Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
This page turns the definition of concavity into the answer format the exam scores. The answer is always an interval, and the evidence is always the sign of f′′ on that interval. An inflection point is a point where the concavity actually changes, so a zero of f′′ is only a candidate until the signs on both sides disagree. When the derivative is supplied as a graph, the concavity of f is read from the slope of that graph, and the inflections of f sit at the extrema of f′. The two sentences that earn the points are "concave down on (a,b) because f′′<0 there" and "a point of inflection at x=c because the concavity changes there".
Figure 15 — One curve, one change of bend, and the sign strip underneath. The left half bends down, the right half bends up, and the inflection is the single point where the sign flips.
Concavity is reported on intervals, and the endpoints of those intervals come from the zeros of f′′ together with the points where f′′ fails to exist. A single point is never concave; it either has a second derivative or it does not.
| What you found | What you still need | Verdict |
|---|---|---|
| f′′(c)=0 | the signs on both sides | inflection only if they differ |
| f′′(c)=0 and the signs match | nothing more | not an inflection |
| f′′ undefined at c | the signs on both sides | inflection if they differ |
| f′(c)=0 and f′′(c)=0 | the signs on both sides | a stationary inflection is possible |
The stationary case is the one that trips students: f(x)=x3 has f′(0)=f′′(0)=0 and a genuine inflection at the origin, while f(x)=x4 has the same two zeros and no inflection at all.
If the graph of f′ is supplied, the concavity of f follows the slope of f′. Where f′ is increasing, f is concave up, and where f′ is decreasing, f is concave down. The inflections of f are therefore the extrema of f′, which is why a question about inflection points can be answered by locating peaks and valleys on a graph of the derivative.
With a table of f′ values, increasing or decreasing entries decide the concavity, and a change of direction between consecutive entries locates an inflection between them. With a table of f values, the second difference plays the same role, and its sign is the discrete version of f′′.
The scored sentence combines three pieces: the interval, the sign, and the name. Writing only the interval drops the reason, and writing only the reason drops the answer. A sign chart is the cheapest way to have all three ready, since each row of the chart is one finished sentence.
Borrowed idea: One chart, three columns. Build f′′, find its zeros, and test one number in each interval between them. Two signs decide everything: the sign names the bend, and a change of sign names an inflection. Nothing else is needed. So why do so many answers lose the point? Because the interval is written without the sign, or the sign is written without the interval, and the guideline scores the pair.
| Zone in the chart | What to record | Sentence it produces |
|---|---|---|
| left of the smallest zero | the sign of f′′ there | "concave down on (−∞,c)" |
| between two zeros | the sign of f′′ there | "concave up on (c,d)" |
| at a zero | whether the sign changed | "an inflection at x=c" |
① Test one number per interval. Two test numbers in the same interval add nothing, and a missed interval loses a point.
② Name the interval before the sign. "Concave up on (0,3) because f′′>0 there" is the full shape of a scoring sentence.
③ Check the zeros of f′′ in order. The intervals you report must tile the domain, or the answer has a gap the reader will find.
Core Archetype (Simplest Form): Find the intervals of concavity and the points of inflection of f(x)=x4−4x3.
Standard Execution:
- Differentiate twice. f′(x)=4x3−12x2 and f′′(x)=12x2−24x=12x(x−2).
- Zeros. x=0 and x=2.
- Signs. f′′(−1)=36>0, f′′(1)=−12<0 and f′′(3)=36>0.
- Translate. Concave up on (−∞,0) and (2,∞), concave down on (0,2), with inflections at x=0 and x=2 since the sign changes at both.
- Coordinates. f(0)=0 and f(2)=−16, so the inflection points are (0,0) and (2,−16).
Exam Variation (Context / Parameter Shift): The graph of f′ is supplied on (−5,6), and its graph is decreasing on (−5,1) and increasing on (1,6) with a minimum at x=1. Where is the graph of f concave up, and where are its points of inflection?
Solution: The concavity of f follows the slope of f′, so f is concave down on (−5,1) and concave up on (1,6). The single point of inflection is at x=1, because that is where f′ stops decreasing and starts increasing.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
On which intervals is f(x)=x3−6x2+5 concave up?
MCQ 2
A sign chart shows f′′>0 on (−3,1), f′′<0 on (1,5), and f′′>0 on (5,∞). How many points of inflection does f have?
MCQ 3
The graph of f′ has a local maximum at x=4. What happens to the graph of f at x=4?
Let f(x)=x4−4x3.
(a) Find f′(x) and f′′(x).
(b) Determine the open intervals on which the graph of f is concave up and those on which it is concave down. Justify your answer.
(c) Find the x-coordinates of all points of inflection of f, and give the coordinates of each.
(d) A student writes that x=0 cannot be an inflection point because f′(0)=0. Explain why that reasoning fails.
| What decides | What to write | |
|---|---|---|
| where is f concave up | f′′>0 there | "concave up on (a,b) because f′′>0" |
| where is f concave down | f′′<0 there | "concave down on (b,c) because f′′<0" |
| where are the inflections | sign of f′′ changes | "a point of inflection at x=c because the concavity changes" |
| what if only f′ is graphed | slope of f′ | increasing f′ means concave up |
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
This page turns the definition of concavity into the answer format the exam scores. The answer is always an interval, and the evidence is always the sign of f′′ on that interval. An inflection point is a point where the concavity actually changes, so a zero of f′′ is only a candidate until the signs on both sides disagree. When the derivative is supplied as a graph, the concavity of f is read from the slope of that graph, and the inflections of f sit at the extrema of f′. The two sentences that earn the points are "concave down on (a,b) because f′′<0 there" and "a point of inflection at x=c because the concavity changes there".
