Unit 5 · Topic 5.8 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 5.8: Sketching Graphs of Functions and Their Derivatives

Q: If I am given only the graph of f', can I find the values of f?

Not from f' alone. f' tells you how f changes, not where it starts, so you need at least one value of f to place the graph. With one value you can sketch the shape; finding other values exactly takes integrals, which Unit 6 teaches.

Q: Is a point where f''(x) = 0 always a point of inflection?

No. f(x) = x4 has f''(0) = 0, but f''(x) = 12x2 is positive on both sides of 0, so the graph is concave up throughout. Check that f'' changes sign.

Q: How do I find concavity from a graph of f' when I do not have f''?

Look at whether the graph of f' is rising or falling. Where it rises, the graph of f is concave up; where it falls, concave down. Which side of the x-axis f' is on does not matter for this.

Q: Why write f' instead of "the derivative" or "it" in a justification?

A reader has two or three graphs in mind at once. "It is positive" could be about f or about f', and those mean different things. Naming the function removes the doubt.

Q: Can f have a relative extremum where f' does not exist?

Yes, if f is continuous there and f' changes sign, as f(x) = |x| does at 0. Look for sign changes of f' at the points where it is 0 and where it does not exist.

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 a function rises, falls, turns and changes its bend from what its first and second derivatives do, whether they come as a graph, a table or a formula.

Key vocabulary

Term	What it means	The exam trap
Key features of a graph	The places and intervals that give a graph its shape: where it increases or decreases, its relative extrema, where it is concave up or down, and its points of inflection.	A feature of $f$ shows up as a different feature of $f^{'}$ , so say which graph you are reading before you name a feature.
Relative (local) extremum	A value $f (c)$ that is the greatest (a relative maximum) or the least (a relative minimum) of the values of $f$ near $c$ .	"Where" asks for the $x$ -value $c$ , and "what value" asks for $f (c)$ ; read which one the question wants.
Concave up and concave down	The graph of $f$ is concave up on an open interval where $f^{'}$ is increasing, and concave down on one where $f^{'}$ is decreasing.	Give the interval and a reason that names $f^{'}$ or $f^{''}$ ; a description of the picture alone is not a justification.
Point of inflection	A point on the graph of a continuous function $f$ where the graph changes from concave up to concave down, or from concave down to concave up.	$f^{''} (c) = 0$ alone does not make one: check that $f^{''}$ really changes sign at $c$ .
Graphical, numerical and analytical representations	A function or its derivative given as a graph, as a table of values, or as a formula.	From a table, claim only what the listed values and the stated facts support, and say which facts you used.

Two graphs lined up on the same x-axis from -3 to 5. Above, f(x) = x cubed over 3 minus x squared minus 3x plus 4 rises to about (-1, 5.67), falls through (1, 0.33) to (3, -5), then rises. Below, its derivative f'(x) = x squared minus 2x minus 3 is 0 at x = -1 and x = 3 and lowest at (1, -4). Dashed lines mark x = -1, 1 and 3 on both graphs.

Part 1 Think first When is the line longest?

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

Predict first

A coffee shop counts the people waiting in line. $t$ minutes after it opens, the length of the line changes at the rate $L^{'} (t) = \frac{t (25 - t)}{50}$ people per minute, for $0 \leq t \leq 40$ . The rate is greatest at $t = 12.5$ , falls to 0 at $t = 25$ , and is negative after that. When is the line longest?

In short. Sketching a function from its derivatives means reading the sign of $f^{'}$ for the direction of $f$ , reading the sign of $f^{''}$ (or whether $f^{'}$ rises or falls) for its bend, and joining the pieces where either one changes.

On an interval where $f^{'} > 0$ the graph of $f$ rises, where $f^{'} < 0$ it falls, and where $f^{'}$ changes sign $f$ has a relative extremum.
On an interval where $f^{'}$ is increasing, or $f^{''} > 0$ , the graph of $f$ is concave up; where $f^{'}$ is decreasing, or $f^{''} < 0$ , it is concave down.
Each key feature of $f$ is a different feature of $f^{'}$ , and you can find it in a graph, a table or a formula.

Part 2 Learn it Three readings of a derivative, each checked as you go

8a. The sign of $f^{'}$ gives the direction of $f$

When you are handed $f^{'}$ instead of $f$ , read its sign first. Where $f^{'} (x) > 0$ , $f$ is increasing. Where $f^{'} (x) < 0$ , $f$ is decreasing. Where $f^{'}$ changes sign, $f$ turns: from positive to negative is a relative maximum of $f$ , and from negative to positive is a relative minimum.

Take the graphs at the top of this page. In the lower panel, the graph of $f^{'}$ is above the $x$ -axis for $x < - 1$ , below it for $- 1 < x < 3$ , and above it again for $x > 3$ . So in the upper panel $f$ rises, falls, then rises, with a relative maximum at $x = - 1$ and a relative minimum at $x = 3$ .

The height of $f^{'}$ does not decide any of this. The lowest point of the graph of $f^{'}$ , at $x = 1$ , is not a turning point of $f$ : $f^{'} (1) = - 4$ , so $f$ is still falling there, at its steepest.

A zero of $f^{'}$ with no sign change is not a turning point either. If $f^{'} (x) = x^{2} (x - 5)$ , then $f^{'}$ is negative just left of 0 and just right of it, because $x^{2}$ is positive on both sides and $x - 5$ is negative. So $f$ keeps decreasing through $x = 0$ , and it turns only at $x = 5$ .

Quick check · Multiple choice

No calculator. The derivative of a function $f$ is $f^{'} (x) = (x + 1) (x - 2)^{2}$ . At which values of $x$ does $f$ have a relative extremum?

8b. Whether $f^{'}$ rises or falls gives the bend of $f$

Direction is half a sketch. The other half is the bend. The graph of $f$ is concave up on an interval where $f^{'}$ is increasing, and concave down on an interval where $f^{'}$ is decreasing. When $f^{''}$ exists, that is the same as $f^{''} > 0$ or $f^{''} < 0$ there.

At the top of the page, the graph of $f^{'}$ falls until $x = 1$ and rises after it. So the graph of $f$ is concave down for $x < 1$ and concave up for $x > 1$ . The point where the bend changes, $x = 1$ , is a point of inflection of $f$ . In the $f^{'}$ panel it is the lowest point. In the $f$ panel it is where the curve falls most steeply.

Put the two readings together and every interval gets one of four shapes: rising and bending up, rising and bending down, falling and bending up, or falling and bending down. A sketch of $f$ joins those pieces at the $x$ -values where $f^{'}$ or $f^{''}$ changes sign, and uses any values of $f$ you are given to place them. Without at least one value of $f$ , the derivatives fix the shape but not the height.

Quick check · Sketch the graph

No calculator. A function $f$ is twice differentiable on $0 \leq x \leq 6$ , and $f (0) = 0$ . Its derivative $f^{'}$ is decreasing on the whole interval, and $f^{'} (3) = 0$ . Sketch the graph of $f$ from $x = 0$ to $x = 6$ . 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.

8c. One feature, three ways to see it

The facts about a function can come as a graph, a table of values or a formula, and each feature of $f$ has a partner feature of $f^{'}$ in all three. A relative maximum or minimum of $f$ is a place where $f^{'}$ changes sign. An interval where $f$ increases is one where $f^{'}$ is positive. A point of inflection of $f$ is a place where $f^{'}$ changes from increasing to decreasing, or the reverse. A concave up stretch of $f$ is one where $f^{'}$ is increasing.

From a formula, find these with algebra: solve $f^{'} (x) = 0$ , then test the sign of $f^{'}$ (and of $f^{''}$ ) between the solutions. From a table, read the signs and trends of the listed values, and claim only what the table and the given facts support. From a graph, read them by eye, then say what you read in calculus terms: " $f$ is decreasing on $- 1 < x < 3$ because $f^{'} (x) < 0$ there," never "it goes down."

Name $f$ , $f^{'}$ or $f^{''}$ every time. With two graphs on the page, "it" leaves a reader guessing which one you mean, and the two say different things.

Quick check · Match

No calculator. Each statement about $f$ follows from one statement about $f^{'}$ . Tap a statement about $f$ , then its partner.

Common mistakes

The mistake	Why it is wrong	Do this instead
Reading the height of the graph of $f^{'}$ as the height of $f$ , so that the highest point of $f^{'}$ is taken for a maximum of $f$ .	$f^{'}$ is the slope of $f$ . Its highest point is where $f$ rises fastest, and $f$ can keep rising long after that.	Read the sign of $f^{'}$ for where $f$ rises and falls, and look for the sign change of $f^{'}$ to find the extrema of $f$ .
Calling every zero of $f^{'}$ a relative extremum.	$f$ turns only where $f^{'}$ changes sign. $f^{'} (x) = x^{2} (x - 5)$ is 0 at $x = 0$ but negative on both sides of it.	Test the sign of $f^{'}$ on each side of every zero before you name a maximum or a minimum.
Reading concavity from the sign of $f^{'}$ , so that every falling stretch of $f$ is called concave down.	The sign of $f^{'}$ gives direction only. The bend comes from whether $f^{'}$ is increasing or decreasing, which is the sign of $f^{''}$ .	Use $f^{'}$ for direction and $f^{''}$ , or the trend of $f^{'}$ , for the bend, and state both for each interval.

Part 3 Watch it done Describing a graph from its derivatives

Watch each step, then do the last one yourself.

Let $f (x) = x^{3} - 3 x^{2} - 9 x + 2$ . Use $f^{'}$ and $f^{''}$ to describe the graph of $f$ : where it rises and falls, its relative extrema, how it bends, and where the bend changes.

Step 1 · Find the first derivative and its zeros
$f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x + 1) (x - 3)$ , so $f^{'} (x) = 0$ at $x = - 1$ and at $x = 3$ .
Step 2 · Read the sign of the first derivative
$f^{'} (- 2) = 15 > 0$ , $f^{'} (0) = - 9 < 0$ and $f^{'} (4) = 15 > 0$ . So $f$ is increasing for $x < - 1$ , decreasing for $- 1 < x < 3$ , and increasing for $x > 3$ . Because $f^{'}$ changes from positive to negative at $- 1$ , $f$ has a relative maximum $f (- 1) = 7$ ; because $f^{'}$ changes from negative to positive at 3, $f$ has a relative minimum $f (3) = - 25$ .
Why this step?
Why does one test value of $f^{'}$ settle the sign on a whole interval?
$f^{'}$ is a polynomial, so it is continuous, and it is 0 only at $- 1$ and 3. A continuous function cannot change sign without passing through 0, so between those zeros $f^{'}$ keeps one sign, and any value you test shows it.
Step 3 · Find the second derivative
$f^{''} (x) = 6 x - 6$ . It is negative at $x = 0$ , since $f^{''} (0) = - 6$ , and positive at $x = 2$ , since $f^{''} (2) = 6$ .
Your step · Calculate

No calculator. Finish it. For $f (x) = x^{3} - 3 x^{2} - 9 x + 2$ , the second derivative is $f^{''} (x) = 6 x - 6$ . At what value of $x$ does the graph of $f$ have a point of inflection?

Your answer

Part 4 Try it Two new derivatives to read

Try it · Spot the error

No calculator. The derivative of a function $g$ is $g^{'} (x) = x^{2} - 6 x$ . A student describes the graph of $g$ . Tap the first line with a mistake.

Try it · Decide and justify

No calculator. A function $f$ is twice differentiable on $0 \leq x \leq 4$ . Its derivative $f^{'}$ is decreasing on $0 \leq x \leq 2$ and increasing on $2 \leq x \leq 4$ , with $f^{'} (0) = 5$ , $f^{'} (2) = 2$ and $f^{'} (4) = 5$ . A classmate says $f$ has a relative minimum at $x = 2$ , because $f^{'}$ has its least value there. 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. The depth of water in a tidal pool is $D (t)$ centimeters, $t$ hours after 6 a.m., for $0 \leq t \leq 8$ . The depth changes at the rate $D^{'} (t) = t^{2} - 8 t + 12$ centimeters per hour.

(a) Find $D^{'} (3)$ . Using correct units, interpret the meaning of this value in the context of the problem.

(b) For $0 < t < 8$ , at what times $t$ does the depth of the water have a relative maximum or a relative minimum? Justify your answer.

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

Model answer

(a) $D^{'} (3) = 9 - 24 + 12 = - 3$ . At 9 a.m. ( $t = 3$ ), the depth of the water is decreasing at a rate of 3 centimeters per hour.

(b) $D^{'} (t) = (t - 2) (t - 6)$ , which is 0 at $t = 2$ and $t = 6$ . $D^{'} (t) > 0$ for $0 < t < 2$ , $D^{'} (t) < 0$ for $2 < t < 6$ , and $D^{'} (t) > 0$ for $6 < t < 8$ . $D$ has a relative maximum at $t = 2$ because $D^{'}$ changes from positive to negative there, and a relative minimum at $t = 6$ because $D^{'}$ changes from negative to positive there.

(c) $D^{''} (t) = 2 t - 8$ , which is negative for $0 < t < 4$ and positive for $4 < t < 8$ . The graph of $D$ is concave up on $4 < t < 8$ because $D^{''} (t) > 0$ there; equally, $D^{'}$ is increasing there.

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

Gets $D^{'} (3) = - 3$ and says the depth is decreasing at 3 centimeters per hour at $t = 3$ , with units.
Finds $t = 2$ and $t = 6$ from $D^{'} (t) = 0$ , and names a relative maximum at $t = 2$ and a relative minimum at $t = 6$ .
Justifies each extremum with the sign change of $D^{'}$ , naming $D^{'}$ rather than "it".
Gives $4 < t < 8$ and justifies it with $D^{''} (t) > 0$ , or $D^{'}$ increasing, on that interval.

Part 6 Check yourself A function known only through a table

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

A function $f$ is twice differentiable for $- 1 \leq x \leq 5$ , with $f (0) = 5$ , $f (2) = 1$ and $f (4) = - 3$ . The table gives the sign of $f^{'}$ and of $f^{''}$ at those points and on the intervals between them.

Signs of the first and second derivatives of f
$x$	$f^{'} (x)$	$f^{''} (x)$
$- 1 < x < 0$	Positive	Negative
$0$	0	Negative
$0 < x < 2$	Negative	Negative
$2$	Negative	0
$2 < x < 4$	Negative	Positive
$4$	0	Positive
$4 < x < 5$	Positive	Positive

Case question 1 · Calculate

No calculator. Based on the table and the given values, what is the relative maximum value of $f$ on $- 1 < x < 5$ ?

Your answer

Case question 2 · Sort it

No calculator. Using the table, tap an interval, then tap the shape of the graph of $f$ on it.

Increasing, concave up

Increasing, concave down

Decreasing, concave up

Decreasing, concave down

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

Check yourself · Fill the statement

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

The graph of $f^{'}$ crosses the $x$ -axis from above to below at $x = 7$ , so $f$ has a at $x = 7$ . The graph of $f^{'}$ falls until $x = 9$ and rises after it, so the graph of $f$ has a at $x = 9$ .

Check yourself · Multiple choice

No calculator. Let $f (x) = x^{4} - 4 x^{3}$ , so $f^{'} (x) = 4 x^{3} - 12 x^{2}$ and $f^{''} (x) = 12 x^{2} - 24 x$ . Which statement about $f$ is true?

Part 7 Keep it Bring back earlier ideas

This lesson stands on the second derivative test from earlier in this unit and on the chain rule from Unit 3. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. A function $g$ is twice differentiable, with $g^{'} (2) = 0$ and $g^{''} (2) = - 5$ . Which statement about $g$ must be true?

Bring it back · Calculate

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

Your answer

In three lines

The sign of $f^{'}$ says where $f$ rises or falls, and a sign change of $f^{'}$ marks a relative extremum of $f$ .
Whether $f^{'}$ rises or falls, which is the sign of $f^{''}$ , says how the graph of $f$ bends, and a change marks a point of inflection.
Each feature of $f$ has a partner feature of $f^{'}$ , whether you are given a graph, a table or a formula.

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 you justify a conclusion about $f$ from a derivative, name the derivative and say what it does: " $f$ has a relative minimum at $x = 4$ because $f^{'}$ changes from negative to positive there." A sentence such as "it goes down" gives no calculus reason, and "it" leaves the reader unsure which function you mean.

Questions students ask

If I am given only the graph of

f^{'}

, can I find the values of

f

Not from $f^{'}$ alone. $f^{'}$ tells you how $f$ changes, not where it starts, so you need at least one value of $f$ to place the graph. With one value you can sketch the shape; finding other values exactly takes integrals, which Unit 6 teaches.

Is a point where

f^{''} (x) = 0

always a point of inflection?

No. $f (x) = x^{4}$ has $f^{''} (0) = 0$ , but $f^{''} (x) = 12 x^{2}$ is positive on both sides of 0, so the graph is concave up throughout. Check that $f^{''}$ changes sign.

How do I find concavity from a graph of

f^{'}

when I do not have

f^{''}

Look at whether the graph of $f^{'}$ is rising or falling. Where it rises, the graph of $f$ is concave up; where it falls, concave down. Which side of the $x$ -axis $f^{'}$ is on does not matter for this.

Why write

f^{'}

instead of "the derivative" or "it" in a justification?

A reader has two or three graphs in mind at once. "It is positive" could be about $f$ or about $f^{'}$ , and those mean different things. Naming the function removes the doubt.

Can

f

have a relative extremum where

f^{'}

does not exist?

Yes, if $f$ is continuous there and $f^{'}$ changes sign, as $f (x) = | x |$ does at 0. Look for sign changes of $f^{'}$ at the points where it is 0 and where it does not exist.

Course alignment, for teachers

AP Calculus AB topic 5.8, 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 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.