Confirming continuity over an interval

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

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

A function is continuous over an interval if every point in the interval satisfies all three conditions for continuity at a point: $f (c)$ is defined, $lim_{x \to c} f (x)$ exists, and $lim_{x \to c} f (x) = f (c)$ . This topic accounts for 3-6% of your total AP Calculus BC exam score, and acts as a foundational justification step for core concepts like the Intermediate Value Theorem, differentiability, and integration.

📘 Definition

Continuity over an interval

$f$ is continuous on $I$ , where $I$ can be open $(a, b)$ , closed $[a, b]$ , half-open, or infinite

All points inside the interval satisfy the three continuity conditions. For closed intervals, endpoints only require one-sided continuity matching the edge of the interval.

Example:

A function is continuous on $[a, b]$ if it is right-continuous at $a$ and left-continuous at $b$ .

2. Continuity of Elementary Functions Over Open Intervals★★☆☆☆⏱ 4 min

Elementary functions (polynomials, rational functions, trigonometric functions, exponentials, logarithms, and root functions) have a key simplifying property: every elementary function is continuous at all points in its domain. This result comes directly from limit laws, so you do not need to check every point individually in an interval.

To confirm continuity of an elementary function over an open interval, you only need to verify that the entire interval is contained within the function's domain. If there are no excluded points (division by zero, negative radicands, non-positive log arguments) inside the interval, the function is continuous over the full interval.

📐 Worked Example

Confirm whether $f (x) = \frac{x ^{2} - 9}{x + 2}$ is continuous on the open interval $(0, 4)$ .

1
Identify that $f (x)$ is a rational (elementary) function, which is continuous everywhere on its domain.
2
Find excluded domain points: denominator equals zero when
$x + 2 = 0 \implies x = -2$
3
The domain of $f$ is $(- \infty, - 2) \cup (- 2, \infty)$ . The entire interval $(0, 4)$ is inside the domain, since $- 2 \in / (0, 4)$ .
4
Conclude: since $f$ is continuous on its domain, $f$ is continuous on $(0, 4)$ .

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

For closed intervals $[a, b]$ , two-sided limits do not exist at the endpoints because the function is only considered on the interval itself. The full definition of continuity on $[a, b]$ requires three conditions:

$f$ is continuous at every point in the open interval $(a, b)$
$f$ is right-continuous at the left endpoint $a$ : $lim_{x \to a^{+}} f (x) = f (a)$
$f$ is left-continuous at the right endpoint $b$ : $lim_{x \to b^{-}} f (x) = f (b)$

This requirement is explicitly tested on AP free-response questions, most often when justifying the use of the Intermediate Value Theorem. Skipping the one-sided endpoint check will cost you justification points.

📐 Worked Example

Confirm whether $f (x) = 16 - x^{2}$ is continuous on the closed interval $[- 4, 4]$ .

1
Check continuity on the open interval $(- 4, 4)$ : for all $x \in (- 4, 4)$ , $16 - x^{2} > 0$ , so all points are in the domain. Since $f$ is elementary, it is continuous on $(- 4, 4)$ .
2
Check right-continuity at $x = - 4$ :
$\lim_{x \to -4^+} \sqrt{16 - x^2} = \sqrt{16 - (-4)^2} = 0 = f(-4)$
3
Check left-continuity at $x = 4$ :
$\lim_{x \to 4^-} \sqrt{16 - x^2} = \sqrt{16 - 4^2} = 0 = f(4)$
4
Conclude: all conditions are satisfied, so $f$ is continuous on $[- 4, 4]$ .

4. Continuity of Piecewise-Defined Functions Over an Interval★★★★☆⏱ 5 min

Piecewise functions use different expressions for different sub-intervals, so the only possible points of discontinuity inside an interval are the breakpoints (points where the expression changes). To confirm continuity over an interval containing breakpoints, follow these steps:

Confirm each individual piece is continuous on its open sub-interval (almost always, each piece is elementary, so only check domain per piece)
Check all three continuity conditions at every breakpoint inside the interval: evaluate the left limit with the left piece, right limit with the right piece, and confirm both equal $f (c)$ at the breakpoint
If the interval is closed, confirm one-sided continuity at the full interval endpoints as usual

📐 Worked Example

Confirm whether $f (x) = {3 x + 2 x^{2} + 4 x < 2 2 \leq x \leq 5$ is continuous on $[0, 5]$ .

1
Check continuity on open sub-intervals: On $(0, 2)$ , $f (x) = 3 x + 2$ is a polynomial (elementary), so continuous. On $(2, 5)$ , $f (x) = x^{2} + 4$ is also a polynomial, so continuous.
2
Check the breakpoint $x = 2$ : left limit is
$\lim_{x \to 2^-} (3x + 2) = 6 + 2 = 8$
3
Right limit is
$\lim_{x \to 2^+} (x^2 + 4) = 4 + 4 = 8$
4
$f (2) = 2^{2} + 4 = 8$ , so $lim_{x \to 2} f (x) = f (2)$ , and $f$ is continuous at $x = 2$ .
5
Check endpoints: Right-continuity at $x = 0$ : $lim_{x \to 0^{+}} (3 x + 2) = 2 = f (0)$ , which holds. Left-continuity at $x = 5$ : $lim_{x \to 5^{-}} (x^{2} + 4) = 29 = f (5)$ , which holds.
6
Conclude: all conditions are satisfied, so $f$ is continuous on $[0, 5]$ .

