Unit 2 · Topic 2.1 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 2.1: Defining Average and Instantaneous Rates of Change at a Point

Q: Is the average rate of change the same as the slope of a line?

Yes, of the line through the two points (a, f(a)) and (a + h, f(a + h)) on the graph. That line is called a secant line, and the graph at the top of this lesson shows two of them.

Q: Why can't I just put h = 0 into the difference quotient?

Because the top and the bottom are both 0, and 00 is not a number. Rewrite the quotient for h 0 until the h in the denominator cancels, then let h approach 0.

Q: Does every function have an instantaneous rate of change at every point?

No. The definition says provided the limit exists. For f(x) = |x| at 0, the averages from the right are 1 and from the left are -1, so the limit does not exist and f'(0) does not exist.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 2: 10 to 15% of the multiple-choice section

What you'll be able to do

Find the average rate of change of a function over an interval with a difference quotient.
Write the derivative of a function at a point as the limit of a difference quotient.

Key vocabulary

Term	What it means	The exam trap
Average rate of change	The change in a function's output over an interval divided by the change in its input.	Keep the two outputs and the two inputs in the same order on the top and the bottom, or the sign comes out wrong.
Difference quotient	An expression such as $\frac{f (a + h) - f (a)}{h}$ or $\frac{f (x) - f (a)}{x - a}$ : a difference of outputs over a difference of inputs.	Before you compute anything, say which two inputs the quotient uses; that is the interval it describes.
Instantaneous rate of change	The rate at one input, defined as the limit of average rates of change over intervals that shrink to that input.	Read whether a question is about an interval or about one input before you choose a formula.
Derivative at a point, $f^{'} (a)$	The limit $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ , which equals $\lim_{x \to a} \frac{f (x) - f (a)}{x - a}$ , when that limit exists.	$f^{'} (a)$ is a rate, a number about how fast $f$ changes at $a$ ; $f (a)$ is the output there, a different number.

Graph of s(t) = t squared for t from 0 to 6. Two secant lines start at the point (3, 9): one ends at t = 5 and has slope 8, the other ends at t = 4 and has slope 7. A dashed line through (3, 9) has slope 6, the value the secant slopes approach as the second point moves toward t = 3.

Part 1 Think first How fast at exactly two seconds?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A stone is dropped from a bridge. After $t$ seconds it has fallen $d (t) = 5 t^{2}$ meters, so 20 meters after 2 seconds and 45 meters after 3. How fast is it falling at the instant $t = 2$ ?

In short. The average rate of change of $f$ from $a$ to $a + h$ is the difference quotient $\frac{f (a + h) - f (a)}{h}$ , and the instantaneous rate of change at $a$ is its limit as $h \to 0$ , written $f^{'} (a)$ .

Both $\frac{f (a + h) - f (a)}{h}$ and $\frac{f (x) - f (a)}{x - a}$ are difference quotients: change in output over change in input.
The instantaneous rate at $a$ is $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ , or equally $\lim_{x \to a} \frac{f (x) - f (a)}{x - a}$ .
That limit is the derivative $f^{'} (a)$ , and it exists only when the limit does.

Part 2 Learn it Three short ideas, each checked as you go

2a. Average rate of change is a difference quotient

The average rate of change of $f$ over an interval is the change in the output divided by the change in the input. A quotient of two differences like this is called a difference quotient.

Calculus writes it two ways. If the interval starts at $a$ and has length $h$ , it runs from $a$ to $a + h$ , and

\frac{f (a + h) - f (a)}{h} .

If the interval runs from $a$ to some other input $x$ , it is

\frac{f (x) - f (a)}{x - a} .

For the graph at the top, $s (t) = t^{2}$ , from $t = 3$ to $t = 5$ : $\frac{s (5) - s (3)}{5 - 3} = \frac{25 - 9}{2} = 8$ meters per second. That is the slope of the line through the two points on the graph.

Quick check · Calculate

No calculator. Let $f (x) = x^{3} - 2 x$ . What is the average rate of change of $f$ over the interval $[1, 3]$ ?

Your answer

2b. Reading the interval off the quotient

Before you work out a difference quotient, read which interval it measures.

In $\frac{f (a + h) - f (a)}{h}$ the number $h$ is the length of the interval, not its far end. With $a = 1$ and $h = 4$ the interval runs from 1 to 5.

In $\frac{f (x) - f (a)}{x - a}$ the two inputs are the endpoints themselves. The order on top matches the order on the bottom, so $x$ can be on either side of $a$ : with $a = 3$ and $x = 1$ the quotient is $\frac{f (1) - f (3)}{1 - 3}$ , the average rate from 1 to 3.

