Confirming Continuity over an Interval

Lesson ~10 min read

In simple terms: In simple terms, this topic is about finding the stretches on a graph where you can draw the function's curve without ever lifting your pencil from the paper.

Why this matters

Imagine you’re planning a cross-country road trip on a brand-new highway system. Some stretches of road, like the ones across the flat plains of Kansas, are perfectly smooth and continuous for hundreds of miles. You can just set your cruise control and go. But other parts, especially in mountainous areas like the Rockies, might have tunnels, bridges, or sections closed for repair.

In calculus, functions are like these highways. Some are continuous everywhere, smooth sailing from negative infinity to positive infinity. Others have specific points where they are "out"—a hole, a jump, or a vertical asymptote you can't cross. Our job is to be the mapmakers, to identify exactly which stretches of the "function highway" are open for travel. We'll learn how to spot the trouble areas and confidently declare the intervals where the path is clear.

Concept overview

flowchart TD
    A[Start: Given a function f(x)] --> B{What type of function is it?};
    B --> C{Polynomial, exp, sin, cos?};
    C --> D[Yes: Continuous on (-inf, inf)];
    B --> E{Rational? f(x) = P(x)/Q(x)};
    E --> F[Yes: Set denominator Q(x) = 0 and solve];
    F --> G[Continuous everywhere EXCEPT at those x-values];
    B --> H{Even Root? f(x) = sqrt(stuff)};
    H --> I[Yes: Set stuff >= 0 and solve];
    I --> J[Continuous on the solution interval];
    B --> K{Logarithm? f(x) = ln(stuff)};
    K --> L[Yes: Set stuff > 0 and solve];
    L --> J;

Core explanation

In the last topic, we looked at continuity with a microscope, examining one single point at a time. Now, we’re zooming out to look at the big picture: entire intervals of a function's graph.

What Does "Continuous on an Interval" Mean?

A function is continuous on an interval if it's continuous at every single point inside that interval.

Think of it like a string of holiday lights. If every single bulb in the string is working, you can say the whole string is lit. But if even one bulb is burnt out, you can't. Similarly, if there's even one point within an interval where the function has a hole, jump, or asymptote, the function is not continuous on that entire interval. The chain is broken.

The good news? You don't have to check every single point one by one. That would be impossible! Instead, we have a powerful shortcut based on the type of function we're dealing with.

The "Well-Behaved" Functions

AP Calculus gives us a list of function types that are famously continuous wherever they are defined. Think of these as your "go-to" tools that you know you can trust.

A function is continuous on its domain.

This is the central idea. For the following function families, their interval of continuity is simply their domain. If you can find the domain, you've found the interval of continuity.

Let's break them down:

Polynomial Functions
f(x) = x^3 - 2x + 5
- These are the champs of continuity. They have no breaks, no holes, no asymptotes, nothing. Their domain is all real numbers.
- Interval of Continuity: (-∞, ∞)
Exponential Functions
f(x) = e^x or f(x) = 2^x
- Like polynomials, these are continuous everywhere.
- Interval of Continuity: (-∞, ∞)
Sine and Cosine Functions
f(x) = sin(x) or f(x) = cos(x)
- These functions wave up and down forever, smoothly and without any interruptions.
- Interval of Continuity: (-∞, ∞)

Functions with Built-in "Caution Signs"

Now for the functions that require a little more thought. They are still continuous on their domains, but their domains are restricted. Our job is to find those restrictions.

Rational Functions
f(x) = P(x) / Q(x)
- The Rule: A rational function is continuous everywhere except where the denominator Q(x) equals zero. Those spots create vertical asymptotes or holes.
- How to find the interval: Set the denominator equal to zero and solve for x. The function is continuous everywhere except for those x-values.
- For example, with f(x) = 1 / (x - 3), the trouble spot is at x = 3. So, the function is continuous on (-∞, 3) and (3, ∞).
Logarithmic Functions
f(x) = ln(x) or f(x) = log_b(x)
- The Rule: You can only take the log of a positive number.
- How to find the interval: Take whatever is inside the logarithm (the "argument") and set it to be greater than zero (> 0). Solve the inequality.
- For g(x) = ln(x - 4), we need x - 4 > 0, which means x > 4. The interval of continuity is (4, ∞).
Radical (Root) Functions
f(x) = sqrt(x) or f(x) = (x)^(1/n)
- The Rule: This depends on whether the root is even or odd.
  - Odd roots (like cube roots) are continuous everywhere, just like polynomials. f(x) = ³√x is continuous on (-∞, ∞).
  - Even roots (like square roots) are the ones to watch. You cannot take an even root of a negative number in the real number system.
- How to find the interval (for even roots): Take whatever is inside the root (the "radicand") and set it to be greater than or equal to zero (≥ 0). Solve the inequality.
- For h(x) = sqrt(x - 2), we need x - 2 ≥ 0, which means x ≥ 2. The interval of continuity is [2, ∞).

Other Trig Functions: tan(x), sec(x), etc.
- Remember that these can be written as ratios, e.g., tan(x) = sin(x) / cos(x). This makes them rational functions!
- They are discontinuous where their denominators are zero. For tan(x), this happens whenever cos(x) = 0, which is at x = π/2, 3π/2, etc.

So, the process is simple: identify the function type, apply its specific domain rule, and you've found your intervals of continuity.

Worked examples

Let's put this into practice. We'll follow our "mapmaking" process to chart the continuous intervals for a few functions.

Example 1

A Rational Function

Problem: Find the interval(s) over which the function f(x) = (x^2 - 9) / (x^2 - x - 6) is continuous.

