Rates of Change
Why this matters
Imagine you’re driving from your home in Dallas to visit a friend in Austin for the weekend. The trip is 200 miles and takes you 4 hours. If someone asked for your speed, you might say, "I averaged 50 miles per hour." That's a great answer, and it’s a perfect real-world example of an average rate of change. You found the total change in distance and divided by the total change in time.
But was your speed exactly 50 mph for the entire trip? Of course not. You slowed down in traffic near Waco, and you sped up a bit on the open highway. The number on your speedometer at any given second—say, 65 mph as you passed a field of bluebonnets—is your instantaneous rate of change.
In this lesson, we'll explore both of these ideas. We'll learn the solid mathematical way to calculate the average rate of change and then, most importantly, we'll learn how to use that skill to get a very close estimate of the rate of change at a single, specific moment.
Concept overview
flowchart TD
A[Start: Given a function f(x) and a point x = c] --> B{Want rate of change AT the point c?};
B --> C[Choose a second point, x₂, very close to c];
C --> D[Calculate the two y-values: f(c) and f(x₂)];
D --> E[Calculate Average Rate of Change: (f(x₂) - f(c)) / (x₂ - c)];
E --> F[Result: The AROC is a good approximation of the instantaneous rate at c];
F --> G{Is the function increasing or decreasing?};
G -- AROC > 0 --> H[Increasing at point c];
G -- AROC < 0 --> I[Decreasing at point c];
Core explanation
What is Average Rate of Change?
At its heart, the average rate of change (AROC) is a concept you’ve known for years: it’s the slope. That’s it! It measures how much a function’s output (y or f(x)) changes for every one unit of change in its input (x).
Remember the slope formula from Algebra 1?
slope = (y₂ - y₁) / (x₂ - x₁)
For a function f(x), we just write it in function notation. The average rate of change between two points, x = a and x = b, is:
Average Rate of Change = (f(b) - f(a)) / (b - a)
This formula gives you the slope of the secant line—the straight line that passes through the two points (a, f(a)) and (b, f(b)) on the function's graph. It tells you the constant rate you would have needed to travel between those two points to achieve the same net change.
From Average to Instantaneous: The Big Idea
So, we can find the average rate of change between two points. But how do we find the rate of change at a single point, like the speedometer reading at one instant? In Precalculus, we don't find it exactly—we approximate it.
And here’s how we do it: we calculate the average rate of change over a very, very small interval around the point we care about.
Imagine you want to know how fast a function is changing at x = -3. You can't use the AROC formula with just one point. But what if you picked a point incredibly close to -3, like x = -3.01? You could then calculate the average rate of change between x = -3.01 and x = -3.
Because the interval is so tiny, the average rate of change over that interval becomes a fantastic approximation for the instantaneous rate of change right at x = -3. The secant line between those two super-close points almost perfectly matches the steepness of the curve at that single spot.
Let's see this in action with a function you'll see in the visuals for this lesson:
f(x) = 0.25x³ - 2x + 2
Suppose we want to approximate the rate of change at point P, where x = -3.
First, let's find the value of the function here:
f(-3) = 0.25(-3)³ - 2(-3) + 2 = 0.25(-27) + 6 + 2 = -6.75 + 8 = 1.25
So, point P is at (-3, 1.25).
Now, let's pick a point very close to x = -3, like x = -3.01.
f(-3.01) = 0.25(-3.01)³ - 2(-3.01) + 2 ≈ 1.2023
Now we use the AROC formula:
AROC = (f(-3) - f(-3.01)) / (-3 - (-3.01))
AROC = (1.25 - 1.2023) / (0.01)
AROC = 0.0477 / 0.01 = 4.77
So, our approximation for the instantaneous rate of change at x = -3 is 4.77. This means that near x = -3, for every 1-unit increase in x, the y-value increases by about 4.77 units.
Comparing Rates of Change
What if we want to know whether the function is changing faster at x = -3 or at another point, say R where x = 2? We can use the same method.
First, find the value at x = 2:
f(2) = 0.25(2)³ - 2(2) + 2 = 0.25(8) - 4 + 2 = 2 - 4 + 2 = 0
So, point R is at (2, 0).
Now, let's approximate the rate of change at x = 2 by using a tiny interval, like [2, 2.01].
f(2.01) = 0.25(2.01)³ - 2(2.01) + 2 ≈ 0.01015
Let's calculate the AROC:
AROC = (f(2.01) - f(2)) / (2.01 - 2)
AROC = (0.01015 - 0) / (0.01)
AROC = 1.015
So, the rate of change at x = 2 is approximately 1.015.
Now we can compare!
