What Is The Average Rate Of Change?
The average rate of change of a function $f$ over an interval $[a, b]$ measures how much the output changes for each unit of change in the input, on average, across that whole interval. It is defined by one formula:
$$\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}$$
The numerator $f(b) - f(a)$ is the total change in output. The denominator $b - a$ is the total change in input. Their ratio is a rate: output change per unit of input change. Both endpoints must lie in the domain of $f$, and $a \neq b$ so the denominator is not zero.
This is exactly the slope of the straight line joining the two endpoints of the curve. That straight line is called the secant line, and computing an average rate of change is the same task as finding slope from two points.
The idea sits at the doorway of calculus. It is the finite, whole-interval version of the rate of change that the derivative later measures at a single instant.
How Do You Calculate The Average Rate Of Change? (Worked Examples)
Each example below states the interval, substitutes into the formula, and keeps the units attached.
Example 1: A quadratic on a clean interval.
Find the average rate of change of $f(x) = x^2$ on $[1, 3]$.
$$\frac{f(3) - f(1)}{3 - 1} = \frac{3^2 - 1^2}{3 - 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4$$
Final answer: the average rate of change is $4$. On $[1, 3]$, the outputs of $x^2$ rise by $4$ units for every $1$ unit of $x$, on average.
Example 2: Average velocity from a position function.
A ball dropped from rest falls a distance $s(t) = 4.9t^2$ metres in $t$ seconds. Find its average velocity on $[0, 2]$.
$$\frac{s(2) - s(0)}{2 - 0} = \frac{4.9(2)^2 - 4.9(0)^2}{2 - 0} = \frac{19.6 - 0}{2} = 9.8 \text{ m/s}$$
Final answer: the average velocity is $9.8$ m/s. Average velocity is simply the average rate of change of position, so the units are metres per second, not a bare number. For the wider treatment of motion, see velocity and acceleration.
Example 3: Average cost per unit.
A workshop's cost to produce $x$ chairs is $C(x) = x^2 + 20x + 100$ dollars. Find the average rate of change of cost as production rises from $x = 10$ to $x = 20$.
$$\frac{C(20) - C(10)}{20 - 10} = \frac{(400 + 400 + 100) - (100 + 200 + 100)}{10} = \frac{900 - 400}{10} = \frac{500}{10} = 50$$
Final answer: the average rate of change of cost is $50$ dollars per chair. Over that batch of ten chairs, each extra chair added about $50$ dollars to total cost, on average.
What Does The Average Rate Of Change Look Like On A Graph?
On a graph the average rate of change is a slope you can see. Mark the two endpoints $(a, f(a))$ and $(b, f(b))$ on the curve, then draw the straight line through them. That line is the secant, and its steepness is the average rate of change.
$$\text{slope of secant} = \frac{\text{rise}}{\text{run}} = \frac{f(b) - f(a)}{b - a}$$
A positive value means the secant climbs left to right, so the function is higher at $b$ than at $a$. A negative value means the secant falls, so the output has dropped across the interval. A value of zero means the two endpoints sit at the same height, even if the curve rose and fell in between.
How Is The Average Rate Of Change Different From The Instantaneous Rate?
The average rate of change describes a whole interval. The instantaneous rate of change describes a single point. The bridge between them is what the whole of differential calculus is built on.
Fix the left endpoint at $x = a$ and let the right endpoint be $a + h$, where $h$ is the width of the interval. The average rate of change becomes the difference quotient:
$$\frac{f(a + h) - f(a)}{h}$$
This is the same average-rate formula, written with the interval width $h$ in the denominator. Now let the interval shrink. As $h$ approaches $0$, the secant line pivots until it just grazes the curve at $a$, becoming the tangent line, and the average rate of change approaches the instantaneous rate of change:
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
That limit is the definition of the derivative. Watch it happen with $f(x) = x^2$ near $x = 1$. The average rate of change on $[1, 1 + h]$ works out to $2 + h$, because
$$\frac{(1 + h)^2 - 1^2}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = 2 + h.$$
Table: The average rate of change of $f(x) = x^2$ on $[1, 1+h]$ as the interval shrinks.
