Unit 5 · Topic 5.4 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 5.4: Using the First Derivative Test to Determine Relative (Local) Extrema

Q: Is every critical point a relative maximum or minimum?

No. At x = 0, f(x) = x3 has f'(0) = 0, but f' is positive on both sides, so f just keeps rising. The sign change of f' is what shows a critical point is an extremum.

Q: Can a relative minimum be larger than a relative maximum?

Yes. Each one compares f only with nearby inputs. For f(x) = x + 4x, the relative maximum value at x = -2 is -4, and the relative minimum value at x = 2 is 4.

Q: What if f' does not exist at the point?

If the input is in the domain of f and f is continuous there, it is still a critical point and the test works the same way. The corner of |x| at 0 is a relative minimum because f' changes from -1 to 1.

Q: Do I need the value of f' at the point to use the test?

No. You need the sign of f' just to the left and just to the right. Any test input between neighboring critical points shows the sign for that whole interval.

Q: How is this different from the Second Derivative Test?

The Second Derivative Test, in Topic 5.7, uses the sign of f'' at a critical point where f' = 0. The First Derivative Test uses only f', and it also handles critical points where f' 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

Use the sign of the first derivative on each side of a critical point to decide whether a function has a relative maximum, a relative minimum or neither there, and justify the conclusion.

Key vocabulary

Term	What it means	The exam trap
Relative (local) extrema	The relative maximum and relative minimum values of a function, each compared only with the values of the function at nearby inputs.	Relative and local mean the same thing; neither word means greatest or least on the whole domain.
Relative maximum	A value $f (c)$ that is at least as large as every value of $f$ at inputs near $c$ .	The maximum is the output $f (c)$ and its location is the input $c$ . Read which one the question asks for.
Relative minimum	A value $f (c)$ that is at most as large as every value of $f$ at inputs near $c$ .	Say which function has it: write $f$ has a relative minimum, not "it" has one.
Critical point	An input in the domain of $f$ at which $f^{'}$ equals 0 or fails to exist.	You find its $x$ -value from $f^{'}$ ; if a question asks for the value of the extremum, that comes from $f$ .
First Derivative Test	A test that classifies a critical point of a continuous function by the sign of $f^{'}$ just to its left and just to its right.	In a justification, state the sign change in words and say which function makes it.

Two stacked graphs. The top graph is f(x) = x to the fourth over 4 minus x cubed plus 4x, with the points (-1, -2.75) and (2, 4) marked. The bottom graph is its derivative f'(x) = (x + 1)(x - 2) squared: negative left of x = -1, zero at x = -1, positive between -1 and 2, touching zero at x = 2 and positive after it.

Part 1 Think first When is the drone highest?

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

Predict first

A survey drone flies over hills. The rate at which its altitude changes, in meters per second, was logged each second:

$t = 0$ : 10.5, $t = 1$ : 6, $t = 2$ : 2.5, $t = 3$ : 0, $t = 4$ : $- 1.5$ , $t = 5$ : $- 2$ , $t = 6$ : $- 1.5$ , $t = 7$ : 0, $t = 8$ : 2.5.

At which time is the drone higher than it is at every nearby time?

In short. The First Derivative Test locates the relative (local) extrema of a function $f$ by reading how the sign of $f^{'}$ changes at each critical point.

The candidates are the critical points: inputs in the domain of $f$ where $f^{'}$ is 0 or does not exist.
What decides each candidate is the sign of $f^{'}$ just to its left and just to its right, not the value of $f^{'}$ at the point.
A justification names $f^{'}$ and the sign change it makes, in words, at that input.

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

4a. The First Derivative Test

A function $f$ has a relative maximum at $c$ when $f (c)$ is at least as large as every value of $f$ at inputs near $c$ , and a relative minimum when $f (c)$ is at most as large. Relative and local mean the same thing, and together these are the relative (local) extrema.

You already know that $f$ increases where $f^{'} > 0$ and decreases where $f^{'} < 0$ . Put that next to a critical point $c$ , an input in the domain of $f$ where $f^{'} (c) = 0$ or $f^{'} (c)$ does not exist, with $f$ continuous at $c$ . The First Derivative Test reads the sign of $f^{'}$ on each side of $c$ :

$f^{'}$ changes from positive to negative at $c$ : $f$ rises and then falls, so $f$ has a relative maximum at $c$ .
$f^{'}$ changes from negative to positive at $c$ : $f$ falls and then rises, so $f$ has a relative minimum at $c$ .
$f^{'}$ has the same sign on both sides: $f$ keeps going the same way through $c$ , so there is no relative extremum at $c$ .

