Confirming Continuity over an Interval - AP Calculus BC

SECTION 1

Historical Context & Motivation

The notion of a function being continuous — behaving without breaks or sudden jumps — has been central to analysis since the early development of calculus. Newton and Leibniz implicitly assumed continuity when computing derivatives and integrals, yet neither provided a formal definition. It was not until the nineteenth century that mathematicians encountered pathological functions (nowhere-differentiable curves, for instance) that demanded precise language for what continuity actually means, both at a point and across an interval.

1687 • Newton's Principia

Newton uses 'fluent' quantities that vary continuously, relying on intuitive continuity without formal definitions.

1817 • Bolzano's Intermediate Value Theorem

Bernard Bolzano provides one of the first rigorous proofs that a continuous function on a closed interval attains every value between its endpoints.

1821 • Cauchy's Cours d'Analyse

Augustin-Louis Cauchy publishes the first systematic ε-δ framework for limits and continuity, formalizing the concept at a point.

1861 • Weierstrass Refines ε-δ

Karl Weierstrass sharpens Cauchy's definitions and introduces uniform continuity, distinguishing pointwise from interval-wide behavior.

Why does confirming continuity over an interval matter for modern calculus students? Theorems you rely on daily — the Intermediate Value Theorem, the Extreme Value Theorem, and the Fundamental Theorem of Calculus — all require continuity on an interval as a hypothesis. If you cannot verify that hypothesis, you cannot invoke the theorem. This lesson equips you to rigorously confirm that a function is continuous on a specified interval.

SECTION 2

Core Principles & Definitions

Before addressing intervals, recall the definition of continuity at a point: a function f is continuous at x = c if and only if (1) f(c) is defined, (2) lim as x → c of f(x) exists, and (3) that limit equals f(c). Continuity over an interval extends this pointwise check to every point in the interval, with special attention to the endpoints when the interval is closed.

Continuity at an Interior Point

For every point c in the open interval (a, b), the two-sided limit of f(x) as x → c must exist and equal f(c).

Right-Continuity at a Left Endpoint

At x = a (left endpoint of [a, b]), the right-hand limit lim x → a⁺ f(x) must equal f(a). The left-hand limit is irrelevant because x = a is the boundary.

Left-Continuity at a Right Endpoint

At x = b (right endpoint of [a, b]), the left-hand limit lim x → b⁻ f(x) must equal f(b). We only approach from within the interval.

Open vs. Closed Intervals

On an open interval (a, b), only two-sided interior checks are needed. On a closed interval [a, b], the one-sided endpoint checks are additionally required.

SECTION 3

Visualizing Continuity & Discontinuity on an Interval

Left panel: a function continuous on the closed interval [a, b] — the curve can be drawn without lifting the pen, and both endpoints are filled (included). Right panel: a jump discontinuity at an interior point c breaks continuity on [a, b].

In the left panel, every point on the curve between a and b satisfies the three-part continuity test: the function value exists, the limit exists, and they agree. The filled circles at a and b represent the one-sided endpoint checks — right-continuous at a, left-continuous at b. In the right panel, the open circle at c indicates that the limit from the left and the limit from the right disagree (a jump discontinuity), violating condition (2) at that interior point. Even though the function is well-behaved everywhere else, this single failure prevents us from declaring continuity on the interval.

SECTION 4

Mathematical Framework

We now formalize the ideas from Section 2. The three definitions below constitute the complete mathematical framework for confirming continuity on open, closed, and half-open intervals.

CONTINUITY AT AN INTERIOR POINT

f is continuous at c ∈ (a, b) ⟺ lim(x→c) f(x) = f(c)

This is the standard three-part test compressed into a single equality: it implicitly requires f(c) to be defined and the two-sided limit to exist.

CONTINUITY ON A CLOSED INTERVAL [a, b]

f continuous on [a, b] ⟺ (∀c ∈ (a,b): lim(x→c) f(x) = f(c)) ∧ lim(x→a⁺) f(x) = f(a) ∧ lim(x→b⁻) f(x) = f(b)

The conjunction (∧) requires all three parts: interior two-sided continuity, right-continuity at the left endpoint, and left-continuity at the right endpoint.

PROPERTIES PRESERVING CONTINUITY

If f and g are continuous at c, then f ± g, f · g, f/g (g(c) ≠ 0), and f ∘ g are continuous at c.

Sums, differences, products, quotients (with nonzero denominator), and compositions of continuous functions remain continuous. This lets you build arguments about complicated functions from simpler ones.

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AP Exam Tip

SECTION 5

Types of Discontinuities & How They Block Interval Continuity

When confirming continuity on an interval, you need to know exactly what can go wrong. There are three primary types of discontinuity, each violating a different part of the continuity definition. Recognizing them lets you target your analysis efficiently: check the points most likely to be problematic rather than blindly testing every value.

Three discontinuity types that prevent interval continuity. A removable discontinuity (left) can theoretically be patched by redefining f(c). A jump discontinuity (center) involves unequal one-sided limits. An infinite discontinuity (right) features a vertical asymptote.

Type	What Fails	Common Cause
Removable	f(c) is undefined or f(c) ≠ lim f(x)	Common factor cancels (e.g., (x²−1)/(x−1) at x = 1)
Jump	lim x→c⁻ f(x) ≠ lim x→c⁺ f(x)	Piecewise definitions with mismatched branches
Infinite	Limit does not exist (unbounded)	Vertical asymptote, e.g., 1/(x − 2) at x = 2

Summary of the three discontinuity types

SECTION 6

Worked Example: Piecewise Function on [−1, 4]

Consider the piecewise function f defined by f(x) = x² + 1 for x < 2, and f(x) = 3x − 1 for x ≥ 2. Determine whether f is continuous on the closed interval [−1, 4].

Worked example

1
Step 1 — Identify the pieces and the potential trouble point

The function has two branches: f(x) = x² + 1 on (−∞, 2) and f(x) = 3x − 1 on [2, ∞). Each branch is a polynomial, hence continuous on its own domain. The only place where continuity might fail is the junction point x = 2, which lies in the interior of [−1, 4].

Potential discontinuity at x = 2
2
Step 2 — Compute the left-hand limit at x = 2

For x → 2⁻ we use the first branch: lim(x→2⁻) (x² + 1) = (2)² + 1 = 5.

lim(x→2⁻) f(x) = 5
3
Step 3 — Compute the right-hand limit at x = 2

For x → 2⁺ we use the second branch: lim(x→2⁺) (3x − 1) = 3(2) − 1 = 5.

lim(x→2⁺) f(x) = 5
4
Step 4 — Evaluate f(2) and compare

Since x = 2 falls in the branch x ≥ 2, f(2) = 3(2) − 1 = 5. Because the left-hand limit, right-hand limit, and function value all equal 5, f is continuous at x = 2.

f(2) = 5 = lim f(x) → continuous at x = 2
5
Step 5 — Check the endpoints of [−1, 4]

At x = −1 (left endpoint), we need right-continuity. Since −1 < 2, f(−1) = (−1)² + 1 = 2 and lim(x→−1⁺) (x² + 1) = 2. These agree. At x = 4 (right endpoint), we need left-continuity. Since 4 ≥ 2, f(4) = 3(4) − 1 = 11 and lim(x→4⁻) (3x − 1) = 11. These agree.

f is continuous on [−1, 4]

SECTION 7

Strategies & Common Pitfalls

Strategy	When to Use	Pitfall to Avoid
Cite function type (polynomial, exponential, etc.)	Non-piecewise functions whose domain covers the interval	Forgetting domain restrictions — e.g., √x is not defined for x < 0
Check piecewise junctions	Any piecewise-defined function	Only checking that limits agree but neglecting to verify f(c) matches
Use algebraic continuity theorems (sums, products, compositions)	Complex expressions built from simple continuous functions	Applying quotient rule without verifying the denominator ≠ 0 on the interval
One-sided limits at endpoints	Closed or half-open intervals	Using a two-sided limit at an endpoint, which is outside the domain

Strategies and associated pitfalls when confirming continuity

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Key takeaway

EXAM STRATEGY

SECTION 8

Connections to Major Theorems

Confirming continuity is rarely an end in itself; it is the gateway hypothesis for the most powerful results in calculus. The table below shows how interval continuity feeds directly into theorems you will use throughout the AP Calculus BC course. Failure to verify continuity before invoking any of these theorems is a logical error that the AP exam specifically tests.

Theorem	Hypothesis	Conclusion
Intermediate Value Theorem (IVT)	f continuous on [a, b]	For any value L between f(a) and f(b), there exists c ∈ (a, b) with f(c) = L.
Extreme Value Theorem (EVT)	f continuous on [a, b]	f attains an absolute maximum and an absolute minimum on [a, b].
Fundamental Theorem of Calculus (FTC)	f continuous on [a, b]	∫ from a to b of f(x) dx = F(b) − F(a), where F′ = f.
Mean Value Theorem (MVT)	f continuous on [a, b], differentiable on (a, b)	∃ c ∈ (a, b) with f′(c) = (f(b) − f(a))/(b − a).

Key calculus theorems requiring continuity on an interval

As you advance into topics like improper integrals, Taylor series convergence, and differential equations in Calculus BC, the requirement for continuity (or piecewise continuity) will recur. Mastering how to confirm continuity now builds the verification habit that will serve you across every major unit.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following conditions is NOT required to confirm that f is continuous on the closed interval [a, b]?

A. f(c) is defined for every c in (a, b)
B. lim(x→a⁺) f(x) = f(a)
C. lim(x→b⁻) f(x) = f(b)
D. lim(x→a⁻) f(x) = f(a)
E. lim(x→c) f(x) = f(c) for every c in (a, b)

Show answer (D)

D is correct. At the left endpoint a of a closed interval [a, b], only the right-hand limit is required: lim(x→a⁺) f(x) = f(a). The left-hand limit lim(x→a⁻) examines values of x less than a, which lie outside the interval and are irrelevant. Option A is required (function values must be defined at interior points). Option B is the right-continuity condition at a — required. Option C is the left-continuity condition at b — required. Option E is the standard interior continuity condition — required.

