To 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. A 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. That’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. So 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? Do 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. And 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. Now 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. So 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. So 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. So 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? Would 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. But 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? You 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. And 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. At 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. It’s passing our horizontal line test, so I would check this

one right over there.

## 2 Comments

## Arandomcheese

You always seem to post a video on the topic I am on in school. It's getting weird!

## Felipe Chang

Nice video!! It seems to be a hard question at first but it's actually quite simple.