Confirming continuity over an interval

AP Calculus AB· AP Calculus AB CED — Limits and Continuity· 14 min read

1. Core Definition of Continuity Over an Interval★★☆☆☆⏱ 3 min

Confirming continuity over an interval means verifying a function meets the formal continuity definition at every point in the interval. This topic makes up 10–12% of Unit 1 exam weight, and is a required first step for justifying major theorems like the Intermediate Value Theorem (IVT) and Extreme Value Theorem (EVT) on AP FRQs.

📘 Definition

Continuity on an interval

$f$ continuous on interval $I$

A function $f$ is continuous on interval $I$ if it meets the continuity condition at every point $c \in I$ . For closed intervals, we add one-sided continuity requirements at endpoints, since the function is not defined outside the interval.

2. Continuity on Open Intervals★★☆☆☆⏱ 4 min

An open interval $(a, b)$ does not include its endpoints, so you only need to verify that $f$ is continuous at every interior point $c$ between $a$ and $b$ . Recall the three conditions for continuity at any interior point:

$f (c)$ is defined
$lim_{x \to c} f (x)$ exists
$lim_{x \to c} f (x) = f (c)$

A key AP exam shortcut: all elementary functions (polynomials, rational, root, exponential, logarithmic, trigonometric) are continuous at every point in their domain. This means you do not need to test every point individually: you only need to confirm that the entire open interval lies within the function's domain. For piecewise functions, the only possible points of discontinuity inside an open interval are the breakpoints where the function's definition changes.

📐 Worked Example

Is $f (x) = \frac{x ^{2} - 9}{x - 3}$ continuous on the open interval $(0, 4)$ ? Justify your answer.

1
First, identify domain restrictions: the denominator is zero at $x = 3$ , so $f (3)$ is undefined.
2
$f(3) = \frac{9-9}{3-3} = \frac{0}{0}$
3
$x = 3$ lies inside the interval $(0, 4)$ . Even though $lim_{x \to 3} f (x) = 6$ exists, this is a removable discontinuity.
4
The first continuity condition fails at $x = 3$ , so $f (x)$ is not continuous on $(0, 4)$ .

3. Continuity on Closed Intervals★★★☆☆⏱ 4 min

A closed interval $[a, b]$ includes both endpoints, so the definition for continuity adds two extra one-sided continuity requirements beyond continuity on the interior $(a, b)$ . Since the function is only defined for $x \in [a, b]$ , we cannot approach the left endpoint $a$ from the left, so we only require right-continuity at $a$ , and left-continuity at the right endpoint $b$ .

📘 Definition

Continuity on a Closed Interval

For a closed interval $[a, b]$ : 1) $f$ is continuous on the open interval $(a, b)$ , 2) $lim_{x \to a^{+}} f (x) = f (a)$ (right-continuous at $a$ ), 3) $lim_{x \to b^{-}} f (x) = f (b)$ (left-continuous at $b$ ). All three conditions are required.

📐 Worked Example

Is $f (x) = 9 - x^{2}$ continuous on the closed interval $[- 3, 3]$ ? Justify your answer.

1
First confirm continuity on the open interval $(- 3, 3)$ : the domain of $f$ is $- 3 \leq x \leq 3$ , so all interior points are in the domain. $f$ is a composition of continuous elementary functions, so it is continuous at all interior points.
2
Check right-continuity at left endpoint $x = - 3$ :
3
$\lim_{x \to -3^+} \sqrt{9 - x^2} = \sqrt{9 - (-3)^2} = 0 = f(-3)$
4
So $f$ is right-continuous at $x = - 3$ . Next check left-continuity at right endpoint $x = 3$ :
5
$\lim_{x \to 3^-} \sqrt{9 - x^2} = \sqrt{9 - (3)^2} = 0 = f(3)$
6
All conditions are satisfied, so $f$ is continuous on $[- 3, 3]$ .

4. Continuity of Piecewise Functions Over Intervals★★★★☆⏱ 3 min

Piecewise functions have different definitions for different subintervals, so they require a targeted check for discontinuities. The only possible points of discontinuity are: (1) breakpoints where the function's definition changes that lie inside your target interval, and (2) points inside any subinterval where the individual piece has a discontinuity (such as a zero denominator).

For any interval that includes a breakpoint inside the interval, you must test all three continuity conditions at that breakpoint by calculating one-sided limits from each side. If the breakpoint is an endpoint of your overall interval, you only check the one-sided limit from inside the interval.

📐 Worked Example

Confirm whether $f (x) = {3 x - 1 x^{2} + 1 x < 2 2 \leq x \leq 5$ is continuous on the interval $[0, 5]$ .

1
List candidate discontinuities on $[0, 5]$ : the only breakpoint is $x = 2$ , which lies inside the interval. Both pieces are polynomials, which are continuous everywhere, so there are no other interior discontinuities.
2
Check the three conditions at $x = 2$ : (1) $f (2) = 2^{2} + 1 = 5$ , so $f (2)$ is defined. (2) Calculate one-sided limits:
3
$\lim_{x \to 2^-} (3x - 1) = 5, \quad \lim_{x \to 2^+} (x^2 + 1) = 5$
4
One-sided limits are equal, so $lim_{x \to 2} f (x) = 5$ exists. (3) $lim_{x \to 2} f (x) = 5 = f (2)$ , so all conditions are satisfied at $x = 2$ .
5
Check endpoint continuity: At $x = 0$ , $lim_{x \to 0^{+}} (3 x - 1) = - 1 = f (0)$ , so $f$ is right-continuous. At $x = 5$ , $lim_{x \to 5^{-}} (x^{2} + 1) = 26 = f (5)$ , so $f$ is left-continuous.
6
All conditions are satisfied, so $f (x)$ is continuous on $[0, 5]$ .

5. Common Pitfalls

Wrong move:

Forgetting to check endpoint continuity when confirming continuity on a closed interval, and only checking interior points.

Why:

Students memorize continuity rules for interior points and ignore the extra endpoint requirement that is mandatory for theorem justifications.

Correct move:

Always add the two one-sided endpoint continuity checks when the interval is closed, and explicitly write the equality $lim_{x \to a^{+}} f (x) = f (a)$ in your justification.

Wrong move:

Claiming a function with a removable discontinuity is continuous on an interval containing that discontinuity, just because the limit exists.

Why:

Students confuse the existence of the limit with the full definition of continuity, which requires the function to be defined at the point.

Correct move:

Always check that $f (c)$ is defined for any point $c$ in the interval, even if the limit at $c$ exists.

Wrong move:

Forgetting to test breakpoints inside the interval for piecewise functions, and only checking that each individual piece is continuous.

Why:

Students assume that because each piece is continuous on its own, the combined function is continuous over the whole interval.

Correct move:

Always list all breakpoints that lie inside your target interval and test continuity at each one explicitly.

Wrong move:

Checking two-sided limits at endpoints of a closed interval, including the side outside the interval.

Why:

Students default to two-sided limits out of habit, even though the function is not defined outside the interval.

Correct move:

Only check right-continuity at the left endpoint and left-continuity at the right endpoint, ignoring behavior outside the interval.

Wrong move:

Claiming a function is continuous on an interval that includes a vertical asymptote, because the asymptote is "almost at the endpoint".

Why:

Students think only interior discontinuities count, but any point in the interval where the function is discontinuous breaks continuity over the whole interval.

Correct move:

If the interval contains any discontinuity (removable, jump, infinite), the function is not continuous over that interval.

6. Quick Reference Cheatsheet

Category	Rule / Formula	Notes
Continuity on Open Interval $(a, b)$	$f$ is continuous at every $c \in (a, b)$	No endpoint checks required; only interior points
Continuity on Closed Interval $[a, b]$	Continuous on $(a, b)$ 2. $lim_{x \to a^{+}} f (x) = f (a)$ 3. $lim_{x \to b^{-}} f (x) = f (b)$	Required for IVT and EVT; AP requires explicit justification of all three conditions
Continuity of Elementary Functions	All elementary functions are continuous on every interval in their domain	Shortcut: No need to test every point, just confirm the interval is within the domain
Piecewise Function Continuity Check	Test continuity at all breakpoints inside the interval, plus endpoints for closed intervals	Only breakpoints can be discontinuities if each individual piece is continuous
Three Conditions at a Point	$f (c)$ is defined 2. $lim_{x \to c} f (x)$ exists 3. $lim_{x \to c} f (x) = f (c)$	Failing any condition means discontinuous at $c$
Removable Discontinuity	Limit exists at $c$ , but $f (c)$ is undefined or not equal to the limit	Even removable discontinuities make the function discontinuous on any interval containing $c$
Infinite Discontinuity	One or both one-sided limits at $c$ are infinite, $f (c)$ undefined	Any interval containing $c$ is not continuous
Jump Discontinuity	One-sided limits exist but are not equal, so two-sided limit does not exist	Any interval containing $c$ is not continuous

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Identify continuous interval
2022 · FRQ
Justify IVT via continuity
2021 · MCQ
Check piecewise continuity

What's Next

Confirming continuity over an interval is the foundational prerequisite for the next core topic in Unit 1: applying the Intermediate Value Theorem, one of the most frequently tested concepts on the AP Calculus AB exam. IVT explicitly requires a function to be continuous on a closed interval to apply, so you cannot earn full points for an IVT justification without first correctly confirming continuity over the interval. Beyond Unit 1, continuity is also a required condition for differentiability, so you will use this skill when checking where a function is differentiable, and when justifying the Mean Value Theorem later in the course. Mastering this topic eliminates a large number of avoidable point losses on both MCQ and FRQ sections.