Question 2: BASIC CALCULATION

Let g(x) = (x² − 9)/(x − 3) for x ≠ 3, and g(3) = 7. On which of the following intervals is g continuous?

A. [0, 5]
B. [3, 6]
C. [4, 6]
D. [1, 3]
E. (−∞, ∞)

Show answer (C)

C is correct. For x ≠ 3, g(x) = (x − 3)(x + 3)/(x − 3) = x + 3, so lim(x→3) g(x) = 6. But g(3) = 7 ≠ 6, creating a removable discontinuity at x = 3. The interval [4, 6] does not contain x = 3, and on that interval g(x) = x + 3, a polynomial, hence continuous. Option A includes x = 3 in its interior — discontinuous there. Option B has x = 3 as its left endpoint; lim(x→3⁺) g(x) = 6 ≠ 7 = g(3), so right-continuity fails. Option D has x = 3 as its right endpoint; lim(x→3⁻) g(x) = 6 ≠ 7 = g(3), so left-continuity fails. Option E includes x = 3 and is therefore incorrect.

Question 3: INTERMEDIATE

Define h(x) = { 2x + k, for x ≤ 1; x² + 3, for x > 1 }. For what value of k is h continuous on [−2, 5]?

A. k = 0
B. k = 1
C. k = 2
D. k = 3
E. k = 4

Show answer (C)

C is correct. Both branches are polynomials, so the only concern is the junction at x = 1. Left-hand limit: lim(x→1⁻) (2x + k) = 2 + k. Right-hand limit: lim(x→1⁺) (x² + 3) = 4. Function value: h(1) = 2(1) + k = 2 + k. For continuity, 2 + k = 4, so k = 2. Option A gives h(1) = 2 ≠ 4. Option B gives h(1) = 3 ≠ 4. Option D gives h(1) = 5 ≠ 4. Option E gives h(1) = 6 ≠ 4. Only k = 2 makes all three quantities equal.

Question 4: APPLIED

A temperature sensor records the temperature T(t) in °C of a chemical reaction, where t is time in minutes. The function T is defined by T(t) = { 20 + 5t, for 0 ≤ t < 4; t² + a, for 4 ≤ t ≤ 10 }. (a) Find the value of a that makes T continuous on [0, 10]. (b) Explain why continuity of T on [0, 10] guarantees that the reaction reaches exactly 50°C at some time in that interval.

Show answer

(a) For continuity at t = 4: lim(t→4⁻) (20 + 5t) = 20 + 20 = 40. We need T(4) = 4² + a = 16 + a = 40, so a = 24. (b) With a = 24, T(0) = 20 and T(10) = 100 + 24 = 124. Since T is continuous on [0, 10] and 50 is between T(0) = 20 and T(10) = 124, the Intermediate Value Theorem guarantees the existence of some c ∈ (0, 10) such that T(c) = 50. [1 pt for computing the left-hand limit = 40. 1 pt for solving a = 24. 1 pt for citing IVT with correct reasoning.]

Question 5: CRITICAL THINKING

Let f be defined on [0, 6] by f(x) = { x sin(1/x), for 0 < x ≤ 6; 0, for x = 0 }. (a) Prove that f is continuous at x = 0. (b) Explain why f is continuous on (0, 6]. (c) Conclude that f is continuous on [0, 6]. (d) The Extreme Value Theorem applies to f on [0, 6]. Explain what it guarantees and discuss why verifying continuity was essential before making that claim.

Show answer

(a) Since −1 ≤ sin(1/x) ≤ 1 for all x ≠ 0, we have −|x| ≤ x sin(1/x) ≤ |x| for x > 0. As x → 0⁺, both −|x| and |x| tend to 0. By the Squeeze Theorem, lim(x→0⁺) x sin(1/x) = 0. Since f(0) = 0, we have lim(x→0⁺) f(x) = f(0), establishing right-continuity at x = 0. [1 pt: Squeeze Theorem; 1 pt: concludes lim = f(0).] (b) On (0, 6], f(x) = x sin(1/x). The functions x and sin(1/x) are both continuous on (0, 6] — x is a polynomial and sin(1/x) is a composition of sin (continuous everywhere) with 1/x (continuous for x ≠ 0). Their product is therefore continuous on (0, 6]. [1 pt.] (c) From (a), f is right-continuous at x = 0. From (b), f is continuous at every point in (0, 6]. At the right endpoint x = 6, left-continuity follows from (b) since f is continuous at every interior point and at 6 itself. Combining these, f is continuous on [0, 6] by definition. [1 pt.] (d) Because f is continuous on the closed interval [0, 6], the Extreme Value Theorem guarantees that f attains an absolute maximum M and an absolute minimum m at some points in [0, 6]. Without first verifying continuity, we could not invoke EVT — if f had a discontinuity (for example, an unbounded oscillation), the function might fail to achieve a maximum or minimum, and the theorem's conclusion would be invalid. [1 pt for EVT statement; 1 pt for justification of why continuity was essential.]