The value $f^{'} (c)$ never decides it. It is 0 or does not exist at every critical point, whether that point turns out to be a maximum, a minimum or neither. The signs on the two sides decide.

Quick check · Sort it

No calculator. On every card, $c$ is a critical point of $f$ and $f$ is continuous at $c$ . Tap a card, then tap what $f$ has at $c$ .

Relative maximum

Relative minimum

Neither

4b. Find every candidate, then test each one

With a formula for $f^{'}$ , the test runs in three steps.

Find the critical points: the inputs in the domain of $f$ where $f^{'} (x) = 0$ or $f^{'} (x)$ does not exist.
The critical points, and any input where $f$ is not defined, cut the number line into intervals. In each interval, pick a test input and find the sign of $f^{'}$ there. A continuous $f^{'}$ cannot change sign without passing through 0, so one test input speaks for the whole interval.
At each critical point, compare the signs on the two sides.

Take $f^{'} (x) = x (x - 3)^{2}$ . It is 0 at $x = 0$ and $x = 3$ . Test inputs: $f^{'} (- 1) = - 16$ , $f^{'} (1) = 4$ , $f^{'} (4) = 4$ . So $f^{'}$ changes from negative to positive at $x = 0$ , a relative minimum. At $x = 3$ it is positive on both sides, so $f$ has no extremum there.

That second point is the one students miss. A squared factor such as $(x - 3)^{2}$ is never negative, so it never flips the sign of $f^{'}$ . Check the signs; do not assume every zero of $f^{'}$ is a turning point.

One more caution: an input outside the domain of $f$ is a cut but never a critical point, even where the sign of $f^{'}$ changes, because $f$ has no value there to be a maximum or a minimum.

Quick check · Calculate

No calculator. A function $f$ is defined for all real numbers, and $f^{'} (x) = (x + 4) (x - 1)^{2} (x - 6)$ . At what value of $x$ does $f$ have a relative minimum?

Your answer

4c. Say why, and name the function

A free-response answer earns the conclusion only with a reason built on the derivative. Write the sign change of $f^{'}$ , in words, at the input:

" $f$ has a relative maximum at $x = 4$ because $f^{'}$ changes from positive to negative at $x = 4$ ."

Two things make that sentence work. It names the functions: $f$ has the maximum, $f^{'}$ changes sign. "It changes sign" leaves a reader unsure which function you mean. And it gives the sign change, not just " $f^{'} (4) = 0$ ", because a zero of $f^{'}$ only makes 4 a critical point.

Often you are given the graph of $f^{'}$ , not of $f$ . Then read the graph for where it is above or below the $x$ -axis. A relative maximum of $f$ is where the graph of $f^{'}$ crosses the axis from above to below. The highest point on the graph of $f^{'}$ is where $f$ climbs fastest, which is not a maximum of $f$ at all.

When the question asks whether $f$ has a relative maximum, a relative minimum or neither, begin with its own words: " $f$ has neither at $x = 2$ because $f^{'}$ does not change sign at $x = 2$ ."

Quick check · Decide and justify

No calculator. A function $f$ is differentiable for all real numbers, and $f^{'} (2) = 0$ . Asked to justify a relative maximum, a student writes: " $f$ has a relative maximum at $x = 2$ because $f^{'} (2) = 0$ ." Is that justification complete?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Calling every zero of $f^{'}$ a relative maximum or minimum.	$f^{'} (c) = 0$ only makes $c$ a critical point. If $f^{'}$ has the same sign on both sides, as at a squared factor, $f$ goes straight through.	Test the sign of $f^{'}$ in every interval and decide each critical point from the signs on its two sides.
Reversing the test: calling positive-to-negative a minimum.	Positive $f^{'}$ means $f$ rises, so positive then negative is rising then falling: a peak.	Say the motion out loud: rising then falling is a maximum, falling then rising is a minimum.
Reading the highest or lowest point on a graph of $f^{'}$ as an extremum of $f$ .	A peak of $f^{'}$ is where $f$ changes fastest. The extrema of $f$ are where the graph of $f^{'}$ crosses the $x$ -axis.	On a graph of $f^{'}$ , look only at where it is above or below the axis, and where it crosses.

