How to Give the Range of a Non Continuous Compound Inequality in Interval Notation

In our article on Continuity, we looked at the criterion that functions need to be continuous at a point. What if we were asked to analyze the continuity of a function over a whole interval instead? One way is to prove that it is continuous at every single point, but that's impossible since there is an infinite set of points in any given interval. In this article, we will look at functions that are continuous over their domain and some theorems related to continuity over an interval.

For the definition of function continuity at a point, see Continuity.

Interval Notation Review

First, let's review a few things about intervals. Remember that intervals can be open or closed, and you can write them in various notations. The most common are interval notation and inequality notation:

Interval Notation	Inequality Notation
\( [a, b] \)	\( a \le x \le b \)
\( (a, b) \)	\( a < x < b \)
\( (-\infty, \infty ) \)	all real numbers, also written as \( \mathbb{R} \)
\( [a, b) \)	\( a \le x < b \)
\( (a,b] \)	\( a < x \le b \)

More details on interval notation can be found in our article Functions

In addition, let's remind ourselves of a few key terms we will be using in this article: Interior points and endpoints.

An endpoint is a point at the left or right end of an interval.

A point \( x \) is in the interior of an interval \( I \) if \( x \in I \) and \( x \) is not an endpoint of \( I \).

Let's do a couple of examples with applications to interval notation.

For the interval \( [2, 3) \) what are the endpoints, and what is in the interior?

Answer:

The endpoints are \( x = 2 \) and \( x = 3 \). Notice that one of the endpoints is in the interval , but the other endpoint is not. Endpoints do not need to be inside the interval.

For the interior, you know that it can't be an endpoint, and it has to be inside the interval. So for this example, any point between 2 and 3 is in the interior, or in other words, points where \( 2 < x < 3 \). This is the same as the interval \( (2,3) \)!

What if your interval is \( (-\infty, \infty ) \)? Does this interval have endpoints? What points are in the interior?

Answer:

The endpoints of an interval need to be numbers; in this case, they aren't (infinity isn't a number), so there are no endpoints.

In fact, any real number is between \( -\infty \) and \( \infty \), so any real number is in the interior. That means the interior is \( \mathbb{R} \).

Continuity of a Function Over an Interval

Since there are endpoints and interior points of intervals, the definition of continuity over an interval needs to take both of them into account. But it is a good idea to use things you already know about continuity at a point and limits from the left and right. So let's start by defining continuity from the left and right.

A function \( f(x) \) is said to be continuous from the right at \( a \) if

\[ \lim\limits_{x \to a^+} f(x) = f(a) . \]

A function is said to be continuous from the left at \( a \) if

\[ \lim\limits_{x \to a^-} f(x) = f(a). \]

For more information on limits from the left and right, see One-Sided Limits.

The problem with defining continuity over an interval is that there are many different kinds of intervals. Sometimes the endpoints are in the interval, and sometimes they are not. So the definition needs to take all of those cases into account! Let's put together a wish list of what should go into the definition:

1. Remember that for a function to have a hope of being continuous at a point, the function needs to be defined at that point. So the first part is ensuring that the interval you care about is in the function's domain.

2. Interior points of the interval are easier since we know we can evaluate the limit of the function there. So the definition needs to say that the function is continuous at any interior point of the interval.

3. You don't know if the interval has a left endpoint that is in the interval or not. In fact, the left endpoint of the interval might not exist as in the example \( ( -\infty, 0] \). So the definition needs to say something like "if the left endpoint is in the interval then the function is continuous from the left there".

4. You also don't know if the interval has a right endpoint that is in the interval, so the definition needs to take care of that case, similar to how it takes care of the left endpoint.

Condensing the wish list down into math-speak gives the following:

Let \( I \) be an interval in the domain of the function \( f(x)\). We say that \( f(x) \) is continuous on the interval \( I \) if all of the following are true:

1. \( f(x) \) is continuous at all interior points of \( I \);
2. if the left endpoint \( a \) of the interval \( I \) is in the interval, then \( f(x) \) is continuous from the right at \( a \); and
3. if the right endpoint \( b \) of the interval \( I \) is in the interval, then \( f(x) \) is continuous from the left at \( b\).

