# Rates of change (average and over equal intervals)

> AP Precalculus · Unit 1: Polynomial and Rational Functions
> Source: https://www.owlsprep.com/study/ap-precalculus-u1-rates-of-change/

How to measure how fast a function changes: the average rate of change over an interval, the rate of change at a point (approximated with averages — no calculus), how to compare rates at two points, and how the sign of a rate describes the way two quantities vary together.

**Prerequisites:** Topic 1.1 — increasing/decreasing, domain and range; Reading points and computing slope on a graph; Arithmetic with fractions and signed numbers

## Learning objectives

- Find the average rate of change of a function over an interval as the change in output divided by the change in input.
- Approximate the rate of change of a function at a point using average rates of change over small intervals.
- Compare the rates of change at two points.
- Use the sign of a rate of change to describe how two quantities vary together.

## Average Rate of Change

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.

**Average rate of change** — 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.

*Notation:* AROC

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

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 $f(x) = x^2 - 2x$ 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.

> **tip**
>
> Keep the order consistent: output-change on top, input-change on the bottom, both from the same interval. $\frac{f(b)-f(a)}{b-a}$ — not $\frac{f(a)-f(b)}{b-a}$.

## Rate of Change at a Point

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.

**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.

> **warning**
>
> This is an **approximation**, not an exact value. Finding the exact rate at a single point requires calculus, which comes *after* this course — here we only estimate with averages.

## Comparing Rates & Reading the Sign

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.

**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 $[0, 4]$.

> **tip**
>
> This connects back to Topic 1.1: a positive rate of change matches an *increasing* function, a negative rate matches a *decreasing* function — now with a number attached that says how fast.

## AP-Style Practice

**Check your understanding**

Test your understanding with this multiple-choice question:

1. The table gives $g(x)$. What is the average rate of change of $g$ over $[1, 5]$?

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

   - $4$
   - $8$
   - $16$
   - $2$

   *Answer:* $4$

   *Why:* Average rate of change over $[1,5]$ uses only the endpoints: $\frac{g(5)-g(1)}{5-1} = \frac{20-4}{4} = \frac{16}{4} = 4$.

**Worked example:** The value $V$ (in thousands of dollars) of a machine after $t$ years is $V(t) = 40 - 3t$. 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.

## Common pitfalls

- **Wrong:** Writing the average rate of change over $[a,b]$ as $\frac{f(a)-f(b)}{b-a}$.
  - Why it fails: Flipping the order of the outputs but not the inputs gives the wrong sign.
  - Correct: 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:** Treating the rate of change at a point as an exact value you can compute directly.
  - Why it fails: Finding the exact rate at a single point requires calculus, which is beyond this course. In precalculus it is only ever approximated.
  - Correct: 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:** Computing the average rate of change as the average of the two outputs, e.g. $\frac{f(a)+f(b)}{2}$.
  - Why it fails: That is the average *value*, not the average *rate of change*. Rate of change is a change divided by a change.
  - Correct: Always divide the change in output by the change in input: $\frac{f(b)-f(a)}{b-a}$.
- **Wrong:** Reporting only the change in output (e.g. "it went up 18") and calling that the rate.
  - Why it fails: A rate must account for how much the input changed too — 18 over 2 seconds is very different from 18 over 6 seconds.
  - Correct: Divide by the change in input so the answer is a rate (output per input), with units.
- **Wrong:** Concluding a function increases on the whole interval just because its average rate of change is positive.
  - Why it fails: A positive average rate only means the net change is positive; the function could dip and recover in between.
  - Correct: A positive average rate tells you the overall/net direction over the interval, not that it rises at every point inside it.

## Cheatsheet

| Concept | Rule | Key Notes |
| --- | --- | --- |
| Average rate of change over $[a,b]$ | $\dfrac{f(b)-f(a)}{b-a}$ | 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 |

## 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).

- [Rates of change in linear and quadratic functions](https://www.owlsprep.com/study/ap-precalculus-u1-rates-of-change-in-linear/)
- [Polynomial functions and rates of change](https://www.owlsprep.com/study/ap-precalculus-u1-polynomial-functions-and-rates-of/)

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