Unit 1Continuity on a Closed Interval

Concept 4.145Built for May 2027 AP Exam

Continuity on a Closed Interval

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Continuity on a closed interval $[a, b]$ is one claim about the inside and two narrower claims at the ends. Inside, every point must pass the ordinary three-part test, because both sides exist there. At $a$ only the right-hand side is available, so the test is rewritten with $\lim_{x \to a^{+}} f(x) = f(a)$ ; at $b$ only the left-hand side is available, so it becomes $\lim_{x \to b^{-}} f(x) = f(b)$ . Nothing is weakened by this. The same equality is checked, just with one side removed because the domain ends. Exams phrase it as "is $f$ continuous on $[0, 3]$ ", and the answer has to distinguish the interior from the two endpoints.

On the AP Exam

Closed-interval continuity is the hypothesis of the theorems that follow, so the exam asks you to establish it rather than to use it. Free response supplies a piecewise or radical function on a stated interval and asks whether continuity holds there, and the scoring splits into an interior check and two endpoint checks. Multiple choice compresses it into one question, usually about a square root at its left endpoint or a piecewise seam inside the interval. The endpoint check is where responses lose points, because the left endpoint has no left side and students either ignore the point entirely or demand a two-sided limit that cannot exist.

What you’ll see: Radicals on their domain, rational functions with a seam inside a stated interval, piecewise rules on $[0, k]$ , and prompts that quote the interval explicitly.
What you’ll do: Check the interior with the ordinary test, check each endpoint with its one-sided version, and state the conclusion in interval language with the endpoint work shown.
Memory hook: Inside both sides, ends one side. The interior gets the full test; the endpoints get half of it, and the half is the half that exists.
Use it on the exam: For MCQs: At a left endpoint, compute the right-hand limit only. If the item offers "does not exist" because the left limit is undefined, that choice is the trap, not the answer. For FRQs: Write three labelled checks, interior and two endpoints, then one sentence closing the interval claim. A single sentence saying "continuous everywhere on the interval" without the endpoint checks is the standard way to drop a point.

Concept Review

The Interval Statement, Broken Into Parts

The phrase "continuous on $[a, b]$ " is an abbreviation for an infinite number of point claims plus two special ones.

Location	Sides available	What must be checked	Why it differs
Interior point $c$ with $a < c < b$	Both	$\lim_{x \to c} f(x) = f(c)$	The ordinary test, unchanged
Left endpoint $a$	Right only	$\lim_{x \to a^{+}} f(x) = f(a)$	There is nothing to the left of $a$ in the domain
Right endpoint $b$	Left only	$\lim_{x \to b^{-}} f(x) = f(b)$	There is nothing to the right of $b$ in the domain
Open interval $(a, b)$	Both everywhere	The test at every interior point	Endpoints are excluded and carry no claim

The endpoint conditions are called one-sided continuity, and they are not a weaker version of continuity. They are the only version that could possibly be true at a point where the domain stops.

Checking a Closed Interval in Three Moves

Read the interval, then read the domain. A function can only be continuous on an interval it is defined on. For a radical, the domain may end exactly at the stated endpoint, which is what makes the right-hand check meaningful.
Check the interior. Any seam inside the interval gets the full three-part test. Points with no seam are continuous by inheritance from polynomial, rational, radical, trigonometric, or exponential pieces on their domains.
Check each endpoint with its one-sided limit. Compute only the side that exists, compare it with the endpoint value, and say which side you used.
Close with the interval sentence. "Since $f$ is continuous at every interior point, right-continuous at $0$ , and left-continuous at $3$ , $f$ is continuous on $[0, 3]$ ." The closing sentence is what the question asked for.

Figure 1 — One interval, three kinds of check.

Caption. At every interior point the neighbourhood is two-sided, so the ordinary test applies. At $a$ the curve is only approached from the right, and at $b$ only from the left; those two points are tested with one-sided limits against the endpoint values, and the dots mark that both endpoint values exist.

