Confirming continuity over an interval — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Continuity at a point definition, one-sided continuity at interval endpoints, testing continuity for piecewise, elementary, and composite functions over open and closed intervals per AP CED.

You should already know: How to evaluate one-sided and two-sided limits, How to find the domain of common functions, How to work with piecewise-defined 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 BC 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?

A function is continuous over an interval if every point in the interval satisfies the three conditions for continuity at a point: $f(c)$ is defined, ${\lim }_{x\to c}f(x)$ exists, and ${\lim }_{x\to c}f(x)=f(c)$. Per the 2019 AP CED, this topic falls within Unit 1: Limits and Continuity, accounting for approximately 3-6% of total exam score weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Most often, it is tested as a foundational justification step for other concepts like the Intermediate Value Theorem, differentiability, or integration. Notation: We write $f$ is continuous on $I$, where $I$ can be an open interval $(a,b)$, closed interval $[a,b]$, half-open $[a,b)$ or $(a,b]$, or infinite interval like $(-\infty ,c]$. Synonyms for confirming continuity include "proving $f$ is continuous on $I$" or "verifying the interval of continuity for $f$". For closed intervals, we require a modified condition for endpoints, since two-sided limits do not exist at the edges of the interval.

2. Continuity of Elementary Functions Over Open Intervals

Elementary functions (polynomials, rational functions, trigonometric functions, exponential functions, logarithms, and root functions) have a key property that simplifies confirming continuity: every elementary function is continuous at all points in its domain. This theorem comes directly from limit laws: for any point $c$ in the domain of an elementary function $f$, ${\lim }_{x\to c}f(x)=f(c)$ by the properties of limits, so all three continuity conditions are automatically satisfied. To confirm continuity of an elementary function over an open interval, you only need to check that the entire interval is contained within the function's domain. If there are no points in the interval that are excluded from the domain (e.g., division by zero, negative radicands, non-positive arguments for logs), the function is continuous over the entire open interval. This saves significant time on exam questions, as you do not need to check every point individually.

Worked Example

Problem: 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 function, which is an elementary function continuous everywhere in its domain.
2. Find points excluded from the domain: the denominator equals zero when $x+2=0⟹x=-2$. The domain of $f$ is $(-\infty ,-2)\cup (-2,\infty )$.
3. Check if the entire interval $(0,4)$ is inside the domain: $-2\notin (0,4)$, so every $x\in (0,4)$ is in the domain of $f$.
4. Conclude: Since $f$ is continuous on its domain, $f$ is continuous on $(0,4)$.

Exam tip: On AP MCQ, you can eliminate options that contain any excluded point (discontinuity) from the domain — you only need to find one discontinuity to rule out the interval.

3. Continuity on Closed Intervals With Endpoints

When confirming continuity on a closed interval $[a,b]$, we need an additional rule for endpoints, since the function is only defined or considered on the interval $[a,b]$, so two-sided limits do not exist at $x=a$ or $x=b$. The full definition of continuity on $[a,b]$ is:

1. $f$ is continuous at every point in the open interval $(a,b)$,
2. $f$ is right-continuous at the left endpoint $a$: ${\lim }_{x\to a^{+}}f(x)=f(a)$,
3. $f$ is left-continuous at the right endpoint $b$: ${\lim }_{x\to b^{-}}f(x)=f(b)$. This requirement is explicitly tested on AP FRQ, most often when justifying the use of the Intermediate Value Theorem, which requires continuity on a closed interval. Skipping the one-sided endpoint check will cost you justification points on the exam.

Worked Example

Problem: Confirm whether $f(x)=\sqrt{16-x^2}$ 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 of the square root function. Since $f$ is elementary, it is continuous on $(-4,4)$.
2. Check right-continuity at $x=-4$: ${\lim }_{x\to -4^{+}}\sqrt{16-x^2}=\sqrt{16-(-4{)}^2}=0=f(-4)$, so right-continuity holds.
3. Check left-continuity at $x=4$: ${\lim }_{x\to 4^{-}}\sqrt{16-x^2}=\sqrt{16-4^2}=0=f(4)$, so left-continuity holds.
4. Conclude: All conditions are satisfied, so $f$ is continuous on $[-4,4]$.

Exam tip: If an FRQ asks you to confirm continuity on $[a,b]$ for IVT, explicitly state that you checked one-sided continuity at the endpoints to earn full credit.

4. Continuity of Piecewise-Defined Functions Over an Interval

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

1. Confirm that each individual piece is continuous on its open sub-interval (almost always, each piece is elementary, so this only requires checking the domain for each piece),
2. Check all three continuity conditions at every breakpoint that lies inside the interval: evaluate the left-hand limit with the left piece, the right-hand limit with the right piece, confirm both are equal to the value of $f$ at the breakpoint,
3. If the interval is closed, confirm one-sided continuity at the endpoints of the full interval as usual. If all breakpoints are continuous and all other conditions hold, the entire function is continuous over the full interval.

Worked Example

Problem: Confirm whether $f(x)=\{\begin{array}{ll}3x+2& x<2\\ x^2+4& 2\le x\le 5\end{array}$ is continuous on $[0,5]$.

1. Check continuity on sub-intervals: On $(0,2)$, $f(x)=3x+2$ 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 ${\lim }_{x\to 2^{-}}(3x+2)=6+2=8$. Right limit ${\lim }_{x\to 2^{+}}(x^2+4)=4+4=8$. $f(2)=2^2+4=8$. So ${\lim }_{x\to 2}f(x)=f(2)$, so $f$ is continuous at $x=2$.
3. Check endpoints: Right-continuity at $x=0$: ${\lim }_{x\to 0^{+}}(3x+2)=2=f(0)$, holds. Left-continuity at $x=5$: ${\lim }_{x\to 5^{-}}(x^2+4)=25+4=29=f(5)$, holds.
4. Conclude: All conditions are satisfied, so $f$ is continuous on $[0,5]$.

