A critical point of a function f is a point in the domain of f where the first derivative equals zero or fails to exist (FUN-1.C.2). All local (relative) extrema occur at critical points, but not every critical point is an extremum, which is why AP Calc requires a follow-up test to classify them.
A critical point is an x-value in the domain of f where f′(x) = 0 or f′(x) does not exist. That's the entire CED definition (FUN-1.C.2), and the exam takes it literally. A horizontal tangent gives you f′ = 0. A corner, cusp, or vertical tangent gives you f′ undefined. Both count.
Here's why critical points matter so much in Unit 5. The CED states that all local (relative) extrema occur at critical points, though not all critical points are local extrema (FUN-1.C.3). In plain terms, critical points are your suspect list. Every max and min is hiding at one of them, but some suspects turn out to be innocent. The classic innocent suspect is f(x) = x³ at x = 0, where f′ = 0 but the function just flattens out and keeps climbing. Finding critical points is step one. Classifying them with the First or Second Derivative Test is step two, and the AP exam grades you on both.
Critical points live in Unit 5 (Analytical Applications of Differentiation) and show up in five separate topics. Topic 5.2 defines them alongside the Extreme Value Theorem (learning objective 5.2.A). Topics 5.4 and 5.7 use them as the input to the First and Second Derivative Tests (5.4.A, 5.7.A). Topic 5.5 makes them half of the Candidates Test, since absolute extrema on a closed interval can only occur at critical points or endpoints (5.5.A). Topic 5.11 turns them into answers for optimization problems (5.11.A). Almost every justification you write in Unit 5 starts with the sentence "f has a critical point at x = c because f′(c) = 0." If you can't find and classify critical points cleanly, the entire unit falls apart.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryFirst Derivative Test (Unit 5)
Critical points tell you where to look; the First Derivative Test tells you what you found. If f′ changes from positive to negative at a critical point, it's a local max. Negative to positive means local min. No sign change means no extremum at all.
Candidates Test (Unit 5)
On a closed interval, the absolute max and min can only live at critical points or endpoints. The Candidates Test just evaluates f at every one of those candidates and picks the biggest and smallest values. Critical points are literally the "candidates" in the name.
Inflection Point (Unit 5)
These run on different derivatives. Critical points come from f′ and flag possible extrema. Inflection points come from f″ changing sign and flag changes in concavity. A point can be both, either, or neither.
Extreme Value Theorem (Unit 5)
EVT guarantees a continuous function on [a, b] actually has an absolute max and min (FUN-1.C.1). Critical points tell you where to hunt for them. EVT is the promise; critical points are the map.
Multiple-choice questions test the definition directly with stems like "How many critical points does f(x) = 2x³ - 9x² + 12x - 5 have?" You take the derivative, set it equal to zero, factor (here f′ = 6(x-1)(x-2), so two critical points), and remember to check where f′ is undefined. FRQs lean on critical points constantly, especially graph-of-the-derivative problems like 2022 FRQ Q3, where you're given the graph of f′ and asked about extrema of f. There, critical points are where the f′ graph crosses zero, and you justify a max or min by describing the sign change of f′. The scoring rubrics reward precise justification language. "f has a relative maximum at x = c because f′ changes from positive to negative at x = c" earns the point; "the graph goes up then down" usually doesn't.
Critical points are about the FIRST derivative (f′ = 0 or undefined) and flag possible maxes and mins. Inflection points are about the SECOND derivative changing sign and flag a change in concavity. They answer different questions. A function can have a critical point with no inflection there, an inflection point with no critical point there, or occasionally both at the same x-value, like f(x) = x³ at x = 0.
A critical point is an x-value in the domain of f where f′(x) = 0 or f′(x) does not exist.
Every local extremum occurs at a critical point, but not every critical point is a local extremum, so you always need a test (First Derivative Test or Second Derivative Test) to classify it.
Don't forget the "f′ undefined" case; corners, cusps, and vertical tangents are critical points too, and MCQs love to hide them.
Absolute extrema on a closed interval can only occur at critical points or endpoints, which is exactly what the Candidates Test checks.
If a continuous function has only one critical point on an interval and it's a local extremum, it's automatically the absolute extremum on that interval (a shortcut the CED explicitly allows in optimization problems).
On FRQs, justify with derivative behavior, like "f′ changes from positive to negative at x = c," not with vague descriptions of the graph's shape.
A critical point is a point in the domain of a function where the first derivative equals zero or fails to exist. It's the official CED definition (FUN-1.C.2), and it's where all local maxes and mins must occur.
No. The CED is explicit that not all critical points are local extrema (FUN-1.C.3). f(x) = x³ has a critical point at x = 0 with no max or min there, because f′ never changes sign. That's why you always follow up with the First or Second Derivative Test.
Critical points come from f′ being zero or undefined and mark possible extrema. Inflection points come from f″ changing sign and mark where concavity flips. Same function, different derivatives, different information.
Yes, as long as the point is in the domain of f. A corner like |x| at x = 0 or a cusp has no derivative there, but it's still a critical point and can absolutely be a max or min. This is a favorite trap on multiple choice.
Take the derivative, set it equal to zero, and solve. Then also check where the derivative is undefined within the domain. For f(x) = 2x³ - 9x² + 12x - 5, f′(x) = 6x² - 18x + 12 = 6(x-1)(x-2), giving critical points at x = 1 and x = 2.