- At
x = -3, the rate of change is about 4.77. - At
x = 2, the rate of change is about 1.015.
Since 4.77 > 1.015, we can conclude that the function is increasing much more steeply at x = -3 than it is at x = 2.
What Do the Signs Mean?
The sign of the rate of change tells you everything about the function's behavior.
- Positive Rate of ChangeThis means as
xincreases,f(x)also increases. On a graph, the function is moving "uphill" from left to right. In our example, both rates (4.77 and 1.015) were positive, and if you look at the graph, the function is indeed increasing at bothx = -3andx = 2. - Negative Rate of ChangeThis means as
xincreases,f(x)decreases. On a graph, the function is moving "downhill." If we had calculated the rate of change atx=0, we would have found a negative value, because the graph dips down in that region.
This is where many students slip up. They do the calculation correctly but forget to interpret the result. The number tells you how fast things are changing, and the sign tells you in what direction. Both are critical.
Worked examples
Approximating Rate of Change from a Function
Problem: Given the function g(t) = -16t² + 80t + 5, which models the height of a baseball (in feet) t seconds after being hit, approximate the speed of the ball at exactly t = 1 second.
Solution:
- 1Identify the GoalWe need the instantaneous rate of change (speed) at
t = 1. We will approximate this using the average rate of change over a very small interval aroundt = 1. Let's use the interval[1, 1.001]. - 2Calculate Function ValuesWe need
g(1)andg(1.001).g(1) = -16(1)² + 80(1) + 5 = -16 + 80 + 5 = 69feet.g(1.001) = -16(1.001)² + 80(1.001) + 5g(1.001) = -16(1.002001) + 80.08 + 5g(1.001) = -16.032016 + 85.08 = 69.047984feet.
- 3Apply the AROC FormulaAROC =
(g(1.001) - g(1)) / (1.001 - 1)AROC =(69.047984 - 69) / 0.001AROC =0.047984 / 0.001AROC =47.984 - 4Interpret the ResultThe average rate of change on this tiny interval is approximately 48. This means that at
t = 1second, the baseball's height is increasing at a rate of about 48 feet per second. This is our approximation for the instantaneous speed.
Comparing Rates of Change from a Table
Problem: The table below shows the population of a suburb of Seattle, measured in thousands, over several years.
| Year (t) | Population P(t) (in thousands) |
|---|---|
| 2015 | 12.5 |
| 2017 | 14.1 |
| 2019 | 16.0 |
| 2021 | 18.2 |
| 2023 | 20.1 |
Was the population growing faster, on average, between 2015-2017 or between 2021-2023?
Solution:
- 1Identify the GoalWe need to compare the average rate of change for two distinct intervals.
- 2Calculate AROC for 2015-2017
- The inputs are
t₁ = 2015andt₂ = 2017. - The outputs are
P(2015) = 12.5andP(2017) = 14.1. - AROC =
(14.1 - 12.5) / (2017 - 2015)AROC =1.6 / 2 = 0.8
- The inputs are
- 3Interpret the First ResultThe rate is 0.8. Since the population is in thousands, this means the population grew at an average rate of 800 people per year from 2015 to 2017.
- 4Calculate AROC for 2021-2023
- The inputs are
t₁ = 2021andt₂ = 2023. - The outputs are
P(2021) = 18.2andP(2023) = 20.1. - AROC =
(20.1 - 18.2) / (2023 - 2021)AROC =1.9 / 2 = 0.95
- The inputs are
- 5Interpret and CompareThe rate is 0.95, meaning an average growth of 950 people per year from 2021 to 2023.
Try it yourself
Practice Problem 1
A company's profit, P, in thousands of dollars, can be modeled by the function P(x) = -x² + 20x - 60, where x is the number of units sold in thousands.
Approximate the instantaneous rate of change of profit when the company sells 8,000 units (i.e., at x = 8). Use the interval [8, 8.01] for your approximation. What does the sign of your answer tell you about the company's profit at this level of sales?
Practice Problem 2
The temperature T (in °F) in Chicago on a winter day is recorded in the table.
| Time (hour past noon) | Temperature (°F) |
|---|---|
| 0 (noon) | 10 |
| 2 | 11 |
| 4 | 7 |
| 6 | 1 |
Calculate the average rate of change of temperature from noon to 2 PM, and from 4 PM to 6 PM. What do the different signs of your answers indicate about the weather trend?
Practice — 8 questions
In simple terms, rates of change help us measure how fast a function's output is changing compared to its input, just like calculating your average speed on a road trip.
- 1.2.A: Compare the rates of change at two points using average rates of change near the points.
