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How do we measure how fast a function changes, both over an interval and at a single point?

Topic 1.2 Rates of Change: compute and interpret the average rate of change of a function over an interval, and estimate the rate of change at a point.

A focused answer to AP Precalculus Topic 1.2, covering average rate of change over an interval, the rate of change at a point, and how to compute and interpret both from graphs, tables and formulas.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Average rate of change
Reading the sign and size
The rate of change at a point
Connecting rates of change to concavity
Try this

What this topic is asking

The College Board (Topic 1.2) wants you to measure how fast a function changes. The average rate of change over an interval is the slope of the secant line joining the two endpoints. The rate of change at a point is what that average rate approaches as the interval shrinks around the point. You must compute both, interpret them with units in context, and connect them to increasing, decreasing and concavity from Topic 1.1.

Average rate of change

The numerator is the change in output and the denominator is the change in input, so the average rate of change carries units of "output per input". For a distance-time function it is an average speed; for a population-time function it is people per year.

Reading the sign and size

A positive average rate of change means the output rose over the interval (the function increased on average), and a negative value means it fell. The size measures steepness: a larger absolute value means a steeper secant and a faster average change. Because the average rate of change depends only on the two endpoints, a function can wiggle in between and still show, say, a positive average rate.

The rate of change at a point

In AP Precalculus you estimate this rate numerically rather than computing it exactly (that exact limit is the derivative, which belongs to calculus). The key idea is that the tangent slope is the limiting case of the secant slope.

Average rate of change and a point estimate

For $f(x) = x^2$ , find the average rate of change over $[2, 5]$ , then estimate the rate of change at $x = 2$ using the interval $[2, 2.1]$ .

step 1 Average rate of change over [2, 5]

$f(5) = 25$ and $f(2) = 4$ , so $\frac{f(5) - f(2)}{5 - 2} = \frac{25 - 4}{3} = \frac{21}{3} = 7$ .

step 2 Set up the narrow interval for the point estimate

To estimate the rate at $x = 2$ , use a short interval $[2, 2.1]$ so the secant slope approximates the tangent slope.

step 3 Compute the narrow average rate

$f(2.1) = 2.1^2 = 4.41$ and $f(2) = 4$ , so $\frac{4.41 - 4}{2.1 - 2} = \frac{0.41}{0.1} = 4.1$ .

step 4 Interpret

The average rate over $[2, 5]$ is $7$ , but the rate right at $x = 2$ is close to $4.1$ . The true instantaneous rate at $x = 2$ is $4$ ; the short interval estimate of $4.1$ is already close, and it would improve as the interval shrinks further.

Connecting rates of change to concavity

Average rates of change over equal-width intervals are the cleanest way to detect concavity. Slice an interval into equal pieces, find the average rate of change on each piece, and compare. If those rates are increasing, the function is concave up; if they are decreasing, it is concave down. This is the same idea as successive differences from Topic 1.1, now expressed as rates so the units make sense in context.

A point worth keeping straight is that the average rate of change is a single number describing an interval, while the rate of change at a point describes one instant. They agree only for linear functions, whose rate of change is constant everywhere. For any curved function they differ, and watching the average rate change as you move the interval is precisely how you sense the function speeding up or slowing down. This distinction between an interval average and a point rate underlies every later topic about how polynomial, rational, exponential and logarithmic models grow.

Try this

Q1. Find the average rate of change of $f(x) = 3x - 4$ over $[0, 10]$ . [1 point]

Cue. $\frac{f(10) - f(0)}{10 - 0} = \frac{26 - (-4)}{10} = \frac{30}{10} = 3$ . For a line, the average rate equals the slope.

Q2. Average rates of change over three equal intervals are $2$ , $5$ , $9$ . Is the function concave up or down? [1 point]

Cue. The rates are increasing, so the function is concave up.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). For

f(x) = x^2 + 1

, what is the average rate of change of

f

over the interval

[1, 4]

? (A)

3

(B)

5

(C)

15

(D)

17

Show worked answer →

The correct answer is (B), $5$ .

The average rate of change is $\frac{f(4) - f(1)}{4 - 1}$ . Compute $f(4) = 16 + 1 = 17$ and $f(1) = 1 + 1 = 2$ , so the average rate of change is $\frac{17 - 2}{4 - 1} = \frac{15}{3} = 5$ . Choice (C) is the change in output without dividing by the change in input, a common slip.

AP 2024 (style)3 marksSection II (free response, calculator allowed). The population

P

of a town, in thousands, is recorded as

P(2010) = 18

P(2015) = 24

, and

P(2020) = 39

. (a) Find the average rate of change of

P

from 2010 to 2015 and from 2015 to 2020. (b) Using your two rates, state whether the population is growing at an increasing or decreasing rate, with justification.

Show worked answer →

A 3-point question on interpreting average rates of change.

(a) From 2010 to 2015 (1 point): $\frac{24 - 18}{2015 - 2010} = \frac{6}{5} = 1.2$ thousand per year. From 2015 to 2020 (1 point): $\frac{39 - 24}{2020 - 2015} = \frac{15}{5} = 3$ thousand per year.
(b) Increasing rate (1 point): the average rate of change rose from $1.2$ to $3$ thousand per year, so the population is growing at an increasing rate over this period.

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Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)