Confirming Continuity over an Interval (AP Calculus AB 1.12)

Topic 1.11 tests continuity one point at a time, but an interval has infinitely many points, so nobody checks them one by one. Instead you use two facts: the basic function families are continuous everywhere they are defined, and functions built from continuous pieces stay continuous. That turns "where is $f$ continuous?" into two smaller questions: where is $f$ defined, and what happens at the points where a piecewise rule changes?

What continuous on an interval means

A function is continuous on an open interval $(a, b)$ when it is continuous at every point of that interval. For a closed interval, the endpoints need care. At $x = a$ the interval contains only points to the right of $a$ , so what $f$ does to the left of $a$ has nothing to do with the interval; often $f$ is not even defined there. So at an endpoint you require only the one-sided version of continuity.

Take $s(x) = \sqrt{9 - x^2}$ , the upper half of the circle of radius 3. It is defined only for $-3 \le x \le 3$ , so $\lim\limits_{x \to 3} s(x)$ cannot be two-sided: there are no inputs above 3. But $\lim\limits_{x \to 3^-} s(x) = 0 = s(3)$ , and similarly at $-3$ from the right, so $s$ is continuous on $[-3, 3]$ . The semicircle is drawn in one stroke, endpoints included, and the definition agrees.

Functions that are continuous on their domains

The reasons come from the limit laws. The functions $x$ and any constant are continuous, and the sum and product laws show that sums and products of continuous functions are continuous, which builds every polynomial. The quotient law handles rational functions, but only where the denominator's limit is not 0. Sine and cosine are continuous everywhere: as an angle changes a little, the point on the unit circle moves a little, so its coordinates change a little. The other trig functions are quotients of these, such as $\tan x = \frac{\sin x}{\cos x}$ , so they are continuous except where the denominator is 0. Exponential functions have unbroken graphs, and logarithms, their inverses, inherit that.

What "on their domains" leaves out is the work in every problem. Each family's domain:

Function	Continuous on
polynomial	all real numbers
$\dfrac{p(x)}{q(x)}$ (rational)	all $x$ with $q(x) \neq 0$
$\sqrt{x}$ and other even roots	$x \ge 0$
$\sqrt[3]{x}$ and other odd roots	all real numbers
$b^x$ , including $e^x$	all real numbers
$\ln x$	$x > 0$
$\sin x$ , $\cos x$	all real numbers
$\tan x$ , $\sec x$	all $x$ except $\dfrac{\pi}{2} + k\pi$ , $k$ an integer
$\cot x$ , $\csc x$	all $x$ except $k\pi$ , $k$ an integer

"Continuous on its domain" is not "continuous everywhere." The function $\frac{1}{x}$ is continuous at every point where it is defined, yet it has an infinite discontinuity at $x = 0$ (topic 1.10), the one point its domain leaves out. So when a question asks where $\frac{\ln x}{x - 3}$ is continuous, the answer is its domain: $\ln x$ needs $x > 0$ and the denominator needs $x \neq 3$ , giving $(0, 3) \cup (3, \infty)$ .

Finding the intervals, piecewise functions included

For a function given by one formula, the intervals of continuity are its domain, written as intervals. List what each part demands: denominators not 0, even roots of nonnegative quantities, logarithms of positive quantities, $\tan$ and $\sec$ away from odd multiples of $\frac{\pi}{2}$ . An included endpoint, like the $-3$ and $3$ of the semicircle, gets a square bracket.

For a piecewise function, there are two separate jobs:

Inside each piece's interval, the piece's formula is continuous wherever it is defined. A point where the formula breaks matters only if that point lies in the piece's interval.
At each boundary, the formulas hand off, and nothing about the families guarantees continuity there. Check the boundary with the three conditions of topic 1.11.

If $f$ breaks at a boundary $b$ , $f$ may still be continuous from one side at $b$ . That decides the brackets: if $f(b)$ comes from the right-hand piece and matches the right-hand limit, then $b$ can be the closed left end of an interval of continuity.

Worked examples

Worked exampleExample 1: a root over a rational expression

Find the intervals on which $f(x) = \dfrac{\sqrt{x + 2}}{x^2 - 1}$ is continuous.

Show solution

List the requirements. The square root needs $x + 2 \ge 0$ , so $x \ge -2$ . The denominator needs $x^2 - 1 \neq 0$ , so $x \neq 1$ and $x \neq -1$ . The numerator is a composition of continuous functions on $x \ge -2$ , the denominator is a polynomial, and a quotient of continuous functions is continuous where the denominator is nonzero. So $f$ is continuous on its domain:

[-2, -1) \cup (-1, 1) \cup (1, \infty).

The bracket at $-2$ is correct because $f$ is continuous from the right there: $\lim\limits_{x \to -2^+} f(x) = \dfrac{0}{3} = 0 = f(-2)$ . The points $\pm 1$ are excluded because $f$ is undefined there; since the numerator is not 0 at either one, they are vertical asymptotes.

Worked exampleExample 2: a composition on a restricted interval

On the interval $[0, 2\pi]$ , where is $g(x) = \ln(\cos x)$ continuous?

Show solution

$g$ is the composition of $\ln u$ , continuous for $u > 0$ , with $u = \cos x$ , continuous everywhere. A composition is continuous wherever the inside is continuous and the outside is continuous at the inside's value, so $g$ is continuous wherever $\cos x > 0$ .

On $[0, 2\pi]$ , $\cos x > 0$ for $0 \le x < \frac{\pi}{2}$ and for $\frac{3\pi}{2} < x \le 2\pi$ ; it equals 0 at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ and is negative between them. So $g$ is continuous on

\left[0, \tfrac{\pi}{2}\right) \cup \left(\tfrac{3\pi}{2}, 2\pi\right].