Figure 15 — One curve, one change of bend, and the sign strip underneath. The left half bends down, the right half bends up, and the inflection is the single point where the sign flips.
Concavity is reported on intervals, and the endpoints of those intervals come from the zeros of f′′ together with the points where f′′ fails to exist. A single point is never concave; it either has a second derivative or it does not.
| What you found | What you still need | Verdict |
|---|---|---|
| f′′(c)=0 | the signs on both sides | inflection only if they differ |
| f′′(c)=0 and the signs match | nothing more | not an inflection |
| f′′ undefined at c | the signs on both sides | inflection if they differ |
| f′(c)=0 and f′′(c)=0 | the signs on both sides | a stationary inflection is possible |
The stationary case is the one that trips students: f(x)=x3 has f′(0)=f′′(0)=0 and a genuine inflection at the origin, while f(x)=x4 has the same two zeros and no inflection at all.
If the graph of f′ is supplied, the concavity of f follows the slope of f′. Where f′ is increasing, f is concave up, and where f′ is decreasing, f is concave down. The inflections of f are therefore the extrema of f′, which is why a question about inflection points can be answered by locating peaks and valleys on a graph of the derivative.
With a table of f′ values, increasing or decreasing entries decide the concavity, and a change of direction between consecutive entries locates an inflection between them. With a table of f values, the second difference plays the same role, and its sign is the discrete version of f′′.
The scored sentence combines three pieces: the interval, the sign, and the name. Writing only the interval drops the reason, and writing only the reason drops the answer. A sign chart is the cheapest way to have all three ready, since each row of the chart is one finished sentence.
Borrowed idea: One chart, three columns. Build f′′, find its zeros, and test one number in each interval between them. Two signs decide everything: the sign names the bend, and a change of sign names an inflection. Nothing else is needed. So why do so many answers lose the point? Because the interval is written without the sign, or the sign is written without the interval, and the guideline scores the pair.
| Zone in the chart | What to record | Sentence it produces |
|---|---|---|
| left of the smallest zero | the sign of f′′ there | "concave down on (−∞,c)" |
| between two zeros | the sign of f′′ there | "concave up on (c,d)" |
| at a zero | whether the sign changed | "an inflection at x=c" |
① Test one number per interval. Two test numbers in the same interval add nothing, and a missed interval loses a point.
② Name the interval before the sign. "Concave up on (0,3) because f′′>0 there" is the full shape of a scoring sentence.
③ Check the zeros of f′′ in order. The intervals you report must tile the domain, or the answer has a gap the reader will find.
Core Archetype (Simplest Form): Find the intervals of concavity and the points of inflection of f(x)=x4−4x3.
Standard Execution:
- Differentiate twice. f′(x)=4x3−12x2 and f′′(x)=12x2−24x=12x(x−2).
- Zeros. x=0 and x=2.
- Signs. f′′(−1)=36>0, f′′(1)=−12<0 and f′′(3)=36>0.
- Translate. Concave up on (−∞,0) and (2,∞), concave down on (0,2), with inflections at x=0 and x=2 since the sign changes at both.
- Coordinates. f(0)=0 and f(2)=−16, so the inflection points are (0,0) and (2,−16).
Exam Variation (Context / Parameter Shift): The graph of f′ is supplied on (−5,6), and its graph is decreasing on (−5,1) and increasing on (1,6) with a minimum at x=1. Where is the graph of f concave up, and where are its points of inflection?
Solution: The concavity of f follows the slope of f′, so f is concave down on (−5,1) and concave up on (1,6). The single point of inflection is at x=1, because that is where f′ stops decreasing and starts increasing.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
On which intervals is f(x)=x3−6x2+5 concave up?
MCQ 2
A sign chart shows f′′>0 on (−3,1), f′′<0 on (1,5), and f′′>0 on (5,∞). How many points of inflection does f have?
MCQ 3
The graph of f′ has a local maximum at x=4. What happens to the graph of f at x=4?
Let f(x)=x4−4x3.
(a) Find f′(x) and f′′(x).
(b) Determine the open intervals on which the graph of f is concave up and those on which it is concave down. Justify your answer.
(c) Find the x-coordinates of all points of inflection of f, and give the coordinates of each.
(d) A student writes that x=0 cannot be an inflection point because f′(0)=0. Explain why that reasoning fails.
| What decides | What to write | |
|---|---|---|
| where is f concave up | f′′>0 there | "concave up on (a,b) because f′′>0" |
| where is f concave down | f′′<0 there | "concave down on (b,c) because f′′<0" |
| where are the inflections | sign of f′′ changes | "a point of inflection at x=c because the concavity changes" |
| what if only f′ is graphed | slope of f′ | increasing f′ means concave up |
Strengthen this topic with AP-style practice, or continue to next topic.