A curve drawn over a closed interval from a to b, with a filled dot at each end and dashed vertical lines marking the endpoints

Three Corrections Worth Memorising

“The left endpoint fails because the left-hand limit does not exist.”
- Correct: the left-hand limit is not part of the definition at $a$ . Only $\lim_{x \to a^{+}} f(x) = f(a)$ is required, and that can hold perfectly well.
“Continuous on an open interval and continuous on a closed interval are the same thing.”
- Correct: they differ at the ends. The closed interval adds two one-sided claims, and those claims are exactly what later theorems rely on.
“A function continuous on $[a, b]$ is continuous everywhere.”
- Correct: the claim is confined to the interval. Outside $[a, b]$ the function may be undefined or badly behaved, and the interval statement says nothing about it.

Topic-Specific Study Method

The Interior-Plus-Two-Ends Check

Borrowed idea: What does continuity mean at the edge of an interval? Inside the interval both sides exist, so the usual three-part test applies unchanged. At the left endpoint, which side is even available? Only the right side, because the function is not defined to the left of $a$ . So the test there asks for the right-hand limit, the value, and their equality. Does that make endpoint continuity weaker? It makes it narrower, not weaker: the same equation with one side removed. Write the endpoint check separately from the interior check, and label which side you used. Readers look for that label, because an endpoint claim without it reads like an interior claim. Continuity on a closed interval is the interior plus two one-sided claims at the ends.

Write the interval and the domain side by side. If the domain is $[2, \infty)$ and the question asks about $[2, 5]$ , say so. The left endpoint of the question coincides with the left endpoint of the domain, and that is the case the one-sided check is built for.
Sweep the interior first. Name the branch types and conclude continuity by inheritance, then treat any seam with the full test. Interior work is usually short and should not consume the time the endpoints need.
Test each endpoint with the side that exists. One line per endpoint: the one-sided limit, the value, the comparison, and the words "right-continuous" or "left-continuous".
Close with the interval sentence, naming all three. Interior, right endpoint, left endpoint, then "therefore continuous on $[a, b]$ ". The closing sentence is the answer; the lines above it are the evidence.

What the item gives	Interior	Left endpoint	Right endpoint	Verdict
$\sqrt{x-2}$ on $[2, 5]$	Continuous	Right-continuous, limit $0 = f(2)$	Left-continuous	Continuous on $[2,5]$
Rational with a seam at $x = 2$ inside $[0,4]$	Seam fails or passes the test	Fine	Fine	Depends entirely on the seam
Piecewise on $[0, 3]$ with a jump at $1$	Fails at $1$	Check separately	Check separately	Not continuous on $[0,3]$
Polynomial on $[-1, 4]$	Continuous	Right-continuous automatically	Left-continuous automatically	Continuous on $[-1,4]$
$\frac{1}{x}$ on $[1, 3]$	Continuous	Right-continuous	Left-continuous	Continuous on $[1,3]$ despite the asymptote at $0$

Training & Error Diagnosis

Core Archetype (Simplest Form): Show that $f(x) = \sqrt{x - 2}$ is continuous on $[2, 5]$ .

Standard Execution:

Interior. For $2 < c < 5$ the radicand is positive, so $f$ is continuous at $c$ by inheritance.

Left endpoint. $\lim_{x \to 2^{+}} \sqrt{x-2} = 0 = f(2)$ , so $f$ is right-continuous at $2$ .

Right endpoint. $\lim_{x \to 5^{-}} \sqrt{x-2} = \sqrt{3} = f(5)$ , so $f$ is left-continuous at $5$ .

Conclusion. All three hold, so $f$ is continuous on $[2, 5]$ .

Exam Variation (Context / Parameter Shift): Let $g(x) = \begin{cases} x + 1, & 0 \le x < 2 \cr 5 - x, & 2 \le x \le 3 \end{cases}$ . Is $g$ continuous on $[0, 3]$ ?

Solution: Interior seams: at $x = 2$ , the left limit is $3$ and the right limit is $3$ , and $g(2) = 3$ , so the seam passes. Each branch is a polynomial and continuous on its own piece. Left endpoint: $\lim_{x \to 0^{+}} g(x) = 1 = g(0)$ , so $g$ is right-continuous at $0$ . Right endpoint: $\lim_{x \to 3^{-}} g(x) = 2 = g(3)$ , so $g$ is left-continuous at $3$ . Therefore $g$ is continuous on $[0, 3]$ .

