Unit 5 · Topic 5.6 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 5.6: Determining Concavity of Functions over Their Domains

Q: Is concave up the same as increasing?

No. Increasing is about the sign of f'; concave up is about f' getting larger. y = x2 is decreasing and concave up for x 0.

Q: Can a point of inflection be where f'' does not exist?

Yes, if f is continuous there and the concavity changes. For f(x) = x1/3, f''(x) = -29x-5/3 does not exist at 0, but it is positive for x 0, so (0, 0) is a point of inflection.

Q: What does concavity mean in a real context?

It says how a rate is changing. If W(t) is the amount of water in a tank and its graph is concave up, the rate W'(t) is increasing: water comes in faster and faster, or, while W' is negative, drains more and more slowly.

Q: If I am given the graph of f' instead of f, how do I read concavity?

Look at whether the graph of f' rises or falls, not whether it is above or below the axis. Where the graph of f' rises, the graph of f is concave up; where it falls, the graph of f is concave down. A peak or a valley of the graph of f' is where the graph of f changes concavity.

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

Justify where the graph of a function is concave up or concave down from whether its derivative is increasing or decreasing, or from the sign of its second derivative, and use the second derivative to locate its points of inflection.

Key vocabulary

Term	What it means	The exam trap
Concave up	The graph of $f$ is concave up on an open interval if $f^{'}$ is increasing on that interval.	Rising and concave up are separate facts about a graph, and each is read from its own derivative.
Concave down	The graph of $f$ is concave down on an open interval if $f^{'}$ is decreasing on that interval.	A graph can be concave down on a stretch where it is still rising, so do not wait for it to fall.
Second derivative, $f^{''}$	The derivative of $f^{'}$ , which provides information about $f$ and its graph, including intervals of upward or downward concavity.	Before you draw a conclusion from a sign, say whether it is the sign of $f^{'}$ or of $f^{''}$ .
Point of inflection	A point on the graph of $f$ , where $f$ is continuous, at which the graph changes concavity.	Some questions ask for the point and some only for its $x$ -coordinate; give what is asked, and name the function in your reason.

Graph of f(x) = x cubed minus 6x squared plus 9x plus 1 from x = 0 to 4.2, with short tangent segments whose slopes fall from 3.75 at x = 0.5 to -3 at x = 2, then rise back to 3.75 at x = 3.5. The point of inflection (2, 3) is marked, with concave down to its left and concave up to its right.

Part 1 Think first Downloads that keep rising, more slowly

Commit to a guess before the lesson. Even a wrong guess makes the idea easier to remember.

Predict first

A new app got 500 downloads on its first day, then 800, 1000 and 1100 on the next three days. The daily count keeps rising, but by less each day: up 300, then 200, then 100. Let $N (t)$ be the total number of downloads after $t$ days. Over these four days, what does the graph of $N$ look like?

In short. The graph of $f$ is concave up on an open interval where $f^{'}$ is increasing and concave down where $f^{'}$ is decreasing, and the sign of $f^{''}$ tells you which.

$f^{''} (x) > 0$ on an open interval means the graph of $f$ is concave up there, and $f^{''} (x) < 0$ means it is concave down.
A point of inflection is a point where the graph of $f$ changes concavity, so it sits where $f^{''}$ changes sign.
An input where $f^{''} (x) = 0$ or $f^{''} (x)$ does not exist is only a candidate until the sign of $f^{''}$ is checked on both sides.

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

2a. Concavity is about how the slope changes

Two graphs can both rise and still look different. One gets steeper as you move right; the other flattens out. The difference is in the derivative.

The graph of $f$ is concave up on an open interval if $f^{'}$ is increasing on that interval. The tangent slopes get larger from left to right, and the graph bends upward like a cup. The graph of $f$ is concave down on an open interval if $f^{'}$ is decreasing there: the slopes get smaller, and the graph bends downward.

Look at the graph at the top. From $x = 0.5$ to $x = 2$ the tangent slopes are 3.75, 0, $- 2.25$ and $- 3$ . Each is smaller than the one before, so $f^{'}$ is decreasing and the graph is concave down there, even on the stretch where $f$ is still rising. After $x = 2$ the slopes climb back through $- 2.25$ , 0 and 3.75, so the graph is concave up.

Concavity says nothing about whether $f$ goes up or down; that is the sign of $f^{'}$ . Concavity is whether $f^{'}$ itself goes up or down. A graph can be decreasing and concave up at once, like the stretch of this one between $x = 2$ and $x = 3$ .

Quick check · Sketch the graph

No calculator. Water flows into an empty tank for 5 minutes. The rate of flow, in liters per minute, stays positive the whole time but decreases from minute to minute. Sketch the volume of water in the tank against time, from 0 to 5 minutes. The first point is fixed at 0.

Drag a point up or down, or tap it and use the arrow keys. Only the shape is checked.

2b. The second derivative reads concavity

Whether $f^{'}$ is increasing or decreasing is a question about the derivative of $f^{'}$ . That derivative is the second derivative, $f^{''}$ .

If $f^{''} (x) > 0$ on an open interval, $f^{'}$ is increasing there, so the graph of $f$ is concave up.
If $f^{''} (x) < 0$ on an open interval, $f^{'}$ is decreasing there, so the graph of $f$ is concave down.

So $f^{''}$ gives the intervals of concavity the way $f^{'}$ gives the intervals where $f$ increases and decreases. Find $f^{''}$ , find where it is 0 or does not exist, and check its sign between those inputs.

For the curve at the top, $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ , so $f^{'} (x) = 3 x^{2} - 12 x + 9$ and $f^{''} (x) = 6 x - 12$ . That is negative for $x < 2$ and positive for $x > 2$ . The graph of $f$ is concave down on $(- \infty, 2)$ and concave up on $(2, \infty)$ , which is what the falling and then rising tangent slopes showed.

When you justify a claim like this, name the function and the derivative you used: "the graph of $f$ is concave up on $(2, \infty)$ because $f^{''} (x) > 0$ there." A sentence such as "it curves up" names no function and gives no reason.

Quick check · Multiple choice

No calculator. Let $f (x) = x^{4} - 6 x^{2}$ . Which gives every open interval on which the graph of $f$ is concave down?

2c. Points of inflection

A point of inflection is a point on the graph of $f$ , where $f$ is continuous, at which the graph changes concavity: from up to down, or from down to up. On the graph at the top it is $(2, 3)$ , where the tangent slopes stop falling and start rising.

Concavity follows the sign of $f^{''}$ , so the second derivative locates these points. The graph changes concavity at $x = a$ when $f^{''}$ changes sign there. The candidates are the inputs where $f^{''} (x) = 0$ or $f^{''} (x)$ does not exist, and each one has to be tested on both sides.

A candidate can fail the test. For $g (x) = x^{4}$ , $g^{''} (x) = 12 x^{2}$ , which is 0 at $x = 0$ . But $g^{''}$ is positive on both sides of 0, so the graph of $g$ is concave up on both sides and $(0, 0)$ is not a point of inflection. A zero of $f^{''}$ tells you where to look, not what you will find.

Write the reason with the function named: "The graph of $f$ has a point of inflection at $x = 2$ because $f^{''}$ changes sign there." Because $f^{''}$ changes sign exactly where $f^{'}$ switches between decreasing and increasing, "because $f^{'}$ changes from decreasing to increasing at $x = 2$ " is an equally good reason.

Quick check · Decide and justify

No calculator. Let $h (x) = x^{5} - 5 x^{4}$ , so $h^{''} (x) = 20 x^{3} - 60 x^{2}$ . Since $h^{''} (0) = 0$ and $h^{''} (3) = 0$ , a classmate says the graph of $h$ has points of inflection at both $x = 0$ and $x = 3$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Reading concavity from the sign of $f^{'}$ , so calling the graph concave down wherever $f$ is decreasing.	The sign of $f^{'}$ says whether $f$ rises or falls. Concavity says whether $f^{'}$ itself rises or falls, which is a different question.	Use the sign of $f^{''}$ , or whether $f^{'}$ is increasing or decreasing, and nothing else.
Calling every input where $f^{''} (x) = 0$ a point of inflection.	The graph has to change concavity there. For $x^{4}$ at 0, $f^{''}$ is 0 but positive on both sides, so nothing changes.	Treat each zero of $f^{''}$ as a candidate, and check the sign of $f^{''}$ on both sides of it.
Justifying with "it curves up" or "the graph bends down."	That names no function and gives no calculus reason, so a reader cannot tell whether you mean $f$ , $f^{'}$ or $f^{''}$ .	Write "the graph of $f$ is concave up on this interval because $f^{''} (x) > 0$ there" (or "because $f^{'}$ is increasing there").

Part 3 Watch it done Concavity and inflection points of a quartic

Open the steps one at a time, then write the conclusion yourself.

Let $f (x) = x^{4} - 4 x^{3} + 10$ . Find the open intervals on which the graph of $f$ is concave up and concave down, and find the $x$ -coordinates of its points of inflection. Justify your answers.

Step 1 · Differentiate twice
$f^{'} (x) = 4 x^{3} - 12 x^{2}$ , so $f^{''} (x) = 12 x^{2} - 24 x = 12 x (x - 2)$ .
Step 2 · Find the candidates
$f^{''} (x) = 0$ at $x = 0$ and $x = 2$ . Since $f^{''}$ is a polynomial, it exists everywhere, so there are no other candidates.
Step 3 · Test the sign of f'' between them
$f^{''} (- 1) = 36 > 0$ , $f^{''} (1) = - 12 < 0$ and $f^{''} (3) = 36 > 0$ . So $f^{''} (x) > 0$ on $(- \infty, 0)$ , $f^{''} (x) < 0$ on $(0, 2)$ , and $f^{''} (x) > 0$ on $(2, \infty)$ .
Why this step?
Why test a value on each side instead of stopping at the two zeros of $f^{''}$ ?
A zero of $f^{''}$ only marks a place where the sign might change. Here it changes at both, but for $x^{4}$ at 0 it does not. Only the signs on each side show whether the concavity actually changes.
Your step · Fill the statement
No calculator. Finish it. Tap a blank, then tap an answer.

From the signs of $f^{''}$ , the graph of $f (x) = x^{4} - 4 x^{3} + 10$ is concave down on , and it has points of inflection at .

Part 4 Try it A new context and four new functions

Try it · Calculate

No calculator. After a dose of a medicine, the amount of it in a patient's blood is $C (t) = - t^{3} + 9 t^{2}$ milligrams, $t$ hours after the dose, for $0 \leq t \leq 6$ . At what time $t$ , in hours, does the graph of $C$ have a point of inflection?

Your answer

Try it · Match

No calculator. Tap a function, then the interval on which its graph is concave up.

Part 5 Explain it Write it as a free-response answer

Free response on the exam is written by hand, so work this on paper. Nothing you type here leaves the page. When you are done, open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. The depth of the water in a reservoir is $D (t)$ feet, $t$ months after January 1, for $0 \leq t \leq 12$ . The depth changes at the rate $D^{'} (t) = t^{2} - 10 t + 21$ feet per month.

(a) Find $D^{''} (4)$ . Using correct units, interpret the meaning of $D^{''} (4)$ in the context of the problem.

(b) On what open interval or intervals in $0 < t < 12$ is the graph of $D$ concave up? Justify your answer.

(c) Find the value of $t$ , for $0 < t < 12$ , at which the graph of $D$ has a point of inflection. Justify your answer.

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

Model answer

(a) $D^{''} (t) = 2 t - 10$ , so $D^{''} (4) = 2 (4) - 10 = - 2$ feet per month per month. At $t = 4$ months, the rate at which the depth of the water changes is decreasing at 2 feet per month per month.

(b) $D^{''} (t) = 2 t - 10 > 0$ when $t > 5$ . The graph of $D$ is concave up on $5 < t < 12$ because $D^{''} (t) > 0$ there.

(c) $t = 5$ . $D^{''} (t) = 2 t - 10$ is negative for $0 < t < 5$ and positive for $5 < t < 12$ , so $D^{''}$ changes sign at $t = 5$ and the graph of $D$ has a point of inflection there.

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

Finds $D^{''} (t) = 2 t - 10$ and gets $D^{''} (4) = - 2$ .
Interprets $D^{''} (4)$ : at $t = 4$ the rate of change of the depth is decreasing at 2 feet per month per month, with those units.
Gives $5 < t < 12$ and justifies it with $D^{''} (t) > 0$ there (or $D^{'}$ increasing there).
Gives $t = 5$ and justifies it with $D^{''}$ changing sign at $t = 5$ .

Part 6 Check yourself A function known by the signs of its derivatives

Answer both case questions from the table.

A function $f$ is twice differentiable for all real numbers. The table gives the sign of $f^{'} (x)$ and of $f^{''} (x)$ on each interval, and the inputs where either one is 0.

Signs of the first and second derivatives of f
$x$	$f^{'} (x)$	$f^{''} (x)$
$x < - 1$	Positive	Negative
$x = - 1$	0	Negative
$- 1 < x < 1$	Negative	Negative
$x = 1$	Negative	0
$1 < x < 3$	Negative	Positive
$x = 3$	0	Positive
$x > 3$	Positive	Positive

Case question 1 · Multiple choice

No calculator. Which gives all the values of $x$ for which the graph of $f$ is concave up?

Case question 2 · Calculate

No calculator. At what value of $x$ does the graph of $f$ have a point of inflection?

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 the graph of $g (x) = x^{3} + 6 x^{2} - 5$ is concave up and where it is concave down. Tap the first line with a mistake.

Check yourself · Put it in order

No calculator. Put the steps for finding the points of inflection of the graph of a function $f$ in order.

Part 7 Keep it Bring back earlier ideas

Concavity is read the way increasing and decreasing were, one derivative later. These two questions bring back the pieces it stands on.

Bring it back · Multiple choice

No calculator. The derivative of a function $f$ is $f^{'} (x) = (x - 2) (x + 5)$ . On which interval or intervals is $f$ decreasing?

Bring it back · Calculate

No calculator. Let $f (x) = 2 x^{3} - 5 x^{2} + 4 x$ . Find $f^{''} (2)$ .

Your answer

In three lines

The graph of $f$ is concave up on an open interval where $f^{'}$ is increasing, and concave down where $f^{'}$ is decreasing.
The sign of $f^{''}$ gives the intervals of concavity: positive for concave up, negative for concave down.
Points of inflection are where the graph changes concavity, found where $f^{''}$ changes sign.

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.

Answer in the question's own words and name the function: "The graph of $f$ is concave down on $a < x < b$ because $f^{''} (x) < 0$ there," or "because $f^{'}$ is decreasing there." A sentence like "it curves down" leaves the reader unsure which function you mean and gives no calculus reason. When you are given the graph of $f^{'}$ , argue from that graph directly: whether it rises or falls tells you the concavity of $f$ .

Questions students ask

Is concave up the same as increasing?

No. Increasing is about the sign of $f^{'}$ ; concave up is about $f^{'}$ getting larger. $y = x^{2}$ is decreasing and concave up for $x < 0$ , and $y = \sqrt{x}$ is increasing and concave down for $x > 0$ .

Can a point of inflection be where

f^{''}

does not exist?

Yes, if $f$ is continuous there and the concavity changes. For $f (x) = x^{1 / 3}$ , $f^{''} (x) = - \frac{2}{9} x^{- 5 / 3}$ does not exist at 0, but it is positive for $x < 0$ and negative for $x > 0$ , so $(0, 0)$ is a point of inflection.

What does concavity mean in a real context?

It says how a rate is changing. If $W (t)$ is the amount of water in a tank and its graph is concave up, the rate $W^{'} (t)$ is increasing: water comes in faster and faster, or, while $W^{'}$ is negative, drains more and more slowly.

If I am given the graph of

f^{'}

instead of

f

, how do I read concavity?

Look at whether the graph of $f^{'}$ rises or falls, not whether it is above or below the axis. Where the graph of $f^{'}$ rises, the graph of $f$ is concave up; where it falls, the graph of $f$ is concave down. A peak or a valley of the graph of $f^{'}$ is where the graph of $f$ changes concavity.

Does the graph have to be flat at a point of inflection?

No. The slope there can be any number. On the graph at the top, the slope at the point of inflection $(2, 3)$ is $- 3$ . In the worked example, the point $(0, 10)$ happens to have slope 0, but at $(2, - 6)$ the slope is $f^{'} (2) = - 16$ .

Course alignment, for teachers

AP Calculus AB topic 5.6, Unit 5: Analytical Applications of Differentiation.

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

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 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 Downloads that keep rising, more slowly

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

2a. Concavity is about how the slope changes

2b. The second derivative reads concavity

2c. Points of inflection

Part 3 Watch it done Concavity and inflection points of a quartic

Part 4 Try it A new context and four new functions

Part 5 Explain it Write it as a free-response answer

Part 6 Check yourself A function known by the signs of its derivatives

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic