Lesson Part 1

Introduction

In your explorations of functions and their graphs, you will have noticed that the graphs of some functions, from start to end, are made of one unbroken curve, whereas others include breaks within their domain.

The following three graphs demonstrate this idea.

Functions Having an Unbroken Curve

Let's first look at the function \(f(x)=(x+2)^2(x-1)(x-4)\) which has an unbroken curve.

This function is a polynomial function with degree four. And we see that the function's graphed from one edge of the grid to the other with no breaks within its domain. If we were to zoom out on this graph, we would continue to see no breaks along this curve.

Functions that contain no breaks along their entire domain are continuous.

As the \(x\)-values of a continuous function vary, the \(y\)-values vary continuously and do not jump from one value to another.

If a function is continuous, then it is possible to draw the function's graph from one end of its domain to the other without removing your pencil from the paper.

If this is not possible, then the graph must contain breaks.

Functions with Breaks

Let's now look at a few functions that contain breaks within their domain. The first function that I want you to consider is a piecewise function.

\begin{align*} f(x) &= \begin{cases} x+3 &\mbox{for } x \lt 1 \\ (x-2)^2+1 &\mbox{for } x\geq 1 \end{cases} \end{align*}

Here, the piecewise function is made of two parts, the first part being linear and the second part being quadratic. When we look at the graph, we notice that the linear function and the quadratic function are not connected when \(x = 1\). The first part of \(f(x)\) has an open circle at \(y=4\) because \(x\) is strictly less than \(1\). The second part has a closed circle at \(y=2\) because \(x\) can be equal to \(1\). We refer to this type of break as a jump discontinuity, since we are jumping from a \(y\) value of \(4\) to a \(y\) value of \(2\) as we move from left to right along \(f(x)\).

The second function is a rational function.

\begin{align*}f(x) &= \dfrac{(x+2)^2}{(x-1)(x-4)}\end{align*}

This rational function has a vertical asymptote at \(x = 1\) and \(x = 4\). In this case, this function has an infinite discontinuity, meaning that the values of the function around the vertical asymptote are approaching positive or negative infinity as we get closer and closer to the vertical asymptote.

A function that is not continuous, at a particular value let's say \(x=a\), is referred to as discontinuous at \(a\).

The point, \(a\), is known as a point of discontinuity.

Definition

Let's now take some time to define what it means to be continuous and what it means to be discontinuous.

A function, \(f(x)\), is continuous at \(x=a\) if \(\displaystyle \lim_{x\to a} f(x)=f(a)\), meaning that as we approach the value \(a\) from the left side and the right side, we are approaching the same value for \(f(x)\) and this value is the same value as \(f(a)\).

For this definition to be satisfied, the following three conditions must hold.

1. \(f(a)\) must exist.
2. \(\displaystyle \lim_{x\to a} f(x)\) must exist.
3. \(\displaystyle \lim_{x\to a} f(x)=f(a)\).

Lesson Part 2

Types of Discontinuity

Removable Discontinuities

Let's now explore through some of the different types of discontinuities. Let's begin by looking at removable discontinuities. You may have seen these in previous courses.

Some functions may contain removable discontinuities (sometimes commonly referred to as holes or missing points) in their graphs.

For example, holes occur when a rational function has an equal factor in its numerator and denominator that can be cancelled.

Here is an example.

Consider the rational function, \(f(x)=\dfrac{(2x-5)(x+3)}{(2x-5)(x-1)}\).

This function has a hole at \(x=\dfrac{5}{2}\) since we see in the equation that the factor \(2x - 5\) is in the numerator and also in the denominator.

On its graph, we see a missing point within the domain of \(f(x)\) at \(x=\dfrac{5}{2}\).

Note that \(\displaystyle \lim_{x\to 5/2}f(x)\) exists, since we can approach the value \(x=\dfrac{5}{2}\) from the left and right sides.

Here, \(\displaystyle \lim_{x\to 5/2} f(x) = \dfrac{11}{3}\).

However, \(f\left (\dfrac{5}{2}\right )\) does not exist because when \(x=\dfrac{5}{2}\), \(f(x)\) has the indeterminate form \(f(x)=\dfrac{0}{0}\). This is the reason why a hole appears in the graph of \(f(x)\).

Vertical Asymptotes

The second kind of discontinuity that I'd like us to look at are vertical asymptotes. Rational functions can also have vertical asymptotes.

Let's consider the function that we just looked at for holes.

The function, \(f(x)=\dfrac{(2x-5)(x+3)}{(2x-5)(x-1)}\), can be simplified, for \(x\neq \dfrac{5}{2}\).

We can cancel the equal factor \((2x-5)\) from the numerator and the denominator of \(f(x)\) to give

\[f(x)=\dfrac{x+3}{x-1}\]

However, we must remember to place a restriction on its domain, so we must explicitly state that \(x\neq \dfrac{5}{2}\).

Once \(f(x)\) has been completely simplified, any factor of \(x-a\) that remains in the denominator will produce a vertical asymptote at \(x=a\).

