x=4[/latex] using the formula [latex]\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}[/latex].
Find the average rate of change of [latex]f\left(x\right)=x-2\sqrt[\leftroot{1}\uproot{2} ]{x}[/latex] on the interval [latex]\left[1,\text{ }9\right].[/latex]
[latex]\frac{1}{2}[/latex]
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.
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]
The average rate of change is [latex]-\frac{1}{9}[/latex] newton per centimeter.
Find the average rate of change of [latex]g\left(t\right)={t}^{2}+3t+1[/latex] on the interval [latex]\left[0,\text{ }a\right].[/latex] The answer will be an expression involving [latex]a.[/latex]
We use the average rate of change formula with input values [latex]a[/latex] and 0.
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]
Find the average rate of change of [latex]f\left(x\right)={x}^{2}+2x-8[/latex] on the interval [latex]\left[5,\text{ }a\right].[/latex]
[latex]a+7[/latex]
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 3 shows examples of increasing and decreasing intervals on a function.
Figure 3 The function [latex]f\left(x\right)={x}^{3}-12x[/latex] is increasing on [latex]\left(-\infty \text{,}\text{ }-\text{2}\right){{\cup }^{\text{}}}^{\text{}}\left(2,\text{ }\infty \right)[/latex] and is decreasing on [latex]\left(-2\text{,}\text{ }2\right).[/latex]
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 whose graph is shown in Figure 4 , the local maximum occurs when [latex]x=-2[/latex] The maximum value is the output value of 16. The local minimum occurs when [latex]x=2[/latex] . The minimum value is the output value of [latex]-16[/latex]
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 5 illustrates these ideas for a local maximum.
Figure 5 Definition of a local maximum
These observations lead us to a formal definition of local extrema.
A function [latex]f[/latex] is an increasing function on an open interval if [latex]f\left(b\right)>f\left(a\right)[/latex] for every two input values [latex]a[/latex] and [latex]b[/latex] in the interval where [latex]b>a.[/latex]
A function [latex]f[/latex] is a decreasing function on an open interva l if [latex]f\left(b\right)<f\left(a\right)[/latex] for every two input values [latex]a[/latex] and [latex]b[/latex] in the interval where [latex]b>a.[/latex]
A function [latex]f[/latex] has a local maximum at a point [latex]b[/latex] in an open interval [latex]\left(a,c\right)[/latex] if [latex]f\left(b\right)[/latex] is greater than or equal to [latex]f\left(x\right)[/latex] for every point [latex]x[/latex] ([latex]x[/latex] does not equal [latex]b[/latex]) in the interval. Likewise, [latex]f[/latex] has a local minimum at a point [latex]b[/latex] in [latex]\left(a,c\right)[/latex] if [latex]f\left(b\right)[/latex] is less than or equal to [latex]f\left(x\right)[/latex] for every [latex]x[/latex] ([latex]x[/latex] does not equal [latex]b[/latex]) in the interval.
Given the function [latex]p\left(t\right)[/latex] in Figure 6 , identify the intervals on which the function appears to be increasing and decreasing.
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. The function appears to be decreasing until [latex]t=1[/latex] and then again from [latex]t=3[/latex] to [latex]t=4.[/latex]
In interval notation , we would say the function appears to be increasing [latex]\left(1,\text{ }3\right)\cup\left(4,\infty \right)[/latex] and decreasing on [latex]\left(-\infty,\text{ }1\right)\cup\left(3,\text{ }4\right).[/latex]
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).
Use technology to graph the function [latex]f\left(x\right)=\frac{2}{x}+\frac{x}{3}.[/latex] Then use features of your graphing utility to estimate the local extrema of the function and to determine the intervals on which the function is increasing.
Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 8 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 [latex](-\infty \text{,}-\text{2}\text{.449)}[/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]±\sqrt{6},[/latex] but determining this requires calculus.)
Use technology to graph the function [latex]f\left(x\right)={x}^{3}-6{x}^{2}-15x+20[/latex] and to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.
Using technology, we find the local maximum of 28 appears to occur at [latex]\left(-1,28\right),[/latex] and the local minimum of -80 appears to occur at [latex]\left(5,-80\right).[/latex] The function is increasing on [latex]\left(-\infty ,-1\right)\cup \left(5,\infty \right)[/latex] and decreasing on [latex]\left(-1,5\right).[/latex]
For the function [latex]f[/latex] whose graph is shown in Figure 9 , find all local maxima and minima.
Observe the graph of [latex]f.[/latex] The graph attains a local maximum at [latex]x=1[/latex] because the highest point in an open interval occurs when [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]x=-1[/latex] because the lowest point in an open interval occurs when [latex]x=-1.[/latex] The local minimum is the y -coordinate at [latex]x=-1,[/latex] which is [latex]-2.[/latex]
We will now return to our toolkit functions and discuss their graphical behavior in Figure 10 , Figure 11 , and Figure 12 .
