Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
The second derivative is the derivative of the slope, so it describes how the curve bends rather than how high it sits. When f′′>0 the slope is increasing, the curve is concave up, and every tangent line lies below the graph. When f′′<0 the slope is decreasing, the curve is concave down, and every tangent line lies above it. That one idea answers three exam questions: the intervals of concavity, the location of inflection points, and the type of extremum at a critical point. The last of these is the second derivative test, and it has a failure mode. When f′′(c)=0 the test says nothing, and a sign chart has to decide instead.
Four of the nine released exams since 2017 contain a part that turns on the sign of f′′, and five separate sub-parts are involved because the 2023 exam asked the question twice in the same paper.
Figure 1 — The sign of f′′ is the sign of the slope's direction. It names the bend and it decides which side of the curve the tangent sits on.
f′′(x)>0 on an interval⟺f′ is increasing there⟺f is concave up there
The chain of equivalences is the reason this topic feels easy once it clicks and confusing before. Nothing here is a new object. The second derivative is a first derivative wearing a different hat, applied to a function you already know how to read. Everything you learned about the sign of f′ applies again one level up.
Three equivalent descriptions of the same geometry, in the order students usually meet them.
| Description | What it says | How to test it |
|---|---|---|
| tangent test | the tangent lies below the curve | f′′>0 on the interval |
| slope test | the slope is increasing as you move right | f′ is increasing there |
| chord test | the curve sags below its own chords | f is concave up there |
Any one of the three is a valid reason on the exam.
The tangent test is the one that appears in textbooks, the slope test is the one that explains why the second derivative is involved at all, and the chord test is the one that survives when the function is only described by a table. When a released rubric says "concave up because f′ is increasing", it is accepting the middle row.
At a critical point c where f′(c)=0, the sign of f′′(c) decides the type.
f′′(c)>0⟹f(c) is a relative minimum
f′′(c)<0⟹f(c) is a relative maximum
f′′(c)=0⟹the test is inconclusive
The first two cases are quick and they appear often on multiple-choice questions, because a single number answers them. The third case is the one that matters on free-response: the test is silent, not negative, and an answer that treats "inconclusive" as "neither" is wrong for the right reason and wrong for the wrong one at the same time.
f′′(c)=0 at a critical point means the slope is momentarily not changing. That is compatible with all three outcomes.
The fallback is always the first derivative test. When f′′(c)=0, write the sign chart for f′ and let the pattern decide. The rubric accepts this substitution without penalty, and it is the reason the two tests are taught as a pair rather than as alternatives.
Borrowed idea: The second derivative is the derivative of the slope, so it reports on the slope's direction rather than on the curve's height. When the slope is increasing, the curve bends upward and every tangent line sits below it. When the slope is decreasing, the curve bends downward and every tangent line sits above it. That single idea produces three separate exam answers: where a function is concave up, where it is concave down, and what type of extremum sits at a critical point. The first two come from the sign of f′′. The third comes from the sign of f′′ evaluated at a point. Then the test fails.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "on what intervals is the graph concave down" | a formula for f | sign chart of f′′ | factor f′′, test intervals, write open intervals |
| "increasing and concave down" | two conditions in one sentence | sign of f′ and sign of f′′ | intersect the two interval lists |
| "classify the critical point" | a critical point and a formula | the value f′′(c) | evaluate, then name the type |
| "is the approximation an overestimate" | a tangent line used to estimate | concavity at the point of tangency | state the sign of f′′, then the direction |
| "does the graph have an inflection point at x=k" | a candidate input | sign of f′′ on both sides | report the change, not the zero |
① Compute f′′ and factor it. An unfactored second derivative hides both its zeros and their multiplicities, and the multiplicity is what decides whether the concavity actually changes.
② Draw the sign chart before writing any sentence. Concavity questions are graded on intervals, and intervals come straight off the chart.
③ When a critical point is involved, evaluate rather than reason. One number, f′′(c), decides the type whenever that number is nonzero.
Core Archetype (Simplest Form): Let f(x)=x3−3x2. Find the intervals of concavity and classify the critical points using the second derivative test.
Standard Execution:
- First derivative. f′(x)=3x2−6x=3x(x−2), so the critical points are x=0 and x=2.
- Second derivative. f′′(x)=6x−6=6(x−1), which is zero at x=1.
- Concavity. f′′<0 on (−∞,1) and f′′>0 on (1,∞), so the graph is concave down then concave up, with an inflection point at x=1.
- Types. f′′(0)=−6<0, so f(0)=0 is a relative maximum. f′′(2)=6>0, so f(2)=−4 is a relative minimum.
Exam Variation (Context / Parameter Shift): Let f(x)=x4−4x3. Find all critical points of f, classify each one, and determine where the second derivative test is inconclusive.
Solution: f′(x)=4x3−12x2=4x2(x−3), so the critical points are x=0 and x=3. Also f′′(x)=12x2−24x=12x(x−2), so f′′(3)=36>0 and f(3)=−27 is a relative minimum. At x=0 we get f′′(0)=0, so the second derivative test is inconclusive there; the sign chart of f′ shows that f′<0 on both (0,3) and (−1,0), so f′ does not change sign at x=0 and there is neither a maximum nor a minimum at that input.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f(x)=x3−3x2. At the critical point x=2, the function f has
MCQ 2
How many inflection points does the graph of f(x)=x4 have?
MCQ 3
The graph of a differentiable function f is concave down on the interval (2,5). Which statement must be true on that interval?
FRQ 4
Let f be the function defined by f(x)=x4−4x3.
(a) Find f′(x) and find all critical points of f.
(b) Classify each critical point. If the second derivative test is inconclusive at any of them, say so and use another test.
(c) Find the open intervals on which the graph of f is concave up and the open intervals on which it is concave down. Give a reason for each answer.
(d) A student claims that because f′′(0)=0, the graph of f must have an inflection point at x=0. Explain whether the claim's reasoning is valid, and state what actually decides the question.
| What to compute | What it tells you | |
|---|---|---|
| "on what intervals is the graph concave up" | sign chart of f′′ | intervals where f′′>0 |
| "point of inflection" | zeros of f′′ plus a sign check | the inputs where concavity changes |
| "classify the critical point" | f′′ at that input | minimum if positive, maximum if negative |
| "is the tangent approximation an overestimate" | f′′ at the point of tangency | concave up gives an underestimate |
| "increasing and concave up" | sign of f′ and of f′′ | the overlap of both interval lists |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
The second derivative is the derivative of the slope, so it describes how the curve bends rather than how high it sits. When f′′>0 the slope is increasing, the curve is concave up, and every tangent line lies below the graph. When f′′<0 the slope is decreasing, the curve is concave down, and every tangent line lies above it. That one idea answers three exam questions: the intervals of concavity, the location of inflection points, and the type of extremum at a critical point. The last of these is the second derivative test, and it has a failure mode. When f′′(c)=0 the test says nothing, and a sign chart has to decide instead.
Four of the nine released exams since 2017 contain a part that turns on the sign of f′′, and five separate sub-parts are involved because the 2023 exam asked the question twice in the same paper.
Figure 1 — The sign of f′′ is the sign of the slope's direction. It names the bend and it decides which side of the curve the tangent sits on.
f′′(x)>0 on an interval⟺f′ is increasing there⟺f is concave up there
The chain of equivalences is the reason this topic feels easy once it clicks and confusing before. Nothing here is a new object. The second derivative is a first derivative wearing a different hat, applied to a function you already know how to read. Everything you learned about the sign of f′ applies again one level up.
Three equivalent descriptions of the same geometry, in the order students usually meet them.
| Description | What it says | How to test it |
|---|---|---|
| tangent test | the tangent lies below the curve | f′′>0 on the interval |
| slope test | the slope is increasing as you move right | f′ is increasing there |
| chord test | the curve sags below its own chords | f is concave up there |
Any one of the three is a valid reason on the exam.
The tangent test is the one that appears in textbooks, the slope test is the one that explains why the second derivative is involved at all, and the chord test is the one that survives when the function is only described by a table. When a released rubric says "concave up because f′ is increasing", it is accepting the middle row.
At a critical point c where f′(c)=0, the sign of f′′(c) decides the type.
f′′(c)>0⟹f(c) is a relative minimum
f′′(c)<0⟹f(c) is a relative maximum
f′′(c)=0⟹the test is inconclusive
The first two cases are quick and they appear often on multiple-choice questions, because a single number answers them. The third case is the one that matters on free-response: the test is silent, not negative, and an answer that treats "inconclusive" as "neither" is wrong for the right reason and wrong for the wrong one at the same time.
f′′(c)=0 at a critical point means the slope is momentarily not changing. That is compatible with all three outcomes.
The fallback is always the first derivative test. When f′′(c)=0, write the sign chart for f′ and let the pattern decide. The rubric accepts this substitution without penalty, and it is the reason the two tests are taught as a pair rather than as alternatives.
Borrowed idea: The second derivative is the derivative of the slope, so it reports on the slope's direction rather than on the curve's height. When the slope is increasing, the curve bends upward and every tangent line sits below it. When the slope is decreasing, the curve bends downward and every tangent line sits above it. That single idea produces three separate exam answers: where a function is concave up, where it is concave down, and what type of extremum sits at a critical point. The first two come from the sign of f′′. The third comes from the sign of f′′ evaluated at a point. Then the test fails.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "on what intervals is the graph concave down" | a formula for f | sign chart of f′′ | factor f′′, test intervals, write open intervals |
| "increasing and concave down" | two conditions in one sentence | sign of f′ and sign of f′′ | intersect the two interval lists |
| "classify the critical point" | a critical point and a formula | the value f′′(c) | evaluate, then name the type |
| "is the approximation an overestimate" | a tangent line used to estimate | concavity at the point of tangency | state the sign of f′′, then the direction |
| "does the graph have an inflection point at x=k" | a candidate input | sign of f′′ on both sides | report the change, not the zero |
① Compute f′′ and factor it. An unfactored second derivative hides both its zeros and their multiplicities, and the multiplicity is what decides whether the concavity actually changes.
② Draw the sign chart before writing any sentence. Concavity questions are graded on intervals, and intervals come straight off the chart.
③ When a critical point is involved, evaluate rather than reason. One number, f′′(c), decides the type whenever that number is nonzero.
Core Archetype (Simplest Form): Let f(x)=x3−3x2. Find the intervals of concavity and classify the critical points using the second derivative test.
Standard Execution:
- First derivative. f′(x)=3x2−6x=3x(x−2), so the critical points are x=0 and x=2.
- Second derivative. f′′(x)=6x−6=6(x−1), which is zero at x=1.
- Concavity. f′′<0 on (−∞,1) and f′′>0 on (1,∞), so the graph is concave down then concave up, with an inflection point at x=1.
- Types. f′′(0)=−6<0, so f(0)=0 is a relative maximum. f′′(2)=6>0, so f(2)=−4 is a relative minimum.
Exam Variation (Context / Parameter Shift): Let f(x)=x4−4x3. Find all critical points of f, classify each one, and determine where the second derivative test is inconclusive.
Solution: f′(x)=4x3−12x2=4x2(x−3), so the critical points are x=0 and x=3. Also f′′(x)=12x2−24x=12x(x−2), so f′′(3)=36>0 and f(3)=−27 is a relative minimum. At x=0 we get f′′(0)=0, so the second derivative test is inconclusive there; the sign chart of f′ shows that f′<0 on both (0,3) and (−1,0), so f′ does not change sign at x=0 and there is neither a maximum nor a minimum at that input.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f(x)=x3−3x2. At the critical point x=2, the function f has
MCQ 2
How many inflection points does the graph of f(x)=x4 have?
MCQ 3
The graph of a differentiable function f is concave down on the interval (2,5). Which statement must be true on that interval?
FRQ 4
Let f be the function defined by f(x)=x4−4x3.
(a) Find f′(x) and find all critical points of f.
(b) Classify each critical point. If the second derivative test is inconclusive at any of them, say so and use another test.
(c) Find the open intervals on which the graph of f is concave up and the open intervals on which it is concave down. Give a reason for each answer.
(d) A student claims that because f′′(0)=0, the graph of f must have an inflection point at x=0. Explain whether the claim's reasoning is valid, and state what actually decides the question.
| What to compute | What it tells you | |
|---|---|---|
| "on what intervals is the graph concave up" | sign chart of f′′ | intervals where f′′>0 |
| "point of inflection" | zeros of f′′ plus a sign check | the inputs where concavity changes |
| "classify the critical point" | f′′ at that input | minimum if positive, maximum if negative |
| "is the tangent approximation an overestimate" | f′′ at the point of tangency | concave up gives an underestimate |
| "increasing and concave up" | sign of f′ and of f′′ | the overlap of both interval lists |
Strengthen this topic with AP-style practice, or continue to next topic.