Functions. Now this would have been covered to an extent in additional mathematics and mathematics, however we cover it to a slightly greater extent in pure. Okay, what exactly is a function?
I want you to think of it like a sort of machine, okay, and in the machine there's an input and an output and in order for the input to get to the output it has to go through this machine. So let's say this machine had an operation that when you put the input in, it operates on that input in order to get the output. So let's say our input was 5 and in the machine it ordered that your input must be multiplied by 5. So if we say our input was 5, in order to get our output we would have to say 5 multiplied by 5 to get 25 as our output.
Okay, so our machine says multiply by 5 if we were to put the input of 2. Okay, to get our output it would be 10. Exactly. So we have an input and an output and what relates the input and output is our machine or function. Okay, so that's kind of what you can think of function is a sort of machine and you have to put something in, basically substitute something in order to get a value called your output.
So it's a relationship. A function defines a relationship between two variables, an independent and a dependent. So it relates an element of a set to exactly one element of another set. And we'll go into this one element of another set in detail, you know, relations and...
right. So you should also know that a function can be represented in different ways. It can be seen in these ways. As a formula, as a graph, a list of values, a set of ordered pairs, and an arrow diagram or also known as a mapping. Okay?
And the notation. y is equal to f of x. All f of x is is a placeholder for a function. So you might say f of x is equal to 2x plus 1. Okay, so that's all that is and you have to say f of x.
So this is not f brackets x. You say it as f of x. And this would be g of x, h of x.
Okay, and this here means that for every x value, there's a unique value of y. Right. Now to understand our topic a little better, of course we go through our fundamental concepts and the first one being ordered pairs.
So an ordered pair is comprised of an x coordinate and a y coordinate written in a fixed order within brackets. So we use these to plot graphs for example. You know that if you had to plot 1 and 2 and you find it on a graph, you would see the x value first being 1 and then the y value second which is 2. and this would be our point.
So that's all an ordered pair is. So our example was, which 1 being the x coordinate and 7 being the y coordinate. Now what if we had a set of these ordered pairs? In set notation, that's what this is.
So and we had a set of them, 1, 7, 2, 7, 3, 7. Okay, so what if we had a set of them? Okay, that is what you call a relation. Okay.
And a relation is not necessarily a function, remember that, we'll be going on to that soon. So an example of a set of ordered pairs would be y is equal to this, and you can see there's a set of ordered pairs here, and y can be described as a relation because it is a set of ordered pairs. Now on a more serious note, I want you to pay attention to these definitions, because in order to truly understand this topic, you must understand these definitions.
So the domain is the set of input values. So we remember the input up here, right? So that's our domain.
Then we have our co-domain, which pay attention to because we need to know the difference between a co-domain and range. Don't mix that up. So the set of output values that could possibly come out of a function, okay?
So possibly come out. No, possibly. it doesn't actually come out because what actually comes out we call that the range all right so try to remember the difference between this because it's not crucial it's just important it's the same thing right anyways mapping relations so relations can also be represented in diagrams called mappings and if you'd also remember from before i mentioned it's called an arrow diagram as well so if we're given this relation, remember set of ordered pairs, relation We can say that, okay, since we have this, we can represent it as a mapping on our diagram, okay? So what we would do is we would say, okay, this is our domain, which is the x coordinates, and that would be elements of the domain.
So all of these here are called elements, 9, 15, 18, these are all elements, right? These are elements of the domain and elements of the co-domain, okay? So, x coordinates would be elements of the domain and y coordinates are elements of the co-domain that the respective x coordinates are marked onto. Okay? So this is kind of confusing, I know, because you need to understand the difference between a co-domain and a range.
However, all I want you to know is that if this side has an arrow connecting to this side, it's a positive range. However, if there's a number like let's say 7 and 8 and you can see they're not connected to anything in the domain, they are not a part of the range. That's why our range is this, is just 12, 18 and 24 because they come out, they actually come out of the function, which is this, okay?
So the domain, then there's the function, let's say the function is something, this is the function here, and it goes through the function, and it comes out of the function. However, since there's no arrows connecting to the codomain from the domain, to this number, these numbers are not actually a part of the range. Recalling what I said about relations, we remember that it is a set of ordered pairs, okay?
And there are types of relations. I want you to remember also that I said not all relations are functions. However, functions are relations.
