What is the difference between a local and an absolute extremum?

A local extremum is the highest or lowest value only in a small neighborhood around a point. An absolute extremum is the single highest or lowest value over the entire interval. A local max found by the First Derivative Test might not be the absolute max, which is why closed-interval problems also require checking endpoints with the Candidates Test.

Do I always need to use the First Derivative Test instead of the Second Derivative Test?

Both work for classifying local extrema. The First Derivative Test is more general because it works even when f'' is zero, undefined, or hard to compute, and it works at points where f' is undefined. On free-response questions, either test earns credit as long as you justify clearly. The Second Derivative Test appears in the neighboring topic on concavity.

Why doesn't f'(x)=0 automatically mean there is a max or min?

A zero derivative only means the tangent line is horizontal. If f' does not actually change sign there — as with f(x)=x^3 at x=0 — the function keeps increasing or decreasing through that point, so no extremum exists. Always confirm a sign change, not just a zero.

How do I write a justification that earns full AP credit?

State the sign behavior of f' explicitly. For example: 'f' changes from positive to negative at x=c, so f has a local maximum at x=c.' For absolute extrema, present a table of f values at all critical points and endpoints and identify the largest and smallest. Vague statements like 'f'=0 here' will not receive full credit.

AP-CALCBC-5.3-5.5

U5.3 First Derivative Analysis: Increasing/Decreasing and Local Extrema

Master AP Calculus BC first derivative analysis: use the sign of f'(x) to find increasing/decreasing intervals, apply the First Derivative Test, and find absolute extrema.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U5.3 First Derivative Analysis: Increasing/Decreasing and Local Extrema, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

The sign of the first derivative tells you everything about whether a function is climbing or falling. When

f'(x) > 0

, the graph rises; when

f'(x) < 0

, it falls. The instant the derivative switches sign, you have found a local maximum or minimum. In this lesson you will learn to build a sign chart for

f'(x)

, apply the First Derivative Test to classify local extrema, and use the Candidates Test to pin down absolute extrema on a closed interval. These are among the most frequently tested skills in Unit 5, appearing in both multiple-choice and free-response questions, often disguised inside a table of values or a graph of

f'

Increasing, Decreasing, and Critical Points

A function

f

is increasing on an interval where

f'(x) > 0

and decreasing where

f'(x) < 0

. This follows from the Mean Value Theorem: a positive derivative forces output values to grow as inputs grow.

To analyze

f

, first find its critical points — the

x

-values in the domain where

f'(x) = 0

f'(x)

does not exist. These are the only places where

f'

can change sign, so they partition the number line into test intervals.

A reliable procedure looks like this:

Step	Action
1	Compute $f'(x)$ and simplify
2	Solve $f'(x)=0$ and find where $f'$ is undefined
3	Mark these critical points on a number line
4	Test the sign of $f'$ in each interval
5	Translate: $+$ means increasing, $-$ means decreasing

A common misconception is that every point where

f'(x)=0

is an extremum. That is false. Consider

f(x)=x^3

: here

f'(x)=3x^2=0

x=0

, but

f'

stays positive on both sides, so

f

keeps increasing. The derivative touching zero without changing sign produces no extremum. Always check the sign on both sides rather than assuming.

The First Derivative Test for Local Extrema

The First Derivative Test classifies each critical point by how

f'

changes sign there. If

f'

changes from positive to negative as

x

increases through

c

, then

f

has a local maximum at

c

. If

f'

changes from negative to positive,

f

has a local minimum at

c

. If

f'

does not change sign, there is no local extremum at

c

Sign of $f'$ around $c$	Conclusion at $x=c$
$+$ then $-$	Local maximum
$-$ then $+$	Local minimum
$+$ then $+$	No extremum
$-$ then $-$	No extremum

On the exam, questions frequently give you the graph of

f'

(not

f

) or a sign chart. Read carefully: you are looking at where

f'

crosses the

x

-axis and changes sign, not where

f'

has its own peaks. A very common trap presents the graph of

f'

and asks where

f

has a local max; students wrongly pick where

f'

is largest. The correct answer is where

f'

crosses from positive to negative.

