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What does it mean for a function to be continuous over an entire interval, and which familiar functions qualify?

Topic 1.12 Confirming Continuity over an Interval: determine intervals on which a function is continuous, using one-sided continuity at endpoints and the continuity of standard function families.

A focused answer to AP Calculus AB Topic 1.12, defining continuity over open and closed intervals, the continuity of polynomial, rational, root, trig, exponential and log families, and one-sided continuity at endpoints.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. From a point to an interval
  3. Continuity of the standard families
  4. Finding the interval of continuity
  5. Building continuous functions from continuous pieces
  6. Why this matters downstream

What this topic is asking

The College Board (Topic 1.12) extends continuity from a single point to an interval. You should know that a function is continuous on an interval when it is continuous at every point inside it (with one-sided continuity at any included endpoint), and that the standard function families are continuous on their entire domains.

From a point to an interval

At an endpoint of a closed interval there is only one side to approach from, so only the appropriate one-sided continuity is required. For instance, x\sqrt{x} is continuous on [0,∞)[0, \infty) because it is right-continuous at 00.

Continuity of the standard families

Finding the interval of continuity

The practical method is: identify the domain of the function, then exclude any points of discontinuity (denominator zeros, jumps in a piecewise definition, or arguments outside a root or log domain). What remains is the set of intervals on which the function is continuous.

Building continuous functions from continuous pieces

A practical reason the standard families matter is that continuity is preserved under the operations you use to build functions. If ff and gg are continuous on an interval, then so are f+gf + g, fβˆ’gf - g, fgfg, and fg\frac{f}{g} (wherever gβ‰ 0g \neq 0), and the composition f(g(x))f(g(x)) is continuous wherever the inside and outside are. This closure means that almost any function written with a single formula - a polynomial over a polynomial, a root of a polynomial, a product of a trig function and an exponential - is automatically continuous on its domain, and you can justify it in one sentence by naming the families and the operations involved. The only places to watch are the domain restrictions (denominator zeros, negative radicands of even roots, non-positive logarithm arguments) and the joins of a piecewise definition, where you must check continuity by hand.

Why this matters downstream

Continuity over an interval is the hypothesis for the big theorems of the course. The Intermediate Value Theorem (Topic 1.16) requires continuity on a closed interval, and later the Extreme Value Theorem and the Mean Value Theorem do too. Knowing that the standard families are continuous on their domains lets you verify these hypotheses quickly. When an exam question opens with "since ff is continuous on [a,b][a, b]", it is signalling that one of these theorems is about to be used, and a full-credit answer states explicitly why the continuity holds - "ff is a polynomial" or "ff is continuous on its domain, which includes [a,b][a, b]" - rather than assuming it. Treating the interval-continuity claim as something to justify, not just assert, is what separates complete responses from partial ones.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice). On which interval is f(x)=1xβˆ’2f(x) = \frac{1}{x - 2} continuous? (A) All real numbers (B) (βˆ’βˆž,2)βˆͺ(2,∞)(-\infty, 2) \cup (2, \infty) (C) [2,∞)[2, \infty) (D) (2,∞)(2, \infty) only
Show worked answer β†’

The correct answer is (B).

The rational function 1xβˆ’2\frac{1}{x-2} is continuous everywhere it is defined, that is, everywhere except where the denominator is zero. The denominator is zero at x=2x = 2, where there is an infinite discontinuity. So ff is continuous on (βˆ’βˆž,2)βˆͺ(2,∞)(-\infty, 2) \cup (2, \infty), every real number except 22.

AP 2023 (style)3 marksSection II (free response, no calculator). Let f(x)=xβˆ’1f(x) = \sqrt{x - 1}. (a) State the domain of ff. (b) On what interval is ff continuous, and what kind of continuity holds at the left endpoint? (c) Justify why ff is continuous on that interval.
Show worked answer β†’

A 3-point interval-continuity question.

(a) (1 point) The radicand must be nonnegative: xβˆ’1β‰₯0x - 1 \geq 0, so the domain is [1,∞)[1, \infty).
(b) (1 point) ff is continuous on [1,∞)[1, \infty); at the left endpoint x=1x = 1 only right-hand continuity is required, since the function is not defined to the left of 11.
(c) (1 point) Square-root functions are continuous on their entire domain, and lim⁑xβ†’1+xβˆ’1=0=f(1)\lim_{x \to 1^+} \sqrt{x-1} = 0 = f(1), confirming continuity at the endpoint from the right.

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