Unit 5 · Topic 5.3 · about 30 minutes · 7 parts · 12 scored checks, 2 of them multiple choice

AP Calculus AB 5.3: Determining Intervals on Which a Function Is Increasing or Decreasing

Q: Do I write the intervals as open or closed?

Open intervals are always safe, because the sign of f' is checked inside them. If f is continuous at an endpoint, f is also increasing on the interval with that endpoint included, so a closed interval is correct too.

Q: If f'(c) = 0, is f increasing at c?

Increasing is a statement about an interval, not one input. f(x) = x3 has f'(0) = 0, yet f'(x) > 0 on both sides of 0 and f is increasing on every interval, including those that contain 0.

Q: What if f' does not exist somewhere?

Split the number line there as well. f' can change sign at an input where it does not exist, so that input is a boundary between intervals. For f(x) = x2/3, f'(x) = 23x-1/3 does not exist at 0, and f is decreasing on (-, 0) and increasing on (0, ).

Q: On the calculator part, can I just graph f and look?

Use the calculator to find where f'(x) = 0, and to check your picture. The answer still needs the reason: name the equation you solved, give the zeros, and state the sign of f' on each interval.

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

Unit 5: 15 to 20% of the multiple-choice section

What you'll be able to do

Use the sign of a function's first derivative to find, and justify, the intervals on which the function is increasing or decreasing.

Key vocabulary

Term	What it means	The exam trap
Increasing on an interval	A function $f$ is increasing on an interval if $f (a) < f (b)$ for any two inputs $a < b$ in that interval.	It describes $f$ over a stretch of inputs, so give the answer as an interval of $x$ -values, not as $y$ -values.
Decreasing on an interval	A function $f$ is decreasing on an interval if $f (a) > f (b)$ for any two inputs $a < b$ in that interval.	Write the interval with the smaller input first, as $(a, b)$ with $a < b$ , even though $f$ falls as you move across it.
First derivative, $f^{'}$	The function whose value $f^{'} (x)$ is the instantaneous rate of change of $f$ at $x$ ; it gives information about $f$ and its graph.	When you write about it, say $f^{'}$ by name; "it" or "the graph" leaves the reader guessing which function you mean.

Two graphs stacked on one x-axis scale from -2.4 to 2.4. Above is f(x) = x cubed - 3x, with its points (-1, 2) and (1, -2) marked. Below is its derivative f'(x) = 3x squared - 3, which crosses the x-axis at x = -1 and x = 1. Dashed vertical lines at x = -1 and x = 1 join the two graphs.

Part 1 Think first A falling reading on a rising tank

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

Predict first

A meter on a water tank shows how fast the amount of water is changing, in gallons per minute. A positive reading means water is being added. Over 6 minutes the reading drops steadily from 12 to 0. During those 6 minutes, what happened to the amount of water in the tank?

In short. On an open interval where $f^{'} (x) > 0$ the function $f$ is increasing, and where $f^{'} (x) < 0$ it is decreasing, so the sign of the first derivative settles which way $f$ moves.

To find the intervals, split the number line where $f^{'} (x) = 0$ , where $f^{'} (x)$ does not exist, and where $f$ is not defined, then test the sign of $f^{'}$ on each piece.
From a graph of $f^{'}$ , read whether it lies above or below the $x$ -axis, not whether it rises or falls.
A justification names $f^{'}$ and its sign on the interval, as in " $f$ is increasing on $(a, b)$ because $f^{'} (x) > 0$ there."

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

3a. The sign of the derivative gives the direction

A function $f$ is increasing on an interval if its outputs go up as $x$ moves right: for any two inputs $a < b$ in the interval, $f (a) < f (b)$ . It is decreasing on an interval if $f (a) > f (b)$ for any $a < b$ there.

The first derivative tells you which, because $f^{'} (x)$ is the rate at which $f$ is changing at $x$ .

If $f^{'} (x) > 0$ for every $x$ in an open interval, then $f$ is increasing on that interval.
If $f^{'} (x) < 0$ for every $x$ in an open interval, then $f$ is decreasing on that interval.

Only the sign of $f^{'}$ matters here. In Think first the meter reading fell from 12 toward 0, but it stayed positive, so the water kept rising, just more slowly. A positive derivative that is getting smaller still means $f$ is going up. A negative derivative that is getting closer to 0 still means $f$ is going down.

