Continuity over an Interval | AP Calculus Topic 1.12 Notes

1. From continuity at a point to an entire interval

A function is continuous on an interval when it is continuous at every point of that interval, using the appropriate one-sided approach at each included endpoint. One interior hole, jump or other failure is enough to prevent continuity on the whole interval.

At an interior point, the assigned value must be defined, the two-sided limit must exist as a finite real number, and they must agree:

lim_{x \to c} f (x) = f (c)

Checking a few sample inputs is not sufficient. Instead, use continuity theorems for known function families to cover ordinary points, then check domain restrictions and piecewise boundaries. Review Topic 1.11: continuity at a point for the full three-condition test.

An interval statement covers every included point

The pencil test is useful intuition, but a graph can hide small holes or jumps. The definition and domain give the justification. Continuity also does not mean smoothness: a corner can be continuous.

2. Open, closed, half-open and infinite intervals

Read the brackets before doing any calculations. Parentheses exclude a finite endpoint; square brackets include it. An included endpoint needs an assigned value and the relevant finite one-sided limit.

Interval notation determines which endpoint checks are needed
Type	Notation	Points included	Continuity requirements
Open	$(a, b)$	$a < x < b$	All interior points. Neither endpoint belongs to the requested interval.
Closed	$[a, b]$	$a \leq x \leq b$	All interior points, right continuity at the left endpoint and left continuity at the right endpoint.
Includes left only	$[a, b)$	$a \leq x < b$	All interior points and right continuity at the left endpoint.
Includes right only	$(a, b]$	$a < x \leq b$	All interior points and left continuity at the right endpoint.
Infinite	$(a, \infty)$ , $(- \infty, b)$ , $(- \infty, \infty)$	Every real input satisfying the interval bounds.	All included real points; check a finite endpoint if it is included. Infinity always uses a parenthesis.

An infinite interval may include its finite boundary, such as $[a, \infty)$ or $(- \infty, b]$ . There is no function value “at infinity” to check. Continuity on an unbounded interval does not require a finite limit at infinity or bounded outputs.

3. The full test on a closed interval

Assume $a < b$ . To confirm continuity on $[a, b]$ , establish three parts.

Interior continuity

lim_{x \to c} f (x) = f (c), a < c < b

Use a valid continuity theorem or evaluate the side limits at any exceptional interior boundary. The limit and the assigned value must both be finite and agree.

Right continuity at the left endpoint

lim_{x \to a^{+}} f (x) = f (a)

Left continuity at the right endpoint

lim_{x \to b^{-}} f (x) = f (b)

The approach is taken within the requested interval. Even if the formula is defined outside it, continuity on that restricted interval only needs the inward approach at its endpoints. For a half-open interval, apply the appropriate endpoint equality only where the finite endpoint is included.

An open endpoint needs no assigned value

A function may approach infinity near an excluded endpoint and still be continuous at every point of the open interval. Including that endpoint changes the question: a finite assigned value and matching finite inward limit are then required.

4. Check the real domain before claiming continuity

A function cannot be continuous on a requested interval if it is undefined at any included input. Record original exclusions before simplifying, then intersect all domain restrictions when operations or compositions are combined.

Common restrictions for real-valued functions
Source of restriction	Condition	Example
Denominator	It must be nonzero.	$\frac{1}{x - 2}$ excludes $x = 2$ .
Even root	The radicand must be nonnegative.	$\sqrt{x - 3}$ requires $x \geq 3$ .
Logarithm	Its argument must be strictly positive.	$\ln (x)$ requires $x > 0$ .
Tangent or secant	Cosine in the denominator cannot vanish.	$\tan (x)$ and $\sec (x)$ exclude $x = \frac{π}{2} + n π$ , for integer $n$ .
Cotangent or cosecant	Sine in the denominator cannot vanish.	$\cot (x)$ and $\csc (x)$ exclude $x = n π$ , for integer $n$ .

Trigonometric inputs are measured in radians. A transformation changes the condition on the input: for a logarithm of an expression, solve for a positive argument; for a square root of an expression, solve for a nonnegative radicand.

Being defined everywhere on an interval is necessary but not sufficient for an arbitrary function. A fully defined piecewise function can still jump. Domain inclusion proves continuity when paired with an applicable continuity theorem, not by itself.

5. Continuous function families and useful operations

Standard polynomial, rational, power, exponential, logarithmic and trigonometric functions are continuous at real inputs in their domains. Use this fact to justify all ordinary points rather than testing them individually.