✓ Quick check

Test your understanding with these AP-style practice questions:

Which of the following intervals is $f (x) = \frac{e ^{x}}{l n ( x + 3 )}$ continuous on?
- A) $(- 3, 0]$
- B) $(0, 2)$
- C) $[- 2, 1]$
- D) $(- 2.5, - 1.5)$
Reveal answer
B —
The domain of $f (x)$ requires $x > - 3$ and $x \neq = - 2$ (since $ln (1) = 0$ at $x = - 2$ ). Only interval $(0, 2)$ contains no excluded points, so $f$ is continuous here.
Let $f (x) = {k x + 3 2 x^{2} - k x < 1 x \geq 1$ for constant $k$ . (a) Find $k$ that makes $f$ continuous at $x = 1$ . (b) Confirm $f$ is continuous on $[- 2, 3]$ with your value of $k$ and justify.
Reveal answer
$k = -\frac{1}{2}$, $f$ is continuous on $[-2, 3]$ —
For continuity at $x = 1$ , set $k + 3 = 2 - k$ , so $k = - \frac{1}{2}$ . Check continuity on open sub-intervals, confirm continuity at $x = 1$ , then check one-sided continuity at endpoints $- 2$ and $3$ : all conditions are satisfied.
The temperature of a chemical reaction is modeled by $T (t) = {25 e^{0.1 t} 25 e^{0.05 t + 0.15} 0 \leq t < 3 3 \leq t \leq 8$ . Confirm if $T (t)$ is continuous on $[0, 8]$ .
Reveal answer
Yes, $T(t)$ is continuous on $[0,8]$ —
Both pieces are exponential (elementary) functions, so they are continuous on their open sub-intervals. At the breakpoint $t = 3$ , both one-sided limits equal $25 e^{0.3} = T (3)$ , and endpoint one-sided continuity conditions hold. The claim of a continuous temperature function is correct.

5. Common Pitfalls

Wrong move:

Claiming $f (x) = \frac{( x - 2 ) ( x + 3 )}{x - 2}$ is continuous on $[0, 5]$ after simplifying to $x + 3$ , which is defined everywhere.

Why:

Students confuse the simplified form with the original function's domain; any point not in the original domain is a discontinuity regardless of cancellation.

Correct move:

Always check the original function's domain first, before simplifying, to find points of discontinuity.

Wrong move:

For a closed interval $[a, b]$ , requiring $lim_{x \to a} f (x) = f (a)$ (two-sided limit) to confirm continuity at the endpoint.

Why:

Two-sided limits do not exist at endpoints when we only consider the function on the interval.

Correct move:

For left endpoint $a$ , check only right-continuity $lim_{x \to a^{+}} f (x) = f (a)$ ; for right endpoint $b$ , check only left-continuity $lim_{x \to b^{-}} f (x) = f (b)$ .

Wrong move:

When checking continuity of a piecewise function at breakpoint $c$ , evaluating both one-sided limits with the same piece that defines $f (c)$ .

Why:

Students assume the function uses the same expression everywhere near the breakpoint, instead of switching expressions at the break.

Correct move:

Label which piece corresponds to $x < c$ and $x > c$ , and evaluate each one-sided limit with the matching piece.

Wrong move:

Claiming $f (x) = ln (x)$ is continuous on $[0, 5]$ because it is continuous on its domain.

Why:

Students forget that $x = 0$ is not in the domain of $ln (x)$ , so the endpoint condition fails.

Correct move:

Always confirm that endpoints of the interval are in the domain of the function before confirming continuity.

Wrong move:

Claiming a composite function $f (g (x))$ is discontinuous on an interval just because $f (x)$ has a discontinuity at a point outside the range of $g (x)$ over the interval.

Why:

Students check discontinuities of the outer function regardless of the output of the inner function over the interval.

Correct move:

A composite function $f (g (x))$ is continuous on an interval if $g (x)$ is continuous on the interval and $f$ is continuous on the range of $g (x)$ over that interval.

6. Quick Reference Cheatsheet

Category	Rule/Condition	Notes
Continuous on open interval $(a, b)$	$\forall c \in (a, b) : lim_{x \to c} f (x) = f (c)$	All elementary functions satisfy this if the interval is inside the function's domain.
Continuous 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 the Intermediate Value Theorem; always check the one-sided endpoint condition.
Elementary function continuity	All elementary functions are continuous on their domain	Applies to polynomials, rationals, trig, exp, log, and root functions.
Composite function continuity	$f (g (x))$ is continuous on $I$ if $g$ is continuous on $I$ and $f$ is continuous on $g (I)$	No extra checks needed if these conditions hold.
Piecewise function continuity	Check continuity at all breakpoints inside $I$ , plus endpoint conditions for closed intervals	Each piece is almost always continuous on its own sub-interval.
Continuity at a breakpoint $c$	$lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x) = f (c)$	Evaluate each one-sided limit with the matching piece for $x < c$ and $x > c$ .
Removable discontinuity	Point not in original domain = discontinuity	Discontinuity exists even if you can simplify the function to remove the hole.

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.

2022 · MCQ
Identify continuous interval
2021 · FRQ
Justify IVT continuity requirement

What's Next

Confirming continuity over an interval is a foundational prerequisite for almost all major topics in AP Calculus BC that come after Unit 1. Immediately next, you will apply this concept to the Intermediate Value Theorem (IVT), which requires a continuous function on a closed interval to guarantee a root or specific output value — without correctly confirming continuity, you cannot correctly apply or justify IVT on free-response questions. Later, this concept is required to relate continuity and differentiability, find intervals of convergence for power series, and apply the Fundamental Theorem of Calculus to definite integrals. Mastering this topic is non-negotiable for a high exam score, as it underpins nearly all calculus justifications.