Common Traps & Failure Signals:

The missing side (Signal: a claim that the endpoint fails because one side is undefined): At an endpoint only one side exists. The missing side is not evidence.
The unexamined seam (Signal: an interval claim with no mention of an interior break): One seam inside the interval is enough to make the whole interval claim false, and the seam must be named.
The overreach (Signal: the words "continuous everywhere" attached to an interval question): The claim is about $[a, b]$ . Nothing outside the interval has been checked or asserted.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Endpoint Blindness: You check the interior and stop.
Symmetry Assumption: Because interior points need two sides, the endpoint gets tested the same way, and a limit that cannot exist is reported as a failure instead of being recognised as the wrong test for that location.
Seam Forgetfulness: You accept an interval claim after checking only the endpoints, leaving the one interior point that actually breaks the conclusion unexamined.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No scored FRQ part in any released exam; it is the hypothesis, never the question
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Continuity at both endpoints plus the interior seam
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Left Endpoint

The function $f(x) = \sqrt{x - 3}$ is defined on $[3, \infty)$ . Which statement about continuity at $x = 3$ is correct?

MCQ 2

The Interval Claim

Let $g(x) = \begin{cases} x + 1, & 0 \le x < 2 \cr 5 - x, & 2 \le x \le 3 \end{cases}$ Which statement is correct?

MCQ 3

Where the Interval Claim Breaks

Let $h(x) = \dfrac{1}{x - 2}$ and consider the interval $[1, 4]$ . Which statement is correct?

FRQ 4

A Closed Interval, Checked Completely

Let $f(x) = \begin{cases} x^{2} + 1, & 0 \le x < 2 \cr 4, & x = 2 \cr 9 - x, & 2 < x \le 4 \end{cases}$

(a) Evaluate $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ .

(b) Determine whether $f$ is continuous at $x = 2$ , using the three-part test.

(c) Check continuity at the endpoints $x = 0$ and $x = 4$ , naming the side used in each case.

(d) Determine whether $f$ is continuous on $[0, 2]$ , and whether it is continuous on $(2, 4]$ . Justify each conclusion.

Cheatsheet

Interval Continuity

	Sides to check	Condition	Sentence to write
Interior point	Both	$\lim_{x \to c} f(x) = f(c)$	"Since both one-sided limits equal $f(c)$ , $f$ is continuous at $c$ ."
Left endpoint $a$	Right only	$\lim_{x \to a^{+}} f(x) = f(a)$	" $f$ is right-continuous at $a$ ."
Right endpoint $b$	Left only	$\lim_{x \to b^{-}} f(x) = f(b)$	" $f$ is left-continuous at $b$ ."
Whole interval	Interior plus both ends	All of the above	"Therefore $f$ is continuous on $[a, b]$ ."
Interval with an excluded point	Not a continuity claim	The point is outside the domain	"The interval claim fails at $x = c$ , which is not in the domain."

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Unit 1Continuity on a Closed Interval

Concept 4.145Built for May 2027 AP Exam

Continuity on a Closed Interval

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Radicals on their domain, rational functions with a seam inside a stated interval, piecewise rules on $[0, k]$ , and prompts that quote the interval explicitly.
What you’ll do: Check the interior with the ordinary test, check each endpoint with its one-sided version, and state the conclusion in interval language with the endpoint work shown.
Memory hook: Inside both sides, ends one side. The interior gets the full test; the endpoints get half of it, and the half is the half that exists.
Use it on the exam: For MCQs: At a left endpoint, compute the right-hand limit only. If the item offers "does not exist" because the left limit is undefined, that choice is the trap, not the answer. For FRQs: Write three labelled checks, interior and two endpoints, then one sentence closing the interval claim. A single sentence saying "continuous everywhere on the interval" without the endpoint checks is the standard way to drop a point.

Concept Review

The Interval Statement, Broken Into Parts

The phrase "continuous on $[a, b]$ " is an abbreviation for an infinite number of point claims plus two special ones.

Location	Sides available	What must be checked	Why it differs
Interior point $c$ with $a < c < b$	Both	$\lim_{x \to c} f(x) = f(c)$	The ordinary test, unchanged
Left endpoint $a$	Right only	$\lim_{x \to a^{+}} f(x) = f(a)$	There is nothing to the left of $a$ in the domain
Right endpoint $b$	Left only	$\lim_{x \to b^{-}} f(x) = f(b)$	There is nothing to the right of $b$ in the domain
Open interval $(a, b)$	Both everywhere	The test at every interior point	Endpoints are excluded and carry no claim

The endpoint conditions are called one-sided continuity, and they are not a weaker version of continuity. They are the only version that could possibly be true at a point where the domain stops.