Part 3 Watch it done The test when $f^{'}$ does not exist

Watch each step, then do the last one yourself.

Let $f (x) = x^{2 / 3} (x - 5)$ , which is continuous for all real numbers. Find where $f$ has a relative maximum and a relative minimum.

Step 1 · Find the derivative
Write $f (x) = x^{5 / 3} - 5 x^{2 / 3}$ . Then $f^{'} (x) = \frac{5}{3} x^{2 / 3} - \frac{10}{3} x^{- 1 / 3} = \frac{5 x - 10}{3 x^{1 / 3}}$ .
Step 2 · Find the critical points
$f^{'} (x) = 0$ where the numerator is 0: $x = 2$ . $f^{'} (x)$ does not exist where the denominator is 0: $x = 0$ , which is in the domain of $f$ . So the critical points are $x = 0$ and $x = 2$ .
Why this step?
Why does $x = 0$ count as a critical point when $f^{'} (0)$ is not 0?
A critical point is an input in the domain of $f$ where $f^{'}$ is 0 or does not exist. $f (0) = 0$ is defined and $f$ is continuous at 0, but $f^{'} (0)$ divides by 0, so it does not exist. Skip it and you lose one of the two extrema.
Step 3 · Test the sign on each interval
$f^{'} (- 1) = \frac{- 15}{- 3} = 5 > 0$ , $f^{'} (1) = \frac{- 5}{3} < 0$ , and $f^{'} (8) = \frac{30}{6} = 5 > 0$ . So $f^{'} > 0$ for $x < 0$ , $f^{'} < 0$ for $0 < x < 2$ , and $f^{'} > 0$ for $x > 2$ .
Your step · Fill the statement
No calculator. Finish it. Tap a blank, then tap an answer.

For $f (x) = x^{2 / 3} (x - 5)$ , $f^{'}$ is positive for $x < 0$ , negative for $0 < x < 2$ and positive for $x > 2$ . So $f$ has a relative maximum at $x =$ and a relative minimum at $x =$ .

Part 4 Try it Two new functions

Try it · Calculate

Graphing calculator required. Let $f$ be a function with $f^{'} (x) = \cos x - \frac{x}{2}$ for $0 < x < 4$ . In radian mode, find the value of $x$ at which $f$ has a relative maximum. Give your answer to three decimal places.

Your answer

Try it · Multiple choice

No calculator. Let $f (x) = x + \frac{4}{x}$ , so $f^{'} (x) = 1 - \frac{4}{x^{2}}$ . Where does $f$ have a relative minimum?

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 5, from free-response Part B (calculator not permitted).

Free-response practice · Question 5 style

No calculator. Water flows into and out of a tank. The amount of water in the tank, in liters, is $W (t)$ , where $t$ is measured in hours, and $W$ is differentiable with

W^{'} (t) = (t - 1) (t - 4)^{2} (6 - t)

for $0 < t < 7$ .

(a) Find $W^{'} (5)$ . Using correct units, interpret the meaning of $W^{'} (5)$ in the context of the problem.

(b) Find each value of $t$ in the interval $0 < t < 7$ at which $W$ has a relative minimum, and each value at which $W$ has a relative maximum. Justify your answers.

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

Model answer

(a) $W^{'} (5) = (4) (1)^{2} (1) = 4$ . At time $t = 5$ hours, the amount of water in the tank is increasing at a rate of 4 liters per hour.

(b) $W^{'} (t) = 0$ at $t = 1$ , $t = 4$ and $t = 6$ . $W^{'} (t) < 0$ for $0 < t < 1$ , $W^{'} (t) > 0$ for $1 < t < 4$ and for $4 < t < 6$ , and $W^{'} (t) < 0$ for $6 < t < 7$ . $W$ has a relative minimum at $t = 1$ because $W^{'}$ changes from negative to positive at $t = 1$ . $W$ has a relative maximum at $t = 6$ because $W^{'}$ changes from positive to negative at $t = 6$ .

(c) Neither. $W^{'} (4) = 0$ , but $W^{'} (t) > 0$ for $1 < t < 4$ and for $4 < t < 6$ , so $W^{'}$ does not change sign at $t = 4$ , and $W$ has neither a relative minimum nor a relative maximum there.