- 1.2.B: Describe how two quantities vary together at different points and over different intervals of a function.
- 1.2.A.1
- The average rate of change of a function over an interval of the function’s domain is the constant rate of change that yields the same change in the output values as the function yielded on that interval of the function’s domain. It is the ratio of the change in the output values to the change in input values over that interval.
- 1.2.A.2
- The rate of change of a function at a point quantifies the rate at which output values would change were the input values to change at that point. The rate of change at a point can be approximated by the average rates of change of the function over small intervals containing the point, if such values exist.
- 1.2.A.3
- The rates of change at two points can be compared using average rate of change approximations over sufficiently small intervals containing each point, if such values exist.
- 1.2.B.1
- Rates of change quantify how two quantities vary together.
- 1.2.B.2
- A positive rate of change indicates that as one quantity increases or decreases, the other quantity does the same.
- 1.2.B.3
- A negative rate of change indicates that as one quantity increases, the other decreases.
flowchart TD
A[Start: Given a function f(x) and a point x = c] --> B{Want rate of change AT the point c?};
B --> C[Choose a second point, x₂, very close to c];
C --> D[Calculate the two y-values: f(c) and f(x₂)];
D --> E[Calculate Average Rate of Change: (f(x₂) - f(c)) / (x₂ - c)];
E --> F[Result: The AROC is a good approximation of the instantaneous rate at c];
F --> G{Is the function increasing or decreasing?};
G -- AROC > 0 --> H[Increasing at point c];
G -- AROC < 0 --> I[Decreasing at point c];
Read what Saavi narrates
Hi everyone, I'm Saavi, and welcome to Shrutam.
Let's talk about a simple idea: speed. Imagine you’re driving from Dallas to Austin. The trip is 200 miles and takes you 4 hours. If someone asked for your speed, you might say, "I averaged 50 miles per hour." That's a perfect real-world example of an average rate of change. You found the total change in distance and divided by the total change in time.
But was your speed exactly 50 mph the whole time? No way. You slowed down for traffic, you sped up on the highway. The number on your speedometer at any given second—that's your instantaneous rate of change.
In this lesson, we're going to master both of these ideas. We'll learn how to calculate the average rate of change, which is really just the slope between two points. Then, we'll use that skill to get a very close estimate of the rate of change at a single, specific moment.
Let's try an example. Imagine a baseball is hit, and its height is modeled by the function g of t equals negative 16 t-squared plus 80 t plus 5. We want to find the ball's speed at exactly one second.
We can't just plug in t equals 1. That would give us the ball's height, which is 69 feet. That's its position, not its speed. To find the speed, we need to find the rate of change.
So, we'll calculate the average rate of change over a tiny interval, like from t equals 1 to t equals 1.001. We already know g of 1 is 69. If you plug 1.001 into the function, you get about 69.048.
Now, we use the slope formula. The change in height is 69.048 minus 69, which is 0.048. The change in time is 1.001 minus 1, which is 0.001.
So, the rate is 0.048 divided by 0.001, which equals 48. The units are feet per second. So, at exactly one second, the ball is rising at about 48 feet per second.
A really common mistake here is just calculating the height and stopping. Remember, rate of change is about how a value is *changing*. It's a measure of steepness or speed, not just a location.
As you work through these problems, always ask yourself: am I finding a position, or am I finding a rate? That one question will save you a lot of points. You've got this.
The value of the function gives you a position or an amount at a single point. Rate of change measures how that value is *changing* over an interval.
Always use the full slope formula: `(f(b) - f(a)) / (b - a)`. Rate of change requires two points and a ratio.
You'll be calculating the "run over rise" instead of "rise over run." This gives you the reciprocal of the correct answer and will have incorrect units (e.g., seconds per foot instead of feet per second).
Always remember "change in y over change in x." The output `f(x)` always goes in the numerator.
Approximating the speed at `t=1` by using the interval `[1, 5]` gives you the *average* speed over 4 seconds, which is likely very different from the instantaneous speed at `t=1`.
To approximate an instantaneous rate, your interval must be very small, like `[1, 1.01]` or `[0.99, 1]`. The smaller the interval, the better the approximation.
The sign tells you the direction of change. A rate of `-50` means the quantity is *decreasing*. A rate of `50` means it's *increasing*. They are completely different situations.
Be meticulous with your calculations. If you get a negative result, ask yourself: "Does this make sense?" If a ball is falling, its height's rate of change should be negative.
A rate of "15" is meaningless. Is it 15 miles per hour? 15 dollars per month? 15 feet per second? The units provide the context that makes the number useful.
Always state the units of the output divided by the units of the input. For example, `(thousands of people) / (year)`.