Average Rate of Change
"We averaged 80 km/h." You've said something like it after every long car
journey, and you computed it without a moment's thought: the odometer read
18{,}240 km when you left and 18{,}480 km
when you arrived, the clock said the trip took 3 hours, so
\text{average speed} = \frac{18{,}480 - 18{,}240}{3} = \frac{240 \text{ km}}{3 \text{ h}} = 80 \text{ km/h}.
Total change divided by time taken. That's it — that's the whole idea of this
page, and you already own it. But look closer at what the number hides. On that trip you sat in
a roadworks jam for twenty minutes, completely stationary — speed
0. Later, on the empty motorway, the speedometer touched
120. The "80" you reported is neither of those; it never appeared on
the speedometer for more than an instant. It is a single number summarising the whole messy,
variable journey: if you had somehow driven at one perfectly steady speed and still
covered 240 km in 3 hours, that speed
would have been 80 km/h.
That is what an average rate of change is, for any changing quantity —
a growing population, a cooling coffee, a rising share price, a draining tank. Take the total
change in the thing you care about, divide by the change in the input (usually time), and you
get the steady rate that would have produced the same overall change.
The definition
For a function f over an interval from
x = a to x = b, the
average rate of change is the change in the output divided by the change in
the input:
\text{average rate of change} = \frac{\Delta f}{\Delta x} = \frac{f(b) - f(a)}{b - a}
The capital Greek delta, \Delta, is mathematics' shorthand for
"change in": \Delta x = b - a is how far the input moved, and
\Delta f = f(b) - f(a) is how far the output moved in response.
Read the fraction aloud and it says exactly what the road trip said: output change per
unit of input change.
And that fraction should look deeply familiar — it is exactly the
slope of a line,
rise over run, just written with function values instead of coordinates. This is not a
coincidence; it's the geometric heart of the whole idea, and it's where we go next.
Two things to keep straight from the start. First, the average rate carries
units: output units per input unit — km per hour, people per year,
degrees per minute. Second, it has a sign: if f
ends lower than it started, \Delta f is negative and so is the
rate — a draining tank has a negative rate of change in litres per minute.
It is the slope of the secant line
Now draw the picture. Mark the two points \big(a, f(a)\big) and
\big(b, f(b)\big) on the graph of f and
join them with a straight line. That line — a line that cuts the curve at two points —
is called a secant line (from the Latin secare, "to cut"), and the
average rate of change is its slope: rise f(b) - f(a) over
run b - a.
Try it live on the parabola f(x) = x^2. Drag the sliders to move the
two endpoints and watch the bold secant line tilt as the interval changes. Three experiments
worth running: put a = 1, b = 3 and note the steepness; slide both
points to the left half of the parabola and watch the slope go negative (the function
is falling there); and — the fun one — set a = -2, b = 2. The secant
goes perfectly flat: the parabola climbed and fell by exactly the same amount,
so its average rate of change over that interval is zero, even though it was never
standing still except for one instant at the bottom.
That flat secant is the graph-world version of the roadworks jam: an average can be tame while
the function underneath is doing wild things. The secant only ever consults the two endpoints.
Worked example: a quadratic, step by step
Take f(x) = x^2 on the interval from x = 1
to x = 3, and follow the recipe deliberately.
Step 1 — evaluate at both endpoints.
f(1) = 1^2 = 1 and f(3) = 3^2 = 9.
Step 2 — form the two changes. The rise is
\Delta f = f(3) - f(1) = 9 - 1 = 8; the run is
\Delta x = 3 - 1 = 2.
Step 3 — divide.
\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{3 - 1} = \frac{8}{2} = 4
Over this stretch the function rises, on average, 4 units of output
for every 1 unit of input. Watch the same computation happen
geometrically — the secant, its run, its rise:
One more, with a negative answer. Same parabola, but from
x = -3 to x = 1:
f(-3) = 9 and f(1) = 1, so
\frac{f(1) - f(-3)}{1 - (-3)} = \frac{1 - 9}{4} = \frac{-8}{4} = -2.
The function fell overall, so the average rate is negative — even though, near the
right-hand end, the parabola had already turned around and was climbing again. Order matters
inside each difference (top and bottom must both run a \to b), but
it doesn't matter which point you call a: swap them and
both differences flip sign, so the fraction survives untouched.
No formula? No problem: rates straight from data
Here is something quietly powerful about \tfrac{f(b) - f(a)}{b - a}:
it never asks what f is. It only needs two values. So it
works just as well on measured data with no formula in sight — which is how it earns its living
in geography, biology, economics and medicine. Suppose a census records a town's population:
| year |
2000 |
2010 |
2020 |
| population |
12{,}000 |
21{,}000 |
24{,}000 |
Average rate of change from 2000 to 2010:
\frac{21{,}000 - 12{,}000}{2010 - 2000} = \frac{9{,}000}{10} = 900 \text{ people per year,}
and from 2010 to 2020:
\frac{24{,}000 - 21{,}000}{2020 - 2010} = \frac{3{,}000}{10} = 300 \text{ people per year.}
Both rates are positive — the town grew in both decades — but comparing them tells a story a
single number can't: growth slowed to a third of its earlier pace. Town planners,
epidemiologists tracking case counts, and analysts reading revenue tables all live on exactly
this comparison: compute the average rate on successive intervals and watch how it changes.
