D.3 Continuous functions; Continuity theorems

To conclude our discussion of continuity, let’s define continuous functions, and introduce some continuity theorems. This screen will be quick: we’re introducing ideas that we’ll need later, but there isn’t a whole lot of practice worth doing now. That said, the concepts themselves are crucial to have in mind.

Continuous functions

The definition of a continuous function is no surprise: a function is a continuous function if it’s continuous at every number a in its domain.

And we state without proof that the following functions are continuous everywhere they are defined:

Continuous functions

polynomials
trig functions
exponentials
root functions (e.g., $\sqrt{x}$)

rational functions (quotient of two polynomials)
inverse trig functions
logarithmic functions

Do not work to memorize this list

There’s no need to memorize this list; instead, just remember that the functions you normally think of are continuous where they are defined. (We emphasize that last phrase because, as we saw on the preceding screen, a function like $f(x) = \dfrac{1}{x-1}$ is of course not continuous at $x=1,$ but is continuous everywhere else.) Note that proving that these functions are all continuous is well beyond the scope of an introductory Calculus course, and we point you once again to an upper division Mathematics course in Analysis if you’re interested in such things!

Continuity theorems

Let’s also state theorems that codify how continuous functions can be put together to make other continuous functions. Essentially, this all works the way you think it should: adding two continuous functions together gives you a new continuous function; multiplying a continuous function by a constant gives you another new continuous function; dividing one continuous function by another gives you yet another continuous function (everywhere the function in the denominator is defined); and so forth.

Theorems regarding combining continuous functions
If f and g are continuous functions at $x=a,$ then so are

\begin{align*}
&f+g, \\[8px] &f-g,
\end{align*}

\begin{align*}
&fg, \\[8px] &\frac{f}{g}, \text{ where }g(a)\ne 0,
\end{align*}

\[Cf, \text{ where $C$ is a constant,}\] \[\text{and }f\big(g(x) \big) \text{: the composite function $f$ of $g$ of $x$}\]

The following Example illustrates the type of question you might be asked about these ideas.

Example 1: Continuity of $f(x) = \sqrt[3]{\dfrac{\cos(x)}{x^2 + 5}}$

Determine where the function $f(x) = \sqrt[3]{\dfrac{\cos(x)}{x^2 + 5}}$ is continuous.

Solution.
The polynomial $x^2 + 5$ is continuous for all real values of x, $(-\infty, \infty)$. Since $x^2 + 5$ never equals zero, we don’t have to worry about the denominator of f equaling zero. ($x^2 + 5$ never equals zero because $x^2$ is never less than 0, so $x^2 + 5$ is never less than 5.)

The function $\cos (x)$ is defined and continuous for $(-\infty, \infty).$

And the cube-root function $\sqrt[3]{\phantom{x}}$ can take on any input value, positive, negative or zero. It is continuous on $(-\infty, \infty).$

Hence f is continuous on $(-\infty, \infty) \; \cmark$.

The list of continuous functions above, plus these theorems, make it easy to state that a particular function is continuous on an interval of interest. And a function’s continuity is a prerequisite to being able to apply other theorems to and draw other conclusions about the function — as we’ll see starting on the next screen. For now, let’s remind you again that we used to say a function had to “be nice and smooth, with no sudden jumps or gaps,” whereas now we can simply say that it is continuous.

With that, here are a few problems to put your new knowledge to use:

Practice Problems: Continuous Functions

Practice Problem #1

Non-calculator question: Answer without using a graphing device of any sort.
Which of the following functions are continuous for all real numbers x?
I. $y=e^{-x}$
II. $y=x^{2/3}$
III. $y= \tan x$ \begin{array}{lllll} \text{(A) I only} && \text{(B) II only} && \text{(C) I and II } && \text{(D) I, II and III } && \text{(E) None} \end{array}

Show/Hide Solution

This question just requires you to know each of these functions well enough to be able to think about where it might be discontinuous. Ideally, a rough graph of each popped into your head when you saw the function’s expression; we have such graphs below.

If for one (or more) of them the relevant picture didn’t pop into your head, then you might reason as follows:

I. $y = e^{-x}$ ✓: There is no value of x this function cannot take on, so its domain is all real numbers. Furthermore, it has no jumps or gaps, and so is continuous.

II. $y=x^{2/3}$ ✓: Remember that the 2/3 in the exponent means “take the cube root and the square the result.” Or, “square the number and then take its cube root”: the order of operations doesn’t matter. Either way, both the squaring process and the cube-root process can take negative numbers as an input, along with zero and of course positive numbers. So its domain is all of the reals. Furthermore, the function doesn’t have any gaps or suddenly leaps in it, so it’s continuous everywhere.