The endpoints $0$ and $2\pi$ are included because $g$ is defined there ( $g(0) = \ln 1 = 0$ ) and one-sided continuity is all an endpoint of the given interval needs.

Worked exampleExample 3: an exam-style piecewise question (no calculator)

Let

f(x) = \begin{cases} 2^x & x \le 2 \\ \dfrac{x + 6}{x} & 2 < x < 5 \\ \dfrac{x + 1}{x - 7} & x \ge 5 \end{cases}

(a) Find all values of $x$ at which $f$ is not continuous, and justify your answer.

(b) Is $f$ continuous on the closed interval $[1, 5]$ ? Justify your answer.

Show solution

(a) First, inside each piece's interval. $2^x$ is continuous everywhere. $\frac{x + 6}{x}$ is continuous except at $x = 0$ , but 0 is not in $(2, 5)$ , so this piece causes no break. $\frac{x + 1}{x - 7}$ is continuous except at $x = 7$ , which is in $[5, \infty)$ , so $f$ is not continuous at $x = 7$ , where $f(7)$ is undefined.

Next, the boundaries.

At $x = 2$ : $f(2) = 2^2 = 4$ ; $\lim\limits_{x \to 2^-} f(x) = 2^2 = 4$ ; $\lim\limits_{x \to 2^+} f(x) = \frac{8}{2} = 4$ . The limit is 4 and equals $f(2)$ , so $f$ is continuous at 2.

At $x = 5$ : $f(5) = \frac{6}{-2} = -3$ ; $\lim\limits_{x \to 5^-} f(x) = \frac{11}{5}$ ; $\lim\limits_{x \to 5^+} f(x) = \frac{6}{-2} = -3$ . The one-sided limits differ, so $\lim\limits_{x \to 5} f(x)$ does not exist and $f$ is not continuous at 5.

$f$ is not continuous at $x = 5$ and $x = 7$ , and continuous everywhere else: on $(-\infty, 5)$ , $[5, 7)$ and $(7, \infty)$ . The bracket at 5 is right because $\lim\limits_{x \to 5^+} f(x) = -3 = f(5)$ , so $f$ is continuous from the right at 5.

(b) No. Continuity on $[1, 5]$ requires $f$ to be continuous from the left at the endpoint 5, but $\lim\limits_{x \to 5^-} f(x) = \frac{11}{5} \neq -3 = f(5)$ . ( $f$ is continuous at every point of $[1, 5)$ , which is not enough.)

Where students lose points

On the exam

Questions ask for the intervals or points where a function is continuous, as multiple choice from a formula, a piecewise rule or a graph. The idea also sits inside free-response questions: the Intermediate Value Theorem (topic 1.16), the Extreme Value Theorem and the Mean Value Theorem all require continuity on a closed interval, and the justification has to state it. For a function given only by a table or graph, you can't prove continuity from the data; the question will say " $f$ is continuous" (or "differentiable," which implies it), and your justification should cite that statement.

Check yourself

Question 1
On which of the following is $f(x) = \dfrac{x - 1}{\ln x}$ continuous?
1. A
  $(0, \infty)$
2. B
  $(0, 1) \cup (1, \infty)$
3. C
  $(1, \infty)$
4. D
  all real $x$ except $x = 1$
Show answer
B
The logarithm needs $x > 0$ , and the denominator $\ln x$ is 0 at $x = 1$ , so the domain is $(0, 1) \cup (1, \infty)$ , and a quotient of continuous functions is continuous on its domain. Choice A forgets that the denominator vanishes at 1; choice D forgets that $\ln x$ is undefined for $x \le 0$ .
Question 2
Which of the following functions is continuous on the closed interval $[-2, 2]$ ?
1. A
  $\dfrac{1}{x - 3}$
2. B
  $\tan x$
3. C
  $\ln(x + 2)$
4. D
  $\sqrt{1 - x^2}$
Show answer
A
$\frac{1}{x - 3}$ is undefined only at 3, which is outside $[-2, 2]$ . $\tan x$ is undefined at $\frac{\pi}{2} \approx 1.571$ and $-\frac{\pi}{2}$ , both inside the interval. $\ln(x + 2)$ is undefined at the endpoint $-2$ , so it can't be continuous from the right there. $\sqrt{1 - x^2}$ is undefined for $|x| > 1$ .
Question 3
The two-sided limit $\lim\limits_{x \to 4} \sqrt{4 - x}$ does not exist, because $\sqrt{4 - x}$ is undefined for $x > 4$ . Explain why $\sqrt{4 - x}$ is still continuous on $[0, 4]$ .
Show answer
At the right endpoint 4, continuity on $[0, 4]$ requires only continuity from the left: $\lim\limits_{x \to 4^-} \sqrt{4 - x} = 0 = \sqrt{4 - 4}$ . At every point of $[0, 4)$ the function is a square root of a positive or zero quantity, and continuous there.
An interval's endpoint is approached only from inside the interval, so the behavior (or absence) of the function beyond 4 is irrelevant. At 0, the left endpoint, $\sqrt{4 - x}$ is continuous from the right (in fact two-sided).
Question 4
Find all intervals on which $k(x) = \dfrac{1}{\sqrt{x^2 - 4x}}$ is continuous.
Show answer
$(-\infty, 0) \cup (4, \infty)$
The square root must be of a nonnegative quantity and, being in the denominator, must not be 0, so $x^2 - 4x = x(x - 4) > 0$ . That product is positive when both factors have the same sign: $x < 0$ or $x > 4$ . The points 0 and 4 are excluded because the denominator is 0 there.

Go deeper

OpenStax, Calculus Volume 1, 2.4 Continuity