Sometimes you will want to look at the entire domain of a function and say whether or not it is continuous on the whole domain.

A function \( f(x) \) is said to be continuous on its domain if it is continuous at every point in its domain.

Sometimes you will see a function which is continuous on the whole real line called continuous everywhere.

Now you can write down the steps to checking if a function is continuous on an interval.

Step 1: Make sure the interval you care about is part of the function's domain.

Step 2: Check the interior of the interval to see if the function is continuous there.

Step 3: Check to see if it is continuous from the right or left as needed at the endpoints of the interval.

Examples of Continuity Over an Interval

First, let's look at some examples of using the definition to see whether or not a function is continuous on an interval.

Is the function \( f(x) = \sqrt{ x-4} \) continuous on the interval \( [0, 7) \)?

Answer:

Step 1: The first step is to check the function's domain. You know that you can't take the square root of a negative number and end up with a real number, so for a point to be in the domain you need \( x - 4 \ge 0 \), or in other words \( x \ge 4 \). Writing that in interval notation, the domain of \( f(x) \) is \( [4, \infty ) \). Since part of the interval \( [0, 7 ) \) isn't in the domain, the function is definitely not continuous on the interval \( [0, 7 ) \) . You don't even need to do the other steps because it has already failed to be continuous.

Is the function \( f(x) = \sqrt{ x-4} \) continuous on the interval \( [4, 7) \)?

Answer:

From the previous example, you already know that the interval \( [4, 7) \) is in the domain of the function, so Step 1 is covered. Since 7 is not in the interval, that means you only need to check two things:

Step 2: Is the function continuous at any point in the interior? The interior of the interval is \( (4, 7) \), so in other words, if you pick a random point \( p \in (4, 7) \), is the function continuous there?

Step 3: Is the function continuous from the right at \( p = 4\)?

Step 2: Now, pick a random point \( p \in (4, 7) \) . You know that this point is in the interior of the interval and that the function is defined there. So using properties of limits and square roots, you get

\[ \lim\limits_{x \to p^+} f(x) = \lim\limits_{x \to p^+} \sqrt{ x-4} = \sqrt{ p-4} = f(p). \]

But was just any old point in the interior of the interval, which means this works for any point in the interior.

Step 3: First, look at the function value at \( p = 4\). You get \( f(4) = \sqrt{4-4} = 0 \). Then using properties of limits and the square root function, taking the limit from the right at \( p = 4\) you get

\[ \lim\limits_{x \to 4^+} f(x) = \lim\limits_{x \to 4^+} \sqrt{ x-4} = 0. \]

Since this is equal to the function value, you know the function is continuous from the right at \( p = 4 \) .

Putting all the steps together, you know that \( f(x) \) is continuous on the interval \( [4, 7) \) .

Intervals of Continuity on a Graph

Looking at a graph of a function, one question you can ask is "what are the intervals where it is continuous"?

The intervals where the function is continuous are called the intervals of continuity.

For the function in the graph, is it continuous on the interval \( (-2, 3] \)?

Check to see if the function is continuous on an interval, StudySmarter Original

Answer:

Notice that the interval does not contain the left endpoint but does contain the right endpoint.

Going through the steps to check for continuity on an interval:

Step 1: The function is defined on the entire interval, so that part is good to go.

Step 2: Now, you need to check points in the interior to make sure the function is continuous there. The interior of \( (-2, 3] \) is the interval \( (-2, 3) \) . If you pick any point in the interior and look at the limit, it is certainly the same as the function value. That means the function is continuous in the interior.

Step 3: So you just need to check that \( f(x) \) is continuous from the left at \( x = 3 \) because it is in the interval. You don't need to check that \( f(x) \) is continuous from the right at \( x = -2 \) because it isn't in the interval. As you can see from the graph,

\[ \lim\limits_{x \to 3^-} f(x) = 7 = f(3), \]

so the function is continuous from the left at \( x = 3 \) .

