Unit 5 · Topic 5.9 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 5.9: Connecting a Function, Its First Derivative, and Its Second Derivative

Q: If I am given the graph of f', should I sketch f first?

No. Argue from the graph of f' directly: whether it is above or below the x-axis for the direction of f, and whether it rises or falls for the concavity of f. A sketch of f can help you see it, but it is not a reason.

Q: Is every zero of f'' a point of inflection?

No. f'' has to change sign there. For f(x) = x4, f''(x) = 12x2 is 0 at x = 0 but positive on both sides, so the graph is concave up on both sides of 0 and has no point of inflection there.

Q: How do I find points of inflection of f from the graph of f'?

Find the relative maxima and minima of the graph of f'. At each one, f' switches between increasing and decreasing, so the graph of f switches concavity.

Q: Can a function be increasing while its graph is concave down?

Yes, wherever f' is positive and decreasing. The function still rises, but by less in each step, like a hiker whose rate of climb is falling but still positive.

Q: Why does it matter whether I write "it" or "f"?

Because f, f' and f'' are three different functions, and on one interval a sentence such as "it is increasing" can be true of one and false of another. Free-response answers are expected to name each function, so a reader never has to guess.

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 a conclusion about where $f$ increases or decreases, has a relative extremum, is concave up or down, or has a point of inflection, from the behavior of $f^{'}$ and $f^{''}$ .

Key vocabulary

Term	What it means	The exam trap
Second derivative, $f^{''}$	The derivative of $f^{'}$ , written $f^{''} (x)$ , $y^{''}$ or $\frac{d^{2} y}{d x^{2}}$ .	Keep track of which graph you are reading: a statement about $f^{''}$ is a statement about how the slopes of $f$ change.
Concave up and concave down	The graph of $f$ is concave up on an open interval where $f^{'}$ is increasing, and concave down where $f^{'}$ is decreasing.	Concavity and direction come from different derivatives, so check each with its own.
Point of inflection	A point on the graph of $f$ where $f$ is continuous and the graph changes concavity.	Look at both sides of a candidate point before you name it.
Relative (local) extremum	A relative maximum or relative minimum: a value of $f$ at least as large, or at least as small, as every value of $f$ at nearby inputs.	Say which function a feature belongs to before you name it a maximum or a minimum.

Three graphs stacked on one x scale from -2.2 to 2.2. Top: f(x) = x cubed minus 3x, with f(-1) = 2 and f(1) = -2. Middle: f'(x) = 3x squared minus 3, which is 0 at x = -1 and x = 1 and -3 at x = 0. Bottom: f''(x) = 6x, which is 0 at x = 0. Dashed vertical lines at x = -1, 0 and 1 run through all three graphs.

Part 1 Think first Still climbing?

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

Predict first

A hiker's watch shows her rate of climb. Between 10:00 and 11:00 in the morning the reading falls steadily, from 300 feet per hour to 100 feet per hour. What happened to her elevation during that hour?

In short. The graphs of $f$ , $f^{'}$ and $f^{''}$ are tied together, because each derivative's sign and direction describe the graph of the function it came from.

The sign of $f^{'}$ gives the direction of $f$ , and a sign change in $f^{'}$ gives a relative extremum of $f$ .
The sign of $f^{''}$ says whether $f^{'}$ is increasing or decreasing, and so whether the graph of $f$ is concave up or concave down.
A sign change in $f^{''}$ , which is a relative extremum of $f^{'}$ , gives a point of inflection of $f$ .

Part 2 Learn it Reading one graph from another

5a. The height of the graph of f' tells you about f

Every value of $f^{'}$ is a slope of the graph of $f$ . So when you are handed the graph of $f^{'}$ , read its height: whether it sits above or below the $x$ -axis. Whether it is rising or falling is a different question, for the next section.

Where the graph of $f^{'}$ is above the $x$ -axis, $f^{'} (x) > 0$ and $f$ is increasing.
Where it is below, $f^{'} (x) < 0$ and $f$ is decreasing.
Where it crosses the $x$ -axis, $f^{'}$ changes sign and $f$ has a relative extremum: a relative maximum where $f^{'}$ changes from positive to negative, a relative minimum where it changes from negative to positive.

Look at the top two graphs in the figure. On $- 1 < x < 1$ the graph of $f^{'}$ is below the axis, and the graph of $f$ falls. At $x = - 1$ the graph of $f^{'}$ crosses from above to below, and $f$ has its relative maximum there.