Use continuity theorems together with the real domain
Family	Typical expression	Domain reminder
Polynomial	$a_{n} x^{n} + \dots + a_{0}$	All real inputs for a polynomial with nonnegative integer powers.
Rational	$\frac{p (x)}{q (x)}$	Exclude zeros of the original denominator.
Exponential	$a^{x}$ or $e^{x}$	All real inputs for a positive base.
Logarithmic	$\ln (x)$ or $\log_{a} (x)$	Positive arguments; the logarithm base is positive and different from one.
Sine and cosine	$\sin (x)$ , $\cos (x)$	All real inputs.
Other trig functions	$\tan (x)$ , $\sec (x)$ , $\cot (x)$ , $\csc (x)$	Exclude inputs at which their defining denominator vanishes.
Power and root functions	$x^{n}$ ; $\sqrt{x}$	Depends on the exponent and the chosen real-valued definition; even roots need nonnegative radicands, while negative powers also exclude zero.

Sums, differences, products and constant multiples preserve continuity where the component functions are continuous. Quotients also require a nonzero denominator. For a composition, the inner output must lie in the outer function's domain, with the inner and outer functions continuous at the relevant inputs.

A concise proof can therefore identify the family, show that the whole interval lies in the domain, and conclude continuity there. For a piecewise definition, justify each branch and separately inspect its boundaries.

6. A step-by-step strategy for the whole interval

Read the interval.
Mark which endpoints are included. Distinguish an open interval, a closed interval, a half-open interval and an unbounded interval.
Find the real domain.
List denominator zeros, root restrictions, logarithm restrictions and branch conditions. An excluded input inside the requested interval immediately prevents continuity on the whole interval.
Cover the interior.
Use standard continuity results on each ordinary branch. Check every included formula-change point and other exceptional input using the definition.
Check included endpoints.
Use the inward one-sided limit at each included finite endpoint, including the single endpoint of a half-open interval.
State and justify the result.
If continuous, name the interval and the supporting property. If not, give the location and reason. If asked for intervals of continuity, split at genuine exclusions or failures and state the convention used.

This is also a decision guide: a known continuous family reduces the problem to its domain; piecewise definitions need branch and boundary checks; an arbitrary unfamiliar function needs additional evidence. An absence of obvious denominator zeros alone is not a universal continuity test.

7. Five complete worked examples

Example 1: a polynomial on a closed interval

f (x) = 2 x^{3} - 5 x + 1

Show continuity on $[- 2, 3]$ . The function is a polynomial, so it is continuous at every real input. The entire requested interval is within its domain. This theorem covers the interior and implies the needed inward endpoint equalities.

lim_{x \to {- 2}^{+}} f (x) = f (- 2) = - 5

lim_{x \to 3^{-}} f (x) = f (3) = 40

Conclusion: the function is continuous on the entire closed interval. A polynomial continuity theorem is enough to justify this without checking infinitely many inputs one by one.

Example 2: a rational function with an excluded endpoint

g (x) = \frac{x + 3}{x - 1}

Consider $(1, \infty)$ . The only denominator zero is one. For every input in this interval, the denominator is positive and nonzero:

x > 1 \Rightarrow x - 1 > 0

Rational functions are continuous wherever their original denominator is nonzero. The requested interval excludes one, so the function is continuous at every point of it. No value at one or finite limit at infinity is required. If the interval included one, the original function would fail the domain check.

Conclusion: continuous on the requested open interval despite its unbounded behavior near the excluded endpoint.

Example 3: a piecewise function on a closed interval

h (x) = {\begin{cases} x^{2} & if x \leq 1 \\ 2 x - 1 & if x > 1 \end{cases}

Check $[0, 3]$ . Both branches are polynomials. They are continuous away from their joining input one, so check that input and the two included interval endpoints.

h (1) = 1^{2} = 1

lim_{x \to 1^{-}} h (x) = lim_{x \to 1^{-}} x^{2} = 1

lim_{x \to 1^{+}} h (x) = lim_{x \to 1^{+}} (2 x - 1) = 1

Both side limits equal the included value. Now confirm the inward endpoint limits.

lim_{x \to 0^{+}} h (x) = 0 = h (0)

lim_{x \to 3^{-}} h (x) = 5 = h (3)

Conclusion: every interior point and both included endpoints pass; the function is continuous on the entire closed interval.

Only the requested closed interval

[0, 3]

is drawn. The branches join at

(1, 1)

; filled endpoint markers show inclusion.

Example 4: a removable hole inside the interval

k (x) = \frac{x^{2} - 4}{x - 2}

The requested interval is $[0, 4]$ . The denominator vanishes at two, which lies inside it. The original function is undefined there, so it cannot be continuous on the whole interval.

\frac{x^{2} - 4}{x - 2} = x + 2, x \neq 2

lim_{x \to 2} k (x) = 4

The missing point is removable because the nearby limit is finite. Simplifying the expression does not fill the hole. Conclusion: not continuous on the requested closed interval as originally defined.