Hence the function \( f(x) \) is continuous on the interval \( (-2, 3] \) .

For the function in the graph, is it continuous on the interval \( (-2, 3] \) ?

Check to see if the function is continuous on the interval using a graph, StudySmarter Originals

Answer:

This function is almost the same as the one in the previous example. In fact, the check to make sure it is continuous in the interior is exactly the same. So it just remains to check the right endpoint of the interval \( (-2, 3] \) . Notice that \( f(3) = 1 \). Also,

\[ \lim\limits_{x \to 3^-} f(x) = 7 . \]

Now the limit from the left at the endpoint is not the same as the function value, so the function is NOT continuous on the interval \( (-2, 3] \) .

For the function in the graph below, find all intervals of continuity.

Finding intervals of continuity from a graph, StudySmarter Originals

Answer:

It is clear looking at the picture that the function is defined everywhere. Even at \( x = 0 \), the function is defined, and \( f(0) = 3 \). In addition, everywhere other than \( x = 0 \) the limit is the same as the function value. So, the only point you need to be concerned about is \( x = 0 \). Since this point is in the domain, you need to check the limit from the left and right:

\[ \lim\limits_{x \to 0^-} f(x) = 3 , \]

and

\[ \lim\limits_{x \to 0^+} f(x) = \infty . \]

Since those two limits are not the same, the function is not continuous at \( x = 0 \) even though it is defined there. So the intervals of continuity are \( (-\infty , 0) \cup ( 0, \infty ) \).

Using the Equation to Find Intervals of Continuity

Naturally, you don't want to graph every function to see where the intervals of continuity are. So let's look at some examples of using the formula of the function to find them.

Find the intervals of continuity for the function

\[ f(x) = \frac{x + 3}{\sqrt{ x^2 - 4}} . \]

Answer:

The steps are exactly the same if you are looking for intervals of continuity.

Step 1: To start looking for intervals of continuity, first, you need to find the domain of the function. The numerator of this function is a nice line, and it is defined everywhere.

So the only problem would be the denominator of the function, which is \( \sqrt{ x^2 - 4} \).

Remember that you can't have a zero in the denominator, and you can't take the square root of a negative number. You can factor this equation to get:

\[ \sqrt{ x^2 - 4} = \sqrt{ (x - 2)(x+ 2)} , \]

and the roots of this equation are at \( x = 2 \) and \( x = -2 \). In addition, it is only defined when

\[ x^2 - 4 \ge 0 , \]

or in other words when

\[ x^2 \ge 4 \]

which means either \( x \le -2 \) or \( x \ge 2 \) must be true. Then putting together the "no zero in the denominator" rule with the "positive numbers inside square roots" rule, you can see that the function \( f(x) \) has the domain \( ( -\infty , -2) \cup (2, \infty ) \).

Step 2: There are no other possible points of discontinuity in the domain other than the ones you have already found, so the function is continuous on the interior of the domain.

Step 3: The domain's endpoints aren't in the domain, so you don't need to do a special check for them.

That means the intervals of continuity for \( f(x) \) are \( ( -\infty , -2) \) and \( (2, \infty ) \).

Find the intervals of continuity for the function

\[ f(x) = \sqrt{ -x^3 -3x^2 + 13x + 15} . \]

Answer:

Step 1: The first step is to find the domain of the function. It helps that the function inside the square root has a factored form and that

\[ -x^3 -3x^2 + 13x + 15 = -(x-3)(x+1)(x+5). \]

The roots of this function are at \( x = 3 \), \( x = -1 \), and \( x = -5 \).

Plugging in test values at each interval between the roots tells us that the function

\[ y = -(x-3)(x+1)(x+5) \]

is positive on the intervals \( (-\infty , -5) \cup (-1, 3) \). So the domain of \( f(x) \) is \( (-\infty , -5] \cup [-1, 3] \) .

More details on how to determine if the function is positive or negative on an interval can be found in the Interval Notation section of our article on Functions

Step 2: The interior of the domain is \( (-\infty , -5) \cup (-1, 3) \) , and there are no possible points of discontinuity there. That means the function is continuous on the interior of the domain.

