Rates of Change — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

A hiking app says you averaged 1.9 miles per hour. You did not walk at that speed every minute: you were faster on the flat start and slower on the climb. That single number is an average rate of change, and comparing such averages over short stretches is how precalculus says how fast a function changes at a moment.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the slope of the line through $(1, 4)$ and $(4, 8)$ ?

$\frac{4}{3}$
$\frac{3}{4}$
$4$
$12$

Show the answer

$\frac{8 - 4}{4 - 1} = \frac{4}{3}$ .

Correct: $\frac{4}{3}$ :
$\frac{3}{4}$ :
$4$ :
$12$ :

2. The outputs of $f$ at $x = 0, 1, 2, 3$ are $2, 5, 10, 17$ . How does the rate of change of $f$ behave?

It is increasing
It is decreasing
It is constant
It is negative

Show the answer

The changes $3, 5, 7$ rise.

Correct: It is increasing:
It is decreasing:
It is constant:
It is negative:

Part 4 · See it

See it first

The graph of f(x) = 4 times the square root of x for 0 ≤ x ≤ 9, with a blue line through (1, 4) and (4, 8). The slope of that line, (8 − 4)/(4 − 1) = 4/3, is the average rate of change of f on [1, 4]. — The average rate of change of $f(x) = 4\sqrt{x}$ on $[1, 4]$ is $\frac{8 - 4}{4 - 1} = \frac{4}{3}$ , the slope of the line through $(1, 4)$ and $(4, 8)$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Pick an interval $[a, b]$ and find $f(a)$ and $f(b)$ the average rate of change is $\frac{f(b) - f(a)}{b - a}$ , in output units per input unit
Draw the line through $(a, f(a))$ and $(b, f(b))$ its slope is that average rate of change
Shrink the interval around a point $x = c$ the average rate of change estimates how fast $f$ changes at $c$
Compare such estimates at two points over equally short intervalsthe larger one tells where $f$ is changing faster

Part 6 · Key ideas

Key ideas

The average rate of change of $f$ on $[a, b]$ is $\frac{f(b) - f(a)}{b - a}$ : the slope of the line through $(a, f(a))$ and $(b, f(b))$ .
Units: output units per input unit, like miles per hour or dollars per year.
Positive: the output ends higher. Negative: it ends lower. Zero: the outputs at the ends are equal.
To compare rates at two points, use short intervals of the same length around each.

Part 7 · Misconception

A common mistake

The wrong idea: The interval with the biggest change in output has the biggest average rate of change.

What actually happens: Divide by the length of the interval first. The hiker gained $4.5$ miles on $[1, 3]$ but only $2.5$ on $[0, 1]$ , yet the rate on $[0, 1]$ is higher: $2.5$ against $2.25$ miles per hour.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Distance on a hike

Distance d(t) a hiker has walked, in miles, t hours after starting
$t$ (hours)	0	1	3	4	6
$d(t)$ (miles)	0	2.5	7.0	8.8	11.2

1. What is the average rate of change of $d$ on $[1, 3]$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\frac{d(3) - d(1)}{3 - 1}$ $= \frac{7.0 - 2.5}{2}$ $= 2.25$ miles per hour: the slope of the line through $(1, 2.5)$ and $(3, 7.0)$ .

Answer: $\frac{9}{4}$ ≈ 2.250 miles per hour

2. On which interval between consecutive table times is the average rate of change of $d$ greatest?

$[1, 3]$
$[0, 1]$
$[4, 6]$
$[3, 4]$

Show the answer

The intervals have different lengths, so compare rates, not changes: $2.5$ , $2.25$ , $1.8$ and $1.2$ miles per hour on $[0, 1]$ , $[1, 3]$ , $[3, 4]$ and $[4, 6]$ .