Step 1: Identify the function type. This is a rational function because it's a polynomial divided by another polynomial.

Step 2: Identify the potential "trouble spots." For rational functions, the trouble comes from the denominator equaling zero. So, we need to find the x-values that make the denominator zero. x^2 - x - 6 = 0

Step 3: Solve for the trouble spots. Let's factor the quadratic. We need two numbers that multiply to -6 and add to -1. That would be -3 and 2. (x - 3)(x + 2) = 0 This gives us two x-values where the function is undefined: x = 3 and x = -2.

Step 4: State the intervals of continuity. The function is continuous everywhere except at x = -2 and x = 3. We write this using interval notation. Imagine a number line with open circles at -2 and 3. The continuous parts are everything to the left of -2, everything between -2 and 3, and everything to the right of 3.

Example 2

A Function with a Square Root

Problem: Determine the interval(s) over which g(x) = sqrt(9 - x^2) is continuous.

Step 1: Identify the function type. This is a radical function with an even root (a square root).

Step 2: Identify the potential "trouble spots." For an even root, the expression inside the radical (the radicand) cannot be negative. It must be greater than or equal to zero. 9 - x^2 ≥ 0

Step 3: Solve the inequality. This is a quadratic inequality. Let's rearrange it: 9 ≥ x^2 or x^2 ≤ 9

Taking the square root of both sides, we have to be careful. This inequality means x must be between -3 and 3, inclusive. -3 ≤ x ≤ 3

Step 4: State the interval of continuity. The function is defined and continuous only where x is between -3 and 3. Because the inequality is ≥, we use square brackets to show that -3 and 3 are included.

This makes sense if you picture the graph: y = sqrt(9 - x^2) is the equation for the top half of a circle with a radius of 3, centered at the origin. The graph exists only from x = -3 to x = 3.

Try it yourself

Ready to be the mapmaker? Give these a try.

Problem 1: Find the interval(s) on which the function h(t) = (t+1) / sqrt(t - 4) is continuous.

Problem 2: A function is defined as f(x) = ln(x^2 - 5x + 6). Over which intervals is this function continuous?

TL;DR

In simple terms, this topic is about finding the stretches on a graph where you can draw the function's curve without ever lifting your pencil from the paper.

Key terms

AP Calculus Continuity Continuous on an interval Domain Rational functions Radical functions Logarithmic functions Interval notation LIM-2.B Calculus help ===END===

You can now…

LIM-2.B: Determine intervals over which a function is continuous.

Essential knowledge (exam-tested)

LIM-2.B.1: A function is continuous on an interval if the function is continuous at each point in the interval.
LIM-2.B.2: Polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous on all points in their domains.

Concept map

flowchart TD
    A[Start: Given a function f(x)] --> B{What type of function is it?};
    B --> C{Polynomial, exp, sin, cos?};
    C --> D[Yes: Continuous on (-inf, inf)];
    B --> E{Rational? f(x) = P(x)/Q(x)};
    E --> F[Yes: Set denominator Q(x) = 0 and solve];
    F --> G[Continuous everywhere EXCEPT at those x-values];
    B --> H{Even Root? f(x) = sqrt(stuff)};
    H --> I[Yes: Set stuff >= 0 and solve];
    I --> J[Continuous on the solution interval];
    B --> K{Logarithm? f(x) = ln(stuff)};
    K --> L[Yes: Set stuff > 0 and solve];
    L --> J;

Read what Saavi narrates

(gentle, warm intro music fades)

Hi there, it's Saavi from Shrutam. Let's talk about a really foundational idea in calculus: continuity over an interval.

Imagine you’re driving from Dallas to Austin. For most of the trip, Interstate 35 is a smooth, continuous road. You can just drive. But what if there’s a sudden bridge closure right at Waco? That one break in the road means you can't say the *entire* route from Dallas to Austin is continuous. You have a continuous stretch from Dallas to Waco, and another one from Waco to Austin, but the whole path is broken.

Functions work the same way. We want to find those smooth, unbroken stretches of road. The big idea is that we don't have to test every single point. We can use what we know about different types of functions to find their continuous intervals quickly.

Let's work through a classic example. Suppose we have the function... f of x equals the fraction, x squared minus 9, divided by, x squared minus x minus 6.

Where is this function continuous?

First, we see it's a rational function... a fraction. The only place these functions can get into trouble is when the denominator becomes zero. So, let's find out where that happens.

We set the denominator, x squared minus x minus 6, equal to zero.
If we factor that quadratic, we're looking for two numbers that multiply to negative 6 and add to negative 1. Those numbers are negative 3 and positive 2.
So, our factored denominator is (x minus 3) times (x plus 2).
Setting that equal to zero gives us our two trouble spots: x equals 3, and x equals negative 2.

These are the two "bridge closures" on our function highway. The function is continuous everywhere else.

So, how do we write that? We use interval notation. The road is open from negative infinity all the way to negative 2. Then it's open again between negative 2 and 3. And finally, it's open from 3 all the way to positive infinity.

So the answer is... open parenthesis, negative infinity, comma, negative 2, close parenthesis... Union... open parenthesis, negative 2, comma, 3, close parenthesis... Union... open parenthesis, 3, comma, infinity, close parenthesis.

Now, a super common mistake here is to try and simplify the function first. The numerator, x squared minus 9, factors into (x minus 3) times (x plus 3). Students see the (x minus 3) on the top and bottom and cancel them. They think the problem at x equals 3 is gone. But it's not! The original function is still undefined there. For continuity, you always, always, always look at the original function's definition.

Keep practicing this process: identify the function type, find the domain restrictions, and state your intervals. You've got this.

(gentle, warm outro music fades in)

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