Step 3: You just need to check that the left or right limits at the domain endpoints are the same as the function values there.

For \( x = -5 \), evaluating gives \( f(-5) = 0 \). Looking at the limit from the left there,

\[ \lim\limits_{x \to -5^-} f(x) = \lim\limits_{x \to -5^-} \sqrt{ -x^3 -3x^2 + 13x + 15} = 0 . \]

Since the two are the same, \( f(x) \) is continuous from the left at \( x = -5 \) . We can do a similar check to show that is continuous from the left at \( x = 3 \).

For \( x = -1 \) you will need to check the limit from the right. So

\[ \lim\limits_{x \to -1^+} f(x) = \lim\limits_{x \to -1^+} \sqrt{ -x^3 -3x^2 + 13x + 15} = 0 = f(-1), \]

which means that is continuous from the right at \( x = -1 \) .

Putting it all together, you have checked the interior of the domain, and the appropriate limits from the left and right at the endpoints of the domain that are actually in the domain, so you know that \( f(x) \) is continuous on its domain. That means the intervals of continuity are \( (-\infty , -5) \) and \( (-1, 3) \).

Proving Continuity Over an Interval

Let's apply the definition of continuity over an interval to some examples.

Take the line \( f(x) = 4x - 3 \). The domain of this function is the whole real line. Is this function continuous everywhere?

Answer:

Rather than trying to do this for every single point in the domain (which would be impossible!) take \( p \) to be a random real number. Using the definition of continuous, you need to check that the limit exists at \( p \) and is the same as the function value there. Checking it gives

\[ \lim\limits_{x \to p} f(x) = \lim\limits_{x \to p} (4x-3) = 4p-3 = f(p), \]

which means that \( f(x) \) is continuous at \( p \). But \( p \) was just a random real number which means this works for any real number! Therefore \( f(x) \) is continuous everywhere.

In fact, you can do exactly the same process as in the previous example for any polynomial, leading to the following theorem.

Theorem: Every polynomial is continuous on the whole real line.

What about rational functions?

Where is the function

\[ f(x) = \frac{ x^2 - 4}{x + 3} \]

continuous?

Answer:

First, you need to decide where the domain of the function is because you certainly don't want to waste time checking points that aren't in the domain. You already know that the domain of rational functions is everywhere except where the denominator is equal to zero. That means the domain of is \( ( -\infty, -3) \cup (-3, \infty ) \). Just like in the previous example, take \( p \) to be a random point in the domain, so \( p \in ( -\infty, -3) \cup (-3, \infty ) \) . Because \( p \) is in the domain, you know that \( p \) isn't the endpoint of the domain (in other words, it isn't \( -3 \) ), so you don't need to check left or right limits. Checking the limit,

\[ \lim\limits_{x \to p} f(x) = \lim\limits_{x \to p} \frac{ x^2 - 4}{x + 3} = \frac{ p^2 - 4}{p + 3} = f(p) . \]

But \( p \) was just a random point in the domain, so the function is continuous on its whole domain, or in other words, it is continuous on \( ( -\infty, -3) \cup (-3, \infty ) \) .

But the process you did in the previous example would work for any rational function. So you can write the following theorem.

Theorem: Every rational function is continuous on its domain.

Continuity Over an Interval - Key takeaways

• Let be an interval in the domain of the function . We say that is continuous on the interval \( I \) if all of the following are true:
  1. \( f(x) \) is continuous at all interior points of \( I \);
  2. if the left endpoint \( a \) of the interval \( I \) is in the interval, then \( f(x) \) is continuous from the right at \( a \); and
  3. if the right endpoint \( b \) of the interval \( I \) is in the interval, then \( f(x) \) is continuous from the left at \( b \).
• A function is said to be continuous on its domain if it is continuous at every point in its domain.
• The intervals where are function is continuous are called the intervals of continuity.
• A function that is continuous on the whole real line is called continuous everywhere.

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Source: https://www.studysmarter.us/explanations/math/calculus/continuity-over-an-interval/

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