Consider the Function F X It is Continuous on the Interval 0 6

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We have previously discussed the continuity at a point of a function and the different types of discontinuities that exist (see this post for more information: Continuity at a Point). In this tutorial, we will address the continuity on an interval of a function.

What is Continuity on an Interval?

A function $f$ is continuous on an interval if it is continuous at every number in the interval.

The following types of functions are continuous at every number in their domains; in other words, they are continuous on their domains.

polynomials (continuous everywhere on $\mathbb{R}$ ).
rational functions (continuous where they are defined).
root functions (continuous where they are defined).
trigonometric functions (continuous where they are defined).
inverse trigonometric functions (continuous where they are defined).
exponential functions (continuous everywhere on $\mathbb{R}$ ).
logarithmic functions (continuous on the domain of positive, real numbers).

Examples

Let's now take a look at a few examples illustrating the concept of continuity on an interval.

Example 1: Finding Continuity on an Interval

Find the interval over which the function $f(x)= 1- \sqrt{4- x^2}$ is continuous.

Here is what this function looks like:

We know that this is a root function which is defined on the domain of real numbers provided that :

$\sqrt{4- x^2}>= 0 \Longleftrightarrow 4-x^2>= 0 \Longleftrightarrow (2-x)(2+x)>= 0$

In addition, we can see that the inequality holds as long as either

$2-x>= 0$ and $2+x>= 0$

$2-x<= 0$ and $2+x<= 0$

The first set of inequalities implies that:

$x<= 2$ and $x>= -2$

which means that:

$x \in [-2, 2]$

The second set of inequalities implies that:

$x>= 2$ and $x<= -2$

but, there is no solution to this.

In conclusion, the interval of continuity of the function discussed above is $x \in [-2, 2]$ .

Example 2: Finding Continuity on an Interval

Determine the interval on which the function $f(x)= \frac{x-3}{x^2+ 2x}$ is continuous.

Let's take a look at the function above:

First of all, this is a rational function which is continuous at every point in its domain.

Secondly, the domain of this function is $x \in \mathbb{R}$ , provided that

$x^2+ 2x \neq 0 \Longleftrightarrow x(x+2) \neq 0 \Longleftrightarrow x \neq 0, x \neq -2$

Therefore, the function is continuous on the interval $(-\infty, -2), (-2, 0)$ and $(0, \infty)$ .

Example 3: Finding Continuity on an Interval

Find the interval of continuity for the following function $f(x)= \begin{cases} 1-x^2, \text{ if } x<= 1 \\ \ln x, \text{ if } x> 1 \end{cases}$

$f(x)= 1-x^2$ is continuous for all $x<= 1$ , since it is a polynomial function.

$f(x)= \ln x$ is a logarithmic function which is continuous for all $x> 0$ . Therefore, $f(x)= \ln x$ is continuous for all $x> 1$ .

You many also be wondering, what happens when $x= 1$ ?

Well, that is a great question. We compute the left and right limits and we see that:

$\displaystyle\lim_{x \to 1^-}{f(x)}= \displaystyle\lim_{x \to 1^-}{(1-x^2)}= 0 = f(0)$

$\displaystyle\lim_{x \to 1^+}{f(x)}= \displaystyle\lim_{x \to 1^+}{(\ln x)}= \ln 1= 0 = f(0)$

As a result,

$\displaystyle\lim_{x \to 1}{f(x)}= 0$

Therefore, $f(x)= \begin{cases} 1-x^2, \text{ if } x<= 1 \\ \ln x, \text{ if } x> 1 \end{cases}$ is continuous on the entire set of real numbers, $(-\infty, \infty)$ .

Practice Problems

Now it's your turn to put into practice the ideas explained in this tutorial. Find the interval of continuity of the following functions:

$f(x)= \sqrt{1+ \frac{1}{x}}$
$f(x)= 2x^4- 3x^2+ 5x- 2$
Determine the interval on which $f(x)= \frac{2x- 5}{x^2+ 3x}$ is continuous.
$f(x)= 5^x- 3$
$f(x)= 5\log_{10}{x}- 12$

Practice Solutions

$x \in \mathbb{R}$ such that $x \in [-1, 0) \cup (0, \infty)$
$x \in \mathbb{R}$
The answer is $x \in (-\infty, -3) \cup (-3, 0) \cup (0, \infty)$
$x \in \mathbb{R}$
$x \in (0, \infty)$

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Source: https://mathleverage.com/continuity-on-an-interval/