Confirming continuity over an interval — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The three-part continuity test at a point, endpoint continuity for closed intervals, continuity of common function families, and testing continuity over open, closed, and half-open intervals per AP CED requirements.

You should already know: How to evaluate one-sided and two-sided limits. The definition of continuity at a single point. How to find the domain of algebraic and transcendental functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Confirming continuity over an interval?

Confirming continuity over an interval is the process of verifying that a function meets the formal definition of continuity at every point contained in the interval. On the AP Calculus AB exam, this topic is part of Unit 1: Limits and Continuity, which accounts for 10–12% of total exam weight. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections: it is often tested as a standalone MCQ, or as a required first step for FRQ justifications of major theorems like the Intermediate Value Theorem (IVT) or Extreme Value Theorem (EVT).

Notation conventions: A function $f$ is continuous on an open interval $(a,b)$ if it is continuous at every point $c\in (a,b)$. For a closed interval $[a,b]$, we add requirements for one-sided continuity at the endpoints, since we cannot evaluate behavior outside the interval. Confirming continuity is sometimes called verifying continuity or checking continuity on an interval in AP exam wording. This topic is foundational for nearly all major calculus theorems that rely on continuity.

2. Continuity on Open Intervals

An open interval $(a,b)$ does not include its endpoints, so to confirm continuity here, you only need to verify that $f$ is continuous at every interior point $c$ between $a$ and $b$. Recall the three conditions for continuity at any interior point $c$:

1. $f(c)$ is defined,
2. ${\lim }_{x\to c}f(x)$ exists,
3. ${\lim }_{x\to c}f(x)=f(c)$.

A key shortcut for AP exams: all elementary functions (polynomials, rational, root, exponential, logarithmic, trigonometric) are continuous at every point in their domain. This means you do not need to test every point individually: you only need to confirm that the entire open interval lies within the function's domain, with no points where the function is undefined or discontinuous. For piecewise functions, the only possible points of discontinuity inside an open interval are the breakpoints where the function's definition changes, so you only need to test those breakpoints.

Worked Example

Is $f(x)=\frac{x^2-9}{x-3}$ continuous on the open interval $(0,4)$? Justify your answer.

1. First, identify domain restrictions: the denominator is zero at $x=3$, so $f(3)$ is undefined.
2. Confirm that $x=3$ lies inside the interval $(0,4)$.
3. Check the first continuity condition at $x=3$: $f(3)=\frac{9-9}{3-3}=\frac{0}{0}$, so $f(3)$ is not defined.
4. Even though ${\lim }_{x\to 3}f(x)=6$ exists (this is a removable discontinuity), the first condition for continuity fails at $x=3$.
5. Conclusion: $f(x)$ is not continuous on $(0,4)$.

Exam tip: On AP MCQ asking if a rational function is continuous on an interval, always check first if the interval contains a root of the denominator — that is the most common trick tested.

3. Continuity on Closed Intervals

A closed interval $[a,b]$ includes both endpoints, so the definition for continuity adds two extra one-sided continuity requirements beyond continuity on the interior $(a,b)$. Since the function is only defined for $x\in [a,b]$, we cannot approach the left endpoint $a$ from the left, so we only require right-continuity at $a$: ${\lim }_{x\to a^{+}}f(x)=f(a)$. Similarly, we require left-continuity at $b$: ${\lim }_{x\to b^{-}}f(x)=f(b)$.

This distinction is extremely important for AP exams: nearly all major calculus theorems (IVT, EVT) explicitly require continuity on a closed interval to apply, so you must confirm the endpoint conditions to earn full justification points on FRQs. Endpoints are often the edge of a function's domain (for example, the left endpoint of a square root function's domain), so checking one-sided continuity there is required.

Worked Example

Is $f(x)=\sqrt{9-x^2}$ continuous on the closed interval $[-3,3]$? Justify your answer.

1. First confirm continuity on the open interval $(-3,3)$: the domain of $f$ is all $x$ where $9-x^2\ge 0$, so $-3\le x\le 3$. Any point $c\in (-3,3)$ is in the domain, and $f$ is a composition of continuous elementary functions, so $f$ is continuous at all interior points.
2. Check right-continuity at the left endpoint $x=-3$: ${\lim }_{x\to -3^{+}}\sqrt{9-x^2}=\sqrt{9-(-3{)}^2}=0=f(-3)$, so $f$ is right-continuous at $x=-3$.
3. Check left-continuity at the right endpoint $x=3$: ${\lim }_{x\to 3^{-}}\sqrt{9-x^2}=\sqrt{9-(3{)}^2}=0=f(3)$, so $f$ is left-continuous at $x=3$.
4. All conditions are satisfied, so $f$ is continuous on $[-3,3]$.