Two features of the graph of $f^{'}$ are easy to misread. A point where it only touches the $x$ -axis is a zero of $f^{'}$ with no sign change, so $f$ has a horizontal tangent there but no extremum. And a low point on the graph of $f^{'}$ is not a low point of $f$ : at $x = 0$ the graph of $f^{'}$ is at its lowest, and $f$ is simply falling at its fastest.

Quick check · Multiple choice

No calculator. The graph of $f^{'}$ , the derivative of $f$ , crosses the $x$ -axis from below to above at $x = 1$ and touches the $x$ -axis at $x = 4$ without crossing it. These are the only zeros of $f^{'}$ . What are the relative extrema of $f$ ?

5b. The slope of the graph of f' is f''

The second derivative $f^{''}$ is the derivative of $f^{'}$ . Its sign says whether the graph of $f^{'}$ is rising or falling, and that decides which way the graph of $f$ bends.

Where $f^{''} (x) > 0$ , $f^{'}$ is increasing and the graph of $f$ is concave up.
Where $f^{''} (x) < 0$ , $f^{'}$ is decreasing and the graph of $f$ is concave down.
Where $f^{''}$ changes sign, $f^{'}$ has a relative extremum and the graph of $f$ has a point of inflection.

In the figure, $f^{''} (x) = 6 x$ is negative to the left of 0 and positive to the right. Over the same stretches the graph of $f^{'}$ falls and then rises, with its lowest point at $x = 0$ , and the graph of $f$ changes from concave down to concave up at $(0, 0)$ .

Direction and concavity are separate facts, read from different derivatives. On $- 2.2 < x < - 1$ the function $f$ is increasing, because $f^{'}$ is positive there, and its graph is concave down, because $f^{'}$ is decreasing there. A positive, shrinking derivative describes a function that is still rising, only more slowly each moment, like the hiker in the morning.

Quick check · Match

No calculator. Each statement about $f$ follows from one statement about $f^{'}$ or $f^{''}$ . Tap a statement about $f$ , then the one it follows from.

5c. Saying which function, and why

A conclusion about $f$ needs a reason taken from $f^{'}$ or $f^{''}$ , with every function named. Write the conclusion, then because, then what the derivative does on the same interval or at the same point.

$f$ is decreasing on $- 1 < x < 1$ because $f^{'} (x) < 0$ on $- 1 < x < 1$ .
$f$ has a relative maximum at $x = - 1$ because $f^{'}$ changes from positive to negative at $x = - 1$ .
The graph of $f$ is concave up on $0 < x < 2$ because $f^{'}$ is increasing on $0 < x < 2$ .

A justification falls short in a few familiar ways. Writing "it" instead of $f$ , $f^{'}$ or $f^{''}$ leaves the reader guessing which graph you mean, and on one interval "it is increasing" can be true of $f^{'}$ and false of $f$ . A single value, such as $f^{'} (c) = 0$ or $f^{''} (c) = 0$ , is not enough for an extremum or a point of inflection; the reason has to be a sign change. And "because the graph goes up" describes a picture without saying anything about a derivative.

When you are given the graph of $f^{'}$ , argue from it directly, using its sign for the direction of $f$ and its rising or falling for the concavity of $f$ . You do not need to sketch $f$ first.

Quick check · Sort it

No calculator. Each sentence is a student's conclusion about a function $g$ , with a reason. Tap a sentence, then say whether its reason justifies the conclusion.

Justifies the conclusion

Does not justify it

Common mistakes

The mistake	Why it is wrong	Do this instead
Reading a feature of the graph of $f^{'}$ as the same feature of $f$ , such as calling a low point of $f^{'}$ a minimum of $f$ , or saying $f$ decreases where $f^{'}$ decreases.	The graph of $f^{'}$ shows slopes of $f$ . Its height gives the direction of $f$ ; its own rising and falling gives the concavity of $f$ .	Before you conclude anything, say which graph you are reading and which property of it (sign, or direction) you are using.
Treating a single zero, $f^{'} (c) = 0$ or $f^{''} (c) = 0$ , as proof of an extremum or a point of inflection.	A derivative can be 0 at a point and keep the same sign on both sides, and then nothing changes there.	Check the sign on each side, and give the sign change as the reason.
Writing "it is increasing" or "because the graph goes up" as a justification.	"It" can mean $f$ , $f^{'}$ or $f^{''}$ , and a description of a picture is not a reason from a derivative.	Name the function and the derivative: " $f$ is increasing on $a < x < b$ because $f^{'} (x) > 0$ on $a < x < b$ ."

Part 3 Watch it done From a formula for f' to the shape of f

Watch each step, then do the last one yourself.

The derivative of a function $f$ is $f^{'} (x) = x^{3} - 6 x^{2} + 9 x$ . Find where $f$ has relative extrema and where the graph of $f$ is concave down, and justify each conclusion.

Step 1 · Find the zeros of the derivative
$f^{'} (x) = x (x^{2} - 6 x + 9) = x (x - 3)^{2}$ , so $f^{'} (x) = 0$ at $x = 0$ and at $x = 3$ .
Step 2 · Read the sign of the derivative
$f^{'} (x) < 0$ for $x < 0$ , and $f^{'} (x) > 0$ for $0 < x < 3$ and for $x > 3$ , because $(x - 3)^{2} > 0$ everywhere except 3. So $f$ has a relative minimum at $x = 0$ because $f^{'}$ changes from negative to positive there, and $f$ has no relative extremum at $x = 3$ because $f^{'}$ does not change sign there.
Why this step?
Why is $x = 3$ not a relative extremum of $f$ , when $f^{'} (3) = 0$ ?
$f^{'}$ is positive just to the left of 3 and just to the right, so $f$ is increasing on both sides. The graph of $f$ levels off for an instant and keeps rising. Only a sign change in $f^{'}$ makes a relative extremum.
Step 3 · Differentiate again
$f^{''} (x) = 3 x^{2} - 12 x + 9 = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3)$ .
Your step · Fill the statement
No calculator. Finish it from $f^{''} (x) = 3 (x - 1) (x - 3)$ . Tap a blank, then tap an answer.

The zeros of $f^{''}$ are and 3, and $f^{''}$ is negative between them. So the graph of $f$ is concave down on that interval because $f^{'}$ is there, and the graph of $f$ has a point of inflection at each zero because $f^{''}$ at each.

Part 4 Try it A new derivative to read

Try it · Calculate

No calculator. The derivative of $g$ is $g^{'} (x) = e^{x} (x - 2)$ . At what value of $x$ does the graph of $g$ have a point of inflection?

Your answer

Try it · Decide and justify

No calculator. Again $g^{'} (x) = e^{x} (x - 2)$ . A classmate says the graph of $g$ is concave down at $x = 1.5$ , because $g^{'} (1.5)$ is negative. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write the justification on paper

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. Water flows into and out of a tank. The depth of the water is $W (t)$ centimeters at time $t$ hours, for $0 \leq t \leq 6$ , and $W^{'} (t) = t^{2} - 6 t + 8$ .

(a) On what open intervals is $W$ increasing? Justify your answer.

(b) For $0 < t < 6$ , at what time $t$ does $W$ have a relative minimum? Justify your answer.

(d) At $t = 5$ , is the depth of the water increasing at an increasing rate or increasing at a decreasing rate? Explain your reasoning, using correct units.

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

Model answer

(a) $W^{'} (t) = (t - 2) (t - 4)$ , which is positive for $0 < t < 2$ and for $4 < t < 6$ , and negative for $2 < t < 4$ . $W$ is increasing on $0 < t < 2$ and on $4 < t < 6$ because $W^{'} (t) > 0$ on those intervals.

(b) $W^{'} (t) = 0$ at $t = 2$ and $t = 4$ . $W^{'}$ changes from negative to positive at $t = 4$ , so $W$ has a relative minimum at $t = 4$ . (At $t = 2$ , $W^{'}$ changes from positive to negative, which gives a relative maximum.)

(c) $W^{''} (t) = 2 t - 6$ , which is negative for $t < 3$ and positive for $t > 3$ . $W^{''}$ changes sign at $t = 3$ , so the graph of $W$ has a point of inflection at $t = 3$ .

(d) $W^{'} (5) = 25 - 30 + 8 = 3$ centimeters per hour and $W^{''} (5) = 2 (5) - 6 = 4$ centimeters per hour per hour. The depth is increasing at 3 centimeters per hour because $W^{'} (5) > 0$ , and that rate is increasing because $W^{''} (5) > 0$ . The depth is increasing at an increasing rate.

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

(a) Gives $0 < t < 2$ and $4 < t < 6$ , with the reason that $W^{'} (t) > 0$ on those intervals.
(b) Gives $t = 4$ , with the reason that $W^{'}$ changes from negative to positive there.
(c) Finds $W^{''} (t) = 2 t - 6$ and gives $t = 3$ , with the reason that $W^{''}$ changes sign there.
(d) Finds $W^{'} (5) = 3$ and $W^{''} (5) = 4$ , both positive, and concludes increasing at an increasing rate, with units of centimeters per hour.

Part 6 Check yourself A sign table

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

A function $h$ is twice differentiable on $- 2 < x < 4$ . The table gives the sign of $h^{'} (x)$ and of $h^{''} (x)$ on each interval, and their values at the points between.

Signs of the first and second derivatives of h
$x$	$h^{'} (x)$	$h^{''} (x)$
$- 2 < x < 0$	Positive	Negative
$0$	0	Negative
$0 < x < 1$	Negative	Negative
$1$	Negative	0
$1 < x < 2$	Negative	Positive
$2$	0	Positive
$2 < x < 4$	Positive	Positive

Case question 1 · Calculate

No calculator. At what value of $x$ does $h$ have a relative maximum?

Your answer

Case question 2 · Multiple choice

No calculator. On which interval is $h$ decreasing and its graph concave up?

Case question 3 · Multiple choice

No calculator. Which statement justifies that the graph of $h$ has a point of inflection at $x = 1$ ?

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

Check yourself · Spot the error

No calculator. A student analyzes $p (x) = x^{4} - 4 x^{3}$ . Tap the first line with a mistake.

Check yourself · Sketch the graph

No calculator. For $0 \leq x \leq 5$ , $f^{'} (x)$ is positive and $f^{'}$ is decreasing, and $f (0) = 0$ . Drag the points to sketch a possible graph of $f$ . 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.

Part 7 Keep it Bring back earlier ideas

This lesson leans on the second derivative test and on finding a second derivative. Pulling them back up is what makes them last.

Bring it back · Decide and justify

No calculator. A function $k$ is twice differentiable, with $k^{'} (4) = 0$ and $k^{''} (4) = 5$ . Does $k$ have a relative maximum at $x = 4$ ?

Which reason supports your answer?

Bring it back · Calculate

No calculator. Let $f (x) = x^{4} - 2 x^{3}$ . Find $f^{''} (- 1)$ .

Your answer

In three lines

Where $f^{'}$ is positive or negative, $f$ is increasing or decreasing, and a sign change in $f^{'}$ is a relative extremum of $f$ .
Where $f^{'}$ is increasing or decreasing, which is the sign of $f^{''}$ , the graph of $f$ is concave up or concave down.
Every justification names $f$ , $f^{'}$ or $f^{''}$ and gives a sign change, not a single zero, for an extremum or an inflection point.

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.

Name the function in every sentence, and give the derivative's behavior on the same interval or at the same point: "the graph of $f$ is concave up on $a < x < b$ because $f^{'}$ is increasing on $a < x < b$ ." For an extremum or a point of inflection, the reason is a sign change, never a single zero.

Questions students ask

If I am given the graph of

f^{'}

, should I sketch

f

first?

No. Argue from the graph of $f^{'}$ directly: whether it is above or below the $x$ -axis for the direction of $f$ , and whether it rises or falls for the concavity of $f$ . A sketch of $f$ can help you see it, but it is not a reason.

Is every zero of

f^{''}

a point of inflection?

No. $f^{''}$ has to change sign there. For $f (x) = x^{4}$ , $f^{''} (x) = 12 x^{2}$ is 0 at $x = 0$ but positive on both sides, so the graph is concave up on both sides of 0 and has no point of inflection there.

How do I find points of inflection of

f

from the graph of

f^{'}

Find the relative maxima and minima of the graph of $f^{'}$ . At each one, $f^{'}$ switches between increasing and decreasing, so the graph of $f$ switches concavity.

Can a function be increasing while its graph is concave down?

Yes, wherever $f^{'}$ is positive and decreasing. The function still rises, but by less in each step, like a hiker whose rate of climb is falling but still positive.

Why does it matter whether I write "it" or "

f

Because $f$ , $f^{'}$ and $f^{''}$ are three different functions, and on one interval a sentence such as "it is increasing" can be true of one and false of another. Free-response answers are expected to name each function, so a reader never has to guess.

Course alignment, for teachers

AP Calculus AB topic 5.9, Unit 5: Analytical Applications of Differentiation.

Skill 2.D: Identify how mathematical characteristics or properties of functions are related in different representations.

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.

What you'll be able to do

Key vocabulary

Part 1 Think first Still climbing?

Part 2 Learn it Reading one graph from another

5a. The height of the graph of f' tells you about f

5b. The slope of the graph of f' is f''

5c. Saying which function, and why

Part 3 Watch it done From a formula for f' to the shape of f

Part 4 Try it A new derivative to read

Part 5 Explain it Write the justification on paper

Part 6 Check yourself A sign table

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic