Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
A critical point of f is an interior input where f′ is zero or where f′ does not exist. That is the whole definition, and it has exactly two doors. Endpoints are not critical points, even when they turn out to hold the absolute maximum. Locating critical points and classifying them are two separate jobs, and the released exams score them separately. Locating needs only the algebra: solve f′=0 and find where f′ fails. Classifying needs a sign pattern, because a zero derivative can mean a maximum, a minimum, or nothing at all. The most common lost point on this topic is not a wrong derivative. It is a correct derivative followed by a claim of "maximum" that was never tested.
Four of the nine released exams since 2017 contain a part that asks you to locate a critical point or to name what happens at one.
Figure 1 — The definition admits two cases and excludes a third. An endpoint can be a candidate for the absolute extremum without ever being a critical point.
c is a critical point of f⟺c is interior to the domain and (f′(c)=0 or f′(c) does not exist)
Everything on this page is a consequence of that sentence. The word "interior" removes the endpoints. The word "or" supplies the second door. Nothing in the definition mentions maxima or minima, which is why a critical point is a candidate and never a conclusion on its own.
The two cases look different on paper and they come from different kinds of curves.
| Case | What the curve is doing | How you find it |
|---|---|---|
| f′(c)=0 | the tangent is horizontal, so the curve pauses | factor f′ and solve |
| f′(c) does not exist | a corner, a cusp, or a vertical tangent | look for absolute values, roots, and piecewise joins |
| endpoint x=a or x=b | the domain simply stops | read the interval, not the derivative |
Only the first two rows are critical points.
The third row matters because the same question often asks for the absolute extremum, and endpoints are candidates there. Keeping the two lists separate in your head prevents the writing error that costs credit: calling x=a a critical point in an answer that is otherwise perfect.
When the exam gives you f′ instead of f, the location step becomes visual.
The distinction between a crossing and a touch is the single most useful thing to look for, because it decides the classification before any arithmetic happens.
Classification is a sentence with three parts: the input, the type, and the evidence. The released rubrics pay for the evidence.
| Pattern of f′ around c | Classification | Sentence that earns it |
|---|---|---|
| + then − | relative maximum | "f′ changes from positive to negative at c" |
| − then + | relative minimum | "f′ changes from negative to positive at c" |
| + then +, or − then − | neither | "f′ does not change sign at c" |
The evidence is always a sign statement, never a value.
Writing "f′(c)=0, so f has a maximum at c" is the classic version of a claim without evidence. Writing "f′(c)=0 and f′ changes from positive to negative at c, so f(c) is a relative maximum" is the same fact with the justification attached, and it is the version that scores.
Borrowed idea: A critical point is a place where the derivative stops giving an answer, because it is either zero or missing. Both cases deserve a look. The definition is deliberately narrow: interior inputs only, and exactly two ways in. Students lose this point by widening the definition rather than by failing the algebra, since they call an endpoint critical or assume that a zero must be a turn. Locate first. Then name. The naming is a separate step that carries its own point.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "find all critical points of g" | a derivative defined by an integral | g′ is the integrand | solve g′=0, then check where g′ fails |
| "does f have a relative maximum at x=6" | a graph of f′ | the sign of f′ on each side of 6 | report the pattern, then the type |
| "at which t does H have a critical point" | a differential equation and a formula | solve the derivative equation on the stated interval | keep only interior solutions |
| "how many critical points does f have" | an absolute value or a root | one input where f′ fails | count, do not classify |
| "critical point of g where g′=f" | a graph of f | zeros of f are the critical points of g | read the zeros off the graph |
① Solve f′=0 first, because it is mechanical. Then scan for the second door: absolute values, odd roots, and piecewise definitions.
② Delete every solution outside the stated interval before you classify anything. A critical point at x=7 is not a critical point of a function studied on [0,6].
③ Attach the sign pattern to every classification. One clause, either "changes from positive to negative" or "does not change sign".
Core Archetype (Simplest Form): Let f(x)=x3−3x. Find all critical points of f and classify each one.
Standard Execution:
- Derivative. f′(x)=3x2−3=3(x−1)(x+1), so f′ exists everywhere.
- Locate. f′(x)=0 at x=−1 and x=1. Both are interior inputs of the domain.
- Classify. f′>0 on (−∞,−1), f′<0 on (−1,1), f′>0 on (1,∞).
- Names. At x=−1 the sign goes + to −, so f(−1)=2 is a relative maximum. At x=1 the sign goes − to +, so f(1)=−2 is a relative minimum.
Exam Variation (Context / Parameter Shift): Let h(x)=∣x−1∣+2. Explain why x=1 is a critical point of h, and classify it.
Solution: For x>1, h′(x)=1; for x<1, h′(x)=−1. The two one-sided slopes disagree at x=1, so h′(1) does not exist. Since x=1 is an interior input of the domain, it is a critical point. The sign of h′ changes from negative to positive there, so h(1)=2 is a relative minimum.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f(x)=∣x−3∣+2. Which statement about the critical points of f is correct?
MCQ 2
The derivative of f satisfies f′(x)=0 only at x=2 and at x=5. Also f′<0 on (0,2), f′>0 on (2,5), and f′<0 on (5,7). At which value of x does f have a relative minimum?
MCQ 3
Let f(x)=x1/3. Which statement about x=0 is correct?
FRQ 4
Let f(x)=x3−3x on the closed interval [−2,2].
(a) Find all values of x in the open interval (−2,2) at which f has a critical point. Show the work that leads to your answer.
(b) Classify each critical point found in part (a) as a relative maximum, a relative minimum, or neither. Give a reason for each answer.
(c) Let g be the function defined by g(x)=f(x)+2x. Find all critical points of g in the interval (−2,2), and classify each one.
(d) Let h(x)=∣x−1∣+2. Explain why x=1 is a critical point of h even though h′(1) does not exist, and classify it.
| What to do first | What to write | |
|---|---|---|
| "find all critical points" | solve f′=0, then hunt the undefined cases | interior inputs only, listed |
| "how many critical points" | count, do not classify | one number |
| "relative maximum, minimum, or neither" | sign of f′ on both sides | pattern, then type |
| "critical point of g where g′=f" | read the zeros of f from the graph | the zeros inside the interval |
| "at which t does H have a critical point" | set the given model for H′ to zero | interior solutions only |
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
A critical point of f is an interior input where f′ is zero or where f′ does not exist. That is the whole definition, and it has exactly two doors. Endpoints are not critical points, even when they turn out to hold the absolute maximum. Locating critical points and classifying them are two separate jobs, and the released exams score them separately. Locating needs only the algebra: solve f′=0 and find where f′ fails. Classifying needs a sign pattern, because a zero derivative can mean a maximum, a minimum, or nothing at all. The most common lost point on this topic is not a wrong derivative. It is a correct derivative followed by a claim of "maximum" that was never tested.
Four of the nine released exams since 2017 contain a part that asks you to locate a critical point or to name what happens at one.
Figure 1 — The definition admits two cases and excludes a third. An endpoint can be a candidate for the absolute extremum without ever being a critical point.
c is a critical point of f⟺c is interior to the domain and (f′(c)=0 or f′(c) does not exist)
Everything on this page is a consequence of that sentence. The word "interior" removes the endpoints. The word "or" supplies the second door. Nothing in the definition mentions maxima or minima, which is why a critical point is a candidate and never a conclusion on its own.
The two cases look different on paper and they come from different kinds of curves.
| Case | What the curve is doing | How you find it |
|---|---|---|
| f′(c)=0 | the tangent is horizontal, so the curve pauses | factor f′ and solve |
| f′(c) does not exist | a corner, a cusp, or a vertical tangent | look for absolute values, roots, and piecewise joins |
| endpoint x=a or x=b | the domain simply stops | read the interval, not the derivative |
Only the first two rows are critical points.
The third row matters because the same question often asks for the absolute extremum, and endpoints are candidates there. Keeping the two lists separate in your head prevents the writing error that costs credit: calling x=a a critical point in an answer that is otherwise perfect.
When the exam gives you f′ instead of f, the location step becomes visual.
The distinction between a crossing and a touch is the single most useful thing to look for, because it decides the classification before any arithmetic happens.
Classification is a sentence with three parts: the input, the type, and the evidence. The released rubrics pay for the evidence.
| Pattern of f′ around c | Classification | Sentence that earns it |
|---|---|---|
| + then − | relative maximum | "f′ changes from positive to negative at c" |
| − then + | relative minimum | "f′ changes from negative to positive at c" |
| + then +, or − then − | neither | "f′ does not change sign at c" |
The evidence is always a sign statement, never a value.
Writing "f′(c)=0, so f has a maximum at c" is the classic version of a claim without evidence. Writing "f′(c)=0 and f′ changes from positive to negative at c, so f(c) is a relative maximum" is the same fact with the justification attached, and it is the version that scores.
Borrowed idea: A critical point is a place where the derivative stops giving an answer, because it is either zero or missing. Both cases deserve a look. The definition is deliberately narrow: interior inputs only, and exactly two ways in. Students lose this point by widening the definition rather than by failing the algebra, since they call an endpoint critical or assume that a zero must be a turn. Locate first. Then name. The naming is a separate step that carries its own point.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "find all critical points of g" | a derivative defined by an integral | g′ is the integrand | solve g′=0, then check where g′ fails |
| "does f have a relative maximum at x=6" | a graph of f′ | the sign of f′ on each side of 6 | report the pattern, then the type |
| "at which t does H have a critical point" | a differential equation and a formula | solve the derivative equation on the stated interval | keep only interior solutions |
| "how many critical points does f have" | an absolute value or a root | one input where f′ fails | count, do not classify |
| "critical point of g where g′=f" | a graph of f | zeros of f are the critical points of g | read the zeros off the graph |
① Solve f′=0 first, because it is mechanical. Then scan for the second door: absolute values, odd roots, and piecewise definitions.
② Delete every solution outside the stated interval before you classify anything. A critical point at x=7 is not a critical point of a function studied on [0,6].
③ Attach the sign pattern to every classification. One clause, either "changes from positive to negative" or "does not change sign".
Core Archetype (Simplest Form): Let f(x)=x3−3x. Find all critical points of f and classify each one.
Standard Execution:
- Derivative. f′(x)=3x2−3=3(x−1)(x+1), so f′ exists everywhere.
- Locate. f′(x)=0 at x=−1 and x=1. Both are interior inputs of the domain.
- Classify. f′>0 on (−∞,−1), f′<0 on (−1,1), f′>0 on (1,∞).
- Names. At x=−1 the sign goes + to −, so f(−1)=2 is a relative maximum. At x=1 the sign goes − to +, so f(1)=−2 is a relative minimum.
Exam Variation (Context / Parameter Shift): Let h(x)=∣x−1∣+2. Explain why x=1 is a critical point of h, and classify it.
Solution: For x>1, h′(x)=1; for x<1, h′(x)=−1. The two one-sided slopes disagree at x=1, so h′(1) does not exist. Since x=1 is an interior input of the domain, it is a critical point. The sign of h′ changes from negative to positive there, so h(1)=2 is a relative minimum.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f(x)=∣x−3∣+2. Which statement about the critical points of f is correct?
MCQ 2
The derivative of f satisfies f′(x)=0 only at x=2 and at x=5. Also f′<0 on (0,2), f′>0 on (2,5), and f′<0 on (5,7). At which value of x does f have a relative minimum?
MCQ 3
Let f(x)=x1/3. Which statement about x=0 is correct?
FRQ 4
Let f(x)=x3−3x on the closed interval [−2,2].
(a) Find all values of x in the open interval (−2,2) at which f has a critical point. Show the work that leads to your answer.
(b) Classify each critical point found in part (a) as a relative maximum, a relative minimum, or neither. Give a reason for each answer.
(c) Let g be the function defined by g(x)=f(x)+2x. Find all critical points of g in the interval (−2,2), and classify each one.
(d) Let h(x)=∣x−1∣+2. Explain why x=1 is a critical point of h even though h′(1) does not exist, and classify it.
| What to do first | What to write | |
|---|---|---|
| "find all critical points" | solve f′=0, then hunt the undefined cases | interior inputs only, listed |
| "how many critical points" | count, do not classify | one number |
| "relative maximum, minimum, or neither" | sign of f′ on both sides | pattern, then type |
| "critical point of g where g′=f" | read the zeros of f from the graph | the zeros inside the interval |
| "at which t does H have a critical point" | set the given model for H′ to zero | interior solutions only |
Strengthen this topic with AP-style practice, or continue to next topic.