Checking a Closed Interval in Three Moves

Read the interval, then read the domain. A function can only be continuous on an interval it is defined on. For a radical, the domain may end exactly at the stated endpoint, which is what makes the right-hand check meaningful.
Check the interior. Any seam inside the interval gets the full three-part test. Points with no seam are continuous by inheritance from polynomial, rational, radical, trigonometric, or exponential pieces on their domains.
Check each endpoint with its one-sided limit. Compute only the side that exists, compare it with the endpoint value, and say which side you used.
Close with the interval sentence. "Since $f$ is continuous at every interior point, right-continuous at $0$ , and left-continuous at $3$ , $f$ is continuous on $[0, 3]$ ." The closing sentence is what the question asked for.

Figure 1 — One interval, three kinds of check.

A curve drawn over a closed interval from a to b, with a filled dot at each end and dashed vertical lines marking the endpoints

Three Corrections Worth Memorising

“The left endpoint fails because the left-hand limit does not exist.”
- Correct: the left-hand limit is not part of the definition at $a$ . Only $\lim_{x \to a^{+}} f(x) = f(a)$ is required, and that can hold perfectly well.
“Continuous on an open interval and continuous on a closed interval are the same thing.”
- Correct: they differ at the ends. The closed interval adds two one-sided claims, and those claims are exactly what later theorems rely on.
“A function continuous on $[a, b]$ is continuous everywhere.”
- Correct: the claim is confined to the interval. Outside $[a, b]$ the function may be undefined or badly behaved, and the interval statement says nothing about it.

Topic-Specific Study Method

The Interior-Plus-Two-Ends Check

Write the interval and the domain side by side. If the domain is $[2, \infty)$ and the question asks about $[2, 5]$ , say so. The left endpoint of the question coincides with the left endpoint of the domain, and that is the case the one-sided check is built for.
Sweep the interior first. Name the branch types and conclude continuity by inheritance, then treat any seam with the full test. Interior work is usually short and should not consume the time the endpoints need.
Test each endpoint with the side that exists. One line per endpoint: the one-sided limit, the value, the comparison, and the words "right-continuous" or "left-continuous".
Close with the interval sentence, naming all three. Interior, right endpoint, left endpoint, then "therefore continuous on $[a, b]$ ". The closing sentence is the answer; the lines above it are the evidence.

What the item gives	Interior	Left endpoint	Right endpoint	Verdict
$\sqrt{x-2}$ on $[2, 5]$	Continuous	Right-continuous, limit $0 = f(2)$	Left-continuous	Continuous on $[2,5]$
Rational with a seam at $x = 2$ inside $[0,4]$	Seam fails or passes the test	Fine	Fine	Depends entirely on the seam
Piecewise on $[0, 3]$ with a jump at $1$	Fails at $1$	Check separately	Check separately	Not continuous on $[0,3]$
Polynomial on $[-1, 4]$	Continuous	Right-continuous automatically	Left-continuous automatically	Continuous on $[-1,4]$
$\frac{1}{x}$ on $[1, 3]$	Continuous	Right-continuous	Left-continuous	Continuous on $[1,3]$ despite the asymptote at $0$

Training & Error Diagnosis

Core Archetype (Simplest Form): Show that $f(x) = \sqrt{x - 2}$ is continuous on $[2, 5]$ .

Standard Execution:

Interior. For $2 < c < 5$ the radicand is positive, so $f$ is continuous at $c$ by inheritance.

Left endpoint. $\lim_{x \to 2^{+}} \sqrt{x-2} = 0 = f(2)$ , so $f$ is right-continuous at $2$ .

Right endpoint. $\lim_{x \to 5^{-}} \sqrt{x-2} = \sqrt{3} = f(5)$ , so $f$ is left-continuous at $5$ .

Conclusion. All three hold, so $f$ is continuous on $[2, 5]$ .

Exam Variation (Context / Parameter Shift): Let $g(x) = \begin{cases} x + 1, & 0 \le x < 2 \cr 5 - x, & 2 \le x \le 3 \end{cases}$ . Is $g$ continuous on $[0, 3]$ ?

Solution: Interior seams: at $x = 2$ , the left limit is $3$ and the right limit is $3$ , and $g(2) = 3$ , so the seam passes. Each branch is a polynomial and continuous on its own piece. Left endpoint: $\lim_{x \to 0^{+}} g(x) = 1 = g(0)$ , so $g$ is right-continuous at $0$ . Right endpoint: $\lim_{x \to 3^{-}} g(x) = 2 = g(3)$ , so $g$ is left-continuous at $3$ . Therefore $g$ is continuous on $[0, 3]$ .

Common Traps & Failure Signals:

The missing side (Signal: a claim that the endpoint fails because one side is undefined): At an endpoint only one side exists. The missing side is not evidence.
The unexamined seam (Signal: an interval claim with no mention of an interior break): One seam inside the interval is enough to make the whole interval claim false, and the seam must be named.
The overreach (Signal: the words "continuous everywhere" attached to an interval question): The claim is about $[a, b]$ . Nothing outside the interval has been checked or asserted.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Endpoint Blindness: You check the interior and stop.
Symmetry Assumption: Because interior points need two sides, the endpoint gets tested the same way, and a limit that cannot exist is reported as a failure instead of being recognised as the wrong test for that location.
Seam Forgetfulness: You accept an interval claim after checking only the endpoints, leaving the one interior point that actually breaks the conclusion unexamined.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No scored FRQ part in any released exam; it is the hypothesis, never the question
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Continuity at both endpoints plus the interior seam
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Left Endpoint

The function $f(x) = \sqrt{x - 3}$ is defined on $[3, \infty)$ . Which statement about continuity at $x = 3$ is correct?

MCQ 2

The Interval Claim

Let $g(x) = \begin{cases} x + 1, & 0 \le x < 2 \cr 5 - x, & 2 \le x \le 3 \end{cases}$ Which statement is correct?

MCQ 3

Where the Interval Claim Breaks

Let $h(x) = \dfrac{1}{x - 2}$ and consider the interval $[1, 4]$ . Which statement is correct?

FRQ 4

A Closed Interval, Checked Completely

Let $f(x) = \begin{cases} x^{2} + 1, & 0 \le x < 2 \cr 4, & x = 2 \cr 9 - x, & 2 < x \le 4 \end{cases}$

(a) Evaluate $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ .

(b) Determine whether $f$ is continuous at $x = 2$ , using the three-part test.

(c) Check continuity at the endpoints $x = 0$ and $x = 4$ , naming the side used in each case.

(d) Determine whether $f$ is continuous on $[0, 2]$ , and whether it is continuous on $(2, 4]$ . Justify each conclusion.

Cheatsheet

Interval Continuity

	Sides to check	Condition	Sentence to write
Interior point	Both	$\lim_{x \to c} f(x) = f(c)$	"Since both one-sided limits equal $f(c)$ , $f$ is continuous at $c$ ."
Left endpoint $a$	Right only	$\lim_{x \to a^{+}} f(x) = f(a)$	" $f$ is right-continuous at $a$ ."
Right endpoint $b$	Left only	$\lim_{x \to b^{-}} f(x) = f(b)$	" $f$ is left-continuous at $b$ ."
Whole interval	Interior plus both ends	All of the above	"Therefore $f$ is continuous on $[a, b]$ ."
Interval with an excluded point	Not a continuity claim	The point is outside the domain	"The interval claim fails at $x = c$ , which is not in the domain."

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Start Topic Practice Next Topic: 1.17 Vertical Asymptotes

Unit 1Continuity on a Closed Interval

Concept 4.145Built for May 2027 AP Exam

Continuity on a Closed Interval

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: Radicals on their domain, rational functions with a seam inside a stated interval, piecewise rules on $[0, k]$ , and prompts that quote the interval explicitly.
What you’ll do: Check the interior with the ordinary test, check each endpoint with its one-sided version, and state the conclusion in interval language with the endpoint work shown.
Memory hook: Inside both sides, ends one side. The interior gets the full test; the endpoints get half of it, and the half is the half that exists.
Use it on the exam: For MCQs: At a left endpoint, compute the right-hand limit only. If the item offers "does not exist" because the left limit is undefined, that choice is the trap, not the answer. For FRQs: Write three labelled checks, interior and two endpoints, then one sentence closing the interval claim. A single sentence saying "continuous everywhere on the interval" without the endpoint checks is the standard way to drop a point.

Concept Review

The Interval Statement, Broken Into Parts

The phrase "continuous on $[a, b]$ " is an abbreviation for an infinite number of point claims plus two special ones.

Location	Sides available	What must be checked	Why it differs
Interior point $c$ with $a < c < b$	Both	$\lim_{x \to c} f(x) = f(c)$	The ordinary test, unchanged
Left endpoint $a$	Right only	$\lim_{x \to a^{+}} f(x) = f(a)$	There is nothing to the left of $a$ in the domain
Right endpoint $b$	Left only	$\lim_{x \to b^{-}} f(x) = f(b)$	There is nothing to the right of $b$ in the domain
Open interval $(a, b)$	Both everywhere	The test at every interior point	Endpoints are excluded and carry no claim

The endpoint conditions are called one-sided continuity, and they are not a weaker version of continuity. They are the only version that could possibly be true at a point where the domain stops.

Checking a Closed Interval in Three Moves

Read the interval, then read the domain. A function can only be continuous on an interval it is defined on. For a radical, the domain may end exactly at the stated endpoint, which is what makes the right-hand check meaningful.
Check the interior. Any seam inside the interval gets the full three-part test. Points with no seam are continuous by inheritance from polynomial, rational, radical, trigonometric, or exponential pieces on their domains.
Check each endpoint with its one-sided limit. Compute only the side that exists, compare it with the endpoint value, and say which side you used.
Close with the interval sentence. "Since $f$ is continuous at every interior point, right-continuous at $0$ , and left-continuous at $3$ , $f$ is continuous on $[0, 3]$ ." The closing sentence is what the question asked for.

Figure 1 — One interval, three kinds of check.

A curve drawn over a closed interval from a to b, with a filled dot at each end and dashed vertical lines marking the endpoints

Three Corrections Worth Memorising

“The left endpoint fails because the left-hand limit does not exist.”
- Correct: the left-hand limit is not part of the definition at $a$ . Only $\lim_{x \to a^{+}} f(x) = f(a)$ is required, and that can hold perfectly well.
“Continuous on an open interval and continuous on a closed interval are the same thing.”
- Correct: they differ at the ends. The closed interval adds two one-sided claims, and those claims are exactly what later theorems rely on.
“A function continuous on $[a, b]$ is continuous everywhere.”
- Correct: the claim is confined to the interval. Outside $[a, b]$ the function may be undefined or badly behaved, and the interval statement says nothing about it.

Topic-Specific Study Method

The Interior-Plus-Two-Ends Check

Write the interval and the domain side by side. If the domain is $[2, \infty)$ and the question asks about $[2, 5]$ , say so. The left endpoint of the question coincides with the left endpoint of the domain, and that is the case the one-sided check is built for.
Sweep the interior first. Name the branch types and conclude continuity by inheritance, then treat any seam with the full test. Interior work is usually short and should not consume the time the endpoints need.
Test each endpoint with the side that exists. One line per endpoint: the one-sided limit, the value, the comparison, and the words "right-continuous" or "left-continuous".
Close with the interval sentence, naming all three. Interior, right endpoint, left endpoint, then "therefore continuous on $[a, b]$ ". The closing sentence is the answer; the lines above it are the evidence.

What the item gives	Interior	Left endpoint	Right endpoint	Verdict
$\sqrt{x-2}$ on $[2, 5]$	Continuous	Right-continuous, limit $0 = f(2)$	Left-continuous	Continuous on $[2,5]$
Rational with a seam at $x = 2$ inside $[0,4]$	Seam fails or passes the test	Fine	Fine	Depends entirely on the seam
Piecewise on $[0, 3]$ with a jump at $1$	Fails at $1$	Check separately	Check separately	Not continuous on $[0,3]$
Polynomial on $[-1, 4]$	Continuous	Right-continuous automatically	Left-continuous automatically	Continuous on $[-1,4]$
$\frac{1}{x}$ on $[1, 3]$	Continuous	Right-continuous	Left-continuous	Continuous on $[1,3]$ despite the asymptote at $0$

Training & Error Diagnosis

Core Archetype (Simplest Form): Show that $f(x) = \sqrt{x - 2}$ is continuous on $[2, 5]$ .