When we look at this function the factor in the denominator is \(x - 1\). Therefore, the simplified function, \(f(x)=\dfrac{x+3}{x-1}\), has a vertical asymptote at \(x=1\).

We say that \(f(x)\) has an infinite discontinuity at \(x=1\) since the graph of \(f(x)\) approaches negative infinity from the left side of the vertical asymptote and positive infinity from the right side of the vertical asymptote.

Jump Discontinuities

The next type of discontinuity that we'll look at are jump discontinuities. We have already seen a jump discontinuity in the introduction of this module. Piecewise functions often have jump discontinuities.

Consider the function,

\begin{align*} f(x) &= \begin{cases} x^2+1 &\text{for } x \leq 1 \\ x &\text{for }x\gt 1 \end{cases} \end{align*}

\(f(x)\) is made up of two different parts, the first part being quadratic and the second part being linear. When we graph the two parts of \(f(x)\), we see that the quadratic part does not connect to the linear part.

Tracing \(f(x)\) from the left side to the right side, we see \(\displaystyle \lim_{x\to 1^-} f(x) =2\).

However, at \(x = 1\), the values of \(f(x)\) jump from \(y=2\) downwards to a value very close to \(y=1\).

Now, we notice on the linear part of \(f(x)\) there is an open circle at the end of the line. We use an open circle to denote a point that is not a part of the function.

Beginning on the left side of the grid, \(f(x)\) is continuous until the point where \(x=1\) and \(f(1)=2\).

We say that there is a jump discontinuity at \(x=1\).

After this point of discontinuity, \(f(x)\) is continuous for \(x\gt 1\).

Restricted Domain

Some functions have a restricted domain, meaning that their domain is not the set of real numbers.

For example, consider the function \(f(x)=\sqrt{x}\). Its domain is \(\{x\in\mathbb{R} \mid x\geq 0\}\).

This function does not exist for \(x\lt 0\); however, it is continuous for \(x\gt 0\).

We say that \(f(x)\) is continuous from the right at \(x=0\) since \(\displaystyle \lim_{x\to 0^+} \sqrt{x} \) exists and equals \(f(0)\).

Similarly, the function \(g(x)=\sqrt{-x}\) is defined only for \(x\leq 0\) and is continuous from the left at \(x=0\).

Oscillating Discontinuities

The last type of discontinuity that we're going to look at is an oscillating discontinuity. There are functions whose \(y\)-values oscillate infinitely often within a finite interval.

Consider the function \(f(x)=\sin \left ( \dfrac{1}{x}\right )\).

As \(x\to 0\), the graph of \(f(x)\) oscillates too often in any interval about \(x=0\) for \(\displaystyle \lim_{x\to 0} f(x)\) to exist.

Also, if we try to evaluate \(f(0)\), we have \(\sin \left (\dfrac{1}{0}\right )\), which is undefined. If we were to zoom in on this graph, we would continue to see \(f(x)\) oscillating around when \(x = 0\).

Lesson Part 3

Functions That Are Continuous Everywhere

There are some known functions that are always continuous for their entire domain. You have seen some of these in your previous learning of mathematics. Let's first look at a polynomial function \(f(x) = x^5 - 2x^4 - 3x^3 + 4x^2 + 4x\). Polynomial and exponential functions are continuous for all \(x\in\mathbb{R}\).

This polynomial function can be traced from one edge of the grid to the other, and so we know it is continuous. Again, if we were to zoom out on this polynomial function, we would see that \(f(x)\) is continuous for all real values of \(x\).

Consider the exponential function \(f(x)=2^x + 1\).

Exponential functions are also continuous for all real values. We can still trace the graph of \(f(x)\) from one edge of the grid to the other without removing our pencil. Again, if we were to zoom out on this grid, we would see that this is always the case.

Notice there is a horizontal asymptote at \(y=1\), but this does not affect the continuity of the function.

The absolute value function is also continuous everywhere for all \(x\in\mathbb{R}\).

We could trace the absolute value function from the left side of the grid to the right side of the grid without removing our pencil from the paper.

Notice that the corner at \(x=0\) does not affect the continuity of the function.

The trigonometric functions \(\sin(x)\) and \(\cos(x)\) are also continuous everywhere for all \(x\in\mathbb{R}\).

Let's look at the graph of \(\sin(x)\).

We can trace this curve from the left side of the grid to the right side, again without removing our pencil from the paper.

Similarly, \(\cos(x)\) is also continuous.

Now, not all trigonometric functions are continuous for all real values of \(x\). Consider the graph of \(\tan(x)\). \(\tan(x)\) has vertical asymptotes within its domain. On your own, explore the graphs of \(\tan(x)\) and the reciprocal trigonometric functions, such as \(\csc(x)\), \(\sec(x)\), and \(\cot(x)\) to determine their continuity. Are these graphs continuous, or do they have discontinuities within their domain?

Quiz

See the quiz in the side navigation.