Recall that the domain of a function is the set of $x$x values on which the function is defined. Graphically, we can think of the domain as all values of $x$x which correspond to a point on the curve.
A function $f\left(x\right)$f(x) is said to be continuous at a point $x=c$x=c if:
i. the function is defined at that point (that is, $c$c is in the domain of $f\left(x\right)$f(x)), and
ii. we can get function values as close as we like to $f\left(c\right)$f(c) by taking a small enough region of domain values around $x=c$x=c.
A function $f\left(x\right)$f(x) is then said to be a continuous function if it is continuous at all points in its domain. Similarly, a function is continuous over an interval if it is continuous at all points in that interval.
If instead there exists a point in the function's domain at which it is not continuous, we say that the function is discontinuous at that point.
Let's now take a look at some different types of continuous functions by inspecting their graphs.
Linear functions (that is, functions of the form $f\left(x\right)=ax+b$f(x)=ax+b) are defined for all real numbers. We can represent this by saying its domain is the interval $\left(-\infty,\infty\right)$(−∞,∞).
Look at the linear function in the graph above. At the domain value $x=2$x=2 we have the function value $f\left(2\right)=3$f(2)=3. We can get other function values as close as we like to $3$3 by taking a small enough region of the domain around $x=2$x=2. Watch the applet below for a demonstration.
There is nothing special about $x=2$x=2 in this case - the same is true for any other point in the domain! So the function above is a continuous function, as are all linear functions.
Other examples of continuous functions which have a domain of $\left(-\infty,\infty\right)$(−∞,∞) include any type of polynomial function (quadratic functions, cubic functions etc) as well as exponential functions.
Some of these types are sketched below:
Many functions are not defined for every real number, but are still continuous at every point in their domain. Let's take a look at some of these types of functions.
A hyperbolic function has a form such as $f\left(x\right)=\frac{a}{bx+c}$f(x)=abx+c. Since the domain variable $x$x appears in the denominator there is a value of $x$x for which the function is not defined, so this value of $x$x is not in the domain of the function.
The hyperbolic function in the graph above has a vertical asymptote at $x=1$x=1. The domain of this function is all of the real numbers except for $x=1$x=1, which can be written as $\left(-\infty,1\right)\cup\left(1,\infty\right)$(−∞,1)∪(1,∞). Now if we exclude $x=1$x=1 from consideration, we can see that the function is continuous over each value in the domain - even if we take an $x$x-value very close to $x=1$x=1, we can still move along the curve (even just a little bit!) in either direction. This means that the hyperbolic function is a continuous function!
A logarithmic function, which takes a form such as $f\left(x\right)=\log\left(ax-b\right)$f(x)=log(ax−b), is only defined when the input of the logarithm is positive.
The logarithmic function shown in the graph above has a vertical asymptote at $x=1$x=1. It is defined on all $x$x values larger than $1$1, and we can represent this domain as $\left(1,\infty\right)$(1,∞). It is continuous at all points in this domain, so it is also a continuous function.
A square root function, which takes a form such as $f\left(x\right)=\sqrt{ax-b}$f(x)=√ax−b, is defined when the input of the square root is not negative.
The function $f\left(x\right)=\sqrt{x+4}-1$f(x)=√x+4−1 has been graphed above. The value under the square root cannot be negative, and (as we can see from the graph) this results in a domain of $\left[-4,\infty\right)$[−4,∞). The function is continuous at all points in this domain (including the endpoint where $x=-4$x=−4), and so it is also a continuous function.
As we have seen so far, most of the functions that we are familiar with are continuous. Not all functions are continuous though! Let's take a look at an example of a step function:
In the graph above, the step function has a step at $x=0$x=0. It takes a constant value (in this case $-1$−1) for $x<0$x<0 and a different constant value (in this case $+1$+1) for $x\ge0$x≥0, and so its domain is $\left(-\infty,\infty\right)$(−∞,∞) (that is, all of the real numbers).
This function has a point of discontinuity at $x=0$x=0 - no matter how small we make our region of domain values around $x=0$x=0, we will always include part of the function that takes the value $-1$−1 and part of the function that takes the value $+1$+1. This means that the step function is not a continuous function.
Note that the function is still continuous over the interval $\left(0,\infty\right)$(0,∞) and over the interval $\left(-\infty,0\right)$(−∞,0).
Consider the function $f\left(x\right)=x^2+7x+10$f(x)=x2+7x+10 drawn below.
What is the domain of $f\left(x\right)$f(x)? Give your answer using interval notation.
Select the largest interval over which the function is continuous.
$\left(-\infty,\infty\right)$(−∞,∞)
$\left(-2,\infty\right)$(−2,∞)
$\left(-\infty,-5\right)$(−∞,−5)
$\left[-5,-2\right]$[−5,−2]
So is the function $f\left(x\right)=x^2+7x+10$f(x)=x2+7x+10 a continuous function over its domain?
No
Yes
Consider the function $f\left(x\right)=\frac{x^2-25}{x-5}$f(x)=x2−25x−5 drawn below.
What is the domain of the function $f\left(x\right)=\frac{x^2-25}{x-5}$f(x)=x2−25x−5?
$\left(-\infty,5\right)\cup\left(5,\infty\right)$(−∞,5)∪(5,∞)
$\left(-\infty,\infty\right)$(−∞,∞)
$\left[5,\infty\right)$[5,∞)
$\left(-\infty,5\right]$(−∞,5]
Select the largest region over which the function is continuous.
$\left(-\infty,5\right)$(−∞,5)
$\left(-\infty,5\right)\cup\left(5,\infty\right)$(−∞,5)∪(5,∞)
$\left(-\infty,\infty\right)$(−∞,∞)
$\left(5,\infty\right)$(5,∞)
So is the function $f\left(x\right)=\frac{x^2-25}{x-5}$f(x)=x2−25x−5 a continuous function over its domain?
No
Yes
Which of the following functions are continuous?
Select all that apply.