Average Rate of Change for Any Function
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.1: “Defining the Derivative” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/3-1-defining-the-derivative |
| Supplementary | OpenStax Calculus Volume 1, Section 1.4: “The Tangent and Velocity Problems” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/1-4-the-tangent-and-velocity-problems |
Try This First
A ball is thrown upward. Its height $s$ (in feet) at time $t$ (in seconds) is given by the table below.
| $t$ (s) | $s(t)$ (ft) |
|---|---|
| 0 | 0 |
| 1 | 48 |
| 2 | 64 |
| 3 | 48 |
| 4 | 0 |
Before reading further:
- Between $t = 0$ and $t = 4$: how much does height change? How many seconds elapse? Can you form a ratio?
- Between $t = 0$ and $t = 1$: form the same ratio. Between $t = 1$ and $t = 2$: same ratio. Do they match?
- What does the ratio represent physically? What would a ratio of 0 mean?
Hold your answers. This lesson names what you just computed and explains why it varies.
Key Idea
You already know how to compute the amount of change: $\Delta y = f(b) - f(a)$. The average rate of change goes one step further: it divides the change in output by the change in input.
$$\text{Average rate of change of } f \text{ on } [a,b] = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}.$$
This ratio answers: “how many units does the output change per one unit of input, on average, over the interval $[a,b]$?” For a linear function, this ratio is the same for every interval -- it is the slope. For a non-linear function, the ratio varies with the interval.
Two Connected Representations
Numerically (from the table above): over $[0, 4]$: $$\frac{\Delta s}{\Delta t} = \frac{s(4) - s(0)}{4 - 0} = \frac{0 - 0}{4} = 0 \text{ ft/s}.$$
A ratio of 0 means the height ended where it started. The ball returned to ground level. The average rate is zero even though the ball was moving the whole time. <!-- slop-ok: literal duration, not a meta frame -->
Over $[0, 1]$: $\dfrac{48 - 0}{1 - 0} = 48$ ft/s. Over $[1, 2]$: $\dfrac{64 - 48}{2 - 1} = 16$ ft/s.
Graphically: on a graph of $s$ vs $t$, draw a straight line through the two points $(a, f(a))$ and $(b, f(b))$. The slope of that straight line equals the average rate of change. This line is called a secant line.
Translation prompt: Plot the ball’s data on a graph. Draw the secant line from $(0, 0)$ to $(4, 0)$. What is its slope? Now draw the secant from $(0, 0)$ to $(1, 48)$. What is its slope? Which secant line is steeper?
Why Rate Varies with Interval (The Key Distinction)
For a linear function like $f(x) = 3x + 2$, the ratio $\Delta f / \Delta x = 3$ for any interval you choose. The slope is constant.
For a non-linear function like the ball’s trajectory, the ratio changes depending on which interval $[a, b]$ you pick. Over $[0,1]$ the ball rises 48 ft/s on average; over $[1,2]$ it rises only 16 ft/s on average; over $[3,4]$ it falls. The average rate of change is a property of the interval, not a single fixed property of the function.
This interval-dependence is not a problem -- it is the feature that makes calculus necessary. The question “how fast is the ball moving at $t = 1$ exactly?” cannot be answered by a single secant line; it requires letting the interval shrink to a single point. That limit is the derivative, and it is the subject of the next several lessons.
Worked Example
For $f(x) = x^2$, find the average rate of change on each interval and describe how it changes.
Intervals: $[0,1]$, $[1,2]$, $[2,3]$, $[0,3]$.
| Interval $[a,b]$ | $f(a)$ | $f(b)$ | $\Delta f$ | $\Delta x$ | Avg rate $\Delta f / \Delta x$ |
|---|---|---|---|---|---|
| $[0,1]$ | 0 | 1 | 1 | 1 | 1 |
| $[1,2]$ | 1 | 4 | 3 | 1 | 3 |
| $[2,3]$ | 4 | 9 | 5 | 1 | 5 |
| $[0,3]$ | 0 | 9 | 9 | 3 | 3 |
Predict first: since $f(x) = x^2$ is increasing and curving upward, each successive unit interval should show a larger rate. Is that what the table shows?
Result: yes -- rates of 1, 3, 5 for consecutive unit intervals. The rate grows as $x$ grows. The average over the whole interval $[0,3]$ is 3, which lies between the smallest (1) and largest (5) unit-interval rates. This is not a coincidence: the average rate over a large interval averages the rates over the sub-intervals.
the average rate of change is a fixed property of the function, independent of the interval. This is the rate-as-fixed-number error. For a non-linear function, “the average rate of change of $f$” is incomplete as a statement -- it must specify the interval. Saying “the average rate of change of $x^2$ is 3” is meaningful only if the interval is $[0,3]$ (or $[1,2]$, or $[-1,2]$, etc.). A different interval gives a different number. The only functions for which average rate is truly interval-independent are linear functions, where rate equals slope.
Predict-Then-Check Practice
Problem 1 (Low floor): For $g(x) = 2x - 5$, find the average rate of change on $[1, 4]$. Predict whether it will match the slope of $g$.
Check your answer
Predict: $g$ is linear with slope 2, so the average rate should be 2 regardless of interval.
Compute: $\dfrac{g(4) - g(1)}{4 - 1} = \dfrac{3 - (-3)}{3} = \dfrac{6}{3} = 2$.
Confirmed. For linear functions, average rate equals slope, and it never varies with the interval.
Problem 2: For $h(t) = \sqrt{t}$, find the average rate of change on $[1, 4]$ and on $[4, 9]$. Does the average rate increase or decrease as the interval moves to the right?
Check your answer
On $[1,4]$: $\dfrac{\sqrt{4} - \sqrt{1}}{4 - 1} = \dfrac{2 - 1}{3} = \dfrac{1}{3} \approx 0.333$.
On $[4,9]$: $\dfrac{\sqrt{9} - \sqrt{4}}{9 - 4} = \dfrac{3 - 2}{5} = \dfrac{1}{5} = 0.2$.
The average rate decreased. The square root function is increasing but its rate of increase is slowing down as the input grows -- a concave-down shape.
Problem 3 (High ceiling): For $f(x) = x^2$, find the average rate of change on $[a, a+h]$ in terms of $a$ and $h$. What happens to this expression as $h \to 0$? What does the limit represent?
Check your answer
$\dfrac{f(a+h) - f(a)}{h} = \dfrac{(a+h)^2 - a^2}{h} = \dfrac{2ah + h^2}{h} = 2a + h$.
As $h \to 0$, this average rate approaches $2a$. The limit of the average rate as the interval shrinks to a point is the instantaneous rate of change -- the derivative $f'(a) = 2a$. This limit process is the definition of the derivative, and it is what separates calculus from algebra.
Ask Why
Two cars drive between the same two cities, 300 miles apart. Car A takes 5 hours; car B takes 6 hours. Compute the average speed of each.
Now: is it possible that car B was traveling faster than car A at some moment during the trip, even though car A finished first? How would you argue this from the average rate concept alone?
(One way to see it: average rate is a summary over the whole interval. It says nothing about any specific moment. Car B might have sped for part of the trip and idled for the rest. The average only constrains the total change.)
Mastery Checklist
Next: Rates of Change (Derivatives) | Also: Secant Lines and the Difference Quotient