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 13 .
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.
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]
For the function [latex]f[/latex] shown in Figure 14 , find all absolute maxima and minima.
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]
Access this online resource for additional instruction and practice with rates of change.
Average rate of change | [latex]\frac{\Delta y}{\Delta x}=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}[/latex] |
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Chapter 3 Functions
Learning objectives.
In this section, you will:
Gasoline costs have experienced some wild fluctuations over the last several decades. Table 1 [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.
[latex]y[/latex] | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 |
[latex]C(y)[/latex] | 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, looking at how much the price changed per year might be more useful. In this section, we will investigate changes such as these.
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 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.
The Greek letter[latex]\text{Δ}\,[/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]\,\text{Δ}f\,[/latex]instead of[latex]\,\text{Δ}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 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.
Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values [latex]\,{x}_{1}\,[/latex] and [latex]\,{x}_{2}.[/latex]
Using the data in Table 1 , find the average rate of change of the price of gasoline between 2007 and 2009.
In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is
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 1 , find the average rate of change between 2005 and 2010.
[latex]\frac{$2.84-$2.31}{5\text{ years}}=\frac{\$0.53}{5\text{ years}}=$0.106\,[/latex] per year.
Given the function [latex]\,g\left(t\right)\,[/latex] shown in Figure 1, find the average rate of change on the interval [latex]\,\left[-1,2\right].[/latex]
At [latex]t=-1,[/latex] Figure 2 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]\,\text{Δ}t=3\,[/latex]is shown by the red arrow, and the vertical change [latex]\text{Δ}g\left(t\right)=-3[/latex] 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
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]
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 Table 2. Find her average speed over the first 6 hours.
(hours) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
( ) (miles) | 10 | 55 | 90 | 153 | 214 | 240 | 292 | 300 |
Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours.
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 [latex]f\left(x\right)={x}^{2}-\frac{1}{x}[/latex] on the interval [latex]\text{[2,}\,\text{4].}[/latex]
We can start by computing the function values at each endpoint of the interval.
Now we compute the average rate of change.
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]
[latex]\frac{1}{2}[/latex]
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.
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]
The average rate of change is [latex]-\frac{1}{9}[/latex] newton per centimeter.
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] in simplest form.
We use the average rate of change formula.
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]
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]in simplest forms in terms
[latex]\,a+7\,[/latex]
Access this online resource for additional instruction and practice with rates of change.
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 3 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 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 whose graph is shown in Figure 4, 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]
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 5 illustrates these ideas for a local maximum.
These observations lead us to a formal definition of local extrema.
A function [latex]\,f\,[/latex] is an increasing function on an open interval if [latex]\,f\left(b\right)>f\left(a\right)\,[/latex] for any two input values [latex]\,a\,[/latex]and[latex]\,b\,[/latex] in the given interval where [latex]\,b>a.[/latex]
A function[latex]\,f\,[/latex] is a decreasing function on an open interval if [latex]\,f\left(b\right)[/latex] < [latex]f\left(a\right)\,[/latex] for any two input values [latex]\,a\,[/latex] and [latex]\,b\,[/latex] in the given interval where [latex]\,b>a.[/latex]
A function [latex]f[/latex] has a local maximum at [latex]\,x=b[/latex] if there exists an interval [latex]\,\left(a,c\right)[/latex] with a<b<c such that, for any [latex]x[/latex] in the interval [latex]\left(a,c\right),[/latex][latex]f\left(x\right)\le f\left(b\right).[/latex] Likewise, [latex]f[/latex] has a local minimum at [latex]x=b[/latex] if there exists an interval [latex]\left(a,c\right)[/latex] with a<b<c such that, for any [latex]x[/latex] in the interval [latex]\left(a,c\right),[/latex][latex]f\left(x\right)\ge f\left(b\right).[/latex]
Given the function [latex]\,p\left(t\right)\,[/latex] in Figure 6, identify the intervals on which the function appears to be increasing.
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]
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).
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.
Using technology, we find that the graph of the function looks like that in Figure 7. 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]
Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 8 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[latex]\,(-\infty \text{,}-\text{2}\text{.449)}\,[/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]\,±\sqrt{6},\,[/latex]but determining this requires calculus.)
Graph the function[latex]\,f\left(x\right)={x}^{3}-6{x}^{2}-15x+20\,[/latex]to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.