Notice the units did their job too — "people per year" fell straight out of dividing a change
in people by a change in years.
Shrinking the interval — a peek over the horizon
A question should be nagging at you. The average over a wide interval blurs out detail (the
traffic jam vanished into the "80"). So what if we made the interval narrower? Wouldn't
the average then describe what's happening near a point more and more faithfully?
Let's try it, on f(x) = x^2, anchoring the left end at
x = 1 and pulling the right end in:
| interval |
width |
average rate of change |
| [1,\ 3] |
2 |
\tfrac{9 - 1}{2} = 4 |
| [1,\ 2] |
1 |
\tfrac{4 - 1}{1} = 3 |
| [1,\ 1.5] |
0.5 |
\tfrac{2.25 - 1}{0.5} = 2.5 |
| [1,\ 1.1] |
0.1 |
\tfrac{1.21 - 1}{0.1} = 2.1 |
| [1,\ 1.01] |
0.01 |
\tfrac{1.0201 - 1}{0.01} = 2.01 |
The averages march in single file: 4,\ 3,\ 2.5,\ 2.1,\ 2.01,\ \dots —
homing in on 2. There's a slick reason. For
f(x) = x^2 the average rate between any two points has a
beautiful closed form:
\frac{b^2 - a^2}{b - a} = \frac{(b - a)(b + a)}{b - a} = a + b.
The average rate of x^2 from a to
b is simply a + b! (Check the table:
1 + 3 = 4, 1 + 2 = 3,
1 + 1.5 = 2.5…) So as b slides toward
1, the average slides toward 1 + 1 = 2,
and nothing can stop it.
Pause on what just happened, because it is the doorway to the rest of calculus. The averages
aren't just wandering — they are converging, as if the function has a definite
steepness at the single point x = 1, and the
shrinking averages are sniffing it out. That limiting value — the "instantaneous" rate of
change, the speedometer reading rather than the trip average — is the derivative,
and chasing it down properly is exactly where this thread of pages is headed, starting with a
tidier notation for the shrinking interval: the
difference
quotient. For now, hold on to the picture: average rate over an interval first;
squeeze the interval later.
Three traps catch nearly every newcomer, and they're all the same trap in disguise:
-
The average rate is computed from the two endpoints — full stop. Between
them, the function can do absolutely anything: your 80 km/h trip average contains a
twenty-minute dead standstill and a 120 km/h sprint, and the arithmetic neither knows nor
cares. Two wildly different functions with the same endpoint values have identical
average rates of change. Never read "average rate 4" as "the function was changing at 4
throughout."
-
It is the secant's slope — not the curve's slope anywhere in particular.
The number describes one straight line drawn between two points. At most individual points
of the interval, the curve itself is steeper or shallower than the secant. (On
x^2 from -2 to
2 the average is 0, yet the curve is
flat at only a single point.) The question "how steep is the curve right here?" is a
genuinely different question — the derivative's question.
-
Units, always: change in f per change in
x. An answer of "-5" for a
draining tank is incomplete; "-5 litres per minute" is the fact.
If an exam answer's units aren't output-unit per input-unit, the fraction was built
upside-down — a good self-check that costs two seconds.
On many motorways you'll pass a sign for an average speed check: a camera
photographs your number plate at the start of the zone, a second camera photographs it again at
the end, and a computer does one piece of maths — the maths of this page. Cameras
10 km apart, timestamps 6 minutes apart:
\frac{\Delta \text{distance}}{\Delta \text{time}} = \frac{10 \text{ km}}{0.1 \text{ h}} = 100 \text{ km/h}.
If the limit is 80, you're getting a ticket — and here's the delicious part — without
anyone ever measuring your speed. No radar, no speedometer reading, just two positions and
two times. How can an average of 100 prove your actual speed ever broke 80? Because a
car's speed can't teleport: to average 100 over the zone while never touching 100, you'd
have to be below it sometimes and… no, wait — below the average sometimes forces you
above it at other times, or the average couldn't come out that high. At some instant
in that zone, your true speedometer speed was at least 100.
That "at some instant the instantaneous rate must equal the average rate" principle is a real
theorem — the Mean Value Theorem, one of the crown jewels of calculus, waiting
for you a few concepts down this path. It needs the derivative to even state properly, but you
have just understood its soul from a speeding ticket. Courts in several countries have accepted
exactly this argument; the mathematics is not the weak point of the prosecution's case.
Watch it explained
Sal Khan introduces the average rate of change of a function over an interval, and shows how it
connects to slope — a good second pass over everything above.