A negative $h$ works the same way: with $a = 4$ and $h = - 1$ the interval runs from 3 to 4.

Quick check · Match

No calculator. Each difference quotient gives the average rate of change of $f$ over one interval. Tap a quotient, then its interval.

2c. From average to instant: the limit

An average needs two different inputs. At a single instant $h = 0$ , and $\frac{f (a + 0) - f (a)}{0}$ divides by zero. So calculus does what you did with the stone in Think first: it lets $h$ shrink toward 0 and asks what the averages approach.

The instantaneous rate of change of $f$ at $x = a$ is

\lim_{h \to 0} \frac{f (a + h) - f (a)}{h} or \lim_{x \to a} \frac{f (x) - f (a)}{x - a},

provided the limit exists. The two limits are equivalent forms of the definition of the derivative, and either one is written $f^{'} (a)$ , read "f prime of a."

On the graph at the top, each secant slope is an average rate, and as the second point slides toward $t = 3$ the slopes approach 6. So $s^{'} (3) = 6$ meters per second.

The limit can fail to exist. Then $f$ has no instantaneous rate of change at $a$ , and $f^{'} (a)$ does not exist.

Quick check · Multiple choice

No calculator. Which expression is $g^{'} (4)$ , the instantaneous rate of change of $g$ at $x = 4$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Reading $h$ in $\frac{f (a + h) - f (a)}{h}$ as the far end of the interval, and so writing $f (h) - f (a)$ on top.	The interval runs from $a$ to $a + h$ : $h$ is its length, and its far end is $a + h$ .	Write the interval first: start at $a$ , end at $a + h$ . The outputs on top are at those two inputs.
Calling a difference quotient with a fixed $h$ the instantaneous rate.	With any $h$ that is not 0, the quotient averages over an interval. Only its limit as $h \to 0$ is the rate at one input.	Write $\lim_{h \to 0}$ in front of the quotient, and say average rate when there is no limit.
Substituting $h = 0$ straight into the difference quotient.	That gives $\frac{0}{0}$ , which is not a number and says nothing about the limit.	Rewrite the quotient for $h \neq 0$ until the $h$ in the denominator cancels, and then find the limit.

Part 3 Watch it done The derivative at a point, from the definition

Watch each step, then do the last one yourself.

Let $f (x) = x^{2} - 4 x$ . Use the definition of the derivative to find $f^{'} (5)$ .

Step 1 · Write the limit
$f^{'} (5) = \lim_{h \to 0} \frac{f (5 + h) - f (5)}{h}$ , and $f (5) = 25 - 20 = 5$ .
Step 2 · Expand the top
$f (5 + h) = (5 + h)^{2} - 4 (5 + h) = 25 + 10 h + h^{2} - 20 - 4 h = 5 + 6 h + h^{2}$ . So the top is $(5 + 6 h + h^{2}) - 5 = 6 h + h^{2}$ .
Step 3 · Simplify the quotient
$\frac{6 h + h^{2}}{h} = 6 + h$ , for every $h \neq 0$ .
Why this step?
Why may you cancel the $h$ when the limit is about $h$ going to 0?
The limit looks at values of $h$ close to 0 but never equal to 0 (Unit 1). For every such $h$ , dividing by $h$ is allowed, so $\frac{6 h + h^{2}}{h}$ and $6 + h$ are equal there and have the same limit.
Your step · Calculate

No calculator. Finish it. For $f (x) = x^{2} - 4 x$ , the difference quotient at 5 simplifies to $6 + h$ . What is $f^{'} (5)$ ?

Your answer

Part 4 Try it Two new functions

Try it · Calculate

No calculator. Let $g (x) = \frac{1}{x}$ . What is the value of $\frac{g (2 + h) - g (2)}{h}$ when $h = 2$ ? Give an exact decimal.

Your answer

Try it · Decide and justify

No calculator. Let $f (x) = | x |$ . A classmate says the instantaneous rate of change of $f$ at $x = 0$ is 0, because the average rate of change over $[- 1, 1]$ is 0. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 3, from free-response Part B (calculator not permitted).

Free-response practice · Question 3 style

No calculator. A model rocket is launched straight up. Its height above the ground, in feet, $t$ seconds after launch is $H (t) = 48 t - 16 t^{2}$ for $0 \leq t \leq 3$ .

(a) Find the average rate of change of $H$ over the interval $0 \leq t \leq 1$ . Show the computation that leads to your answer. Using correct units, explain the meaning of this value in the context of the problem.

(b) Write, but do not evaluate, a limit expression that gives the instantaneous rate of change of the rocket's height at $t = 1$ .

