Restricting function domain to make invertible
Articles,  Blog

Restricting function domain to make invertible


Voiceover:It’s useful to
make an invertible function from one that is not invertible by restricting it’s domain. The graph of f(x) is shown below. To which intervals could we restrict f(x) to get
an invertible function? So in order for a
function to be invertible, it’s inverse relationship
has to be a function as well. What do I mean by that? Well let’s show that this one right over here is not invertible. And there’s a couple of ways
you could think about it. F is clearly a function, it
passes the vertical line test. Vertical line test, you take
a vertical line anywhere in this graph and it doesn’t
intersect more than one time. So here it only intersects once. Why does that help us think about whether this is a function? Well imagine if it did intersect. Imagine if the function was
a circle right over here. So imagine if the vertical line did intersect the
function twice like this. Well then what would f of 6 be? Is it 2 or is it negative 2? F of 6 needs to map to a unique value, you need to know what f of 6 is and here you wouldn’t now. So that’s the vertical line test for testing whether
something is a function but you could also apply
a horizontal line test to test whether it’s inverse, the inverse relationship is a function. Or whether this is invertible. One way to think about it is if you swap the x and y values here, the graph of this function is going to look something like this. The graph of this function is going to look something like this. So let me see, actually it’s
not going to look like that. It’s going to look, it’s
going to look like this. It’s going to look like this, let me draw this as neatly as I can. I think you get the idea. So like that and and like that. So it’s going to look like that if you were to take the
inverse relationship, if you were to swap the x and y’s and now this clearly would not
pass the vertical line test. That clearly would not pass
the vertical line test. Or another way to think about it is we could just apply
a horizontal line test, we could apply a horizontal line test, let me clean it up a little bit, to our original function to
test whether it is invertible. The horizontal line
test, it clearly fails. Now once again, why does that show that that’s not invertible? Actually let me do it on
a value that’s easier to, even here we see a horizontal line is intersecting the function
there, there, there and there. Now why is that a problem? Well, let’s just remind ourselves what a function and an
inverse function do. They’re just mappings from the… The function is a mapping from the domain to the range. So let’s say this is the domain. Domain. And this is the range. This is the range. And so our function is clearly, let’s just take 2 of these values. Our function takes 2 in the domain and maps it to 2 in the range. So the function does that, f of 2 is 2. We see that right over here. And it takes 6 and it also maps it to 2. F of 6 is also, is also, equal to 2. Now this is fine for f being a function but it introduces a problem when we’re trying to take the
inverse relationship. The inverse relationship all of a sudden, if i’m trying to set up f inverse, should f inverse, what should
f inverse of 2 be equal to? Should it be equal to 6? So should it be equal to 6? Or should it be equal or
should it be equal to 2? And so this is the problem. When you have multiple
elements in the domain mapping to the same element in the range, that’s okay for being a function but then when you’re trying to find to map from that element to the range to those multiple elements in the domain, that is no longer a function. This is not invertible. So essentially when we
look at the intervals that we could restrict through to get an invertible function
it essentially would be intervals where the thing would pass that horizontal line test. Where over that interval, every member of that domain, that restricted domain, maps to a only, there’s
a one to one mapping between each member of that domain and a member of the new restricted range, or potentially restricted range. So let’s think about these, these, let’s think about these
different restricted domains. So if we think about from negative 2 to 2, so from negative 2 to 2
including negative 2 and 2. So this is a closed interval. We have brackets here. Actually all of these
are closed intervals. Notice it does not pass
the horizontal line test. F of negative 2 and f of 2 both map to 2 and so we don’t know what
f inverse of 2 should be. And you could put a horizontal
line right over there, makes it very clear that this is not passing the horizontal line test. So let me clear this up a little. So let’s rule out the first interval. Now the second one, between 0 and 4. 0 and 4. Once again, not passing
the horizontal line test. Rule that out. Negative 8 to 8. Well that doesn’t even restrict the domain relative to what we had before so that’s not going to work. 6 to 8, now this is interesting. Between 6 and 8, you do
have a 1 to 1 mapping between elements of the domain
and elements of the range. Notice between 6 and 8
it does, it does pass, it does pass the horizontal line test. So this is a legitimate
restriction of the domain. So if we restricted the
domain of the function to just that, it would be invertible. It would be invertible. In fact, we could draw it’s inverse. It’s inverse would look like this. So, f of 6 is 2, so f
inverse of 2 should be 6 and f of 8 is 0 so f
inverse of 0 should be 8. So it’s inverse will look like this, it’s inverse would look like this which is clearly a reflection
over the line, y equals x. Clearly a reflection over that line you see it over here, it’s a reflection. And this, this inverse
function is clearly a function. It passes the vertical line
test for the same reason this passes the horizontal line test. So this is really good. So let’s see negative 8 to negative 6. Well, yeah that’s the same
thing on the other side. So negative 8 to negative 6, once again over this interval we do pass
the horizontal line test. So that looks good as well. And then 0 to 2. From 0 to 2, yeah it does
look like over that interval we pass the horizontal line test. So that looks good as well.

Leave a Reply

Your email address will not be published. Required fields are marked *