And these functions are either one-to-one relations or many-to-one relations. These two are not functions, okay? Because if we remember the definition, it relates an element of a set to exactly one element of another set. And that's why these are not functions.
So a function can only be known as one-to-one relation or many-to-one relation. Now what if there was a way in order for us to determine whether or not a relation was a function? And I'll tell you, there is.
And this is called the vertical line test. So we use this when we want to know if a relation is a function or not. So if we are given the graph of a relation, we can use the vertical line test to determine whether or not the relation is a function.
I think the hardest part in this is knowing what vertical is. So vertical is like this, okay? It's like this, not like this.
This is horizontal. This is vertical. So remember that, okay? So the steps would be, the first thing is to draw a vertical line parallel to the y-axis. So obviously, it'll be parallel to y-axis, right?
Because y-axis is like this and it's parallel, you know. I don't know why I put that there but yeah you have to draw a vertical line that intersects the graph and the line remember the line can be drawn anywhere on the graph okay so it has to intersect the graph or cut the graph and then you can determine whether or not the line cuts the graph more than once so in this example here you can see I drew a line down here it can be drawn here here here anyway I just drew it here right You can draw it anywhere you like as long as it cuts the graph at one point for us to then do our test. And the test is to see if it cuts the graph more than once.
So the relation is in fact a function if the vertical line intersects the graph only once. And we can see at this point it does actually only intersect the graph only once and hence it is a function. 1-to-1 functions, or in fancy language, injector functions, but in English, 1-to-1 functions, okay? So a function can only be considered 1-to-1 if and only if every element of the domain is mapped onto only one element of the codomain. So if you'd remember in the mapping or the error diagram I put above in the types of relations, I showed you a 1-to-1, and you can see that only one element of the domain was actually marked onto one element of the codomain.
Okay? And that's what we call a one-to-one function or an injective function. So there are two ways in which we can show whether or not a function is in fact injective. The first way is the algebraic method, and in this method we would simply have to substitute a and b into the function, equate them to each other, and see if somehow we can get to a is equal to b. And we'll do an example, so no worries about that right now.
And the second way, I would recommend this way highly. because it's much easier, is a horizontal line test. Okay? Why on earth does Rahima have a sun in the middle of a video? I will tell you.
Because this sun represents a horizon and this is a horizontal line. Now you can see the relation, right? This is how you're going to remember the difference between a horizontal line and a vertical line.
Okay? It's crucial. And I know, let's keep it a secret, but I know people like you and I have trouble remembering these things. So this is how we remember, okay?
This horizon, you can see that the horizon is actually this way, okay? And then our sun comes out of it, yeah? So the horizon is our horizontal line, okay? And that's how we remember it. So our horizontal line test.
is if a line drawn parallel to the x-axis cuts the graph of the function only once, then the function is in fact injected. Okay, so we're going to do this test in the example below. And I know I said parallel to the x-axis, so you know that the x-axis is this way, the y-axis is this way. So of course, if it's parallel, it'll be a horizontal line or a horizontal line, right? So let's move on to our example.
Our example says to show that f is 1 to 1. f is equal to 3x minus 5. So algebraically, how are we going to say it and how are we going to approach this? We're going to first say, make a statement, and this is exactly what we're trying to get. So it's kind of like a proof, right?
And if it doesn't work out, then we know that it is not injective. So if f is a 1 to 1 function, then f is equal to f. will then get us to this point a is equal to b.
So the first things first, substitute a and b into the function above and we say we get 3a minus 5 and 3b minus 5 of course. It's simple right? Then we say that we're going to equate them to each other and we did.
Now we can cross off the 5s on both sides of course and then if we divide both sides by 3 we're going to get a is equal to b. And that's exactly how you would do the algebraic method. Of course it's not as simple as this, but of course, basis of everything, this is where it comes from and this is what you're going to have to apply. Now our horizontal line test. Okay, we first have to sketch the graph of f of x.
Okay, so to do that of course you input certain x values and then you get the y values based on that. So you can give random x values and just get the y values based on it. Okay, and you can draw a graph. I got this graph here and I then draw my horizontal line And it has to intersect the graph Okay, so it's just like the vertical line test. It must intersect the graph at one point any point Um, you can draw it.
I just drew it here, but you can draw it at any point Okay, so since the line cuts the graph only once the function is one to one However, if it were to cut more than once, it would not be an injective function. Now moving onto onto functions, also known as surjective functions in fancy language. So for a given function to be onto, every element of the core domain must be mapped onto at least one element of the domain. So this means that the range and core domain are the same.