Exam tip: When justifying that IVT applies on a FRQ, you must explicitly state that $f$ is continuous on the closed interval, including confirming endpoint continuity, to earn full justification points.

4. Continuity of Piecewise Functions Over Intervals

Piecewise functions have different definitions for different subintervals, so they require a targeted check for discontinuities. The only possible points of discontinuity are: (1) breakpoints where the function's definition changes, which may lie inside your target interval, and (2) points inside any subinterval where the individual piece has a discontinuity (such as a zero denominator).

For any interval that includes a breakpoint, you must test all three continuity conditions at that breakpoint by calculating one-sided limits from each side of the breakpoint to confirm the two-sided limit exists. If the breakpoint is an endpoint of your overall interval, you only check the one-sided limit from inside the interval, following the same rule as closed interval endpoint continuity.

Worked Example

Confirm whether $f(x)=\{\begin{array}{ll}3x-1& x<2\\ x^2+1& 2\le x\le 5\end{array}$ is continuous on the interval $[0,5]$.

1. First list candidate discontinuities on $[0,5]$: the only breakpoint is $x=2$, which lies inside the interval. Both pieces are polynomials, which are continuous everywhere, so there are no other interior discontinuities.
2. Check the three conditions at $x=2$: (1) $f(2)=(2{)}^2+1=5$, so $f(2)$ is defined. (2) Left-hand limit: ${\lim }_{x\to 2^{-}}(3x-1)=6-1=5$. Right-hand limit: ${\lim }_{x\to 2^{+}}(x^2+1)=4+1=5$. One-sided limits are equal, so ${\lim }_{x\to 2}f(x)=5$ exists. (3) ${\lim }_{x\to 2}f(x)=5=f(2)$, so all conditions are satisfied at $x=2$.
3. Check endpoint continuity: At $x=0$, ${\lim }_{x\to 0^{+}}(3x-1)=-1=f(0)$, so right-continuous. At $x=5$, ${\lim }_{x\to 5^{-}}(x^2+1)=26=f(5)$, so left-continuous.
4. All conditions are satisfied, so $f(x)$ is continuous on $[0,5]$.

Exam tip: For piecewise functions, always test continuity at every breakpoint that lies inside your target interval — these are the most common sites of discontinuity tested on AP exams.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to check endpoint continuity when confirming continuity on a closed interval, and only checking interior points. Why: Students memorize continuity rules for interior points and ignore the extra requirement for endpoints that is mandatory for theorem justifications. Correct move: Always add the two one-sided endpoint continuity checks when the interval is closed, and explicitly write the equality ${\lim }_{x\to a^{+}}f(x)=f(a)$ in your justification.
• Wrong move: Claiming a function with a removable discontinuity is continuous on an interval containing that discontinuity, just because the limit exists. Why: Students confuse the existence of the limit with the full definition of continuity, which requires the function to be defined at the point. Correct move: Always check that $f(c)$ is defined for any point $c$ in the interval, even if the limit at $c$ exists.
• Wrong move: Forgetting to test breakpoints inside the interval for piecewise functions, and only checking that each individual piece is continuous. Why: Students assume that because each piece is continuous on its own, the combined function is continuous over the whole interval. Correct move: Always list all breakpoints that lie inside your target interval and test continuity at each one explicitly.
• Wrong move: Checking two-sided limits at endpoints of a closed interval, including the side outside the interval. Why: Students default to two-sided limits out of habit, even though the function is not defined outside the interval. Correct move: Only check right-continuity at the left endpoint and left-continuity at the right endpoint, ignoring behavior outside the interval.
• Wrong move: Claiming a function is continuous on an interval that includes a vertical asymptote, because the asymptote is "almost at the endpoint". Why: Students think only interior discontinuities count, but any point in the interval where the function is discontinuous breaks continuity over the whole interval. Correct move: If the interval contains any discontinuity (removable, jump, infinite), the function is not continuous over that interval.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following intervals is the function $f(x)=\ln (x-2)+\frac{1}{x-5}$ continuous on? A) $(0,5)$ B) $(2,5)$ C) $[2,5)$ D) $(2,10)$