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

Gets $W^{'} (5) = 4$ and says the amount of water is increasing at 4 liters per hour at $t = 5$ hours.
Finds $W^{'} (t) = 0$ at $t = 1$ , 4 and 6, and gives the sign of $W^{'}$ on each interval between them.
Says $W$ has a relative minimum at $t = 1$ because $W^{'}$ changes from negative to positive there.
Says $W$ has a relative maximum at $t = 6$ because $W^{'}$ changes from positive to negative there.
Says neither at $t = 4$ because $W^{'}$ does not change sign there, naming $W^{'}$ , not "it".

Part 6 Check yourself A sign chart

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

A function $h$ is continuous for all real numbers. Its derivative $h^{'}$ is 0 at $x = - 3$ and $x = 2$ , does not exist at $x = 0$ , and is never 0 anywhere else. The table gives the sign of $h^{'}$ everywhere else.

Sign of h'(x)
$x$	$h^{'} (x)$
$x < - 3$	positive
$- 3 < x < 0$	negative
$0 < x < 2$	positive
$x > 2$	positive

Case question 1 · Multiple choice

No calculator. At which values of $x$ does $h$ have a relative minimum?

Case question 2 · Calculate

No calculator. How many relative extrema, maxima and minima together, does $h$ have?

Your answer

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

Check yourself · Spot the error

No calculator. A student uses the First Derivative Test on $f (x) = x^{3} - 3 x^{2} - 9 x$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. Each function $f$ is defined for all real numbers and has the derivative shown. Tap a derivative, then what it tells you about a relative maximum of $f$ .

Part 7 Keep it Bring back earlier ideas

Two earlier ideas this lesson leans on. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. For which function does the Extreme Value Theorem guarantee both a maximum value and a minimum value on the closed interval $[- 1, 1]$ ?

Bring it back · Calculate

No calculator. Let $y = (x^{2} + 1)^{3}$ . Find $\frac{d y}{d x}$ at $x = 1$ .

Your answer

In three lines

The candidates for relative extrema are the critical points, where $f^{'}$ is 0 or does not exist.
If $f^{'}$ changes from positive to negative, $f$ has a relative maximum; negative to positive, a relative minimum; no change, neither.
Justify with the sign change of $f^{'}$ at the input, naming $f$ and $f^{'}$ .

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 a question asks whether $f$ has a relative maximum, a relative minimum or neither, start your answer with the question's own words and finish with the sign change of $f^{'}$ : " $f$ has a relative minimum at $x = 1$ because $f^{'}$ changes from negative to positive at $x = 1$ ." Name $f$ and $f^{'}$ every time; "it" can mean either.

Questions students ask

Is every critical point a relative maximum or minimum?

No. At $x = 0$ , $f (x) = x^{3}$ has $f^{'} (0) = 0$ , but $f^{'}$ is positive on both sides, so $f$ just keeps rising. The sign change of $f^{'}$ is what shows a critical point is an extremum.

Can a relative minimum be larger than a relative maximum?

Yes. Each one compares $f$ only with nearby inputs. For $f (x) = x + \frac{4}{x}$ , the relative maximum value at $x = - 2$ is $- 4$ , and the relative minimum value at $x = 2$ is 4.

What if

f^{'}

does not exist at the point?

If the input is in the domain of $f$ and $f$ is continuous there, it is still a critical point and the test works the same way. The corner of $| x |$ at 0 is a relative minimum because $f^{'}$ changes from $- 1$ to 1.

Do I need the value of

f^{'}

at the point to use the test?

No. You need the sign of $f^{'}$ just to the left and just to the right. Any test input between neighboring critical points shows the sign for that whole interval.

How is this different from the Second Derivative Test?

The Second Derivative Test, in Topic 5.7, uses the sign of $f^{''}$ at a critical point where $f^{'} = 0$ . The First Derivative Test uses only $f^{'}$ , and it also handles critical points where $f^{'}$ does not exist.

Course alignment, for teachers

AP Calculus AB topic 5.4, Unit 5: Analytical Applications of Differentiation.

Skill 3.D: Apply an appropriate mathematical definition, theorem, or test.

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 When is the drone highest?

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

4a. The First Derivative Test

4b. Find every candidate, then test each one

4c. Say why, and name the function

Part 3 Watch it done The test when f′ does not exist

Part 4 Try it Two new functions

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A sign chart

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic

Part 3 Watch it done The test when $f^{'}$ does not exist