Within that range, it is continuous separately on these two intervals:

[0, 2) and (2, 4]

Their union is a disconnected set, not a single interval spanning the hole. Assigning a new value of four at two would produce a continuous extension over the whole closed interval.

The open point

(2, 4)

lies inside the requested interval. Endpoint continuity cannot compensate for a missing interior value.

Example 5: the domain of a logarithmic composition

f (x) = \ln (3 x)

The logarithm requires a strictly positive argument. Solve the domain condition first:

3 x > 0 \Leftrightarrow x > 0

The inner linear function is continuous, and the natural logarithm is continuous for positive arguments. Their composition is therefore continuous on its full real domain:

(0, \infty)

Any subinterval contained in this domain also works. Zero is excluded; including it would violate the domain condition.

8. Piecewise boundaries and parameter questions

Check each branch on its own assigned region, then inspect every formula-change input inside the requested interval. At an interior join, continuity requires the two finite side limits to equal the actual included value. Read equality signs in the branch conditions carefully.

lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x) = f (c)

If a formula-change point is an endpoint of the requested interval, use only the inward approach relevant to continuity on that interval. Values on the other side are not part of that restricted question.

Extension: select a branch parameter

F (x) = {\begin{cases} k x & if x < 1 \\ x + 1 & if x \geq 1 \end{cases}

lim_{x \to 1^{-}} F (x) = k, lim_{x \to 1^{+}} F (x) = F (1) = 2

Choosing $k = 2$ makes the join continuous. Since the branches are polynomials with no other restrictions, this choice gives continuity on every real interval. In general, solve the matching equation, then recheck domain and endpoint conditions; a parameter candidate is not a substitute for the complete test.

Continue with Topic 1.13: removing discontinuities for more repair and parameter examples.

9. Find maximal intervals and preserve exclusions

When asked for maximal open intervals of continuity, locate the exceptional inputs, sort them and split the real line. Check that the function is continuous throughout every reported piece. A smaller subinterval can also be valid, but “maximal” asks for pieces that cannot be extended as open intervals without crossing a failure.

A canceled factor still splits the original domain

R (x) = \frac{x + 1}{x^{2} - 1}

\frac{x + 1}{x^{2} - 1} = \frac{1}{x - 1}, x \neq - 1, x \neq 1

The original domain excludes both denominator zeros. The exclusion at negative one is a removable hole, while one is a pole. The maximal open intervals for the original function are:

(- \infty, - 1), (- 1, 1), (1, \infty)

A union of these intervals describes the domain but is not one connected interval. Do not join across a canceled exclusion unless the problem explicitly defines a continuous extension there.

If the question allows endpoint inclusion, state that convention. At a jump, a point assigned to one branch may belong to a one-sided interval of continuity even though it is not continuous through the jump. Ordinary “maximal open intervals” reporting avoids that ambiguity.

10. Point continuity versus interval continuity

The same definition applied at one point or throughout a region
Aspect	At a point	On an interval
Scope	Check the assigned value and the relevant finite limit at one target.	Every included point must satisfy the appropriate continuity condition.
Interior limits	Usually two-sided at an interior domain point.	Two-sided throughout the interior, including piecewise joins.
Endpoints	A domain-aware point check may itself be one-sided.	Use inward one-sided limits at each included finite endpoint, including a half-open interval endpoint.
Proof	Give a value, establish a finite limit and compare them, or cite a suitable theorem.	Use theorems across ordinary regions and explicitly check all exceptional points and included endpoints.
Failure	One target may fail.	One included failure prevents continuity on the whole requested interval.

Interval notation is part of the mathematical answer. A parenthesis can intentionally leave out a troublesome endpoint; a bracket cannot be added just because the function is continuous nearby.

11. Tips, decision checks and AP Calculus explanations

A clear explanation identifies the requested interval, the real domain, a continuity property for ordinary points and any required boundary checks. If continuity fails, name a specific included input and the failed condition.

For a standard family: establish that the entire interval lies in its domain, then cite the appropriate continuity theorem.
For a rational function: solve for every original denominator zero and compare it with the interval.
For roots and logarithms: solve the radicand or argument inequality, then compare domains.
For piecewise functions: show one-sided calculations at interior joins and inward calculations at included endpoints.
For a graph: inspect every included input for holes, mismatched values, jumps or unbounded behavior. A finite-resolution picture may miss exceptional points.

Typical questions

On what intervals is the function continuous? Justify the answer.
Is the function continuous on a specified closed interval? Use the definition.
Find parameters that make a piecewise function continuous on the interval.
Use a graph to identify intervals of continuity.

Use complete limit notation with the function being approached, not a bare limit symbol. Show enough reasoning to support the requested conclusion; the precise scoring requirements depend on the individual question.

