Continuity on an Interval — AP® Calculus AB

Part 1 · Hook

Why this matters

A hiking trail is only usable if every step of it is passable, not just the start and the finish. Calculus asks the same of a function on an interval: it must be continuous at every point along the way. Luckily, most functions you know come with that guarantee on their whole domain.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $f$ is continuous at $c$ when

$f(c)$ is defined, $\lim_{x \to c} f(x)$ exists, and they are equal
$f(c)$ is defined
$\lim_{x \to c} f(x)$ exists
$f(c) = 0$

Show the answer

All three conditions are needed.

Correct: $f(c)$ is defined, $\lim_{x \to c} f(x)$ exists, and they are equal:
$f(c)$ is defined:
$\lim_{x \to c} f(x)$ exists:
$f(c) = 0$ :

2. What is the domain of $\ln x$ ?

$x > 0$
$x \ge 0$
every real number
$x \ge 1$

Show the answer

A logarithm is defined only for positive inputs.

Correct: $x > 0$ :
$x \ge 0$ :
every real number:
$x \ge 1$ :

3. Where is $\frac{1}{x - 4}$ undefined?

$x = 4$
$x = -4$
$x = 0$
$x = 1$

Show the answer

The denominator is 0 at x = 4.

Correct: $x = 4$ :
$x = -4$ :
$x = 0$ :
$x = 1$ :

Part 4 · See it

See it first

The graph of f(x) = x/2 + 1 for x < 1 and f(x) = 1/(3 − x) for x ≥ 1, x not 3. The left piece is a line ending at an open circle at (1, 1.5). The right piece starts at a closed dot at (1, 0.5), rises without bound as x approaches 3 from the left, and comes up from below on the other side of the dashed vertical asymptote x = 3. Brackets under the axis mark the intervals where f is continuous: (−∞, 1), [1, 3) and (3, ∞). — A function and the intervals on which it is continuous. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A function is continuous at every point of an open intervalit is continuous on that interval: its graph has no break there
The interval is closed, $[a, b]$ at the endpoints only the one-sided limit from inside counts: $\lim_{x \to a^+} f(x)$ $= f(a)$ and $\lim_{x \to b^-} f(x)$ $= f(b)$
Polynomial, rational, root, exponential, logarithmic and trigonometric functionsare continuous on their domains, so you only check where they are undefined
Continuous functions are combined by sums, products, quotients or compositionthe result is continuous wherever it is defined; only break points of piecewise rules need a separate check

Part 6 · Key ideas

Key ideas

Continuous on an interval: continuous at every point of it. At a closed endpoint, only the one-sided limit from inside must match the value.
Polynomials, $e^x$ , $\sin x$ and $\cos x$ are continuous on $(-\infty, \infty)$ .
Rational functions, roots, $\ln x$ and $\tan x$ are continuous on their domains: avoid zeros of denominators, negative inputs to even roots, inputs to $\ln$ that are not positive, and zeros of $\cos x$ for $\tan x$ .
Sums, products, quotients (denominator not 0) and compositions of continuous functions are continuous.
For a piecewise function, each piece is handled by the rules above; check each break point with the three conditions.

Part 7 · Misconception

A common mistake

The wrong idea: If $f$ is continuous on $(1, 5)$ and on $[5, 8]$ , it is continuous on $(1, 8]$ .

What actually happens: Continuity on $[5, 8]$ only guarantees continuity from the right at 5. The left-hand limit at 5 can still differ from $f(5)$ , so $f$ may jump there. Check the shared point with all three conditions.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $f$ on $[-3, 5]$

1. On which of these intervals is $f$ continuous?

$[0, 2)$
$[-3, 0]$
$[-3, 2)$
$(0, 5]$

Show the answer

On $(0, 2)$ the curve is unbroken. At the left endpoint 0, the right-hand limit is 3 and $f(0) = 3$ , so $f$ is continuous from the right there. That is all a closed left endpoint needs.

Correct: $[0, 2)$ : Unbroken inside, and $\lim_{x \to 0^+} f(x)$ $= 3$ $= f(0)$ at the closed endpoint.
$[-3, 0]$ : At the right endpoint 0, the left-hand limit is 2 but $f(0) = 3$ , so $f$ is not continuous from the left there.
$[-3, 2)$ : This interval contains the jump at $x = 0$ in its interior.
$(0, 5]$ : $f(2)$ is undefined, and 2 is inside this interval.

2. At $x = 0$ , the function $f$ is continuous

from the left, but not from the right
from the right, but not from the left
from both the left and the right
from neither the left nor the right

Show the answer

$\lim_{x \to 0^+} f(x)$ $= 3$ $= f(0)$ , so $f$ is continuous from the right. $\lim_{x \to 0^-} f(x)$ $= 2 \ne f(0)$ , so it is not continuous from the left.

from the left, but not from the right: The left piece heads to the open circle at height 2, not to $f(0) = 3$ .
Correct: from the right, but not from the left: The closed dot $(0, 3)$ sits where the right piece starts.
from both the left and the right: The left side arrives at 2, not at $f(0) = 3$ .
from neither the left nor the right: From the right, the curve arrives exactly at $f(0) = 3$ .

Table

Four functions

Four functions, each continuous on its domain
Function	Formula
$p$	$p(x) = \sqrt{x - 1}$
$q$	$q(x) = \ln(x + 2)$
$r$	$r(x) = \frac{x + 1}{x^2 - 4}$
$t$	$t(x) = \tan x$

3. Which is the largest interval on which $p$ is continuous?

$(1, \infty)$
$[0, \infty)$
$[1, \infty)$
$(-\infty, \infty)$

Show the answer

$\sqrt{x - 1}$ is defined for $x \ge 1$ . At the endpoint 1, $\lim_{x \to 1^+} p(x)$ $= 0$ $= p(1)$ , so it is continuous from the right, and the interval includes 1.

$(1, \infty)$ : This leaves out 1, where $p$ is continuous from the right. It is not the largest.
$[0, \infty)$ : $p(0) = \sqrt{-1}$ is not a real number.
Correct: $[1, \infty)$ : Defined and continuous for $x \ge 1$ , with one-sided continuity at 1.
$(-\infty, \infty)$ : $p$ is undefined for $x < 1$ .

4. Let $f(x) = e^x$ for $x < 0$ and $f(x) = x + 1$ for $x \ge 0$ . Where is $f$ continuous?

On $(-\infty, \infty)$
On $(-\infty, 0)$ and $(0, \infty)$ , not at 0
On $[0, \infty)$ , and nowhere to the left of 0
On $(-\infty, 0]$ , and nowhere to the right of 0

Show the answer

Each piece is continuous on its own interval. At 0: $\lim_{x \to 0^-} e^x = 1$ , $\lim_{x \to 0^+} (x + 1)$ $= 1$ , and $f(0) = 1$ . So $f$ is continuous at 0 as well.

Correct: On $(-\infty, \infty)$ : Both pieces are continuous, and they meet at $(0, 1)$ , which is $f(0)$ .
On $(-\infty, 0)$ and $(0, \infty)$ , not at 0: $e^0 = 1$ and $0 + 1 = 1$ : the pieces meet, so 0 is not a break.
On $[0, \infty)$ , and nowhere to the left of 0: $e^x$ is continuous for $x < 0$ .
On $(-\infty, 0]$ , and nowhere to the right of 0: $x + 1$ is continuous for $x > 0$ .

5. At how many real numbers is $f(x) = \frac{x - 1}{x^3 - x}$ not continuous?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f$ is a rational function, so it is continuous except where $x^3 - x$ $= x(x - 1)(x + 1)$ $= 0$ : at $x = -1$ , $0$ and $1$ . That is 3 points (removable at 1, infinite at 0 and $-1$ ).

Answer: 3

6. A function $f$ is continuous on $(1, 5)$ and on $[5, 8]$ . Is $f$ continuous on $(1, 8]$ ?

Not necessarily: $\lim_{x \to 5^-} f(x)$ might differ from $f(5)$
Yes: together the two intervals cover $(1, 8]$
Yes: $f(5)$ is defined because $5$ is in $[5, 8]$
No: two touching intervals of continuity force a break at 5

Show the answer

Continuity on $[5, 8]$ only says $f$ is continuous from the right at 5. Continuity at 5 also needs $\lim_{x \to 5^-} f(x)$ $= f(5)$ , which nothing guarantees; $f$ could jump at 5.

Correct: Not necessarily: $\lim_{x \to 5^-} f(x)$ might differ from $f(5)$ : The left side at 5 is not covered by either fact, so a jump at 5 is possible.
Yes: together the two intervals cover $(1, 8]$ : Covering the points is not enough; the left-hand limit at 5 must match $f(5)$ .
Yes: $f(5)$ is defined because $5$ is in $[5, 8]$ : A defined value is one condition; the two-sided limit at 5 must equal it.
No: two touching intervals of continuity force a break at 5: Two such intervals need not break; for example, $f(x) = x$ is continuous on both intervals and on $(1, 8]$ .

Part 9 · Summary

Summary

A function is continuous on an interval when it is continuous at every point of it, with one-sided continuity at closed endpoints. The familiar function families are continuous on their domains, and sums, products, quotients and compositions keep that. So you only need to check where a function is undefined and where a piecewise rule switches.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of ff on [−3,5][-3, 5]

Four functions

Summary

What comes next

Connections

The graph of $f$ on $[-3, 5]$