III. $y = \tan x = \dfrac{\sin x}{\cos x}$ ✗: This function is undefined everywhere $\cos x = 0,$ so at $\pi/2,$ $3\pi/2$ and so forth, and so is discontinuous at each of those locations. It is thus not continuous for all real numbers x.

Hence I and II are continuous, while III is not $\implies \;\text{ (C)} \quad \cmark $

If you’d like to track your learning on our site (for your own use only!), simply log in for free with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.

It’s all free: our only aim is to support your learning via our community-supported and ad-free site.

[hide solution]

Practice Problem #2 (university exam question)

Consider the function $f(x) = \sqrt{|x-1|}$.

(a) What is the domain of $f(x)$? The range?

(b) Graph the function. Make sure to label the axes.

(c) Is $f(x)$ an even function, an odd function, or neither? Explain.

(d) Find $\displaystyle{\lim_{x\to 1}f(x)}$ if it exists. If not, explain why not.

(e) Is $f(x)$ continuous at $x=1$? Explain.

Show/Hide Solution

Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) DetailSolution (d) DetailSolution (e) Detail

(a) Domain:$ (-\infty, \infty)$. Range: $[0, \infty)$

(b) See detailed solution.

(c) Neither. (See detailed solution for explanation.)

(d) 0

(e) Yes. (See detailed solution for explanation.)

To determine the function’s domain, we recognize that we cannot have a negative number under the square-root sign. But the absolute-value signs return a non-negative value regardless of the value of $x$. Hence the domain is all real numbers: $(-\infty, \infty)$. $\quad \cmark$ $f(x) = 0$ when $x=1$. For all other values of $x$, the square-root will return a positive value. Hence the range is $y \ge 0$, which we can also write as $[0, \infty)$. $\quad \cmark$

See the figure.

You should be able to quickly sketch $y = \sqrt{x}$ for yourself, and then recognize that $y = \sqrt{x-1}$ just shifts the graph 1 to the right (putting the zero at $x = 1$). The absolute value symbols then “mirror” the graph to the left. At $x=0$, $y = 1$, so the curve must pass through that point.

To check whether a function is even or odd, we see what happens when we compute $f(-x)$. If $f(-x) = f(x)$ then the function is even; if $f(-x) = -f(x)$ then the function is odd. For our function: $$f(-x) = \sqrt{|-x-1|} = \sqrt{|x+1|}$$ Since $f(-x)$ equals neither $f(x)$ nor $-f(x)$, the function is neither even nor odd. We can also reach this conclusion from the graph in part (b): for an even function, the curve is symmetric about the $y$-axis ($y = 0$); for an odd function, it is anti-symmetric. This curve is symmetric about $y = 1$ instead, and so the function is neither even nor odd.

\begin{align*} {\lim_{x\to 1}f(x)} &= \lim_{x\to 1}\sqrt{|x-1|} \\ &= \sqrt{|1-1|} = 0 \\ &= \sqrt{|0|} = 0 \quad \cmark \end{align*}

$\displaystyle{\lim_{x\to 1}f(x)} = f(1)$, and so $f(x)$ is continuous at $x=1$. (In fact the function is continuous everywhere.) $\quad \cmark$

[hide solution]

On the next screen we conclude our study of limits and continuity by introducing the Intermediate Value Theorem.

For now, what questions or thoughts do you have about continuous functions, or continuity in general? Head on over to the Forum and join the discussion!

The Upshot

The functions in the box at the top of this screen are all continuous functions where they are defined. They include all of the ones you know well: polynomials, root functions, trig functions, exponentials, logarithms, and so forth.
Functions made by combining those functions (adding them, multiplying them, composing one with another, …) are also continuous.

We'd appreciate your feedback! 😊

How helpful?

That's great to hear, thanks! If you have any requests or suggestions for us, we'd love to hear them. (If not, please just hit 'Submit.')

Thanks for your input. If you'd like to tell us, what would have made the material more helpful for you? (If not, please just hit 'Submit.')

Thanks very much for your input. Our aim is to help you do well! If you'd like, please tell us what would have been more useful, and we'll use that to improve. (If not, please just hit 'Submit.')

We're sorry you didn't find this helpful and appreciate your letting us know. Our aim is to help you do well! If you'd like, please tell us what would have been more useful, and we'll use that to improve. (If not, please just hit 'Submit.')

What are your thoughts or questions?

0 Comments

Inline Feedbacks

View all comments