Exam tip: Circle or label the breakpoint before you start to avoid forgetting to check it, which is a common missed point on exams.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming $f(x)=\frac{(x-2)(x+3)}{x-2}$ is continuous on $[0,5]$ because simplifying gives $x+3$, which is defined everywhere. Why: Students confuse 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. Why: Students forget that two-sided limits do not exist at endpoints when we only consider the function on the interval. Correct move: For the left endpoint $a$, only check right-continuity ${\lim }_{x\to a^{+}}f(x)=f(a)$; for the right endpoint $b$, only check left-continuity ${\lim }_{x\to b^{-}}f(x)=f(b)$.
• Wrong move: When checking continuity of a piecewise function at a breakpoint $c$, evaluating both left and right 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 the one-sided limit with the matching piece every time.
• 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. 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following intervals is $f(x)=\frac{e^x}{\ln (x+3)}$ continuous on? A) $(-3,0]$ B) $(0,2)$ C) $[-2,1]$ D) $(-2.5,-1.5)$

Worked Solution: First, $f(x)$ is a quotient of elementary functions, so it is continuous everywhere it is defined. The domain requires two conditions: (1) $x+3>0⟹x>-3$, (2) $\ln (x+3)\ne 0⟹x+3=1⟹x=-2$, so $x\ne -2$. Now check each option: A includes $x=-2$, so it is invalid. C includes $x=-2$, so it is invalid. D includes $x=-2$, so it has a discontinuity inside the interval. Only B has no excluded points in the interval, so $f$ is continuous here. Correct answer: $\boxed{B}$

Question 2 (Free Response)

Let $f(x)=\{\begin{array}{ll}kx+3& x<1\\ 2x^2-k& x\ge 1\end{array}$ for a constant $k$. (a) Find the value of $k$ that makes $f$ continuous at $x=1$. (b) Using your value of $k$ from (a), confirm that $f$ is continuous on the closed interval $[-2,3]$. Justify your answer. (c) Explain why any other value of $k$ makes $f$ discontinuous on $[-2,3]$.

Worked Solution: (a) For continuity at $x=1$, we require ${\lim }_{x\to 1^{-}}f(x)={\lim }_{x\to 1^{+}}f(x)=f(1)$. Calculate left limit: ${\lim }_{x\to 1^{-}}(kx+3)=k+3$. Right limit: ${\lim }_{x\to 1^{+}}(2x^2-k)=2-k$. Set equal: $k+3=2-k⟹2k=-1⟹k=-\frac{1}{2}$.

(b) With $k=-\frac{1}{2}$, check: 1. On $(-2,1)$, $f(x)=-\frac{1}{2}x+3$ is a polynomial, so continuous. On $(1,3)$, $f(x)=2x^2+\frac{1}{2}$ is a polynomial, so continuous. 2. At $x=1$, we confirmed continuity in part (a). 3. Endpoints: ${\lim }_{x\to -2^{+}}f(x)=-\frac{1}{2}(-2)+3=1+3=4=f(-2)$, so right-continuous at $x=-2$. ${\lim }_{x\to 3^{-}}f(x)=2(9)+\frac{1}{2}=18.5=f(3)$, so left-continuous at $x=3$. All conditions are satisfied, so $f$ is continuous on $[-2,3]$.

(c) If $k\ne -\frac{1}{2}$, then ${\lim }_{x\to 1^{-}}f(x)\ne {\lim }_{x\to 1^{+}}f(x)$, so ${\lim }_{x\to 1}f(x)$ does not exist. Since $x=1$ is inside $[-2,3]$, $f$ is discontinuous at $x=1$, hence not continuous on the entire interval.

Question 3 (Application / Real-World Style)

The temperature of a chemical reaction over the first 8 hours after mixing is modeled by the function $T(t)=\{\begin{array}{ll}25e^{0.1t}& 0\le t<3\\ 25e^{0.05t+0.15}& 3\le t\le 8\end{array}$, where $T(t)$ is temperature in degrees Celsius, and $t$ is time in hours. A chemist claims the temperature is a continuous function of time over the entire 8-hour reaction. Confirm or reject the chemist's claim.

Worked Solution: First, check continuity on open sub-intervals: On $(0,3)$, $T(t)=25e^{0.1t}$ is an exponential elementary function, so continuous. On $(3,8)$, $T(t)=25e^{0.05t+0.15}$ is also an exponential elementary function, so continuous. Next, check the breakpoint at $t=3$: Left limit: ${\lim }_{t\to 3^{-}}25e^{0.1(3)}=25e^{0.3}$. Right limit: ${\lim }_{t\to 3^{+}}25e^{0.05(3)+0.15}=25e^{0.15+0.15}=25e^{0.3}=T(3)$, so continuous at $t=3$. Check endpoints: ${\lim }_{t\to 0^{+}}T(t)=25e^0=25=T(0)$, and ${\lim }_{t\to 8^{-}}T(t)=25e^{0.05(8)+0.15}=25e^{0.55}=T(8)$, so endpoint conditions hold. Conclusion: The chemist's claim is correct; the temperature is continuous over the entire 8-hour reaction, meaning there is no abrupt jump in temperature at any point during the experiment.

7. Quick Reference Cheatsheet

8. What's Next

Confirming continuity over an interval is the 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 FRQ. 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. All of these topics rely on the ability to correctly confirm continuity on an interval quickly, so mastering this topic is non-negotiable for exam success.

• Intermediate Value Theorem
• Differentiability and continuity
• Interval of convergence of power series
• Fundamental Theorem of Calculus