The claim is about every input in the interval. Knowing $f^{'} (3) > 0$ tells you the rate at one input only. Knowing $f (1) < f (5)$ does not show $f$ increases from 1 to 5 either: $f$ could dip and recover in between. You need the sign of $f^{'} (x)$ across the whole interval.

Quick check · Sort it

No calculator. Each card gives a fact about a function $f$ on the interval named. Sort each card by what it lets you conclude about $f$ on that interval.

$f$ is increasing there

$f$ is decreasing there

Not enough to tell

3b. Finding the intervals from a formula

To find where $f$ increases and decreases, first find where $f^{'}$ could change sign. That can happen only where $f^{'} (x) = 0$ , where $f^{'} (x)$ does not exist (the critical points from 5.2), or where $f$ itself is not defined. Those inputs split the number line into intervals, and on each one $f^{'}$ keeps a single sign, so one test input per interval is enough.

Take $f (x) = x^{4} - 8 x^{2}$ . Then $f^{'} (x) = 4 x^{3} - 16 x = 4 x (x + 2) (x - 2)$ , which is 0 at $x = - 2$ , $0$ and $2$ and exists everywhere. Test one input in each of the four intervals:

Sign of $f^{'} (x) = 4 x (x + 2) (x - 2)$
Interval	Test input	$f^{'}$ there	$f$ on the interval
$(- \infty, - 2)$	$x = - 3$	$f^{'} (- 3) = - 60 < 0$	decreasing
$(- 2, 0)$	$x = - 1$	$f^{'} (- 1) = 12 > 0$	increasing
$(0, 2)$	$x = 1$	$f^{'} (1) = - 12 < 0$	decreasing
$(2, \infty)$	$x = 3$	$f^{'} (3) = 60 > 0$	increasing

Here the signs happen to alternate, but do not count on it. A squared factor such as $(x - 3)^{2}$ is never negative, so $f^{'}$ can touch 0 at $x = 3$ and keep the same sign on both sides. Test every interval.

Quick check · Multiple choice

No calculator. The derivative of a function $f$ is $f^{'} (x) = (x + 1) (x - 3)^{2}$ . On which interval or intervals is $f$ increasing? Choose the answer that names all of them.

3c. Reading the graph of the derivative, and saying why

Often you are given the graph of $f^{'}$ , not of $f$ . Then one feature matters: is the graph of $f^{'}$ above or below the $x$ -axis? Above means $f^{'} (x) > 0$ , so $f$ is increasing. Below means $f^{'} (x) < 0$ , so $f$ is decreasing.

Whether the graph of $f^{'}$ is rising or falling answers a different question, about how $f^{'}$ itself changes. It does not tell you which way $f$ goes. This is the mistake from Think first again, in picture form.

The figure at the top shows both graphs. The lower one, $f^{'}$ , is below the $x$ -axis between $x = - 1$ and $x = 1$ , and that is exactly where the upper graph, $f$ , comes down. Where $f^{'}$ is above the axis, $f$ goes up.

When you justify a conclusion, say which function you mean and give the sign of the derivative:

" $f$ is increasing on $- 3 < x < - 1$ because $f^{'} (x) > 0$ there."
not "it is increasing because the graph goes up."

The word "it" and the phrase "the graph" leave a reader unsure whether you mean $f$ or $f^{'}$ . Write $f$ and $f^{'}$ by name.

Quick check · Decide and justify

No calculator. On the interval $0 < x < 3$ , the graph of $g^{'}$ , the derivative of $g$ , falls steadily from $g^{'} (0) = 5$ to $g^{'} (3) = 1$ . Is $g$ increasing or decreasing on $0 < x < 3$ ?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Reading where the graph of $f^{'}$ rises or falls as where $f$ rises or falls.	Whether $f$ goes up or down depends on the sign of $f^{'}$ . A falling graph of $f^{'}$ can still be above the $x$ -axis.	On a graph of $f^{'}$ , look only at whether it is above or below the $x$ -axis.
Splitting the number line only where $f^{'} (x) = 0$ .	$f^{'}$ can also change sign where it does not exist, or where $f$ is not defined. For $f (x) = x^{2 / 3}$ , $f^{'} (x) = \frac{2}{3} x^{- 1 / 3}$ does not exist at 0 and is negative to its left and positive to its right.	List every input where $f^{'} (x) = 0$ , where $f^{'} (x)$ does not exist, and where $f$ is not defined, and split at all of them.
Assuming the sign of $f^{'}$ alternates at every zero.	At a zero from a squared factor, such as $(x - 3)^{2}$ , $f^{'}$ touches 0 and keeps its sign.	Test one input in every interval rather than filling in signs by pattern.

