- #1

samir

- 27

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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.

At the very basic level, I understand this notion of discontinuity. But I am looking to understand this better, at a deeper level. Because I know I will need to know this well as I start to explore piece-wise functions.

"Did you lift your pencil while drawing the graph", the teacher asked.

"Yes?... I'm sure I did at some point", the student responded.

"There you go then! It's discontinuous!" The teacher turned around and went away.

Why do we bother drawing graphs? I want to be able to tell if a function is discontinuous without using a graph. Can't we tell if it is discontinuous unless we draw a graph? It is very unsettling to know that my pencil is smarter than me! (Nerd) Perhaps the next time I'm in doubt I will ask my pencil for the correct answer.

The simplest kind of discontinuity might be this thing called "point discontinuity". So let's start with that.

The following function is given.

$$\displaystyle f(x)=\begin{cases}1, & x=3 \\x^{2}, & \text{all other real x values} \\ \end{cases}$$

Is this function defined or undefined? Is it well-defined or ill-defined? Is it continuous or discontinuous?

If a function is discontinuous at a single point, does that mean that the function is also undefined at that point? Would that depend on the domain and rule of the function? What I'm having hard time with is understanding the difference between a function that's undefined at a point and a function that's discontinuous at a point. Can a function be discontinuous at a point and still be defined at that point? Can it be both discontinuous and undefined at a point? Perhaps a Venn diagram or some other diagram would tell more than words do alone. But let's start with a graph of the function above.

View attachment 5460

The point $f(3)=1$ is part of the function definition. So do we say that this function is defined at $x=3$? Do we say that it is defined but discontinuous at $x=3$?

Is there a difference in having a function that's undefined at a point, and having a function that's not a function at all (but a relation)? A function that is undefined at a point is interpreted as if the function doesn't exist at that point?

How do we tell which part of the function rule to use in the example above for $x=3$?

$$3 \to 1$$

$$x^2 | x=3 \to 3^2 \to 9$$

$$f(3)=9 \text{ or } f(3)=1$$

Which one is it? How do we decide between the two? My calculator says 1! But why?

I know this is not true:

$$f(3)=9 \text{ and } f(3)=1$$

We can't have two different output values for the same input value. So what does this tell us? The function doesn't exist at $x=3$?

Is the function discontinuous in its domain? It appears to be defined for all real numbers.

What if we limit the domain of the function to an interval before or after the $x=3$ point?

$$Domain = \left\{x \in\Bbb{R} | x \gt 3\right\}$$

$$Codomain = \left\{y \in\Bbb{R} | y \gt 9\right\}$$

Alternatively:

$$Domain = \left\{x \in\Bbb{R} | x \lt 3\right\}$$

$$Codomain = \left\{y \in\Bbb{R} | y \lt 9\right\}$$

Would it be continuous in one of these domains? Despite what it says in the function rule?

Let's have a look at this one. This function is only slightly changed. The point discontinuity in this case overlaps the graph of the function. As can be seen by the graph below.

$$\displaystyle f(x)=\begin{cases}9, & x=3 \\x^{2}, & \text{all other real x values} \\ \end{cases}$$

View attachment 5461

At $x=3$ the function value is 9. But at the same time, the function value for the point is also 9.

$$3 \to 9$$

$$x^2 | x=3 \to 3^2 \to 9$$

$$f(3)=9 \text{ or } f(3)=9$$

$$f(3)=9 \text{ and } f(3)=9$$

In this example, either of the two pieces of the function rule can be used to arrive at the same function value. So this function is continuous then?

Both in this example, and the previous example, I would refer to these points as points of inclusion. That is to say they are in the domain of real numbers. It doesn't say anything in the definition what the input value is not allowed to be. It's more as a consequence of the function rule that we might say that certain input values are not appropriate?

Here is another example, where we have a point of exclusion. That is to say, the point needs to be excluded from the domain by the very definition. We are not allowed to divide by zero.

$$f(x)=\frac{x^2(x-2)}{x-2}$$

$$f(2)=\frac{2^2(2-2)}{2-2}$$

$$f(2)=\frac{4(0)}{0}$$

$$f(2)=\frac{0}{0}$$

The function doesn't exist at $x=2$? But it does exist at all other points?

The function doesn't exist at all, at any x if its domain is defined to be all real numbers? It takes only one point in the domain at which the function is undefined for the entire function to be undefined or not to exist at all? What does that say about discontinuity? Do we even have to consider discontinuity if the function doesn't exist?

What if we changed the domain so that it excludes the point that breaks the universe (division by zero)?

If $x=2$ is not allowed, we can simplify the function rule.

$$f(x)=\frac{x^2(x-2)}{x-2}$$

$$f(x)=x^2$$

$$\displaystyle f(x)=\begin{cases}x^2, & x\neq 2 \\undefined, & x=2 \\ \end{cases}$$

View attachment 5463

What are the criteria for a function to be discontinuous at a single point? What methods do we have to determine if it is discontinuous at a point? Can we examine the limits at the given point do determine if the function is discontinuous at that point? How would the limit need to be?