A closed interval [a, b] is the set of all real numbers from a to b, including both endpoints, written with square brackets. In AP Calculus, it's where you confirm continuity (Topic 1.12) and check the hypotheses of theorems like the Mean Value Theorem (Topic 5.1).
A closed interval is a set of real numbers that includes both of its endpoints. You write it with square brackets, so [a, b] means every x with a ≤ x ≤ b, endpoints included. Compare that to the open interval (a, b), which uses parentheses and leaves the endpoints out.
In AP Calculus, the closed interval isn't just notation. It changes what "continuous" means at the edges. A function is continuous on [a, b] if it's continuous at every point inside the interval, and at the endpoints it only needs one-sided continuity. At x = a, the limit from the right has to equal f(a); at x = b, the limit from the left has to equal f(b). That's why f(x) = √(x + 4) is continuous on [-4, 6] even though it doesn't exist to the left of -4. The endpoint only asks for a one-sided limit, and it delivers.
Closed intervals show up in three CED learning objectives. In Topic 1.12 (LO 1.12.A), you determine where a function is continuous, and EK LIM-2.B.1 says continuity on an interval means continuity at each point in it, with one-sided limits handling closed endpoints. In Topic 2.4 (LO 2.4.A), the closed-vs-open distinction matters because differentiability needs a two-sided limit of the difference quotient, which endpoints can't provide. And in Topic 5.1 (LO 5.1.A), the Mean Value Theorem's hypotheses are written precisely in interval language. Per EK FUN-1.B.1, f must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). That mismatch isn't a typo. Continuity can be checked one-sided at endpoints, but differentiability can't, so the theorem demands less at the edges. If you can read that distinction, you can earn justification points that students who treat [a, b] and (a, b) as interchangeable lose.
Keep studying AP Calculus Unit 5
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view galleryOpen Interval (Units 1, 2 & 5)
The open interval (a, b) is the closed interval with the endpoints chopped off. The famous pairing is in the Mean Value Theorem, which requires continuity on the closed [a, b] but differentiability only on the open (a, b). The MVT's guaranteed point c also lives strictly inside the open interval, never at an endpoint.
Continuity over an Interval (Unit 1)
Topic 1.12 is where the closed interval earns its keep. Confirming continuity on [a, b] means a two-sided limit test at every interior point but only a one-sided limit test at each endpoint. That's why a square root function can be continuous on a closed interval that starts exactly where its domain starts.
Mean Value Theorem and Rolle's Theorem (Unit 5)
Both theorems are stated over a closed interval. MVT guarantees some c in (a, b) where f'(c) equals the average rate of change over [a, b], and Rolle's is the special case where f(a) = f(b). FRQ justifications live or die on stating the closed-interval continuity hypothesis correctly.
Differentiability and Continuity (Unit 2)
Differentiability at a point requires a two-sided limit of the difference quotient, which is exactly what an endpoint of a closed interval can't give you. This is why FRQs say a function is "differentiable on the closed interval [-6, 5]" carefully, and why theorem hypotheses use the open interval for differentiability.
Closed interval language appears constantly in FRQ stems. The 2017 FRQ Q3 starts with "the function f is differentiable on the closed interval [−6, 5]," the 2019 FRQ Q3 defines a continuous f on the closed interval −6 ≤ x ≤ 5, and the 2021 FRQ Q4 defines f on [−4, 6] and builds an accumulation function from it. Your job is to use that given information as theorem fuel. If f is continuous on [a, b] and differentiable on (a, b), you can invoke the MVT; if f is continuous on a closed interval, you can invoke the Intermediate Value Theorem or the Candidates Test for absolute extrema. In multiple choice, expect questions like "Is f(x) = √(x + 4) continuous over [-4, 6]?" or "Which condition is required to confirm continuity over a closed interval [a, b]?" The trap answer always involves forgetting that endpoints only need one-sided limits.
A closed interval [a, b] includes both endpoints; an open interval (a, b) excludes them. The difference matters most in theorem hypotheses. The MVT requires continuity on the CLOSED interval but differentiability only on the OPEN interval, because continuity can be checked one-sided at an endpoint while differentiability needs a full two-sided limit. Writing "differentiable on [a, b]" when the problem only gives you (a, b) is a classic justification error.
A closed interval [a, b] includes both endpoints and uses square brackets, while an open interval (a, b) excludes them and uses parentheses.
Continuity on a closed interval means continuity at every interior point plus one-sided continuity at each endpoint, so the limit from inside the interval must equal the function value there.
The Mean Value Theorem requires continuity on the closed interval [a, b] but only differentiability on the open interval (a, b), and you must state both hypotheses to earn justification points.
Polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous at every point in their domains, so checking continuity on a closed interval often reduces to checking the interval sits inside the domain.
Endpoints of a closed interval can be continuous without being differentiable, because differentiability demands a two-sided limit that an endpoint can't provide.
FRQ stems regularly hand you a closed interval (like [−6, 5] in 2017 and 2019), and that phrasing is your green light to apply theorems like MVT, IVT, or the Candidates Test.
A closed interval [a, b] is the set of all real numbers x with a ≤ x ≤ b, including both endpoints a and b. It's written with square brackets and is the standard setting for continuity checks (Topic 1.12) and theorem hypotheses like the MVT (Topic 5.1).
Both, for different things. The MVT requires f to be continuous on the closed interval [a, b] and differentiable on the open interval (a, b), and the guaranteed point c falls strictly inside the open interval.
Yes. Continuity at endpoints only needs a one-sided limit, but differentiability needs a two-sided limit of the difference quotient. Also, functions like f(x) = |x| are continuous everywhere but not differentiable at x = 0 because the left and right limits of the difference quotient disagree.
Yes. At the endpoint x = -4, you only need the limit from the right to equal f(-4) = 0, and it does. Every point in [-4, 6] is in the domain of the function, so it's continuous on the whole closed interval.
Square brackets [a, b] mean the endpoints a and b are included (closed interval); parentheses (a, b) mean they're excluded (open interval). On the exam, mixing these up in an MVT justification is one of the most common ways to lose a reasoning point.