Continuity on an Interval — AP® Calculus AB Notes

Checking three conditions at one point is fine for one point. But the big theorems ahead ask whether a function is continuous on a whole interval, which contains infinitely many points. This page shows how to answer that quickly: know which functions are continuous where, and check only the few places where something can go wrong.

Continuous on an interval

A function is continuous on an open interval $(a, b)$ when it is continuous at every point of it. On a closed interval $[a, b]$ there is one extra detail. At the endpoints, the function may not even be defined outside the interval, so we ask only about the side inside it:

at the left endpoint $a$ : $\lim_{x \to a^+} f(x)$ $= f(a)$ (continuous from the right);
at the right endpoint $b$ : $\lim_{x \to b^-} f(x)$ $= f(b)$ (continuous from the left).

So $\sqrt{x}$ is continuous on $[0, 4]$ : there is nothing to the left of 0, but $\lim_{x \to 0^+} \sqrt{x}$ $= 0$ $= \sqrt{0}$ .

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, ∞). — **Figure 1.** This function is continuous on $(-\infty, 1)$ , on $[1, 3)$ and on $(3, \infty)$ . LevlPrep original diagram.

In the figure, $f$ jumps at $x = 1$ , but the closed dot belongs to the right piece. So $f$ is continuous from the right at 1, and the middle interval can include 1: it is $[1, 3)$ . It cannot include 3, where $f$ is undefined.

Functions that are continuous on their domains

You do not have to check every point by hand. These families are continuous at every point where they are defined:

Family	Example	Continuous on
polynomials	$x^3 - 2x + 7$	$(-\infty, \infty)$
rational functions	$\frac{x + 1}{x^2 - 4}$	every $x$ where the denominator is not 0
roots	$\sqrt{x - 1}$	where the radicand is $\ge 0$ (even roots)
exponentials	$e^x$ , $2^x$	$(-\infty, \infty)$
logarithms	$\ln x$	$(0, \infty)$
sine and cosine	$\sin x$ , $\cos x$	$(-\infty, \infty)$
tangent	$\tan x$	wherever $\cos x \ne 0$ , such as $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Combinations keep continuity. Sums, differences and products of continuous functions are continuous. A quotient is continuous wherever its denominator is not 0. A composition $f(g(x))$ is continuous wherever $g$ is continuous and $f$ is continuous at the value $g(x)$ .

Finding where a function is continuous

Worked example 1. Where is $q(x)$ $= \ln(x + 2) + \frac{1}{x - 3}$ continuous?

$\ln(x + 2)$ needs $x + 2 > 0$ , so $x > -2$ ; there it is continuous.

$\frac{1}{x - 3}$ is continuous except at $x = 3$ .

The sum is continuous where both parts are: on $(-2, 3)$ and on $(3, \infty)$ .

Worked example 2. Let $f(x) = e^x$ for $x < 0$ and $f(x) = x + 1$ for $x \ge 0$ . Where is $f$ continuous?

Each piece is continuous on its own interval ( $e^x$ and $x + 1$ are continuous everywhere). The only question is the break point $x = 0$ .

$f(0) = 0 + 1 = 1$ . $\lim_{x \to 0^-} e^x$ $= e^0$ $= 1$ and $\lim_{x \to 0^+} (x + 1)$ $= 1$ . So $\lim_{x \to 0} f(x)$ $= 1$ $= f(0)$ .

All three conditions hold at 0, so $f$ is continuous on $(-\infty, \infty)$ .

A short procedure

Identify each part of the function and its family.
List the trouble spots: zeros of denominators, inputs where an even root or a logarithm is undefined, zeros of $\cos x$ under a $\tan x$ , and break points of piecewise rules.
Away from the trouble spots, the function is continuous by the families and the combination rules.
At each break point of a piecewise rule, check the three conditions.
Write the answer as intervals, with a bracket at an endpoint only when the function is defined there and continuous from inside.

Why does this matter so much? The Unit 1 finale, and several theorems in later units, start with "if $f$ is continuous on $[a, b]$ ". To use them, you will name the reason $f$ is continuous on that closed interval, usually with a sentence from this page: " $f$ is a polynomial, so it is continuous on $[1, 4]$ ."