# Average Rate of Change

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Objectives
• Define the Average Rate of Change over an interval
• Understand why AROC only makes sense to examine over a fixed, finite interval
• Learn how AROC can be used to approximate instantaneous slope
• Briefly look at kinematics graphs and interpret AROC in context of a displacement or velocity graph
Lesson Description

In order to further study the instantaneous slope of a function, it will be helpful to first study the average slope of a function over a specified interval. This lesson defines and works with this concept, which is commonly referred to as Average Rate of Change.

Practice Problems

Practice problems and worksheet coming soon!

## A Graphical Average

When we have a list of numbers, finding the list's "straight average" is a familiar process we've been using since grade school. Other types of averages exist, although they tend to be less intuitive and we use them less frequently. In this lesson, we'll define what it means to consider a function's average rate of change.There are two kinds of function averages we will care about in Calculus. The first one is the topic of this lesson, aptly named Average Rate of Change, which we can think of as the function's average slope on an interval. Later on, using some true Calculus tools, we will also consider how to obtain and analyze average function values ». Both of these concepts only make sense when examining a fixed, finite interval. Additionally, both will make the most sense in context of a word problem or specific situation.

## A Real Life Example

Here's a simplified, estimated graph of the value of the Dow Jones Industrial stock index from 1990 to 2015. In context, if we find the average rate of change on an interval, say from 2004 to 2007, then we are finding the average rate at which the Dow was growing. If we found the average value of the Dow, we would be finding the actual Dow index value that we would say existed on average, given that some days the index value increases, and other days it decreases.Take a look at the graph above. Like many functions we work with, this graph has non-constant slope - sometimes the slope is positive, and sometimes it's negative. However, the slope between two points is what we can consider to be the average rate of change between those two points. So, between 2004 and 2007, the Dow changed from 6,000 to 11,000 (approximately, by eyeballing the graph). This is a change of $11000-6000$ or $5000$, and it happened over 3 years. Therefore, the average change is $5000/3$, or about 1,667. The average rate of change helps us look at the end result of how something changed during a set interval on average, regardless of what values and fluctuations happened in between, because we can say that the Dow grew 1,667 points each year on average.On the other hand, the average value of the Dow between 2004 and 2007 would depend on whatever actually happened in between the years. If the Dow spent most of the time being above the average rate of change line, then its average value would be higher than if it only spent one week being above that line.Here is an example of the difference:  These two scenarios have the same AROC, but different average values. The first one would have a larger average value, since it spent more time being higher, when compared with the pattern we see in the second figure.Again, we'll study average function values » in the future, since we'll need to use some core Calculus tools.

## AROC Definition

As we stated informally with the Dow example, the Average Rate of Change can be thought of as the slope between the two endpoints of the interval in question.
Define: Average Rate of ChangeLet $f(x)$ be a function with defined points at $x=a$, $x=b$, and all points in between, where $a \lt b$. The Average Rate of Change (AROC) of $f(x)$ on the interval $[a,b]$ is$$\mathrm{AROC} \, = \frac{f(b)-f(a)}{b-a}$$
Notice how the definition of the AROC is literally the definition of the slope of the function between $x=a$ and $x=b$. This is how we can and should think of AROC.
You Should Know Knowing that the "average rate of change" over an interval is simply the slope between the start and end of the interval might seem like an odd way to define an average, but in context this makes a lot of sense. For example, in Physics, if we consider the graph of a moving object's position ($y$-axis) over time ($x$-axis), the slope ($y/x$) is measured in distance over time, or in other words, speed. Therefore, the AROC in that case makes intuitive sense in that, for a given starting position, the total distance travelled, and the time it took, the object in any scenario has the same average speed over that time regardless of the various speeds it could have along the way.  Let's say these graphs represent cars that travelled for 8 minutes, starting at time 0. In the first figure, the car started speeding up, then slowed down, but was driving forward the whole time (evidenced by the fact that the distance only ever increased as time went on). In the second figure, the car sped up, then made no progress for a short time, then actually went backward briefly (evidenced by the fact that the graph moved downward) before moving forward again until it reached the 5 mile mark. But since both objects traveled 5 miles in 8 minutes, both cars have the same average speed.
Pro Tip It is often useful to look at a graph with units and realize that the units of each axis will give us the units of the slope of the graph. E.g. if we graph cumulative homeruns versus number of games played for your favorite Major League slugger, the slope of that graph would be homeruns per game, and the interpretation of that number would depend on the interval you use (the slope over a week, a season, or whole-career would give you the average homeruns per game in that week, season, or all-time, respectively).