And there are two ways that which we can show whether or not a function is surjective. Similarly to injective functions, there are two ways. The first way is the algebraic method and the second way is the graphical method. And this will also be the horizontal line test.
So parallel to the x-axis, right? Or horizon, yeah. Okay, so let's go to the algebraic method.
So for every y value in the codomain, there's a corresponding x value in the domain and we have to show that algebraically and to do that we would make x the subject of the formula and prove that for all y values there is an x value. Moving on to the graphical method, also our horizontal line test. If a line drawn parallel to the x-axis, or our horizontal line, cuts the graph at least once, at least once, so that means it can cut more than once, right? Then the function is subjective, okay? So let's look at this example.
Prove that the function is onto given that they give us a change, right? So the function is f is equal to 3x plus 4. And the first way is our algebraic method. So like I said, what we would do is make x the subject of the formula and from there prove that for all y values there's an x value. So let y be an element of the codomain. That's the first thing we do.
So we would say y is equal to 3x plus 4. From there, we make x the subject of the formula, and from this, we can see that for every y value in the codomain, there will exist some corresponding value of x in the domain, right? So because of this, we can see that, and therefore f of x is surjective. Now, the way I always recommend, if you have a graph, always just use a horizontal line test. It's much easier, unless they specify that you must use the algebraic method.
So the first thing to do is sketch the graph of course, because if you don't have a graph there will be nothing to draw a line on, right? So sketch your graph and make sure it's correct. Draw a line parallel to the x-axis or horizontal line that intersects the graph at any point.
It can be there, there, there, I just chose here at random, doesn't matter. And you can see that it cuts the graph at least once, right? So since the line is a graph, at least once the function is onto or subjective. Bijective functions. We know that bi means two.
So bijective, two functions, okay? A bijective function is a function that is both injective and subjective. So these two functions, right? So if we can prove that it is both injective and subjective, that one function, we know that... it is a bijective function.
So if the question says prove that this function is bijective, what we would have to do is two things. The first one, prove that it's injective, and the second one, prove that it is surjective. And to do that, we would use our methods above.
The first one, algebraic, and the second one, horizontal line test. The favorite, right? So these two, of course, you have to know the differences, right? So this one is cuts at least once, this one is cuts only once, and of course, they're different methods algebraically for onto and one-to-one, right?
So know the differences, know how to do that, and when they ask you about bijective, just do those two things and have your concluding statement. Since the function is both injective and surjective, it is a bijective function. Inverse functions. In order for a function to have an inverse, the function must be bijective and as we recall, bijective means that the function must be both injective and surjective. So the inverse of a function can be seen as the reverse of a function and we must be able to find the inverse in any given form.
The notation. The inverse of a function f is denoted by f with a minus 1 in the top right hand corner. So let's just say they give us f of x and they asked us to find the inverse.
The inverse would be f minus 1 x. Similarly, for g of x, simply put a minus 1 and that would be the inverse notation. Properties of the inverse. The reason why this is mentioned is for your understanding.
Because in order for us to truly understand what we are doing in math, we need to go down to the basics, right? So this is here because in some questions you might have to apply them or just be able to understand the whole concept of inverse functions. So the domain of f inverse is the range of f1.
So remember I said the inverse can be seen as the reverse of a function. You can see within these two steps that it's kind of like that, right? So the range of f inverse is the domain of f.
Remember domain and range is what actually is mapped onto it and the codomain is different. Remember learn the difference. between those two, right? So the graph of F inverse is the reflection of the graph of F in the line y is equal to x.
So let's say I had a Cartesian plane, y and x, and what I mean by the line y is equal to x is a line across. And this can be kind of controversial, not controversial, but confusing, because the line y is equal to x can vary. I am assuming that the scale of the y and x here are the same and that's why it's just a straight diagonal line through the middle, right? However, the line y is equal to x, okay, that keeps disappearing, right?
So we have the line y is equal to x here and this is like this because I'm assuming that this is 1, this is 1, this is 2, this is 2 in the same scale. right? However, the line y is equal to x can be like this, like this, at any different angle, depending on the scale.
So always be quite meticulous when it comes to that. Be very careful, right? So what I did, all I would have to do to find the inverse of f is reflect it.