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $\frac{H (1) - H (0)}{1 - 0} = \frac{(48 - 16) - 0}{1} = 32$ feet per second. Over the first second after launch, the rocket's height increased at an average rate of 32 feet per second.

(b) $\lim_{h \to 0} \frac{H (1 + h) - H (1)}{h}$ , or equally $\lim_{t \to 1} \frac{H (t) - H (1)}{t - 1}$ .

(c) $H (1 + h) = 48 + 48 h - 16 (1 + 2 h + h^{2}) = 32 + 16 h - 16 h^{2}$ and $H (1) = 32$ , so the quotient is $\frac{16 h - 16 h^{2}}{h} = 16 - 16 h$ for $h \neq 0$ . Its limit as $h \to 0$ is 16, so at $t = 1$ the height is increasing at 16 feet per second.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes the difference quotient $\frac{H (1) - H (0)}{1 - 0}$ and gets 32.
Says the height increased at an average of 32 feet per second over the first second, with units.
Writes a limit of a difference quotient at $t = 1$ , with $h \to 0$ or $t \to 1$ .
Simplifies to $16 - 16 h$ before taking the limit, and reports 16 feet per second.

Part 6 Check yourself A car pulling away

The two case questions use the same table. Answer them from it.

A car pulls away from a stop sign. Its distance from the sign, $p (t)$ meters, $t$ seconds later, is $p (t) = 2 t^{2} + t$ for the first few seconds. Some values are in the table.

Distance of the car from the stop sign
$t$ (seconds)	$p (t)$ (meters)
0	0
1	3
2	10
3	21
4	36

Case question 1 · Calculate

No calculator. What is the car's average rate of change of distance, in meters per second, from $t = 1$ to $t = 3$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Which expression gives the car's rate of change of distance at the instant $t = 2$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds $f^{'} (2)$ for $f (x) = x^{2} + 1$ from the definition. Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

The quotient $\frac{f (x) - f (6)}{x - 6}$ is the average rate of change of $f$ over the interval with endpoints 6 and . Its limit as $x$ approaches 6, when that limit exists, is written .

Part 7 Keep it Bring back earlier ideas

This lesson stands on two ideas from Unit 1. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. Why can't an average rate of change be computed over the single point $x = 2$ , an interval from 2 to 2?

Bring it back · Calculate

No calculator. Find $\lim_{x \to 3} \frac{x^{2} - 9}{x - 3}$ .

Your answer

In three lines

A difference quotient, change in output over change in input, gives the average rate of change over an interval.
As the interval shrinks to one input, the limit of the difference quotient is the instantaneous rate of change.
That limit is the derivative $f^{'} (a)$ , and it exists only when the limit exists.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question asks for an average rate of change, write the quotient with both outputs and the interval length before any arithmetic, and attach units such as feet per second. When it asks you to express an instantaneous rate, write the limit sign and $h \to 0$ (or $x \to a$ ); a quotient without the limit is an average.

Questions students ask

Which form should I use, the one with

h

or the one with

x

They give the same limit, so use whichever makes the algebra easier. The $h$ form is usually easier when you expand $f (a + h)$ ; the $x$ form is usually easier when you can factor $x - a$ out of the top.

Is the average rate of change the same as the slope of a line?

Yes, of the line through the two points $(a, f (a))$ and $(a + h, f (a + h))$ on the graph. That line is called a secant line, and the graph at the top of this lesson shows two of them.

Why can't I just put

h = 0

into the difference quotient?

Because the top and the bottom are both 0, and $\frac{0}{0}$ is not a number. Rewrite the quotient for $h \neq 0$ until the $h$ in the denominator cancels, then let $h$ approach 0.

Does every function have an instantaneous rate of change at every point?

No. The definition says provided the limit exists. For $f (x) = | x |$ at 0, the averages from the right are 1 and from the left are $- 1$ , so the limit does not exist and $f^{'} (0)$ does not exist.

Can an average rate of change be negative?

Yes. If the output goes down while the input goes up, the change in output is negative and so is the quotient, as for $\frac{1}{x}$ from 2 to 4.

Course alignment, for teachers

AP Calculus AB topic 2.1, Unit 2: Differentiation: Definition and Fundamental Properties.

Skill 2.B: Identify mathematical information from graphical, numerical, analytical, and/or verbal representations.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first How fast at exactly two seconds?

Part 2 Learn it Three short ideas, each checked as you go

2a. Average rate of change is a difference quotient

2b. Reading the interval off the quotient

2c. From average to instant: the limit

Part 3 Watch it done The derivative at a point, from the definition

Part 4 Try it Two new functions

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A car pulling away

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic