Algebra Help

Continuous Functions and Discontinuities

Graph of y = sin(x)

A function that is continuous is a function whose graph has no breaks in it; i.e. it is a continuous curve. Generally speaking, a function is continuous if you can draw its graph without picking up your pencil. Notice, on the graph of y = sin(x), that the function is completely connected at all points. Many functions, however, will have isolated points where they are not connected. These problem points are called discontinuities. There are three types of discontinuities:

 


Discontinuity 1: Asymptotic Discontinuities

Graph of f(x) = (x+4)/(x-1)(x+8)

In the function f(x) = (x+4)/(x-1)(x+8) we know that the domain is limited to all real numbers except 1 and -8. Often, the most interesting points in a function are the problematic points, and indeed, we can see in the graph that the function behaves very strangely at the holes in the domain.

It may be revealing to look at the values of the function approaching one of the problematic points, x = 1:

Values of y = (x+4)/(x-1)(x+8)
x
y
0.25 -0.687
0.5 -1.059
0.75 -2.171
1.0 undefined
1.25 2.270
1.5 1.158
1.75 0.786

 

As we approach x = 1 from the left side, with increasing x values, the y value drops to lower and lower negatives. The graph shows us that the y value seems to approach -infinity from the left as the x-value approaches x = 1, marked by the dotted line. This represents a discontinuity, since the function is not connected over the dotted line. Specifically, this type of discontinuity is called an asymptotic discontinuity. The dotted lines represent asymptotes; they are values for which the function never takes a value, yet still approaches. The asymptotes we see with this function are called vertical asymptotes because they are vertical lines. There are also horizontal asymptotes. To see horizontal asymptotes, try graphing the function f(x) = (2x)/(x+4). It should have a horizontal asymptote at y = 2. In order to figure out where horizontal asymptotes occur in a function, check out the section on limits.

In general, asymptotes occur when a function approaches infinity at a specific value of x or y. If a function has values on both sides of an asymptote, then it cannot be connected, so it must have a discontinuity at the asymptote. We look for asymptotes at points where the denominator is zero because: when the denominator gets close to zero and becomes very small, it makes the value of the function very large. For example, if we look at the fraction 5/.1, i.e. divide 5 by 0.1, we get that it equals 50. If we make the denominator smaller, the value of the fraction gets larger: 5/.01 = 500, 5/.00001 = 500000. This tells us that the closer to zero that the denominator is, the larger the value of the fraction. In other words, we've demonstrated lim 1/x = infinity as x approaches 0


Discontinuity 2: Point Discontinuities

Point discontinuities are also called removable discontinuities or removable singularities.

Sometimes we come across functions that are defined differently for a certain point. Consider the function f(x) = x^2, except at x=3, where f(3) = 1. We defined the value of the function to be 1 at the point x = 3, yet, the rest of the function is dictated by f(x) = x^2. We can see in the graph that the function is continuous except for the tiny hole in the curve at x = 1. It is discontinuous at a single point, and this discontinuity is called a point discontinuity.

In general, point discontinuities occur when a function is defined specifically for an isolated x-value. However, this does not guarantee a point discontinuity. For example, if we change our function slightly to f(x) = x^2, except at x = 3, where f(3) = 9 it becomes continuous. This is because we have defined the value of the function at f(3) precisely to be the value of the function f(x) = x^2 at x = 3. In this case, we did not define the value at = 3 to be different from what it would be if the function were f(x) = x^2. Then there is no discontinuity. Compared to our last function with a point discontinuity, we moved the point back up to the function to "plug" up the hole, and it is now continuous. Always remember, if a function is defined like this, to check if the isolated point is a point discontinuity or just a trick.

Point Discontinuities also arise when our function has a denominator that can be equal to zero, but that part of the denominator can also be cancelled out with a like term in the numerator. Consider the function f(x) = (x^2)(x-2)/(x-2). If we try to find the value of the function at x = 2, we end up getting 0/0. 0/0 represents an undefined number - i.e. the function does not exist at that point. However, if we restrict the function to a domain that does not include x = 2, we can simply cancel out the (x-2)/(x-2) and be left with f(x) = x^2. This leaves us to define the function as f(x) = x^2, except at x = 2, where it is undefined. We have effectively removed the discontinuity to show that the function behaves exactly like f(x) = x^2, except at x = 2, where it is undefined.

In conclusion, point discontinuities also occur when we can cancel a term in the denominator and the numerator. They occur at the values for which the cancelled term is equal to zero. In our example, we removed x - 2, and x - 2 = 0 at x = 2. If we were to remove sin(x), we would have point discontinuities at integer multiples of π, since sin(π) = sin(2π) = sin(3π) = sin(nπ) = 0 for any integer n.


Discontinuity 3: Jump Discontinuities

Graph of piecewise function

Jump discontinuities are also called simple discontinuities, or continuities of the first kind.

Just as we can define a function at a specific point, we can also define a function in specific regions. Consider the function f(x) = x^2 for x<=2, 2-x for x>2 For all intents and purposes, we can consider it two seperate functions - one that is defined for x less than or equal to 1 and another function that is defined for x greater than 1. It is useful to think of it this way when we eventually use integration, differentiation, and other such mathematical tools. However, as it is written, f is a single function, called a piecewise function, since it is defined piece-by-piece. Note that the function adheres to our definition of being continuous.

Graph of piecewise function

Now, suppose we change our function slightly to f(x) = x^2 for x<=1, 6-x for x>1. The two pieces now have a different value at x = 1, and we can see in the graph that our function f seems to "jump" from one branch to the other. Note this this jump makes the function discontinuous. We refer to this as a jump discontinuity.

 

 

Notice that the function's discontinuity is entirely dependent on the value of the two branches of the function. Because of this, we can't just look at a piecewise function and immediatlely see if there is a jump discontinuity. Look at the following animation, which essentially moves the linear branch of the piecewise function down, and note that it is only continuous at one point. In this demonstration, we graph the function f(x) = x^2 for x<=1, or c-x for x>1 and the animation changes the value of c.

Jump discontinuities occur where the function approaches two different values from either side of the discontinuity. In our example, on the right side of x = 1, the function is approaching the value f(1) = 5. On the left side of x = 1, the function is approaching the value f(1) = 1. Thus, it has a jump discontinuity. Formally, we can check this by checking if the left-hand limit and the right-hand limit of the function correspond to the same value at a given point. For more information about left-hand and right-hand limits, please check out the limits page.


Example 1

We have to check the following function for discontinuities: f(x) = (x+4)/x for x less than or equal to 2, x^3 + 1 for x greater than 2

We begin by looking at the first branch of the function. First, note that it has an x in its denominator. This tells us that it has a problem in its domain at x = 0. As the values for x get very small, the value of the function approaches infinity, so we have an asymptotic discontinuity at x = 0. We also have to check if the branches correspond to the same value at x = 2. The first branch yields (2+4)/2 = 3 and the second branch yields 2^3 + 1 = 9. We can see the functions approach different values at x = 2 so there is a jump discontinuity at x = 2. There are no other discontinuities.

Graph of Example function


Exercises

Determine whether the following functions are continuous. If they are not, determine where their discontinuities are and classify them as asymptotic, point, or jump discontinuities.

1. f(x) = x^4 + 2x^3 - 8x + 1

2.f(x) = 2x + 4 for x<=0, -x - 3 for x>0

3.f(x) = x(x+5)(x+3)/(x+5)(x+1)

4.f(x) = x/(x+3) for x<-1, -1/2 for x>=-1

5.f(x) = x^2 + 3 for all x except x = 0, at x = 0 f(0) = 4