##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Domain Restrictions and Functions Defined Piecewise - Concept

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

An important concept in the study of functions, especially piece-wise defined functions, is that of domain restrictions. **Domain restrictions** allow us to create functions defined over numbers that work for our purposes. Piecewise defined functions are the composition of multiple functions with domain restrictions that do not overlap. Some functions are restricted from values that make them undefined.

I want to talk about domain restrictions. Here's an example. I have a function, a Quadratic function the quantity 1+x times the quantity 5-x, restricted to the domain x is between 0 and 5. Now normally a quadratic function is defined for all real numbers. We would make it a restriction like this if we just wanted to keep it to a certain interval of numbers like between 0 and 5 and this is often done when you're doing mathematical modelling, you're trying to make the function for example model revenue. You know the amount of money. You want to make sure that that always ends up positive or that the input is always positive something like that. Well anyway let's graph this quadratic function and see what the restricted domain does to the graph.

Now first of all, quadratics are really easy to graph. First thing I like to do is to plot the intercepts. The x intercept is going to be -1 0 and 5 0, so here's -1 0, here's 5 0. And since it's easy enough to find, let's find the y intercept. We get that by plugging x=0 in so we get one times 5, 5. That's 4 that's 6, 5 is right there. And also it's pretty easy to find the vertex. Remember if you have the x intercepts, the vertex is going to be exactly halfway between them. So I can just average the two x intercept values -1 and 5 and I'll get the x value of the vertex. So that's going to be 4 over 2 which is 2. So the vertex is somewhere on this line x=2. I just need to plug 2 into my function, and I get 1+2 or 3, 5-2 times 3. I get 9, so 2 9 is going to be my vertex. I go up to 8 here, so this is 6. That's about that's about a unit so that's 9. Alright let me put that point here.

Alright, and so now I'm ready to draw my graph. It'll look something like this and notice, I stop here at x=5 because the domain restriction tells me to stop there and then when I graph to the left I don't want to go past this point because this is where x=0. So I'm just going to graph back to this point. And that's the entire graph of my quadratic restricted to the domain x between 0 and 5 right? So, I just graph here and in fact I should even erase that point.

Let's take a look at another example. I have y equals negative one half x plus 5 restricted to the domain x is less than 8. Now this is just a line so it's pretty easy to graph. To get a good graph of a line, all you really need to do is find the two intercepts. If they are two. In this case 0 5 is the y intercept so that's 3, 0 5 would be about here, and the x intercept I'll get by solving the equation 0 equals negative one half x plus 5. Alright. You kind of tell by looking at what you would need x to be, right? You'd need this quantity to be -5 so you need x to be 10. So 10 0 is going to be your x intercept. And if I connect those points, I'll get a nice line. So let me do that.

Okay. Well that's the line y equals negative a half x plus 5 but it, it's not correct for the domain restriction. The domain restriction says graph this line only for x less than 8, so I don't want to go past 8. So I need to erase all this stuff. And because I can't have x=8. It's less than 8 not less than or equal to, I want a little hole here, a little circle showing that I'm not including 8. And so this would be the graph of the line y equals negative one half plus x+5 for x less than 8.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete