Rates of change (average and over equal intervals)

AP Precalculus· Unit 1 · Polynomial & Rational Functions· 12 min read

1. Average Rate of Change★★☆☆☆⏱ 4 min

In Topic 1.1 we asked whether a function's output rises or falls. Now we measure how fast — the rate of change. The most basic version is the average rate of change over an interval.

📘 Definition

Average rate of change

AROC

Over an interval $[a, b]$ , the average rate of change of $f$ is the change in output divided by the change in input. It is the single constant rate that would produce the same net change in output over that interval.

\text{average rate of change} = \frac{f(b) - f(a)}{b - a}

Graphically, this is the slope of the line joining the two endpoints $(a, f (a))$ and $(b, f (b))$ — the secant line. It carries units of output per input (for example, meters per second).

📐 Worked Example

Find the average rate of change of $f (x) = x^{2} - 2 x$ over the interval $[1, 4]$ .

1
Evaluate the function at the endpoints:
2
$f(1) = 1 - 2 = -1, \qquad f(4) = 16 - 8 = 8$
3
Divide the change in output by the change in input:
4
$\frac{f(4) - f(1)}{4 - 1} = \frac{8 - (-1)}{3} = \frac{9}{3} = 3$
5
On average, the output rises 3 units for every 1 unit of input across $[1, 4]$ .

📐 Worked Example

A car's distance $d$ (meters) is recorded every 2 seconds. Find the average rate of change (average speed) over each equal interval.

$t$ (s)	$0$	$2$	$4$	$6$
$d$ (m)	$0$	$18$	$40$	$66$

1
Each interval has the same width (2 s), so just divide each change in distance by 2:
2
$[0,2]:\ \tfrac{18-0}{2}=9 \quad [2,4]:\ \tfrac{40-18}{2}=11 \quad [4,6]:\ \tfrac{66-40}{2}=13$
3
The average speeds are 9, 11, and 13 m/s. All are positive (the car keeps moving forward), and they are getting larger — the car is speeding up across these intervals.

2. Rate of Change at a Point★★☆☆☆⏱ 3 min

What about the rate at a single point, not across a whole interval? In precalculus we do not use calculus for this — instead we approximate it.

📘 Definition

Rate of change at a point

The rate at which the output would change at that exact point. We approximate it with the average rate of change over a small interval containing the point. The smaller the interval, the closer the estimate.

📐 Worked Example

A function $h$ is recorded near $x = 2$ . Estimate the rate of change of $h$ at $x = 2$ .

$x$	$1.9$	$2.0$	$2.1$
$h (x)$	$5.2$	$6.0$	$7.0$

1
Use the average rate of change over the small interval $[1.9, 2.1]$ around $x = 2$ :
2
$\frac{h(2.1) - h(1.9)}{2.1 - 1.9} = \frac{7.0 - 5.2}{0.2} = \frac{1.8}{0.2} = 9$
3
So the rate of change of $h$ at $x = 2$ is approximately 9. A smaller interval around $x = 2$ would give a closer estimate — but it is always an estimate built from average rates.

3. Comparing Rates & Reading the Sign★★★☆☆⏱ 4 min

Two more uses of rates. First, once you can estimate a rate at a point, you can compare two points. Second, the sign of a rate tells you the direction in which the two quantities vary together.

📘 Definition

Sign of a rate of change

A positive rate of change means the two quantities move in the same direction — as the input increases, the output increases. A negative rate means they move in opposite directions — as the input increases, the output decreases. A larger size of the rate means faster change.

\text{positive rate} \Rightarrow \text{same direction} \qquad \text{negative rate} \Rightarrow \text{opposite directions}

📐 Worked Example

A cup of coffee cools: its temperature $T$ (°C) at time $m$ (minutes) is $T (0) = 90$ , $T (4) = 74$ , $T (10) = 58$ . (a) Find the average rate of change over $[0, 4]$ and over $[4, 10]$ . (b) What does the sign mean? (c) Over which interval is the temperature changing faster?

1
(a) Divide each change in temperature by the change in time:
2
$[0,4]:\ \frac{74-90}{4-0}=\frac{-16}{4}=-4 \quad\quad [4,10]:\ \frac{58-74}{10-4}=\frac{-16}{6}\approx -2.7$
3
(b) Both rates are negative — as time increases, temperature decreases. The two quantities move in opposite directions (the coffee cools).
4
(c) The first interval has the larger size, $∣ - 4∣ > ∣ - 2.7∣$ , so the temperature is changing faster over $[0, 4]$ .

4. AP-Style Practice★★☆☆☆⏱ 5 min

✓ Quick check

Test your understanding with this multiple-choice question:

The table gives $g (x)$ . What is the average rate of change of $g$ over $[1, 5]$ ?
$x$ $1$ $3$ $5$
$g (x)$ $4$ $10$ $20$
- $4$
- $8$
- $16$
- $2$

$x$	$1$	$3$	$5$
$g (x)$	$4$	$10$	$20$

📐 Worked Example

The value $V$ (in thousands of dollars) of a machine after $t$ years is $V (t) = 40 - 3 t$ . Find the average rate of change of $V$ over $[0, 6]$ , and interpret the sign in context.

1
Evaluate the endpoints: $V (0) = 40$ , $V (6) = 40 - 18 = 22$ .
2
$\frac{V(6)-V(0)}{6-0} = \frac{22-40}{6} = \frac{-18}{6} = -3$
3
The rate is $- 3$ thousand dollars per year. The negative sign means the value decreases as time increases — the machine loses $3{,}000 of value per year, on average.

📐 Worked Example

A function $f$ is given near $x = 5$ : $f (4.8) = 12.4$ , $f (5.2) = 15.6$ . Estimate the rate of change of $f$ at $x = 5$ .

1
Use the average rate of change over the small interval $[4.8, 5.2]$ around $x = 5$ :
2
$\frac{f(5.2)-f(4.8)}{5.2-4.8} = \frac{15.6-12.4}{0.4} = \frac{3.2}{0.4} = 8$
3
The rate of change of $f$ at $x = 5$ is approximately 8 — a positive value, so near $x = 5$ the output rises as the input rises.

5. Common Pitfalls

Wrong move:

Writing the average rate of change over $[a, b]$ as $\frac{f ( a ) - f ( b )}{b - a}$ .

Why:

Flipping the order of the outputs but not the inputs gives the wrong sign.

Correct move:

Keep both changes in the same order: $\frac{f ( b ) - f ( a )}{b - a}$ — end minus start on top, end minus start on the bottom.

Wrong move:

Treating the rate of change at a point as an exact value you can compute directly.

Why:

Finding the exact rate at a single point requires calculus, which is beyond this course. In precalculus it is only ever approximated.

Correct move:

Approximate the rate at a point with the average rate of change over a small interval around it; a smaller interval gives a closer estimate.

Wrong move:

Computing the average rate of change as the average of the two outputs, e.g. $\frac{f ( a ) + f ( b )}{2}$ .

Why:

That is the average value, not the average rate of change. Rate of change is a change divided by a change.

Correct move:

Always divide the change in output by the change in input: $\frac{f ( b ) - f ( a )}{b - a}$ .

Wrong move:

Reporting only the change in output (e.g. "it went up 18") and calling that the rate.

Why:

A rate must account for how much the input changed too — 18 over 2 seconds is very different from 18 over 6 seconds.

Correct move:

Divide by the change in input so the answer is a rate (output per input), with units.

Wrong move:

Concluding a function increases on the whole interval just because its average rate of change is positive.

Why:

A positive average rate only means the net change is positive; the function could dip and recover in between.

Correct move:

A positive average rate tells you the overall/net direction over the interval, not that it rises at every point inside it.

6. Quick Reference Cheatsheet

Concept	Rule	Key Notes
Average rate of change over $[a, b]$	$\frac{f ( b ) - f ( a )}{b - a}$	change in output ÷ change in input; the slope of the line through the two points
"Same net change" idea	the constant rate giving the same overall change	a straight-line stand-in for the interval
Rate of change at a point	≈ average rate over a small interval around it	precalc = approximation, not calculus; smaller interval → closer
Compare two points	estimate the rate at each, then compare	steeper graph → greater rate
Positive rate	output and input change in the same direction	quantities move together (both up or both down)
Negative rate	output and input change in opposite directions	one goes up while the other goes down
Size of the rate	larger magnitude → faster change	the sign gives direction, the size gives speed

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · AP Precalculus
Compute average rate of change from a table or graph
2024 · AP Precalculus
Interpret and compare rates of change in a context

What's Next

You can now measure how fast a function changes — on average over an interval, and (approximately) at a point — and read the direction from the sign. Next, Topic 1.3 looks at what these rates of change look like for two important families: linear functions (constant rate) and quadratic functions (a rate that changes in a regular way).