Unit 1 · Topic 1.12

1.12 Confirming Continuity over an Interval

A function is continuous on an interval when it is continuous at every point of that interval. Most familiar functions are continuous everywhere they're defined, so the real job is to find the few trouble spots and check them.

Key terms

continuous on an interval
domain
endpoints
rational function
piecewise-defined function

Continuous on open and closed intervals

f is continuous on an open interval (a, b) if it is continuous at every x between a and b.

For a closed interval [a, b], you also need the endpoints. At the left endpoint, only the right side matters: lim (x→a⁺) f(x) = f(a). At the right endpoint, only the left side matters: lim (x→b⁻) f(x) = f(b). This one-sided continuity is all an endpoint needs.

Functions that are continuous on their domains

These function families are continuous at every point of their domain:

Polynomials: continuous for all real x.
Rational functions: continuous except where the denominator is 0.
Roots: √x is continuous for x ≥ 0 (one-sided at 0); ∛x is continuous for all x.
Exponentials (eˣ, aˣ): continuous for all x.
Logarithms (ln x): continuous for x > 0.
sin x and cos x: continuous for all x. tan x and sec x break where cos x = 0; cot x and csc x break where sin x = 0.

Function	Where it is continuous
1/(x − 4)	All x except 4
√(x − 1)	x ≥ 1
ln x	x > 0
tan x	All x except π/2 + kπ (k an integer)

Building bigger functions

Sums, differences, products and constant multiples of continuous functions are continuous. A quotient is continuous wherever its denominator isn't 0. A composition f(g(x)) is continuous wherever g is continuous and f is continuous at g(x).

So to find where a combined function is continuous, find where each piece is defined, then remove any zeros of denominators. For piecewise functions, also check each break point with the three conditions from 1.11.

Why intervals of continuity matter

Several big theorems start with “if f is continuous on [a, b].” The Intermediate Value Theorem (1.16), the Extreme Value Theorem (5.2) and the Mean Value Theorem (5.1) all fail without it. So before using any of them, check the interval for the trouble spots: zeros of denominators, logs of numbers ≤ 0, even roots of negatives, and break points of piecewise functions.

If none of those trouble spots lies in the interval, you can state that f is continuous there and name the reason.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Finding the intervals of continuity
Where is f(x) = ln(x) / (x² − 9) continuous?
Show the solutionHide the solution
1. Step 1: ln x is continuous only for x > 0.
2. Step 2: The denominator x² − 9 = (x − 3)(x + 3) is 0 at x = 3 and x = −3. Only x = 3 matters, since −3 isn't in x > 0 anyway.
3. Step 3: Everywhere else in x > 0, f is a quotient of continuous functions with a nonzero denominator.
Answer: f is continuous on (0, 3) and (3, ∞).
Example 2
A piecewise function on all real numbers
h(x) = 2x + 1 for x ≤ 1, and h(x) = x² + 2 for x > 1. Is h continuous for all real numbers?
Show the solutionHide the solution
1. Step 1: Each piece is a polynomial, so h is continuous on (−∞, 1) and (1, ∞). Only x = 1 needs checking.
2. Step 2: h(1) = 2(1) + 1 = 3 (the first piece includes x = 1).
3. Step 3: lim (x→1⁻) h(x) = 3 and lim (x→1⁺) h(x) = 1 + 2 = 3.
4. Step 4: The limit exists and equals h(1).
Answer: Yes. The pieces are continuous, and at x = 1 both one-sided limits equal h(1) = 3, so h is continuous for all real x.
Example 3
Trap: a trouble spot inside the interval
A student says tan x is continuous on [0, π] because trig functions are continuous on their domains. Is the student right?
Show the solutionHide the solution
1. Step 1: tan x = sin x / cos x is continuous wherever cos x ≠ 0.
2. Step 2: cos x = 0 at x = π/2, which is inside [0, π].
3. Step 3: So π/2 isn't in the domain of tan x, and tan x has a vertical asymptote there.
Answer: No. tan x is continuous on [0, π/2) and (π/2, π], but not on all of [0, π]. “Continuous on its domain” only helps if the whole interval is in the domain.

Common mistakes

Forgetting domain restrictions from logs and even roots. ln x needs x > 0 and √x needs x ≥ 0.
Saying a function like √x is not continuous on [0, 4] because there's nothing to the left of 0. Endpoints only need one-sided continuity.
Listing x = −3 as a break for ln(x)/(x² − 9). It's outside the domain, so it isn't part of any interval of continuity.

On the exam

Many theorems on the exam start with “f is continuous on [a, b].” If a problem tells you f is differentiable, that also means f is continuous (2.4).
In free response, you can justify continuity of a familiar function on an interval by naming its type, like “f is a polynomial, so it is continuous on [1, 5].”

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