$[1, 3]$ : The distance grows most here ( $4.5$ miles), but over $2$ hours: $2.25$ miles per hour, less than on $[0, 1]$ .
Correct: $[0, 1]$ : Right: $2.5$ miles in $1$ hour is $2.5$ miles per hour, the greatest of $2.5$ , $2.25$ , $1.8$ and $1.2$ .
$[4, 6]$ : $\frac{11.2 - 8.8}{2} = 1.2$ miles per hour, the smallest.
$[3, 4]$ : $\frac{8.8 - 7.0}{1} = 1.8$ miles per hour.

3. Using the table, which is the best estimate of how fast the hiker is walking at $t = 3.5$ hours?

$1.8$ miles per hour
$2.0$ miles per hour
$8.8$ miles per hour
$1.87$ miles per hour

Show the answer

To estimate the rate at one time, use the average rate of change over a short interval containing it: $\frac{d(4) - d(3)}{4 - 3}$ $= 1.8$ miles per hour.

Correct: $1.8$ miles per hour: Right: $[3, 4]$ is the shortest interval in the table that contains $3.5$ , and its average rate is $1.8$ .
$2.0$ miles per hour: This is $\frac{7.0}{3.5}$ , which mixes a distance at $t = 3$ with the time $3.5$ ; use the average rate over $[3, 4]$ .
$8.8$ miles per hour: $8.8$ is the distance at $t = 4$ , not a rate.
$1.87$ miles per hour: That is the average over the whole hike, $[0, 6]$ ; a short interval around $3.5$ gives a better estimate.

Graph

Concentration of a medicine

4. The concentration of a medicine in the blood is $C(t) = \frac{12t}{t + 2}$ milligrams per liter, $t$ hours after a dose. Find the average rate of change of $C$ on $[2, 6]$ .

Calculator allowed. Give the answer correct to three decimal places, in milligrams per liter per hour.

Show the answer

$C(2) = 6$ and $C(6) = 9$ , so $\frac{C(6) - C(2)}{6 - 2}$ $= \frac{3}{4}$ $= 0.75$ milligrams per liter per hour.

Answer: 0.750 milligrams per liter per hour

5. The concentration of a medicine in the blood is $C(t) = \frac{12t}{t + 2}$ milligrams per liter, $t$ hours after a dose. Use the average rate of change on $[1, 1.1]$ to estimate how fast $C$ is changing at $t = 1$ . Give your answer to three decimal places.

Calculator allowed. Give the answer correct to three decimal places, in milligrams per liter per hour.

Show the answer

$C(1) = 4$ and $C(1.1)$ $= \frac{13.2}{3.1}$ $= 4.258064...$ (store it). $\frac{C(1.1) - C(1)}{0.1}$ $= 2.580645...$ , so about $2.581$ milligrams per liter per hour.

Answer: 2.581 milligrams per liter per hour

6. The concentration of a medicine in the blood is $C(t) = \frac{12t}{t + 2}$ milligrams per liter, $t$ hours after a dose. Which interval's average rate of change gives the best estimate of how fast $C$ changes at $t = 4$ ?

$[0, 8]$
$[4, 10]$
$[3.9, 4.1]$
$[2, 6]$

Show the answer

The shorter the interval around a point, the closer the average rate of change is to how fast the function changes at that point.

$[0, 8]$ : A long interval mixes the fast early change with the slow later change.
$[4, 10]$ : This interval is long and starts at $4$ , so it reflects mostly later, slower change.
Correct: $[3.9, 4.1]$ : Right: a short interval around $t = 4$ uses outputs very close to the one at $t = 4$ .
$[2, 6]$ : Better than a long interval, but $[3.9, 4.1]$ is much shorter around $t = 4$ .

Part 9 · Summary

Summary

The average rate of change of

f

on

[a, b]

is

\frac{f(b) - f(a)}{b - a}

, the slope of the line through

(a, f(a))

and

(b, f(b))

, in output units per input unit. Over a short interval around a point it estimates how fast

f

changes at that point, and comparing those estimates tells where

f

changes faster.

Part 10 · Up next

What comes next

Part 11 · Connections