Unit 1 · Topic 1.12 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 1.12: Confirming Continuity over an Interval

Q: Is a function with a hole continuous on an interval that contains the hole?

No. At a hole either the function has no value, or its value sits off the curve and does not equal the limit, so the first or the third condition fails there, and one such point is enough. It is continuous on intervals on either side of the hole.

Q: What happens at the endpoints of a closed interval?

Only inputs inside the interval count, so at the left endpoint the function needs a value and a right-hand limit equal to it, and at the right endpoint a left-hand limit. That is how x is continuous on [0, 4].

Q: Does a continuous graph have to be smooth, with no corners?

No. A corner is not a break. |x| has a sharp corner at 0, but it has a value there that equals its limit, so it is continuous on every interval.

Q: If f is continuous on (0, 2) and on (2, 5), is it continuous on (0, 5)?

Not necessarily. Neither interval contains 2, so you know nothing about f at 2 yet. Check x = 2 with the three conditions before you join the intervals.

Q: Can I just say a function is continuous because it is a rational function?

Only at the inputs in its domain. A rational function is continuous at every point where its denominator is not 0, so the reason has to say that the interval avoids those zeros.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 1: 10 to 15% of the multiple-choice section

What you'll be able to do

Find the intervals on which a function is continuous, using the domains of the basic function families and checking every input where a piecewise formula changes.

Key vocabulary

Term	What it means	The exam trap
Continuous on an interval	A function is continuous on an interval if it is continuous at each point in the interval.	A graph that looks smooth can still hide a single missing point. Give a reason that covers every input, not a few that you tried.
Domain	The set of inputs at which a function has a value.	Read the domain from the formula you are given, before you simplify it, and find it for the whole formula, not just one part.
Piecewise-defined function	A function that uses different formulas on different intervals of its domain.	Each formula applies only on its own interval, so read the condition beside it before you use it.

Graph of f(x) = (x squared minus 1) over (x squared minus 3x plus 2) for x from -2 to 5. An open dot at (1, -2) marks a hole where f(1) does not exist, and a dashed vertical line at x = 2 marks a vertical asymptote. Everywhere else the graph is unbroken, from (-2, 0.25) to (5, 2).

Part 1 Think first Does the light meter ever lose the reading?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A light meter is carried in a straight line toward and away from a lamp. A model says the brightness it reads at a distance of $d$ meters from the lamp is $I (d) = \frac{400}{d^{2}}$ lux. Over which stretch of distances does the model give a reading at every distance, with no sudden jump anywhere?

In short. A function is continuous on an interval when it is continuous at each point of the interval, and polynomial, rational, power, exponential, logarithmic and trigonometric functions are continuous at every point of their domains.

One input inside an interval where continuity fails is enough to make the function not continuous on that interval.
For a function built from the basic families, the intervals of continuity come from its domain: find the inputs where it has no value and split there.
For a piecewise-defined function, check each formula on its own interval, then test every input where the formula changes with the three conditions for continuity.

Part 2 Learn it Three short ideas, each checked as you go

12a. Continuous at every point of the interval

In Topic 1.11, $f$ is continuous at one input $c$ when $f (c)$ exists, $\lim_{x \to c} f (x)$ exists, and the two are equal.

A function is continuous on an interval if it is continuous at each point in the interval. One failure anywhere inside the interval is enough to lose it.

The graph at the top is $f (x) = \frac{x^{2} - 1}{x^{2} - 3 x + 2}$ . Neither $f (1)$ nor $f (2)$ exists, so $f$ is not continuous at either input. Everywhere else the graph runs without a break. So $f$ is continuous on $(- 2, 1)$ , on $(1, 2)$ and on $(2, 5)$ . It is not continuous on $[0, 3]$ , which holds both bad inputs, or on $(0.5, 1.5)$ , which holds only one.

Two habits follow. To show that a function is continuous on an interval, give a reason that covers every point of it, not a few sample inputs. To show that it is not, one input inside the interval where continuity fails is enough.

At an endpoint of a closed interval, only inputs inside the interval count: the function needs a value there equal to its one-sided limit from inside, the right-hand limit at the left end and the left-hand limit at the right end. The same goes where a domain stops: only the side with values counts. So $\sqrt{x}$ is continuous on $[0, 4]$ , and at 0 itself, even though it has no values to the left of 0.

Quick check · Sort it

No calculator. A function $f$ is defined for all real numbers and is continuous at every real number except $x = - 1$ and $x = 3$ . Tap an interval, then tap whether $f$ is continuous on it.

Continuous on this interval

Not continuous on this interval

12b. The basic families are continuous on their domains

No one checks every point of an interval one at a time. Instead, calculus leans on one fact: polynomial, rational, power, exponential, logarithmic and trigonometric functions are continuous at all points in their domains. Sums, differences, products, quotients and compositions of them are continuous wherever they are defined, by the limit theorems of Topic 1.5.

So for a function built from these families, the work is finding the domain:

a polynomial, $e^{x}$ , $\sin x$ or $\cos x$ : every real number;
a quotient: leave out each input where the denominator is 0;
$\sqrt{u}$ : you need $u \geq 0$ , so an input where $u = 0$ is kept;
$\ln u$ : you need $u > 0$ , so an input where $u = 0$ is left out;
$\tan x = \frac{\sin x}{\cos x}$ : leave out each zero of $\cos x$ .

Then split the domain at the inputs you left out. Take $r (x) = \frac{x + 3}{x^{2} - 16}$ . Its denominator is 0 at $x = - 4$ and $x = 4$ , so $r$ is continuous on $(- \infty, - 4)$ , on $(- 4, 4)$ and on $(4, \infty)$ . And $q (x) = \ln (5 - x)$ needs $5 - x > 0$ , so it is continuous on $(- \infty, 5)$ .

The fact covers the domain and nothing more. At an input outside the domain the function has no value, so it cannot be continuous there.

Quick check · Multiple choice

No calculator. On which of these intervals is $g (x) = \frac{\ln x}{x - 4}$ continuous?

12c. Piecewise functions: each formula, then each seam

A piecewise-defined function uses different formulas on different intervals. Check it in two stages.

First, each formula on its own open interval, using the families from the last idea. A formula counts only where it is used, so a zero of its denominator outside that interval does not matter.

Second, each input where the formula changes. There the families say nothing, so use the three conditions from Topic 1.11: the value at the input, both one-sided limits, and whether all three agree.

p (x) = {\begin{matrix} \frac{8}{x - 6}, & x < 2 \\ x^{2} - 6, & x \geq 2 \end{matrix}

For $x < 2$ , $\frac{8}{x - 6}$ is rational, and its denominator is 0 only at $x = 6$ , an input this formula is never used for. For $x > 2$ , $x^{2} - 6$ is a polynomial. At $x = 2$ : $p (2) = 4 - 6 = - 2$ , $\lim_{x \to 2^{-}} p (x) = \frac{8}{2 - 6} = - 2$ and $\lim_{x \to 2^{+}} p (x) = 4 - 6 = - 2$ . All three agree, so $p$ is continuous at 2, and $p$ is continuous on $(- \infty, \infty)$ .

If the one-sided limits had disagreed, $p$ would still be continuous on $(- \infty, 2)$ and on $[2, \infty)$ , because $p (2)$ comes from $x^{2} - 6$ and matches the right-hand limit, but not on any interval with 2 strictly inside it.

Quick check · Calculate

No calculator. Let $f (x) = \frac{x + 2}{(x - 4) (x - 5)}$ for $x < 1$ and $f (x) = \frac{6}{x - 3}$ for $x \geq 1$ . At how many real numbers is $f$ not continuous?

