1.12 Confirming Continuity over an Interval

1.12 Confirming Continuity over an Interval

Topic 1.11 defined continuity at a single point: $f(a)$ is defined, $\lim_{x\to a}f(x)$ exists, and the two are equal. Most results in calculus, such as the Intermediate Value Theorem, need more than that—continuity on a whole interval. This page shows how to confirm it without testing infinitely many points one at a time.

Continuity on an interval

The pointwise definition scales directly to intervals:

LIM-2.B.1: A function is continuous on an interval if the function is continuous at each point in the interval.

This means that a claim such as “$f$ is continuous on $[0,4]$” is a global claim supported by pointwise reasoning. You must know that every point from $0$ through $4$ passes the continuity test, including the endpoints under the appropriate one-sided interpretation.

A powerful shortcut is supplied by LIM-2.B.2:

LIM-2.B.2: Polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous on all points in their domains.

For example, $p(x)=x^4-2x+7$ is continuous everywhere because it is a polynomial. The rational function

$$ r(x)=\frac{x+1}{x-2} $$

is continuous wherever it is defined, so it is continuous on $[0,1]$, but not continuous on $[0,3]$ because $x=2$ lies in that interval and makes the denominator zero.

Interval endpoints: closed, open, and half-open

At an interior point, continuity uses the two-sided limit. At an endpoint, only values from inside the interval matter.

Interval	Points to check	Endpoint interpretation
$[a,b]$	Every interior point plus $a$ and $b$	Use $\displaystyle\lim_{x\to a^+}f(x)=f(a)$ and $\displaystyle\lim_{x\to b^-}f(x)=f(b)$
$(a,b)$	Every point strictly between $a$ and $b$	No endpoint values are included
$[a,b)$	Every point from $a$ through, but not including, $b$	Check $a$ from the right; do not check $b$ as a point in the interval
$(a,b]$	Every point after $a$ through $b$	Do not check $a$; check $b$ from the left

Worked example: one function, three interval claims

Consider

$$ g(x)=\ln(x+2). $$

Its domain is $x>-2$. Because logarithmic functions are continuous on all points in their domains by LIM-2.B.2, $g$ is continuous on $[0,5]$, $(0,5)$, and $[0,5)$. At the included endpoint $x=0$, the relevant statement is

$$ \lim_{x\to 0^+}\ln(x+2)=\ln 2=g(0). $$

The endpoint $x=5$ requires no check for $[0,5)$ because $5$ is not part of that interval.

Now consider

$$ h(x)=\frac{1}{x+1}. $$

It is continuous on $[-2,0]$ only if its domain contains every point in that interval. But $x=-1$ lies in $[-2,0]$, and $h(-1)$ is undefined. Thus $h$ is not continuous on $[-2,0]$. It is continuous on $[-2,-1)$ and on $(-1,0]$, because the excluded point is not part of either interval.

Misconception check — “Endpoints always require two-sided limits.” For continuity on $[a,b]$, the endpoint $a$ is tested from the right and $b$ from the left. A two-sided limit would ask about values outside the interval, which are irrelevant to continuity over that interval.

AP reasoning: claim, evidence, conclusion

The relevant AP practices are Implementing Mathematical Processes, Connecting Representations, Justification, and Communication and Notation. In practice, a strong response identifies the function type or problematic point, states the domain or endpoint limit being used, and concludes with the exact interval claim.

A complete justification might read: “The function $r(x)=\frac{x+1}{x-2}$ is rational and therefore continuous on its domain. Since $2\notin[0,1]$, $r$ is continuous on $[0,1]$.” That reasoning is stronger than “the graph has no breaks,” because it names the theorem and checks the interval’s domain.

Retrieval check

For $F(x)=\sqrt{x-1}$, decide whether $F$ is continuous on $[1,9]$. What type of endpoint limit is needed at $x=1$? The answer is yes: $F$ is a power function continuous throughout its domain $[1,\infty)$, and at $x=1$ continuity is confirmed with the right-hand limit $\displaystyle\lim_{x\to1^+}F(x)=F(1)=0$.