Standard Execution:

Interior. For $2 < c < 5$ the radicand is positive, so $f$ is continuous at $c$ by inheritance.

Left endpoint. $\lim_{x \to 2^{+}} \sqrt{x-2} = 0 = f(2)$ , so $f$ is right-continuous at $2$ .

Right endpoint. $\lim_{x \to 5^{-}} \sqrt{x-2} = \sqrt{3} = f(5)$ , so $f$ is left-continuous at $5$ .

Conclusion. All three hold, so $f$ is continuous on $[2, 5]$ .

Exam Variation (Context / Parameter Shift): Let $g(x) = \begin{cases} x + 1, & 0 \le x < 2 \cr 5 - x, & 2 \le x \le 3 \end{cases}$ . Is $g$ continuous on $[0, 3]$ ?

Solution: Interior seams: at $x = 2$ , the left limit is $3$ and the right limit is $3$ , and $g(2) = 3$ , so the seam passes. Each branch is a polynomial and continuous on its own piece. Left endpoint: $\lim_{x \to 0^{+}} g(x) = 1 = g(0)$ , so $g$ is right-continuous at $0$ . Right endpoint: $\lim_{x \to 3^{-}} g(x) = 2 = g(3)$ , so $g$ is left-continuous at $3$ . Therefore $g$ is continuous on $[0, 3]$ .

Common Traps & Failure Signals:

The missing side (Signal: a claim that the endpoint fails because one side is undefined): At an endpoint only one side exists. The missing side is not evidence.
The unexamined seam (Signal: an interval claim with no mention of an interior break): One seam inside the interval is enough to make the whole interval claim false, and the seam must be named.
The overreach (Signal: the words "continuous everywhere" attached to an interval question): The claim is about $[a, b]$ . Nothing outside the interval has been checked or asserted.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Endpoint Blindness: You check the interior and stop.
Symmetry Assumption: Because interior points need two sides, the endpoint gets tested the same way, and a limit that cannot exist is reported as a failure instead of being recognised as the wrong test for that location.
Seam Forgetfulness: You accept an interval claim after checking only the endpoints, leaving the one interior point that actually breaks the conclusion unexamined.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No scored FRQ part in any released exam; it is the hypothesis, never the question
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; Continuity at both endpoints plus the interior seam
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

The Left Endpoint

The function $f(x) = \sqrt{x - 3}$ is defined on $[3, \infty)$ . Which statement about continuity at $x = 3$ is correct?

MCQ 2

The Interval Claim

Let $g(x) = \begin{cases} x + 1, & 0 \le x < 2 \cr 5 - x, & 2 \le x \le 3 \end{cases}$ Which statement is correct?

MCQ 3

Where the Interval Claim Breaks

Let $h(x) = \dfrac{1}{x - 2}$ and consider the interval $[1, 4]$ . Which statement is correct?

FRQ 4

A Closed Interval, Checked Completely

Let $f(x) = \begin{cases} x^{2} + 1, & 0 \le x < 2 \cr 4, & x = 2 \cr 9 - x, & 2 < x \le 4 \end{cases}$

(a) Evaluate $\lim_{x \to 2^{-}} f(x)$ and $\lim_{x \to 2^{+}} f(x)$ .

(b) Determine whether $f$ is continuous at $x = 2$ , using the three-part test.

(c) Check continuity at the endpoints $x = 0$ and $x = 4$ , naming the side used in each case.

(d) Determine whether $f$ is continuous on $[0, 2]$ , and whether it is continuous on $(2, 4]$ . Justify each conclusion.

Cheatsheet

Interval Continuity

	Sides to check	Condition	Sentence to write
Interior point	Both	$\lim_{x \to c} f(x) = f(c)$	"Since both one-sided limits equal $f(c)$ , $f$ is continuous at $c$ ."
Left endpoint $a$	Right only	$\lim_{x \to a^{+}} f(x) = f(a)$	" $f$ is right-continuous at $a$ ."
Right endpoint $b$	Left only	$\lim_{x \to b^{-}} f(x) = f(b)$	" $f$ is left-continuous at $b$ ."
Whole interval	Interior plus both ends	All of the above	"Therefore $f$ is continuous on $[a, b]$ ."
Interval with an excluded point	Not a continuity claim	The point is outside the domain	"The interval claim fails at $x = c$ , which is not in the domain."

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