12. Common mistakes and how to avoid them

Ignoring brackets. Decide which endpoints belong to the interval before evaluating limits.
Checking endpoints only for closed intervals. A half-open interval also has one included endpoint to check.
Using the outward approach. The left endpoint needs a right-hand limit; the right endpoint needs a left-hand limit.
Assuming all piecewise functions join. Each included interior boundary needs matching finite side limits and the correct value.
Missing domain restrictions. Keep original denominator exclusions and solve root and logarithm conditions.
Calling rational functions continuous everywhere. They are continuous where their original denominators are nonzero.
Equating “defined” with “continuous.” Domain inclusion needs a supporting continuity result or definition check.
Using a few sample values as proof. Continuity is required at every included point.
Including infinity with a square bracket. Infinity is not a real endpoint value.
Reporting a union as one interval. Separate pieces remain disconnected if a missing input lies between them.

For classification help, review Topic 1.10: discontinuity types. For the underlying notation, see defining limits.

13. Four practice problems with full explanations

Identify the domain, the relevant interior joins and the included endpoints before opening the solution. All four original practice problems are preserved.

1. A polynomial on a closed interval

f (x) = x^{3} - 2 x + 5

Requested interval: $[- 1, 4]$ .

Check the interval test and explanation

A polynomial is continuous at every real input. The whole requested interval lies in its domain, and the continuity theorem includes the needed inward endpoint equalities. Therefore it is continuous on the closed interval.

2. A rational function on an open interval

g (x) = \frac{x + 1}{x - 3}

Requested interval: $(3, \infty)$ .

Check the interval test and explanation

The only denominator zero is three, and that input is excluded from the requested interval. At every included input the denominator is nonzero. By rational continuity, the function is continuous on this open interval; no endpoint value at three is needed.

3. A piecewise function with an interior jump

h (x) = {\begin{cases} x^{2} & if x < 2 \\ 4 x & if x \geq 2 \end{cases}

Requested interval: $[0, 5]$ .

Check the interval test and explanation

lim_{x \to 2^{-}} h (x) = 4

lim_{x \to 2^{+}} h (x) = 8 = h (2)

Two lies inside the requested interval. Both side limits are finite but unequal, so this is a jump and the function is not continuous on the whole interval. Continuity of the individual polynomial branches does not remove the interior mismatch.

4. A square root with an included domain endpoint

k (x) = \sqrt{x - 1}

Requested interval: $[1, 10]$ .

Check the interval test and explanation

lim_{x \to 1^{+}} k (x) = 0 = k (1)

lim_{x \to 10^{-}} k (x) = 3 = k (10)

The real domain starts at one and includes the entire requested interval. Root continuity covers its interior. Both included endpoints have matching finite inward limits and assigned values, so the function is continuous on the closed interval.

The plot shows only

[1, 10]

. Its filled markers are the requested interval endpoints; the full real-domain curve continues beyond ten. At one, only the right-hand approach is available.

14. Quick reference for endpoint checks

Endpoint equalities require finite assigned values and finite limits
Requested interval	Interior requirement	Included endpoint requirement
$(a, b)$	Continuous at every interior point.	Neither endpoint is included.
$[a, b]$	Continuous at every interior point.	$lim_{x \to a^{+}} f (x) = f (a)$ and $lim_{x \to b^{-}} f (x) = f (b)$ .
$[a, b)$	Continuous at every interior point.	$lim_{x \to a^{+}} f (x) = f (a)$ .
$(a, b]$	Continuous at every interior point.	$lim_{x \to b^{-}} f (x) = f (b)$ .
$[a, \infty)$	Continuous at all included real inputs above the finite endpoint.	Right continuity at the included finite endpoint; no value at infinity.

One failed included point is enough for a “not continuous on the whole interval” conclusion. A positive conclusion must cover every included point, usually through a theorem plus checks at exceptional boundaries.

15. Why interval continuity matters and what comes next

Interval continuity is a hypothesis in several central calculus theorems. Identify the actual interval and verify the relevant conditions before using a conclusion.

Topic 1.13, removing discontinuities: find finite limiting values that can repair holes or choose compatible branch parameters.
Topic 1.16, Intermediate Value Theorem: continuity on a closed interval guarantees intermediate output values.
Unit 2, derivatives: differentiability at an interior point implies continuity there; continuity alone does not guarantee differentiability.
Unit 5, Mean Value Theorem: requires continuity on the closed interval and differentiability on its interior.
Unit 6, integration: standard Fundamental Theorem of Calculus results use stated continuity hypotheses. Continuity is a useful sufficient condition for integrability; some discontinuous functions are integrable too.

Previous: Topic 1.11 — Continuity at a point →Next: Topic 1.13 — Removing discontinuities →All Unit 1 study notes →Unit 1 practice test collection →

Keep the overall method simple: read the interval, preserve the domain, justify ordinary points, test joins and check every included endpoint.

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