Restricting domain of trig function to make invertible | Trigonometry | Khan Academy
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Restricting domain of trig function to make invertible | Trigonometry | Khan Academy


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.

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