The local maximum appears to occur at[latex]\,\left(-1,28\right),\,[/latex]and the local minimum occurs at[latex]\,\left(5,-80\right).\,[/latex]The function is increasing on[latex]\,\left(-\infty ,-1\right)\cup \left(5,\infty \right)\,[/latex]and decreasing on[latex]\,\left(-1,5\right).[/latex]
For the function [latex]\,f\,[/latex] whose graph is shown in Figure 9, find all local maxima and minima.
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]
Access this online resource for additional instruction and practice with increasing intervals, decreasing intervals, local maxima, and local minima values of a function.
We will now return to our toolkit functions and discuss their graphical behavior in Figure 10, Figure 11, and Figure 12.
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 13.
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.
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]
For the function [latex]\,f\,[/latex] shown in below, find all absolute maxima and minima.
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]
Access this online resource for additional instruction and practice in finding the absolute maximum and minimum of a function given the graph.
Average rate of change | [latex]\frac{\Delta y}{\Delta x}=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}[/latex] |
Yes, the average rate of change of all linear functions is constant.
The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.
For the following exercises, find the average rate of change of each function on the interval specified for real numbers [latex]\,b\,[/latex] or [latex]\,h[/latex] in simplest form.
[latex]4\left(b+1\right)[/latex]
[latex]4x+2h[/latex]
[latex]\frac{-1}{13\left(13+h\right)}[/latex]
[latex]3{h}^{2}+9h+9[/latex]
[latex]4x+2h-3[/latex]
For the following exercises, consider the graph of [latex]\,f\,[/latex] shown in Figure below.
[latex]\frac{4}{3}[/latex]
For the following exercises, use the graph of each function to estimate the intervals on which the function is increasing or decreasing.
increasing on[latex]\,\left(-\infty ,-2.5\right)\cup \left(1,\infty \right),\,[/latex]decreasing on[latex]\,\left(-2.5,\text{ }1\right)[/latex]
increasing on [latex]\,\left(-\infty ,1\right)\cup \left(3,4\right),\,[/latex] decreasing on[latex]\,\left(1,3\right)\cup \left(4,\infty \right)[/latex]
For the following exercises, consider the graph shown in the Figure below.
local maximum:[latex]\,\left(-3,\text{ }60\right),\,[/latex] local minimum:[latex]\,\left(3,\text{ }-60\right)\,[/latex]
For the following exercises, consider the graph in the Figure below.
absolute maximum at approximately [latex]\,\left(7,\text{ }150\right),\,[/latex] absolute minimum at approximately [latex]\,\left(-7.5,\text{ }-220\right)[/latex]
| |
---|---|
1998 | 201 |
1999 | 219 |
2000 | 233 |
2001 | 243 |
2002 | 249 |
2003 | 251 |
2004 | 249 |
2005 | 243 |
2006 | 233 |
2000 | 87 |
2001 | 84 |
2002 | 83 |
2003 | 80 |
2004 | 77 |
2005 | 76 |
2006 | 78 |
2007 | 81 |
2008 | 85 |
a. –3000; b. –1250
For the following exercises, find the average rate of change of each function on the interval specified.
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.
Local minimum at[latex]\,\left(3,-22\right),\,[/latex]decreasing on[latex]\,\left(-\infty ,\text{ }3\right),\,[/latex]increasing on[latex]\,\left(3,\text{ }\infty \right)\,[/latex]
Local minimum at[latex]\,\left(-2,-2\right),\,[/latex]decreasing on[latex]\,\left(-3,-2\right),\,[/latex]increasing on[latex]\,\left(-2,\text{ }\infty \right)[/latex]
Local maximum at[latex]\,\left(-0.5,\text{ }6\right),\,[/latex]local minima at[latex]\,\left(-3.25,-47\right)\,[/latex]and[latex]\,\left(2.1,-32\right),\,[/latex]decreasing on[latex]\,\left(-\infty ,-3.25\right)\,[/latex]and[latex]\,\left(-0.5,\text{ }2.1\right),\,[/latex]increasing on[latex]\,\left(-3.25,\text{ }-0.5\right)\,[/latex]and[latex]\,\left(2.1,\text{ }\infty \right)\,[/latex]
Based on the calculator screenshot, the point [latex]\,\left(1.333,\text{ }5.185\right)\,[/latex] is which of the following?
[latex]b=5[/latex]
2.7 gallons per minute
Use the graph to estimate the average decay rate from [latex]\,t=5\,[/latex] to [latex]\,t=15.[/latex]
approximately –0.6 milligrams per day
College Algebra Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.
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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 1 did not change by the same amount each year, so the rate of change was not constant.
Find the average rate of change of g(x) = - 403 + 5 from x = - Ito Enter an integer or decimal number (more.] Message instructor about this question int Version Assessment - Assignment 1.3: Rates of Change and Behavior of Graphs Quei You've already done this problem.
To find the average rate of change, we divide the change in the output value by the change in the input value. Average rate of change = Change in output Change in input = Δy Δx = y2 − y1 x2 − x1 = f(x2) − f(x1) x2 − x1. The Greek letter Δ (delta) signifies the change in a quantity; we read the ratio as "delta- y over delta- x ...
Example 1.3.2. Given the function g(t) shown here, find the average rate of change on the interval [0, 3]. Solution. At t = 0, the graph shows g(0) = 1. At t = 3, the graph shows g(3) = 4. The output has changed by 3 while the input has changed by 3, giving an average rate of change of: 4 − 1 3 − 0 = 3 3 = 1.
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.
2 Section 1.3 Rates of Change and Behavior of Graphs Example 1 Using the cost-of-gas function from earlier, find the average rate of change between 2007 and 2009 From the table, in 2007 the cost of gas was $2.64. In 2009 the cost was $2.14. The input (years) has changed by 2. The output has changed by $2.14 - $2.64 = -0.50.
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 3 shows examples of increasing and decreasing intervals on a function.
Given the function g(t) g ( t) shown here, find the average rate of change on the interval [0, 3]. Solution. At t = 0, the graph shows g(0) = 1 g ( 0) = 1. At t = 3, the graph shows g(3) = 4 g ( 3) = 4. The output has changed by 3 while the input has changed by 3, giving an average rate of change of:
Section 1.3 Rates of Change and Behavior of Graphs 39 Example 5 The magnetic force F, measured in Newtons, between two magnets is related to the distance between the magnets d, in centimeters, by the formula 2 2 ( ) d F d. Find the average rate of change of force if the distance between the magnets is increased from 2
This page titled 1.3: Rates of Change and Behavior of Graphs is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
Question: Assignment 1.3: Rates of Change and Behavior of Graphs Score: 110/120 11/12 answered Question 6 > 1 Given f(x) = find the average rate of change of f(x) on the interval (10,10 + h). Your answer will be an expression involving h. 2 + 2' Question Help: Video Message instructor D Post to forum Submit Question Jump to Answer
Definition. 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 output values f (x1) f ( x 1) and f (x2) f ( x 2) divided by the change ...
The output has changed by 3 while the input has changed by 3, giving an average rate of change of: 1 3 3 3 0 4 1 Example 3 (video example here) ... Section 1.3 Rates of Change and Behavior of Graphs 41 Based on these estimates, the function is increasing on the intervals ( f , 2.449) and
This video reviews how to calculate the average rate of change of a function on a given interval. It demonstrates #5 in section 1.3 from the PreCalculus text...
1 Section 1.3 Rates of Change and Behavior of Graphs Learning objectives: in this section we 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.
Section 1.3 Rates of Change and Behavior of Graphs 35 Example 1 Using the cost-of-gas function from earlier, find the average rate of change between 2007 and 2009 From the table, in 2007 the cost of gas was $2.64. In 2009 the cost was $2.14. The input (years) has changed by 2. The output has changed by $2.14 - $2.64 = -0.50.
Math. Precalculus. Precalculus questions and answers. HW03 - Assignment 1.3: Rates of Change and Behavior of Graphs Score: 1.5/11 Save progress Done 8 Vo 4/11 answered x Question 4 > B9 Details Score on last try: 0 of 1 pts. See Details for more. > Next question You can retry this question below 5 Find the average rate of change of g) = 70 + on ...
Question: Assignment 1.3: Rates of Change and Behavior of Graphs Score: 4/27 3/11 answered Done 6 VO : Question 5 < > B0/1 pt 599 Details 1 Given f(x) find the average rate of change of f(x) on the interval 2 + 12 [10, 10+ h]. Your answer will be an expression involving h. Question Help: Video Submit Question Jump to Answer ction gra
Question: Assignment 1.3: Rates of Change and Behavior of GraphsAssignment 1.3: Rates of Change and Behavior of GScore: 15120,312 answeredFind the average rate of change of g(x)=4x4+3x2 on the interval -4,3.Question Help: video ... Rates of Change and Behavior of Graphs. Assignment 1. 3: Rates of Change and Behavior of G. Score: 1 5 1 ...
Precalculus questions and answers. Assignment 1.3: Rates of Change and Behavior of Graphs Score: 50/806/8 answered The function f (x)=2x3−33x2+168x+7 has one local minimum and one local maximum. Use a graph of the function to estimate these local extrema. This function has a local minimum at x= with output value: and a local maximum at x ...
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.
giving an average rate of change of: 1 3 3 3 0 ... Section 1.3 Rates of Change and Behavior of Graphs 25 Example 9 Using the cost of gasoline function from the beginning of the section, find an interval on which the function appears to be decreasing. Estimate any local extrema using the