Restricting domains of functions to make them invertible

0:00To which intervals could we restrict f of x is equal to cosine of x minus pi over four, so that f of x is invertible? And they show us what cosine of x minus pi over four, what it looks like, the graph of it. So lets just think about what it means for a function to be invertible.

0:17A function is a mapping from a set of elements that we would call the domain, so let me, my pen is a little off today so lets see if it works ok. So this right here is our domain. And this over here is our range. And a function maps from an element in our domain, to an element in our range.

0:41That's what a function does. Now the inverse of the function maps from that element in the range to the element in the domain. So that over there would be f inverse. If that's the direction of the function, that's the direction of f inverse. Now, a function is not invertible, one of the situations in which a function is not invertible you could have a function where two elements of the domain map to the same element of the range.

1:08So both of these elements map to that element of the range, so both of these are the function, but then, if this is the case, you're not going to be able to create a function that maps the other way, because if you input this into the inverse function, where do you go?

1:23Do you go to that element of the domain? Or do you go to that element right over there? So one way to think about it is, you need a one to one mapping. For each element of the range, there's only one element of the domain that gets you there. Or another way to think about it, you could try to draw a horizontal line on the graph of the function and see if it crosses through the function more than once.

1:45And you could see that this is indeed the case for this function right over here. If I did a horizontal line, right over here. Now why is this the issue? Well this is showing, actually let me show a number that's a little bit easier to look at. So lets say I drew the horizontal line right over here.

2:01Now why is this horizontal line an issue? Well, showing us the part of the domain that's being graphed here, that there's several points that map to the same element of the range. They're mapping to 0.5. 0.5, this value right over here. When you take f of that is equal to 0.5, f of this is equal to 0.5, f of this right over here is 0.5.

2:29So if you have multiple elements of your domain mapping to the same element of the range, then the function will not be invertible for that domain. So really what we're going to do is, we're going to try to restrict the domain so that, for that domain, if I were to essentially apply this, what I call the horizontal line test, I'd only intersect the function once.

2:49So lets look at the graph of the function once. So lets look at these choices. So the first one is an open set from -5 pi over four. -5 pi over four, so that's pi, that's negative pi, and another fourth of pi. So that's, I think, starting right over here, going all the way to negative 1/4th pi.

3:11So that's this domain right over here. Let me do this in a new color. So that's this, and this does not include the two endpoints. So here I can still apply the horizontal line and, in that domain, I can show that there's two members of the domain that are mapping to the same element in the range, and so if I'm trying to construct the inverse of that, what would this element, which I guess is -0.6, what would that f inverse of -0.6 be?

3:44Would it be this value here? Or would it be this value here? I would rule this one out. Let's see, negative pi to pi. I'll do this in this color right over here. Negative pi to pi. This is a fairly, so once again, this is enclosed, so we're including the two boundaries, we're including negative pi and pi in the domain.

4:10But once again over that interval, I could apply my horizontal line here and notice, or actually I could even apply the original one that I did in blue, and notice there's multiple elements of the domain that map to, say 0.5. So what would f inverse of 0.5 be?

4:28You can't construct a function where it maps only to one element of the domain, so we could rule this one out right as well. Now, negative 1/2 pi to positive 1/2 pi. So negative 1/2 pi, so let me, I'm running out of colors. So negative 1/2 pi to positive 1/2 pi, This one is interesting, if I apply a horizontal line there, there, there, so lets see, but if I apply a horizontal line right over here I do intersect the function twice, so I have two members of this domain mapping to the same element of the range, so I can rule that one out as well.

5:11And I'm left with one last choice. I'm hoping this one will work out. So 1/2 pi, it's an open set, so 1/2 pi, right over there, to five pi over four. So that's pi and another 1/4th, so that's right over there. And lets see, this is, if I were to look at the graph here, it seems like it would pass the horizontal line test.

5:35At any point here I could make a horizontal line over that domain. Actually, let me do it for the whole domain. So you see that. For the whole domain. And I'm only intersecting the function once. So, for every element of the range that we're mapping to, there's only one element in our domain that is mapping to it.

5:54It's passing our horizontal line test, so I would check this one right over there.