Confirming Continuity
over an Interval

Master the tools to verify that a function has no breaks across an entire interval: from point-by-point checks to powerful theorems.

AP Calculus BC

What Does Continuity Mean?

A function \( f \) is continuous at a point \( x = c \) if you can draw through that point without lifting your pencil. Formally, three conditions must hold simultaneously:

1

\( f(c) \) is defined: the function has a value at \( c \).

2

\( \displaystyle\lim_{x \to c} f(x) \) exists: the left-hand and right-hand limits agree.

3

\( \displaystyle\lim_{x \to c} f(x) = f(c) \): the limit equals the function value.

Continuous vs. Discontinuous

Why Check an Entire Interval?

Many calculus theorems: the Intermediate Value Theorem, the Extreme Value Theorem, and the Mean Value Theorem: require \( f \) to be continuous on a closed interval \( [a, b] \). A single break invalidates these results.

IVT

If \( f \) is continuous on \( [a,b] \), then \( f \) takes every value between \( f(a) \) and \( f(b) \).

EVT

If \( f \) is continuous on \( [a,b] \), then \( f \) attains an absolute max and min on that interval.

MVT

If \( f \) is continuous on \( [a,b] \) and differentiable on \( (a,b) \), there exists \( c \) with \( f'(c) = \frac{f(b)-f(a)}{b-a} \).

Formal Definition

For an open interval \( (a, b) \), we only need condition 1: continuity at every interior point. The endpoint conditions are dropped because the endpoints are not in the domain.

Endpoint One-Sided Limits

Open vs. Closed Intervals

	Open \((a, b)\)	Closed \([a, b]\)
Endpoints included?	No	Yes
Interior check	Two-sided limits at every \( c \in (a,b) \)	Two-sided limits at every \( c \in (a,b) \)
Left endpoint \( a \)	Not required	\( \lim_{x \to a^+} f(x) = f(a) \)
Right endpoint \( b \)	Not required	\( \lim_{x \to b^-} f(x) = f(b) \)
Used by IVT, EVT?	No	Yes

Functions That Are Always Continuous

The following function families are continuous on their entire domain. If the interval lies within the domain, you can cite the function type to confirm continuity without computing limits:

Polynomials

Continuous on \( (-\infty, \infty) \).

Rational

Continuous where the denominator \( \neq 0 \).

Trig

\( \sin x, \cos x \): all reals. \( \tan x \): excludes \( \frac{\pi}{2} + n\pi \).

Exponential / Log

\( e^x \): all reals. \( \ln x \): on \( (0, \infty) \).

Algebra of Continuous Functions

If \( f \) and \( g \) are continuous at \( c \), then the following are also continuous at \( c \):

1

Sum / Difference: \( f(x) \pm g(x) \)

2

Product: \( f(x) \cdot g(x) \)

3

Quotient: \( \dfrac{f(x)}{g(x)} \) provided \( g(c) \neq 0 \)

4

Composition: \( f(g(x)) \) if \( f \) is continuous at \( g(c) \)

5

Scalar multiple: \( k \cdot f(x) \) for any constant \( k \)

Step-by-Step Checklist

1

Identify the interval. Is it open \( (a,b) \) or closed \( [a,b] \)? Half-open?

2

Classify the function. Is it a polynomial, rational, trig, exponential, or piecewise function?

3

Check the domain. Ensure the function is defined at every point in the interval (no division by zero, no negative inside an even root, no log of non-positive).

4

Check interior points. For piecewise functions, verify continuity at each piece boundary inside the interval.

5

Check endpoints (if closed). Verify the appropriate one-sided limit equals the function value at each endpoint.

Example 1: Polynomial

Confirm that \( f(x) = 3x^3 - 2x + 5 \) is continuous on \( [-1, 4] \).

Example 2: Rational Function

Is \( g(x) = \dfrac{x^2 + 1}{x - 3} \) continuous on \( [0, 2] \)?

1

Type: Rational function: continuous on its domain.

2

Domain restriction: \( x - 3 = 0 \Rightarrow x = 3 \). The function is undefined at \( x = 3 \).

3

Check interval: \( x = 3 \notin [0, 2] \), so \( g \) is defined everywhere in \( [0, 2] \).

Example 3: Piecewise Function

Confirm continuity of \( h(x) \) on \( [0, 6] \):

1

Each piece: \( x^2 + 1 \) and \( 3x + 1 \) are polynomials, so each is continuous on its sub-interval.

2

Boundary at \( x = 3 \): Left-hand limit: \( \lim_{x \to 3^-}(x^2+1) = 10 \). Right-hand limit: \( \lim_{x \to 3^+}(3x+1) = 10 \). And \( h(3) = 3^2 + 1 = 10 \).

3

Result: Both one-sided limits equal \( h(3) = 10 \), so \( h \) is continuous at \( x = 3 \). Combined with polynomial continuity on each piece, \( h \) is continuous on \( [0, 6] \). ✓

Example 4: Composition

Is \( f(x) = \ln(\sin x) \) continuous on \( (0, \pi) \)?

1

Inner function: \( \sin x \) is continuous everywhere. On \( (0, \pi) \), \( \sin x > 0 \).

2

Outer function: \( \ln(u) \) is continuous for \( u > 0 \). Since \( \sin x > 0 \) on \( (0, \pi) \), the input to \( \ln \) is always positive.

3

Composition rule: The composition of two continuous functions is continuous. Therefore \( f(x) = \ln(\sin x) \) is continuous on \( (0, \pi) \). ✓

Common Mistakes

✕ Ignoring Endpoints

For a closed interval \( [a,b] \), you must check one-sided limits at \( a \) and \( b \). Checking only interior points is insufficient.

✕ Forgetting Domain Restrictions

A rational function may look continuous, but if the denominator equals zero somewhere in the interval, continuity fails.

✕ Skipping Piecewise Boundaries

Each sub-piece may be continuous, but you must verify that the pieces agree at every transition point inside the interval.

Summary of Strategies

Function Type	Strategy	Watch Out For
Polynomial	Always continuous on \( \mathbb{R} \)	Nothing: automatically continuous
Rational	Continuous on domain	Zeros of denominator in the interval
Trig	Continuous on domain	\( \tan, \sec, \csc, \cot \) have asymptotes
Exponential / Log	Continuous on domain	\( \ln(x) \) requires \( x > 0 \)
Piecewise	Check each piece + boundaries	Limits must match \( f(c) \) at transitions
Composition	Use composition rule	Range of inner must lie in domain of outer

Question 1 of 5

Which of the following is not one of the three conditions for continuity at a point \( x = c \)?

Continuity requires the function to be defined, the limit to exist, and the limit to equal the function value. Differentiability (\( f'(c) \) exists) is a stronger condition: it implies continuity but is not required for it.

Question 2 of 5

On which interval is \( f(x) = \dfrac{1}{x - 2} \) continuous?

The function \( \frac{1}{x-2} \) is undefined at \( x = 2 \). Intervals \( [0,3] \), \( [1,4] \), and \( (-1,2] \) all contain \( x = 2 \). Only \( [3, 5] \) avoids the discontinuity entirely.

Question 3 of 5

Let \( g(x) = \begin{cases} 2x + 1 & \text{if } x < 2 \\ 7 & \text{if } x = 2 \\ x^2 + 1 & \text{if } x > 2 \end{cases} \). Is \( g \) continuous at \( x = 2 \)?

\( \lim_{x \to 2^-}(2x+1) = 5 \) and \( \lim_{x \to 2^+}(x^2+1) = 5 \), so the two-sided limit is \( 5 \). But \( g(2) = 7 \neq 5 \). The limit exists and the function is defined, but they don't match: condition 3 fails.

Question 4 of 5

To confirm that \( f \) is continuous on the closed interval \( [a, b] \), which additional condition(s) must be checked beyond interior continuity?

At endpoints, we use one-sided limits: the right-hand limit at \( a \) and the left-hand limit at \( b \). Two-sided limits (option C) don't apply at endpoints because the function may not be defined outside \( [a,b] \).

Question 5 of 5

The function \( f(x) = \sqrt{4 - x^2} \) has domain \( [-2, 2] \). Which statement correctly confirms its continuity?

\( f(x) = \sqrt{4 - x^2} \) is a composition: \( \sqrt{u} \) (continuous for \( u \geq 0 \)) composed with \( u = 4 - x^2 \) (a polynomial, so continuous everywhere). On \( [-2, 2] \), \( 4 - x^2 \geq 0 \), so the input to \( \sqrt{} \) is always valid. The composition rule confirms continuity on \( [-2, 2] \).

Confirming Continuityover an Interval

What Does Continuity Mean?

Continuous vs. Discontinuous

Why Check an Entire Interval?

Formal Definition

Endpoint One-Sided Limits

Open vs. Closed Intervals

Functions That Are Always Continuous

Algebra of Continuous Functions

Step-by-Step Checklist

Example 1: Polynomial

Example 2: Rational Function

Example 3: Piecewise Function

Example 4: Composition

Common Mistakes

Summary of Strategies

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Confirming Continuity
over an Interval