Unit 5Second Derivative Test Conditions

Concept 4.190Built for May 2027 AP Exam

Second Derivative Test Conditions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The second derivative test classifies a critical number by the way the curve bends at that number. If $f'(c)=0$ and $f''(c)>0$ , the curve is a cup at $c$ and the point is a local minimum. If $f''(c)<0$ , the curve is a cap and the point is a local maximum. If $f''(c)=0$ , the test returns no verdict at all, and the sign change of $f'$ has to decide. The test never finds the critical numbers, and it never touches endpoints. Treat it as a shortcut rather than a definition: every problem it can settle can also be settled by the sign chart of $f'$ , and most lost points come from running it at a number that was never critical in the first place.

On the AP Exam

What you’ll see: A critical number already located, and a stem that asks for a classification with a reason, sometimes with the words "use a derivative test" leaving the choice of test to you.
What you’ll do: Evaluate $f''$ at the critical number, read the sign, and translate that sign into a name for the point.
Memory hook: Positive bends up like a valley, negative bends down like a hill. The sign of $f''$ names the shape, and the shape names the point.
Use it on the exam: For MCQs: The distractors reverse the sign, so a positive second derivative is offered as a maximum. Read the sign twice before naming anything. For FRQs: On 2024 Q3(b) the chief reader report records that students who chose this test often failed to differentiate implicitly for $f''$ , so the test is only as good as the derivative you feed it.

Concept Review

The Test in One Line

The test has three inputs and one output. The inputs are a critical number, the value of $f''$ there, and the assumption that $f''$ exists on a neighbourhood of that number.

Condition at the critical number $c$	What the bend says	Verdict
$f'(c)=0$ and $f''(c)>0$	the curve bends up, a cup	local minimum at $c$
$f'(c)=0$ and $f''(c)<0$	the curve bends down, a cap	local maximum at $c$
$f'(c)=0$ and $f''(c)=0$	the bend says nothing	test fails, use the sign of $f'$
$f'(c)\neq 0$	not a critical number	test does not apply

Why the Bend Decides

Figure 11 — Three panels, three different verdicts. A cup at the critical number makes it a minimum, a cap makes it a maximum, and a flat inflection leaves the test silent.

At a critical number the tangent is horizontal, so the only question left is what happens on either side. A curve bending upward leaves the horizontal tangent sitting at the bottom of a dip, and a curve bending downward leaves it at the top of a rise. That is the whole argument, and it never mentions the values of $f$ anywhere else.

Three panels side by side: a cup with a horizontal tangent at its bottom labeled local minimum, a cap with a horizontal tangent at its top labeled local maximum, and a flat inflection where the second derivative vanishes labeled no verdict

When the Test Fails

A zero of $f''$ at a critical number is not a verdict. The function $f(x)=x^{3}$ has $f'(0)=0$ and $f''(0)=0$ , and $x=0$ is neither a maximum nor a minimum. The function $f(x)=x^{4}$ has the same two zeros at the origin and $x=0$ is a minimum. Same test values, opposite answers, so the test genuinely cannot decide and the sign chart must.

What the Test Cannot Reach

Three situations sit outside the test. Endpoints of a closed interval are never critical numbers, so no derivative test classifies them. A critical number where $f''$ does not exist cannot be fed to the test. A critical number where $f'$ itself fails to exist, such as the corner of $\lvert x\rvert$ at the origin, is also out of range.

Choosing Between the Two Tests

Situation	Better test	Reason
$f''$ is easy and $f''(c)\neq 0$	second derivative test	one evaluation settles it
$f''$ is messy or vanishes	first derivative test	the sign chart never fails
the derivative is given as a graph	first derivative test	the sign change is visible
the function is given as a table	neither, compare values	no derivative is available

The first derivative test is the safer default. The second derivative test is the faster one, and it is worth using only when the second derivative is cheap to compute and clearly nonzero.

Common Misconception Check

Myth 1: "A positive second derivative guarantees a minimum." Truth: It guarantees a minimum only at a critical number. At a non-critical point it describes the bend and nothing about extrema.
Myth 2: "If $f''(c)=0$ then $c$ is an inflection point." Truth: A zero of $f''$ is a candidate. The concavity has to change across it, which is the topic of 5.15.
Myth 3: "The test also covers the endpoints." Truth: Endpoints are compared by value, which is 5.7 and 5.8. The test has nothing to say about them.

Topic-Specific Study Method

Locate First, Then Classify

Borrowed idea: The test classifies, it never locates. Solve $f'(x)=0$ first, and keep only the solutions inside the stated domain. Then evaluate $f''$ at each one and read the two signs together. A sign chart is still the fallback when $f''$ vanishes or refuses to exist. So what does a clean write-up look like when time is short? One sentence per critical number, with the value of $f''$ quoted inside it.

What you have	What you do first	What you write
a formula for $f$	solve $f'(x)=0$	" $f''(3)=12>0$ , so a local minimum"
a formula for $f'$ only	find where $f'$ changes sign	"the sign change decides, not $f''$ "
a graph of $f'$	locate the zeros of $f'$	"the slope of $f'$ is the concavity of $f$ "
a table of values	compare the neighbours	"the values fall then rise, so a minimum"

① Quote the number. " $f''(3)=12>0$ " scores, and " $f''$ is positive" alone often does not, because the value is the evidence.

② Say which test you used. When the stem says "use a derivative test", name it, then show the sign that decided.

③ Never skip the domain check. A critical number outside the stated interval is not a candidate, no matter what the test says about it.

Training & Error Diagnosis

Core Archetype (Simplest Form): Classify the critical numbers of $f(x)=x^{3}-3x^{2}-9x+5$ .

Standard Execution:

Locate. $f'(x)=3x^{2}-6x-9=3(x-3)(x+1)$ , so the critical numbers are $x=-1$ and $x=3$ .

Differentiate again. $f''(x)=6x-6$ .

Evaluate. $f''(-1)=-12<0$ and $f''(3)=12>0$ .

Name. The curve bends down at $x=-1$ , so $f$ has a local maximum there with $f(-1)=10$ ; it bends up at $x=3$ , so $f$ has a local minimum there with $f(3)=-22$ .

Exam Variation (Context / Parameter Shift): A twice-differentiable function $H$ models the depth of water, and $H'(t)=0$ at $t=\pi$ with $H''(\pi)<0$ . A student writes that $t=\pi$ must be an inflection because the depth stops changing. Identify the error and give the correct classification.

Solution: The error is treating a zero of the first derivative as a change of concavity. A zero of $H'$ is a critical number, and $H''(\pi)<0$ says the curve bends down there, so $t=\pi$ is a local maximum of the depth. The flat tangent marks the top of the rise, not a change of bend.

Common Traps & Failure Signals:

The sign reversed (Signal: $f''>0$ reported as a maximum): the cup was read backwards.
The zero trusted (Signal: " $f''(c)=0$ , so it is an inflection"): a failed test was treated as a verdict.
The endpoint classified (Signal: an endpoint offered as a local minimum): the test was applied where no derivative condition holds.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Locate Skipped: You evaluated $f''$ before you had the critical numbers, so a sign was attached to a point that was never a candidate, and the write-up reads as if any negative value marks a maximum.
Zero Trusted: You reached $f''(c)=0$ and wrote a conclusion anyway, which means you read the table of conditions as if every row had a verdict and the third row did not.
Value Unquoted: You named the point without quoting a number.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; One released BC set names a derivative test without saying which one, and that is the only place where the second derivative is available in closed form to settle the type
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; Counted where the scored justification is the sign of $f''$ at a critical number rather than the sign change of $f'$
FRQ Years Appeared: 2026 (Q2C ii)

Practice Examples

MCQ 1

Reading Two Signs

For a function $f$ , $f'(2)=0$ and $f''(2)=-4$ . What does the second derivative test conclude at $x=2$ ?

MCQ 2

When the Test Says Nothing

For $f(x)=x^{3}$ , both $f'(0)=0$ and $f''(0)=0$ . Which statement is correct?

MCQ 3

The Fallback

At a critical number $c$ , $f''(c)=0$ , and $f'$ changes from negative to positive across $c$ . What is true at $c$ ?

FRQ 1 • Classify and Justify

Let $f(x)=x^{3}-3x^{2}-9x+5$ .

(a) Find $f'(x)$ and $f''(x)$ .

(b) Find all critical numbers of $f$ .

(c) Use the second derivative test to classify each critical number, quoting the value of $f''$ that decides.

(d) The second derivative vanishes at $x=1$ . Explain what the test says there and whether $x=1$ is an extremum.

Cheatsheet