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Giving $\ln (x - 1)$ the domain $x \geq 1$ , or $\sqrt{x - 1}$ the domain $x > 1$ .	A logarithm needs its input above 0, so $\ln 0$ does not exist. A square root needs its input at least 0, and $\sqrt{0} = 0$ does exist.	Write the condition first, $x - 1 > 0$ for the logarithm and $x - 1 \geq 0$ for the root, then solve it.
Canceling a common factor and then calling the function continuous where that factor was 0.	The canceled formula agrees with the original only where the original is defined. At the canceled zero the original still has no value.	Read the domain from the formula you were given, before any simplifying.
Checking every formula of a piecewise function but never the input where the formula changes.	Two formulas that are each continuous can still meet at different heights.	At each seam, find the value and both one-sided limits, and compare them.

Part 3 Watch it done Intervals of continuity for a quotient with a root

Watch each step, then do the last one yourself.

Let $g (x) = \frac{\sqrt{x + 4}}{x - 1}$ . Find the intervals on which $g$ is continuous.

Step 1 · Name the families
$\sqrt{x + 4}$ is a power function composed with the polynomial $x + 4$ , and $x - 1$ is a polynomial. A quotient of such functions is continuous at every input where it is defined.
Step 2 · Find the domain
The root needs $x + 4 \geq 0$ , so $x \geq - 4$ . The denominator needs $x - 1 \neq 0$ , so $x \neq 1$ . The domain is $x \geq - 4$ with $x \neq 1$ .
Why this step?
Why is $x = - 4$ kept, when $x = 1$ is thrown out?
At $x = - 4$ the root is $\sqrt{0} = 0$ , so $g (- 4) = \frac{0}{- 5} = 0$ exists, and only inputs to its right are in the domain, so $g$ needs only its right-hand limit there. At $x = 1$ the formula divides by zero, so $g (1)$ does not exist and $g$ cannot be continuous at 1.
Step 3 · Split the domain
The domain is cut at the one input left out inside it, $x = 1$ , which leaves one interval to the left of 1 and one to the right.
Your step · Fill the statement
No calculator. Finish it. Tap a blank, then tap an answer.

The domain of $g (x) = \frac{\sqrt{x + 4}}{x - 1}$ splits at 1 into two intervals that together hold every input of the domain: to the left of 1 and to the right of 1. $g$ is continuous on each of them.

Part 4 Try it Two new functions

Try it · Calculate

No calculator. Let $h (x) = \tan x$ . At how many values of $x$ in the interval $0 \leq x \leq 2 π$ is $h$ not continuous?

Your answer

Try it · Decide and justify

No calculator. Let $f (x) = \frac{e^{x}}{x^{2} + 1}$ . A classmate says $f$ is continuous on every interval of real numbers. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. A chemist heats a mixture, and after 5 minutes turns the heater down. The temperature of the mixture, in degrees Celsius, $t$ minutes after the heater is switched on is modeled by

T (t) = {\begin{matrix} 20 + 6 t, & 0 \leq t \leq 5 \\ 50 + 10 \ln (t - 4), & 5 < t \leq 12 \end{matrix}

(a) Explain why $T$ is continuous on the interval $5 < t \leq 12$ .

(b) Find $T (5)$ , using correct units. Using the definition of continuity, determine whether $T$ is continuous at $t = 5$ . Justify your answer.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) For $5 < t \leq 12$ , $t - 4 > 1 > 0$ , so $\ln (t - 4)$ is defined at every $t$ in the interval. A logarithmic function is continuous at every point of its domain, and $T (t) = 50 + 10 \ln (t - 4)$ there, so $T$ is continuous on $5 < t \leq 12$ (from the left at $t = 12$ ).

(b) $T (5) = 20 + 6 (5) = 50$ degrees Celsius. $\lim_{t \to 5^{-}} T (t) = \lim_{t \to 5^{-}} (20 + 6 t) = 50$ and $\lim_{t \to 5^{+}} T (t) = \lim_{t \to 5^{+}} (50 + 10 \ln (t - 4)) = 50 + 10 \ln 1 = 50$ . So $\lim_{t \to 5} T (t) = 50 = T (5)$ , and $T$ is continuous at $t = 5$ .

(c) Yes. On $0 \leq t < 5$ , $T (t) = 20 + 6 t$ is a polynomial, so $T$ is continuous there (from the right at $t = 0$ ). $T$ is continuous at $t = 5$ by part (b) and on $5 < t \leq 12$ by part (a). So $T$ is continuous at each point of $0 \leq t \leq 12$ , and therefore on the interval.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Says $t - 4 > 0$ for $5 < t \leq 12$ , so $\ln (t - 4)$ is defined there, and that a logarithmic function is continuous on its domain.
Gets $T (5) = 50$ degrees Celsius.
Finds both one-sided limits at $t = 5$ , each equal to 50, so $\lim_{t \to 5} T (t) = 50$ .
Concludes $T$ is continuous at $t = 5$ because $\lim_{t \to 5} T (t) = T (5)$ .
Answers yes in (c), because $T$ is continuous at every point of $0 \leq t \leq 12$ : the linear formula on $0 \leq t < 5$ , part (b) at $t = 5$ , and part (a) after it.

Part 6 Check yourself A function in two pieces

The two case questions use the same function. Answer them from the table.

A function $f$ is given by two formulas. The table says which formula gives $f (x)$ on which inputs. Where a formula has no value, $f$ has none either.

The two formulas for f
Inputs	$f (x)$
$x < 0$	$\frac{x + 6}{x + 3}$
$x \geq 0$	$2 + \ln (x + 1)$

Case question 1 · Multiple choice

No calculator. At which values of $x$ is $f$ not continuous?

Case question 2 · Calculate

No calculator. On how many of these four intervals is $f$ continuous: $[- 4, - 2]$ , $[- 2, 6]$ , $(- 3, 0]$ and $(- 5, - 3]$ ?

Your answer

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds where $m (x) = \frac{x^{2} - x - 6}{x^{2} - 9}$ is continuous. Tap the first line with a mistake.

Check yourself · Match

No calculator. Tap a function, then the largest interval or intervals on which it is continuous.

Part 7 Keep it Bring back earlier ideas

Two earlier ideas this lesson stands on. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. Let $f (x) = x^{2} + 1$ for every $x \neq 2$ , and let $f (2) = 3$ . Which statement explains why $f$ is not continuous at $x = 2$ ?

Bring it back · Calculate

No calculator. Suppose $\lim_{x \to 4} f (x) = 3$ and $\lim_{x \to 4} g (x) = - 2$ . Find $\lim_{x \to 4} ([f (x)]^{2} - 3 g (x))$ .

Your answer

In three lines

A function is continuous on an interval when it is continuous at each point of the interval; one failure inside is enough to lose it.
The basic families are continuous on their domains, so find the domain and split it at the inputs left out.
For a piecewise function, check each formula where it is used, then each seam with the value and both one-sided limits.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When you justify that a function is continuous on an interval, name the family each part belongs to and say why every input of the interval is in its domain. At an input where a piecewise formula changes, write the value, both one-sided limits and the comparison before you state a conclusion.

Questions students ask

Is a function with a hole continuous on an interval that contains the hole?

No. At a hole either the function has no value, or its value sits off the curve and does not equal the limit, so the first or the third condition fails there, and one such point is enough. It is continuous on intervals on either side of the hole.

What happens at the endpoints of a closed interval?

Only inputs inside the interval count, so at the left endpoint the function needs a value and a right-hand limit equal to it, and at the right endpoint a left-hand limit. That is how $\sqrt{x}$ is continuous on $[0, 4]$ .

Does a continuous graph have to be smooth, with no corners?

No. A corner is not a break. $| x |$ has a sharp corner at 0, but it has a value there that equals its limit, so it is continuous on every interval.

f

is continuous on

(0, 2)

and on

(2, 5)

, is it continuous on

(0, 5)

Not necessarily. Neither interval contains 2, so you know nothing about $f$ at 2 yet. Check $x = 2$ with the three conditions before you join the intervals.

Can I just say a function is continuous because it is a rational function?

Only at the inputs in its domain. A rational function is continuous at every point where its denominator is not 0, so the reason has to say that the interval avoids those zeros.

Course alignment, for teachers

AP Calculus AB topic 1.12, Unit 1: Limits and Continuity.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.