Worked Solution: First, find the domain of $f(x)$: $\ln (x-2)$ requires $x-2>0$, so $x>2$. The rational term $\frac{1}{x-5}$ requires $x\ne 5$. So the domain of $f$ is $(2,5)\cup (5,\infty )$. Check each option: Option A includes $x\le 2$ where $f$ is undefined, so A is wrong. Option C includes $x=2$ where $f(2)=\ln (0)$ is undefined, so C is wrong. Option D includes $x=5$ where $f(5)$ is undefined, so D is wrong. Option B is entirely within the domain of $f$, and $f$ is an elementary function, so it is continuous on $(2,5)$. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\{\begin{array}{ll}3x+k& x<1\\ x^2+2x& x\ge 1\end{array}$, defined for all real numbers. (a) Find the value of $k$ (if any) that makes $f$ continuous at $x=1$. (b) Using your value of $k$ from part (a), confirm whether $f$ is continuous on the closed interval $[0,3]$. Justify your answer. (c) Explain why $f$ satisfies the conditions to apply the Intermediate Value Theorem on $[0,3]$ with this $k$.

Worked Solution: (a) For continuity at $x=1$, we need ${\lim }_{x\to 1^{-}}f(x)={\lim }_{x\to 1^{+}}f(x)=f(1)$. Calculate the right-hand limit: ${\lim }_{x\to 1^{+}}(x^2+2x)=1+2=3=f(1)$. Set the left-hand limit equal: ${\lim }_{x\to 1^{-}}(3x+k)=3+k=3$, so $k=0$.

(b) With $k=0$, $f$ is continuous at the interior breakpoint $x=1$. Both pieces are polynomials, so they are continuous at all other interior points. Check endpoints: At $x=0$, ${\lim }_{x\to 0^{+}}3x=0=f(0)$, so $f$ is right-continuous at $0$. At $x=3$, ${\lim }_{x\to 3^{-}}(x^2+2x)=9+6=15=f(3)$, so $f$ is left-continuous at $3$. All conditions are satisfied, so $f$ is continuous on $[0,3]$.

(c) The Intermediate Value Theorem requires the function to be continuous on the closed interval $[a,b]$. We confirmed in part (b) that $f$ is continuous on $[0,3]$, so all conditions for IVT are satisfied.

Question 3 (Application / Real-World Style)

The volume of water in a reservoir over a 12-week dry season is modeled by $V(t)=\frac{1200}{t+2}+40e^{-0.1t}$, where $V(t)$ is measured in thousands of cubic meters, and $t$ is time in weeks, for $0\le t\le 12$. An engineer claims that the volume of water changes continuously over the entire 12-week season, with no sudden jumps in volume. Is the engineer's claim correct? Justify your answer.

Worked Solution: First, check the domain of $V(t)$ on $[0,12]$: $V(t)$ is a sum of a rational function $\frac{1200}{t+2}$ and an exponential function $40e^{-0.1t}$. The denominator of the rational term is zero only at $t=-2$, which is outside the interval $[0,12]$, so all points in the interval are in the domain. Check endpoint continuity: At $t=0$, ${\lim }_{t\to 0^{+}}V(t)=\frac{1200}{2}+40e^0=600+40=640=V(0)$, so $V$ is right-continuous at $0$. At $t=12$, ${\lim }_{t\to 12^{-}}V(t)=\frac{1200}{14}+40e^{-1.2}\approx 97.2=V(12)$, so $V$ is left-continuous at $12$. All interior points are in the domain, and $V$ is a combination of continuous elementary functions, so $V$ is continuous at all interior points. Conclusion: The engineer's claim is correct; the volume of water changes continuously over the entire 12-week dry season with no sudden jumps.

7. Quick Reference Cheatsheet

8. What's Next

Confirming continuity over an interval is the foundational prerequisite for the next core topic in Unit 1: applying the Intermediate Value Theorem, one of the most frequently tested concepts on the AP Calculus AB exam. IVT explicitly requires a function to be continuous on a closed interval to apply, so you cannot earn full points for an IVT justification without first correctly confirming continuity over the interval. Beyond Unit 1, continuity is also a required condition for differentiability, so you will use this skill when checking where a function is differentiable, and when justifying the Mean Value Theorem later in the course. Mastering this topic eliminates a large number of avoidable point losses on both MCQ and FRQ sections.

• Intermediate Value Theorem
• Continuity at a Point
• Differentiability and Continuity
• Extreme Value Theorem