Interval $[1, 1+h]$ | Width $h$ | Average rate of change $2 + h$ |
|---|---|---|
$[1, 2]$ | $1$ | $3$ |
$[1, 1.5]$ | $0.5$ | $2.5$ |
$[1, 1.1]$ | $0.1$ | $2.1$ |
$[1, 1.01]$ | $0.01$ | $2.01$ |
$\to$ single point | $\to 0$ | $\to 2 = f'(1)$ |
The average rates march straight toward $2$, which is exactly $f'(1) = 2(1)$. The instantaneous rate is the limit the average rates are heading for. This shrinking-interval picture is the heart of limits and derivatives.
Table: Average versus instantaneous rate of change, side by side.
Feature | Average rate of change | Instantaneous rate of change |
|---|---|---|
Interval | Whole interval $[a, b]$ | A single point $x = a$ |
Formula | $\dfrac{f(b) - f(a)}{b - a}$ | $\displaystyle\lim_{h \to 0}\dfrac{f(a+h) - f(a)}{h}$ |
Geometry | Slope of the secant line | Slope of the tangent line |
Needs | Two points | A limit (calculus) |
Everyday name | Average speed over a trip | Speedometer reading right now |
Does Some Instantaneous Rate Ever Equal The Average Rate?
Yes, and a theorem guarantees it. The Mean Value Theorem says that if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there is at least one point $c$ inside the interval where the instantaneous rate equals the average rate:
$$f'(c) = \frac{f(b) - f(a)}{b - a}, \qquad a < c < b$$
In plain terms, at some moment your instantaneous speed matched your average speed for the trip. If your average was $60$ km/h, then at some instant the speedometer read exactly $60$.
Check it on $f(x) = x^2$ over $[1, 3]$, where the average rate of change was $4$. Set $f'(c) = 4$. Since $f'(x) = 2x$, solving $2c = 4$ gives $c = 2$, which does lie in $(1, 3)$.
So the tangent at $x = 2$ is parallel to the secant across $[1, 3]$. The full statement and its proof live at the Mean Value Theorem page.
Why Does The Average Rate Of Change Formula Work?
The formula is not a rule to memorise so much as the definition of a rate written out.
A rate is a ratio of changes. Speed is distance per time; price growth is dollars per year; a slope is rise per run. Every rate divides how much one quantity changed by how much another changed. The average rate of change is that ratio for any function $f$ against its input.
Dividing by $b - a$ makes it per-unit. The numerator alone, $f(b) - f(a)$, only tells you the total change. Dividing by the input change $b - a$ scales that total down to a change per single unit of input, so intervals of different widths can be compared fairly.
The secant line carries the average. A straight line has one constant slope. Replacing the wiggly curve between $a$ and $b$ with the straight secant is exactly the act of averaging: the secant is the single steady rate that would carry you from the same start to the same finish.
Seen this way, "average rate of change," "slope of the secant," and "total change divided by input change" are three names for one quantity.
Who Shaped The Idea Of Average And Instantaneous Rates?
The step from a slope between two points to a slope at one point took decades, and it began before calculus had a name.
Two later chapters completed the picture:
Isaac Newton (1642–1727, England) and Gottfried Wilhelm Leibniz (1646–1716, Germany) independently turned the shrinking-secant idea into the derivative, giving the instantaneous rate its own symbols and rules.
Augustin-Louis Cauchy (1789–1857, France) later defined the limit precisely, so "the average rate as the interval shrinks to zero" became a rigorous statement rather than an intuition.
Where Is The Average Rate Of Change Used In The Real World?
Any time a total change is spread over an interval, the average rate of change is doing the work.
Physics: average velocity is displacement over elapsed time, and average acceleration is change in velocity over time.
Economics and business: average cost per unit, average revenue growth per quarter, and average marginal change all divide a total change by the range that produced it.
Biology and medicine: an average population growth rate, or the average rate a drug concentration falls between two blood tests, summarises change over a period.
Chemistry: the average reaction rate is the change in a reactant's concentration divided by the time interval measured.
Geography and engineering: the average grade of a road or a wheelchair ramp is its rise over its run, a slope between two endpoints.
Across every one of these, the same ratio turns two measurements into a single, comparable rate.
What Are The Most Common Mistakes With The Average Rate Of Change?
These four errors account for most lost marks on this topic, and each matches a question real students raise on r/learnmath, AP review notes, and course handouts.
Confusing the average rate with the instantaneous rate.
Where it slips in:
A student is asked for the average rate of change on $[a, b]$ but differentiates $f$ and plugs in a single point, or the reverse.
Don't do this:
Do not treat the two as interchangeable. The average rate uses two points and needs no calculus; the instantaneous rate uses a limit at one point.
The correct way:
For an interval, use $\dfrac{f(b) - f(a)}{b - a}$ with both endpoints. For a single instant, use the derivative $f'(x)$.
Dividing by $b$ instead of $b - a$, or mismatching the order.
Where it slips in:
A student writes $\dfrac{f(b) - f(a)}{b}$, or computes $\dfrac{f(b) - f(a)}{a - b}$ so the numerator and denominator run in opposite directions.
Don't do this:
Do not divide by a single endpoint, and do not let the top and bottom subtract in opposite orders.
The correct way:
Divide the output change by the input change, keeping the same order top and bottom: $\dfrac{f(b) - f(a)}{b - a}$.
Dropping the units or the sign.
Where it slips in:
A student reports "$9.8$" for an average velocity with no units, or writes a positive number when the function is decreasing across the interval.
Don't do this:
Do not strip a rate down to a bare number. The units (m/s, dollars per unit) and the sign carry real meaning.
The correct way:
Attach the units from the context, and keep the sign: a negative average rate of change signals that the output fell over the interval.
Assuming the Mean Value Theorem point is the midpoint.
Where it slips in:
A student learns that some $c$ has $f'(c)$ equal to the average rate and assumes $c$ is always the middle of the interval.
Don't do this:
Do not place $c$ at the midpoint by default. The theorem only guarantees $c$ lies somewhere in $(a, b)$.
The correct way:
Solve $f'(c) = \dfrac{f(b) - f(a)}{b - a}$ for $c$. It happens to be the midpoint for a quadratic, but not in general.
Practice Problems On The Average Rate Of Change
Work each one, then check against the answer. Answers are verified.
Find the average rate of change of $f(x) = x^2$ on $[2, 5]$.
(Answer: $\dfrac{25 - 4}{3} = 7$.)Find the average rate of change of $f(x) = 3x + 4$ on $[0, 10]$.
(Answer: $\dfrac{34 - 4}{10} = 3$; a line's average rate equals its slope.)A car's position is $s(t) = t^2 + 2t$ km after $t$ hours. Find its average velocity on $[1, 4]$.
(Answer: $\dfrac{24 - 3}{3} = 7$ km/h.)Find the average rate of change of $f(x) = \dfrac{1}{x}$ on $[1, 4]$.
(Answer: $\dfrac{\frac{1}{4} - 1}{3} = -\dfrac{1}{4}$; negative, since $f$ is decreasing.)Find the average rate of change of $f(x) = x^2 - 4x$ on $[0, 4]$.
(Answer: $\dfrac{0 - 0}{4} = 0$; equal heights at both ends.)For $f(x) = x^2$ on $[2, 6]$, find the Mean Value Theorem point $c$ where $f'(c)$ equals the average rate.
(Answer: average rate $= \dfrac{36 - 4}{4} = 8$, so $2c = 8$ gives $c = 4$.)
Where Should You Go Next After The Average Rate Of Change?
The average rate of change opens straight into the machinery of differential calculus, and several natural doors lead on from here.
The difference quotient. The average-rate formula written with interval width $h$, the exact expression you take a limit of to reach the derivative.
Definition of the derivative. Watch the average rate become the instantaneous rate as the interval shrinks to zero.
The derivative as a rate of change. The point-by-point rate that the average rate approximates, with its physics and economics readings.
If your child is meeting rates of change for the first time, a live Bhanzu trainer teaches the average rate of change from the secant picture up, so the derivative feels like the natural next step, in the Bhanzu math program.
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