When writing free-response justifications, you must explicitly state the sign change. Saying only "

f'(2)=0

" earns no credit. Write something like: "

f'

changes from positive to negative at

x=2

, so

f

has a local maximum there." Justification language is graded strictly.

Absolute Extrema and the Candidates Test

On a closed interval

[a,b]

, the Extreme Value Theorem guarantees that a continuous function attains both an absolute maximum and an absolute minimum. The Candidates Test finds them efficiently: the absolute extrema can only occur at critical points inside the interval or at the endpoints.

The procedure is:

Step	Action
1	Find all critical points in $(a,b)$
2	List these plus both endpoints $a$ and $b$
3	Evaluate $f$ at every candidate
4	Largest value is the absolute max; smallest is the absolute min

Notice you evaluate

f

itself here, not

f'

. A frequent error is comparing derivative values instead of function values. The First Derivative Test tells you the type of a local extremum, but the Candidates Test simply compares actual outputs.

The closed-interval requirement matters. On an open interval or an unbounded domain, an absolute extremum may not exist, and you would need limits or additional reasoning. On the AP exam, when a problem specifies a closed interval, the Candidates Test is almost always the intended method. Show your candidate values in an organized table for full credit — the reader wants to see every function value you compared.

Key terms

Critical point.: An $x$ -value in the domain of $f$ where $f'(x)=0$ or $f'(x)$ does not exist. Extrema can occur only here or at endpoints.
Increasing function.: A function is increasing on an interval when $f'(x) > 0$ throughout, so larger inputs give larger outputs.
Decreasing function.: A function is decreasing on an interval when $f'(x) < 0$ throughout, so larger inputs give smaller outputs.
First Derivative Test.: A method that classifies a critical point as a local max, local min, or neither based on how the sign of $f'$ changes there.
Local (relative) extremum.: A point where $f$ has a value at least as large (max) or small (min) as all nearby points.
Absolute (global) extremum.: The single largest or smallest value of $f$ over an entire interval or domain.
Candidates Test.: On a closed interval, the method of evaluating $f$ at all critical points and endpoints to locate absolute extrema.
Sign chart.: A number line marked with critical points and the sign of $f'$ in each subinterval, used to read off increasing/decreasing behavior.

Worked example

Let

f(x) = x^3 - 6x^2 + 9x + 2

on the closed interval

[0,4]

. Find the intervals where

f

is increasing and decreasing, classify all local extrema, and determine the absolute maximum and minimum on

[0,4]

First compute the derivative:

f'(x) = 3x^2 - 12x + 9

. Factor:

f'(x) = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

.

Set

f'(x)=0

to find critical points:

x=1

and

x=3

. Both lie in

[0,4]

. The derivative is a polynomial, so it is defined everywhere.

Build a sign chart. Test

x=0

3(0-1)(0-3)=3(-1)(-3)=9>0

, so

f

is increasing on

(0,1)

. Test

x=2

3(2-1)(2-3)=3(1)(-1)=-3<0

, so

f

is decreasing on

(1,3)

. Test

x=4

3(4-1)(4-3)=3(3)(1)=27>0

, so

f

is increasing on

(3,4)

.

Apply the First Derivative Test. At

x=1

f'

changes from positive to negative, so

f

has a local maximum. At

x=3

f'

changes from negative to positive, so

f

has a local minimum.

Now the Candidates Test for absolute extrema on

[0,4]

. Candidates are

x=0,1,3,4

. Evaluate:

f(0)=2

;

f(1)=1-6+9+2=6

;

f(3)=27-54+27+2=2

;

f(4)=64-96+36+2=6

.

Compare values: the largest is

6

, attained at both

x=1

and

x=4

, so the absolute maximum is

6

. The smallest is

2

, attained at both

x=0

and

x=3

, so the absolute minimum is

2

Practice questions

The derivative of a function

f

is given by

f'(x) = (x+2)(x-5)

. At which value of

x

does

f

have a local maximum?

