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### Course: Precalculus (Eureka Math/EngageNY) > Unit 4

Lesson 5: Topic C: Inverse trigonometric functions- Intro to arcsine
- Intro to arctangent
- Intro to arccosine
- Evaluate inverse trig functions
- Restricting domains of functions to make them invertible
- Domain & range of inverse tangent function
- Using inverse trig functions with a calculator
- Solving sinusoidal equations of the form sin(x)=d
- Solving cos(θ)=1 and cos(θ)=-1
- Solve sinusoidal equations (basic)
- Solve sinusoidal equations
- Trig word problem: solving for temperature
- Sinusoidal models word problems

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# Restricting domains of functions to make them invertible

Sal is given the graph of a trigonometric function, and he discusses ways in which he can change the function to make it invertible. Created by Sal Khan.

## Want to join the conversation?

- Don't you use the vertical line test?(29 votes)
- No. The vertical line test just ensures that the relation is a function, not that it is invertible. To make sure it is one-to-one, you need to use the horizontal line test.(123 votes)

- I don't have a solid background about functions. Not yet at least. Do you think I can get through this lesson or should I do some Algebra and then come back?(21 votes)
- Start with algebra 1 then move on to algebra 2 and once you master both then come to precalculus. I recommend doing the high school geometry and if you can then trigonometry along the way for some variety. This is what I’ve been doing for the past 4 years along with physics, chemistry, and astronomy. If any of these are hard then there is a whole list of “Get ready for
*_____*” courses to help out. So if algebra 1 is hard then either go to pre algebra or get ready for algebra 1 and same goes for high school geometry if it is too hard then do basic geometry or get ready for geometry. Oh and I forgot to mention, if you prefer studying grade by grade then you have that option too by taking the grade courses which teach you everything you will learn in that year.(26 votes)

- how do you know that the set is open or closed line(11 votes)
- A set can be open, closed, or part open or part closed, depending on whether its endpoints are included.

Using PARENTHESES when writing the interval in question indicates "OPEN" (in other words, the endpoint is NOT included).

Using BRACKETS when writing the interval in question indicates "CLOSED" (in other words, the endpoint IS included).

Using parentheses AND brackets when writing the interval in question indicates that only one of the endpoints is included.

Hope this answers your question!(21 votes)

- Hello, So can I say, I Must find an interval with one direction/slop (negative or positive, but not both in the same interval) to make possible the reversal .

Forgive my approximate English, I'm French

thanks.(8 votes)- Exactly. In more precise mathematical language, we can say that we are finding an interval for which our function is monotonic. The function is invertible within that interval.(4 votes)

- What is the difference between [intervals] and (intervals) - intervals with brackets and parentheses.(2 votes)
- [Intervals] with brackets are
*closed*intervals, that is they*include*the endpoints.

For example [0,1] is the same as 0 ≤ x ≤ 1

(Intervals) with parentheses are*open*intervals, that is they do not include the endpoints.

For example (0,1) is the same as 0 < x < 1

We can also combine the two ideas (the brackets don't have to match).

So (0,1] is the same as 0 < x ≤ 1(10 votes)

- Can someone please explain to me the concept of the vertical line test?(1 vote)
- The vertical line test determines whether or not the graph of a relation (set of points in the coordinate plane) is a function.

If every vertical line in the coordinate plane passes through no more than one point on the graph, then the graph is a function (because this would mean that every input would have no more than one output).

On the other hand, if at least one vertical line in the coordinate plane passes through at least two points on the graph, then the graph is not a function (because such a vertical line would show an example of at least two outputs for the same input).

Have a blessed, wonderful day!(7 votes)

- How can I find the intervals without a graph?(4 votes)
- what's a vertical line test?(2 votes)
- If you have a graph, the vertical line test is a way to visually see if a graph is a function or not. If you can draw a vertical line anywhere in the graph and only pass thru one point on the graph, then you have a function. If a vertical line can pass thru more than one point, this means you have different X-values with the same Y-value. So, the graph can't be a function.

Sal introduces this concept in the video at this link: https://www.khanacademy.org/math/algebra/algebra-functions/recognizing-functions-ddp/v/graphical-relations-and-functions

Hope this helps.(3 votes)

- why not (pi/4, 5pi/4)?(2 votes)
- Your answer is also correct. But it's limited to the options given only and your option isn't among them. There are infinite solutions to this. You could think of more.

Keep up the great effort :)(2 votes)

- By using the restricted domain mentioned in the video, wouldn't some of the range be cut out? Because in class, I learned that the range of the inverse function helps determine which angle the ratio corresponds to. Or does the range values not matter when inverting sine, cosine, and tangent functions?(2 votes)

## Video transcript

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.