Part 3 Watch it done Intervals of increase and decrease, start to finish

Watch each step, then do the last one yourself.

Let $f (x) = x^{3} - 3 x^{2} - 9 x + 4$ . Find the intervals on which $f$ is increasing and the intervals on which $f$ is decreasing. Justify your answer.

Step 1 · Differentiate and factor
$f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x^{2} - 2 x - 3) = 3 (x + 1) (x - 3)$ .
Step 2 · Find where the sign can change
$f^{'} (x) = 0$ at $x = - 1$ and $x = 3$ . A polynomial's derivative exists everywhere, so the intervals are $(- \infty, - 1)$ , $(- 1, 3)$ and $(3, \infty)$ .
Step 3 · Test one input in each interval
$f^{'} (- 2) = 15 > 0$ , $f^{'} (0) = - 9 < 0$ and $f^{'} (4) = 15 > 0$ .
Why this step?
Why is one test input enough for a whole interval?
$f^{'}$ is a polynomial, so it is continuous, and it is not 0 anywhere inside the interval. By the Intermediate Value Theorem it would have to pass through 0 to change sign, so it keeps the sign of the one value you tested.
Step 4 · State where f increases, with the reason
$f$ is increasing on $(- \infty, - 1)$ and on $(3, \infty)$ because $f^{'} (x) > 0$ on those intervals.
Your step · Fill the statement
No calculator. Finish it. Write the conclusion about where $f (x) = x^{3} - 3 x^{2} - 9 x + 4$ is decreasing. Tap a blank, then tap an answer.

$f$ is decreasing on because for every $x$ in that interval.

Part 4 Try it Two new derivatives

Try it · Calculate

Graphing calculator required. The derivative of a function $f$ is $f^{'} (x) = 2 x - 3 \cos x$ . On the interval $0 < x < 3$ , $f$ is decreasing on $(0, c)$ and increasing on $(c, 3)$ . Find $c$ , with your calculator in radian mode. Give your answer to three decimal places.

Your answer

Try it · Match

No calculator. Each formula gives the derivative $f^{'} (x)$ of a different function $f$ . Tap a derivative, then the interval or intervals on which that $f$ is decreasing.

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 2, from free-response Part A (graphing calculator required).

Free-response practice · Question 2 style

Graphing calculator required. Water flows into a storage tank at the rate $E (t) = 4 + 4 \sin (0.9 t)$ gallons per hour and drains out of the tank at the rate $D (t) = 2 + 0.4 t$ gallons per hour, for $0 \leq t \leq 8$ , where $t$ is measured in hours. Let $W (t)$ be the amount of water in the tank, in gallons, at time $t$ . Use your calculator in radian mode.

(a) Find $W^{'} (2)$ . Is the amount of water in the tank increasing or decreasing at time $t = 2$ ? Give a reason for your answer.

(b) Find all intervals of time in $0 < t < 8$ during which the amount of water in the tank is decreasing. Justify your answer.

(c) The rate $E (t)$ at which water flows in is decreasing for $2 < t < 3$ . A worker says that this means the amount of water in the tank is decreasing for $2 < t < 3$ . Explain why the worker is not correct.

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

Model answer

(a) $W^{'} (t) = E (t) - D (t)$ , so $W^{'} (2) = E (2) - D (2) = 5.095$ gallons per hour. The amount of water is increasing at $t = 2$ because $W^{'} (2) > 0$ .

(b) $W^{'} (t) = E (t) - D (t) = 0$ at $t = 3.642$ and $t = 7.231$ (solved on the calculator). $W^{'} (t) < 0$ for $3.642 < t < 7.231$ , for example $W^{'} (5) = - 3.910$ , and $W^{'} (t) > 0$ on the rest of $0 < t < 8$ . So the amount of water is decreasing on $3.642 < t < 7.231$ because $W^{'} (t) < 0$ there.

(c) Whether the amount of water rises or falls depends on the sign of $W^{'} (t) = E (t) - D (t)$ , not on whether $E$ is rising or falling. For $2 < t < 3$ , $W^{'} (t) > 0$ (it is positive at $t = 2$ and does not reach 0 until $t = 3.642$ ), so water flows in faster than it drains and the amount of water is increasing there.

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

Writes $W^{'} (2) = E (2) - D (2)$ and gets 5.095 gallons per hour.
Says the amount is increasing at $t = 2$ because $W^{'} (2) > 0$ .
Sets $E (t) - D (t) = 0$ and finds $t = 3.642$ and $t = 7.231$ .
Gives $3.642 < t < 7.231$ as the interval of decrease, with the reason $W^{'} (t) < 0$ there.
Explains that $W^{'} (t) = E (t) - D (t) > 0$ for $2 < t < 3$ , so the amount increases even though $E$ decreases.

Part 6 Check yourself Reading a derivative from a table

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

A function $f$ is differentiable on $- 4 < x < 5$ . The graph of its derivative $f^{'}$ consists of line segments joining the points in the table, one after the next.

Points on the graph of $f^{'}$
$x$	$- 4$	$- 2$	$0$	$3$	$5$
$f^{'} (x)$	$2$	$2$	$- 2$	$- 2$	$2$

Case question 1 · Multiple choice

No calculator. On which interval or intervals is $f$ decreasing? Choose the answer that names all of them.

Case question 2 · Decide and justify

No calculator. On the interval $- 4 < x < - 2$ the graph of $f^{'}$ is a horizontal segment. Which describes $f$ on that interval?

Which reason supports your answer?

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

Check yourself · Spot the error

No calculator. A student finds where $f (x) = x + \frac{4}{x}$ is decreasing. Tap the first line with a mistake.

Check yourself · Calculate

No calculator. The function $f (x) = x^{2} - k x + 7$ , where $k$ is a constant, is decreasing on $(- \infty, 3)$ and increasing on $(3, \infty)$ . What is the value of $k$ ?

Your answer

Part 7 Keep it Bring back earlier ideas

This lesson leans on critical points from 5.2 and on the derivative rules from Unit 3. Pulling them back up is what makes them last.

Bring it back · Calculate

No calculator. Let $f (x) = x^{4 / 3} - 4 x^{1 / 3}$ , which is defined for every real $x$ . How many critical points does $f$ have?

Your answer

Bring it back · Calculate

No calculator. Let $h (x) = (3 x - 1)^{4}$ . What is $h^{'} (1)$ ?

Your answer

In three lines

Where $f^{'} (x) > 0$ on an open interval, $f$ is increasing there, and where $f^{'} (x) < 0$ , $f$ is decreasing.
Split the number line where $f^{'}$ is 0 or does not exist, or where $f$ is not defined, and test the sign of $f^{'}$ on every piece.
Justify with the sign of $f^{'}$ , by name, on the interval, never with the direction of a graph of $f^{'}$ .

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 asked where $f$ is increasing or decreasing, show $f^{'} (x)$ (or name the $f^{'}$ you were given), state where $f^{'} (x) = 0$ or does not exist, and finish with a sentence that names $f^{'}$ and its sign on the interval, such as " $f$ is decreasing on $(2, 5)$ because $f^{'} (x) < 0$ there." On the calculator part, give the zeros of $f^{'}$ you found to three decimal places and say which equation you solved.

Questions students ask

Do I write the intervals as open or closed?

Open intervals are always safe, because the sign of $f^{'}$ is checked inside them. If $f$ is continuous at an endpoint, $f$ is also increasing on the interval with that endpoint included, so a closed interval is correct too.

f^{'} (c) = 0

, is

f

increasing at

c

Increasing is a statement about an interval, not one input. $f (x) = x^{3}$ has $f^{'} (0) = 0$ , yet $f^{'} (x) > 0$ on both sides of 0 and $f$ is increasing on every interval, including those that contain 0.

What if

f^{'}

does not exist somewhere?

Split the number line there as well. $f^{'}$ can change sign at an input where it does not exist, so that input is a boundary between intervals. For $f (x) = x^{2 / 3}$ , $f^{'} (x) = \frac{2}{3} x^{- 1 / 3}$ does not exist at 0, and $f$ is decreasing on $(- \infty, 0)$ and increasing on $(0, \infty)$ .

Why can't I say "the graph is going up"?

Because it does not say which graph, and a graph going up is an observation, not a reason from calculus. Name the function and give the sign of its derivative: " $f$ is increasing on $(2, 5)$ because $f^{'} (x) > 0$ there."

On the calculator part, can I just graph

f

and look?

Use the calculator to find where $f^{'} (x) = 0$ , and to check your picture. The answer still needs the reason: name the equation you solved, give the zeros, and state the sign of $f^{'}$ on each interval.

Course alignment, for teachers

AP Calculus AB topic 5.3, Unit 5: Analytical Applications of Differentiation.

Skill 2.E: Describe the relationships among different representations of functions and their derivatives.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.