$x=-2$
$x=0$
$x=5$
$x=2$

Answer: $x=-2$

The critical points are

x=-2

and

x=5

. For

x<-2

, both factors are negative, so

f'>0

. Between

-2

and

5

, the product is negative, so

f'<0

. Thus

f'

changes from positive to negative at

x=-2

, giving a local maximum. At

x=5

f'

changes from negative to positive, a local minimum.

Let

g(x) = x^4 - 8x^2 + 3

on the closed interval

[-1,3]

. Find all local extrema using the First Derivative Test, then determine the absolute maximum and minimum on the interval.

Answer: Local minimum at $x=2$ with value $-13$ ; local maximum at $x=0$ with value $3$ ; absolute maximum is $3$ at $x=0$ , absolute minimum is $-13$ at $x=2$ .

Compute

g'(x)=4x^3-16x=4x(x^2-4)=4x(x-2)(x+2)

. Critical points are

x=-2,0,2

, but only

x=0

and

x=2

lie in

[-1,3]

. Sign analysis: on

(-1,0)

g'<0

(decreasing); on

(0,2)

g'<0

as well? Test

x=1

4(1)(1-2)(1+2)=4(1)(-1)(3)=-12<0

, decreasing. Test

x=-0.5

4(-0.5)(-2.5)(1.5)=7.5>0

, increasing. So

f'

goes

+

-

x=0

: local max,

g(0)=3

. Test

x=2.5

4(2.5)(0.5)(4.5)=22.5>0

, increasing, so

f'

goes

-

+

x=2

: local min,

g(2)=16-32+3=-13

. Candidates

x=-1,0,2,3

g(-1)=1-8+3=-4

g(0)=3

g(2)=-13

g(3)=81-72+3=12

. Absolute max is

12

x=3

; absolute min is

-13

x=2

. Note the absolute max occurs at an endpoint, not the local max — this is why the Candidates Test is essential.

Suppose

f

is continuous on

[0,6]

and its derivative satisfies

f'(x)>0

(0,2)

f'(x)<0

(2,5)

, and

f'(x)>0

(5,6)

. Where does

f

attain a local minimum, and why can you not immediately name the absolute minimum?

Answer: $f$ has a local minimum at $x=5$ ; the absolute minimum requires comparing $f(0)$ , $f(5)$ , and the endpoints because a local min may not be the global min.

x=5

f'

changes from negative to positive, giving a local minimum by the First Derivative Test. At

x=2

f'

changes positive to negative, a local maximum. To find the absolute minimum on

[0,6]

you must evaluate

f

at the critical point

x=5

and at the endpoints

x=0

and

x=6

, then compare actual function values. Without those numerical values, you cannot conclude which is smallest — the endpoint

f(0)

could be lower than

f(5)

FAQ

What is the difference between a local and an absolute extremum?: A local extremum is the highest or lowest value only in a small neighborhood around a point. An absolute extremum is the single highest or lowest value over the entire interval. A local max found by the First Derivative Test might not be the absolute max, which is why closed-interval problems also require checking endpoints with the Candidates Test.
Do I always need to use the First Derivative Test instead of the Second Derivative Test?: Both work for classifying local extrema. The First Derivative Test is more general because it works even when $f''$ is zero, undefined, or hard to compute, and it works at points where $f'$ is undefined. On free-response questions, either test earns credit as long as you justify clearly. The Second Derivative Test appears in the neighboring topic on concavity.
Why doesn't f'(x)=0 automatically mean there is a max or min?: A zero derivative only means the tangent line is horizontal. If $f'$ does not actually change sign there — as with $f(x)=x^3$ at $x=0$ — the function keeps increasing or decreasing through that point, so no extremum exists. Always confirm a sign change, not just a zero.
How do I write a justification that earns full AP credit?: State the sign behavior of $f'$ explicitly. For example: ' $f'$ changes from positive to negative at $x=c$ , so $f$ has a local maximum at $x=c$ .' For absolute extrema, present a table of $f$ values at all critical points and endpoints and identify the largest and smallest. Vague statements like ' $f'=0$ here' will not receive full credit.

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