## Finite Intervals Only

Like all average values we know and use in the real world, we need to be working on a finite space. AROC just plain doesn't make sense to take an average over an infinite interval (usually, though exceptions apply that we will not discuss). AROC also doesn't exactly make sense at a single point, as this becomes a tangent slope, and while tangent slopes do have a clean and common interpretation, they can no longer be considered "average rates".Practically speaking, there are two reasons why we're only ever going to analyze AROC over finite intervals.1) ContextIf we are working on a word problem, then the situation will either implicitly or explicitly impose an upper and lower bound on the problem. Take our Dow Jones example - the implicit lower bound for time is the day that the Dow Jones stock index was created. We can't analyze AROC from a starting point before that day. Likewise, the Dow data does not roll infinitely into the future either. On whatever day you're analyzing the data, the future-most data you could possibly have is that very day's stock value. While there's a very good chance you'll wake up tomorrow and the Dow will still exist, we do not know the future values, and thus do we have an upper bound on the data.2) Common SenseLet's say we weren't working on a word problem, but we were asked to find the AROC of a function $f(x)$, say $f(x)=x^3$. If we tried to compute or interpret the AROC from $(-\infty,k)$ or $(k, \infty)$, or any other interval involving infinity, the whole process falls apart. We couldn't even get an answer. Recall that AROC is computed identically to how we calculate slope, and if I'm looking for an AROC from $(-\infty,k)$ then I would compute it as$$\frac{f(k) - f(\infty)}{k-(-\infty)}$$But what is $f(-\infty)$? This is not a real number that we can compute (this is knowledge we have from studying polynomials in Algebra Two, and also reviewed recently in the lesson on end behavior and limits at infinity »). Since we can't compute $f(-\infty)$, we cannot execute the AROC computation.We conclude that every AROC calculation we will ever be asked to analyze will be done on a set interval, from a starting value to an end value.

## What is AROC Used For?

Real-world applications range anywhere from economics to sales to biology, in any scenario where the average rate of growth has a meaning that is worth studying for one reason or another. The concept of the average change being more important than the actual day-to-day fluctuations is appealing because it's not only simpler to compute, but simpler to understand.Getting back to the land of pure mathematics, we can use the average rate of change of a function as an approximation to the instantaneous slope at a point - and instantaneous slope is something that, in a Calculus course, we have a great deal of interest in studying. We may also be asked to simply know how to calculate AROC and to do so for a given function.Straight AROC CalculationHere's an example of a problem that asks for an AROC computation straight up.

Example 1Find the average rate of change of the function $f(x) = x^3 - 3x^2 + 2$ on the interval from $x=-1$ to $x=3$.$\blacktriangleright$ To find an AROC, we only need to calculate the slope between the point that lies on the start of the interval and the point that lies on the end of the interval.$$f(-1) = -2$$$$f(3) = 2$$Therefore, the interval from $x=-1$ to $x=3$ starts at the coordinate $(-1,-2)$ and ends at the coordinate $(3,2)$. A picture is not at all necessary, but it may help proof our work. Here's the graph: To compute the AROC, find the slope of the line between those two points.$$\mathrm{AROC} \,\, = \frac{f(3)-f(-1)}{3-(-1)}$$$$=\frac{2-(-2)}{3-(-1)} = \frac{4}{4} = 1$$Therefore the AROC for this scenario is $1$. We can see that this slope looks reasonable from the graph as well, which again, wasn't at all necessary to answer the question, but was presented as a double-check tool.Using AROC to Approximate Tangent SlopeThe AROC is, in short, the slope between two endpoints of a given interval length. But what happens if we define a very narrow interval? This figure shows the AROC secant line of a function on the interval $[5,5.5]$. If we wanted an approximation to the instantaneous slope at $x=5$ (often called a tangent slope), we can see that we wouldn't be too far off using this AROC as an approximation, though it is definitely not exactly the instantaneous slope. Furthermore, if we were to find the AROC between $x=5$ and $x=5.1$, we should expect an even better approximation. The closer we get the end point of the interval to the start point, the closer we are to obtaining an instantaneous slope approximation via the AROC calculation. However, we cannot exactly let the end point move into the same place as the start point. For example, if we do the AROC between $x=5$ and $x=5$, we would get$$\frac{f(5)-f(5)}{5-5} = \frac{0}{0}$$Therefore we can only use this AROC approach to get an approximation. We'll use limits in the next lesson » to approach the instantaneous slope problem more rigorously, and find an exact method to obtaining instantaneous slope.For now, we may simply be asked to approximate tangent slopes using AROC, as we are in the following example.

Example 2Approximate the instantaneous slope of the function $f(x) = \sqrt{x+2}$ at $x=2$, using an interval length of $0.3$.$\blacktriangleright$ Although the words "average rate of change" and "AROC" are absent from these instructions, that is exactly what the intention is. Let's find the AROC of $f(x)$ between $x=2$ and $x=2.3$.$$\mathrm{AROC} \,\, = \frac{f(2.3)-f(2)}{2.3-2}$$$$=\frac{2.0736-2}{2.3-2} \approx 0.2455$$
You Should Know Some teachers expect you to estimate quantities like this more rigorously. For example, one way we could get a better estimate is to find the AROC between $x=1.85$ and $x=2.15$, which has an interval length of $0.3$ but is instead centered around the estimate location.It is also possible that your teacher wants you to do this twice, once from $x=2$ to $x=2.3$ and again from $x=1.7$ to $x=2$, and average the two results together. This is another way of approximating with $x=2$ as the center location rather than an endpoint. Generally speaking, the first way I did the problem should be fine unless you are either told otherwise in the directions or told by your instructor that he or she expects something specific. "Centered" estimates are more work but yield better approximations.

## Mr. Math Makes It Mean

On a given interval, the average rate of change is given by a formula. However, it is possible to be given a function and an interval starting point and be asked to find the interval endpoint that would give us a specified AROC. This is a harder question to answer because it requires a thorough understanding of AROC and the formula, but if we are prepared for it, it is like many questions in math - solving a formula for a particular term.

Example 3For $f(x) = 2x^3 - 5x$, find $b$ such that the average rate of change of $f(x)$ on the interval from $x=2$ to $x=b$ is $99$, where $b>2$.$\blacktriangleright$ Use the AROC formula, since we are indeed working with average rate of change, and plug in the information you know. What we don't know, we'll solve for.\begin{align} \mathrm{AROC} \,\,\,\, = 99 & = \frac{f(b) - f(2)}{b - 2}\\ 99 & = \frac{\left[2b^3 - 5b\right] - }{b - 2} \end{align}At this point, you may have a freak-out, especially if you aren't allowed a calculator, because cubics are generally not solvable by hand. However, you aren't dead in the water yet - polynomial long division will save us here! (There's a sentence that's never before been written)\begin{align} 2b^3 - & 5b - 6 \div (b - 2)\\ & = 2b^2 + 4b + 3 \end{align}I know, I know, polynomial long division? Out of left field? There's a reason this is a mean question. However, we are left with a quadratic, and quadratics are fully solvable either by factoring or at worst with the quadratic formula.$$99 = 2b^2 + 4b + 3$$$$\Rightarrow 2b^2 + 4b - 96 = 0$$$$2(b+8)(b-6)=0$$Since we are looking for an interval that starts at $x=2$ and ends at an $x$ value greater than $2$, we know the answer to this question is $x=6$.

Just to be clear, most teachers won't ask for this. If you're in AP BC calc, however, you should be ready for anything!

## Put It To The Test

As discussed, the two major tasks we might be expected to perform with the concept of Average Rate of Change is 1) simply to calculate an AROC for a given function over a given interval 2) estimate the instantaneous slope of a function at a given point, using a specified interval length. We've seen both of these ideas in this lesson - and now we'll put them to the test.

Example 4Determine the Average Rate of Change of the function $f(x) = \ln(x+2)$ between $x=-1$ and $x=4$.
Show solution
$\blacktriangleright$ All we need to do to obtain the AROC is find the coordinates at each interval endpoint, and then take the slope between those points.$$x=-1 \,\, \Rightarrow f(-1) = \ln(1) = 0$$$$x=4 \,\, \Rightarrow f(4) = \ln(6) \approx 1.7918$$Therefore,$$\mathrm{AROC} \,\, = \frac{f(4)-f(-1)}{4-(-1)}$$$$=\frac{1.7918-0}{4-(-1)} \approx 0.35835$$Therefore, the Average Rate of Change of $f(x)=\ln(x+2)$ on the interval from $x=-1$ to $x=4$ is approximately $0.35835$.

Example 5Using the Average Rate of Change, estimate the instantaneous rate of change of $g(x) = 0.3x^4-x^3+0.4x+2$ at the point $x=1$ with an interval length of $0.2$, on each side of the $x$ value.
Show solution
$\blacktriangleright$ As we discussed earlier, if your instructor's implicit or explicit instructions call for an approximation, be aware of exactly what they expect. In this case, the instructions specifically ask for the AROC on the interval on each side, so first let's calculate the AROC of the function on the interval $x=0.8$ to $x=1$, and then the AROC on the interval from $x=1$ to $x=1.2$, and average the two results together.First, the AROC from $x=0.8$ to $x=1$.$$\mathrm{AROC} \,\, = \frac{g(1)-g(0.8)}{1-0.8}$$$$=\frac{1.43 - 1.93088}{1-0.8} \approx -2.5044$$Therefore, the AROC of $g(x)$ on the interval from $x=0.8$ to $x=1$ is approximately $-2.5044$.Now let's compute the AROC on the interval from $x=1$ to $x=1.2$.$$\mathrm{AROC} \,\, = \frac{g(1.2) - g(1)}{1.2-1}$$$$=\frac{1.37408-1.43}{1.2-1} \approx -0.2796$$Finally, the average of these two results is $(-2.5044 + -0.2796)/2 \approx -1.392$. This is really close to the true value of $-1.4$, but notice how each of the two pieces that we averaged together were not really close at all to the approximated result. This particular curve was very sensitive to the interval length we used, and we didn't pick one small enough that each side gave a decent approximation, though the average of each side yielded a good approximation.

Example 6For $f(x) = \frac{x^2}{4} - 3x$, find $b$ such that the average rate of change of $f(x)$ on the interval from $x=1$ to $x=k$ is $-\frac{1}{4}$, where $k > 1$.
Show solution
$\blacktriangleright$ This is an uncommon question that we discussed above (Mister Math Makes It Mean). However, all we need to do is keep calm and use the AROC formula.$$\mathrm{AROC} \,\,\,\, - \frac{1}{4} = \frac{f(k)-f(1)}{k - 1}$$$$-\frac{1}{4} = \frac{\frac{k^2}{4} - 3k - \left(-\frac{11}{4}\right)}{k - 1}$$This is an equation that would benefit greatly from multiplying both sides by $4$.$$4 \cdot \left[ -\frac{1}{4} \right] = 4 \cdot \left[ \frac{\frac{k^2}{4} - 3k - \left(-\frac{11}{4}\right)}{k - 1} \right]$$$$-1 = \frac{k^2 - 12k + 11}{k-1}$$You could solve this via polynomial long division, but since it's quadratic, it's easy enough to manipulate and solve directly.$$-k + 1 = k^2 - 12k + 11$$$$0 = k^2 - 11k + 10$$$$0 = (k-10)(k-1)$$Since we're looking for $k$ that is greater than $1$, we know it is $k=10$ that we are looking for.

Lesson Takeaways
• Understand what AROC means conceptually, and know how to find AROC for a given function over a specified interval.
• Know concisely how to compute an AROC given a function and a specified interval
• Know what to do if our instructions ask us vaguely to approximate a tangent slope (use a small interval and calculate AROC in this case)
• Be privy to your teachers' expectation for AROC approximation of instantaneous slope. Some teachers expect a two-sided approach, while others expect a faster, simpler, one-sided computation.
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