So you see here, I would just make the reflection, this would be the mirror line. You know when you have to do like symmetry and stuff like that and they would say okay well mirror this, it's very similar to, it is that. This would be our line of symmetry right, that's our mirror line.
That's all you have to do okay. And the last but not least, probably least, no not least, the fourth and last one is actually a composite function and all it's saying is that this function, the inverse inside of the function is actually equal to the kind of opposite right. the function itself inside the inverse. So all we did was switch them around, switch this and this around, put them...
If we were to input them inside of each other, they would be the same thing, right? Now moving on to this part, we must be able to find the inverse in any given form. This part, so finding the inverse of a function, there are four ways in which we'll be covering today.
The first one is given a mapping diagram. So we were given the function and we can see we have a domain and a co-domain as well as the inverse. Now what do you notice here? I'll tell you.
The domain is now the co-domain of the inverse, right? And what did I say about this? It's quite similar to this, right?
So notice, well, this is the range because of course they connected to one element of the domain, right? So all I did was switch these elements around and to be able to get our inverse and it's as simple as that. Now the one you are very familiar with because this is the one you had to do before.
Given a formula, if we were to find the inverse of this f of x equal to x squared plus 3, let's recap. I know you must remember it but still it's always good to do some revision. There are three steps. The first one being let y is equal to f of x.
Let y equal to the function. So y is equal to x squared plus 3. the second step is to interchange x and y. And all I mean by that, wherever you see y, put an x on here, wherever you see x, put y.
So it'll be x is equal to y squared plus 3. Step 3 is make y the subject of the formula. And when you do that, and when you find y, you transpose for y, that's actually the inverse function. So when we got y, it was equal to the square root of 3 minus x. And that was actually...
just the inverse function, right? Quick note though, you can switch around step 2 and 3. It's whatever you feel more comfortable with, you will end up with the same result, right? Now, we slightly touched on this before, but once more, if we were given a graph, we would find the mirror line y is equal to x.
Remember, depending on the scale can be different to this one, right? So this is just assuming the scale is the same, right? However, if it was was different, the mirror line can be at different angles. And what you would have to do is mirror it.
Mirror it. So this one from here, just use it, just flip it over and you get this. For, let's say we had a line, right? And we had to mirror this line. What we would do is we can look for two coordinates, this one and this one perhaps.
And we say, okay, so this one could be minus 7 minus, sorry. minus 7 comma minus 3 or something, right? Of this point.
It doesn't matter. It can be anything. What we would have to do to find this point on this side is simply switch these around.
So it will be minus 3, minus 7. Okay, so that's what I'm saying about this. When drawing the inverse graph, switch the x and y coordinates and plot the graph, okay? And for a straight line, as I was saying, it would be so simple in comparison to this, because you can't just find two points and then, I mean you can, but I don't think it would be accurate, right?
So the best thing you can do is probably find three points and try to connect them and make it look like this. And for a line, you would just have to find two points. and connect them, a straight line.
That's the beauty about straight lines. So given a set of ordered pairs, this is the last and the simplest actually, all you would have to do is switch the x and y values around to get our inverse. Composite functions.
You can think of composite functions like a function inside of another function because technically that is exactly what it is. So you'll be substituting one function into another and it can get slightly confusing. However, let's break it down. So let's look at the notation first. We can see that we have f being a function f and we have to place g of x inside of it because that's how the notation works, right?
So to read this notation, we can see that there are two functions and this one is actually inside of this one, right? And it can also be written as this. So this is the same thing as that, just different ways to write it.
All right. So all this here means and this means is where g of x is first substituted. So because this is inside on the right of f, what we would do do is substitute g of x first into f of x and then the rest is evaluated normally right so let's go over that one more time this g of x this function will be substituted into this function and all that means is wherever you saw an x value in the function f you would put g of x all right so let's go over it in an example hopefully to make it a bit simpler. So, the example states, given f of x and g of x, find the composite function f of g of x, right? So, we have f of x, we have g of x, and it says f of g of x.
We know that that means this. So we have to place g of x inside f of x, right? So to do that, wherever we see, we look at these two functions and we focus in on f of x because we need to place this into this.
To do that... that wherever we see x in f of x we place the entire function g of x into it so we would end up with this and that was actually the most difficult part being able to identify what we have to substitute into what okay and that was it from there we just simplified to be able to get this there are many ways you can simplify that but we got to the conclusion now if we were to switch it around G of f of x right now instead of substituting it substituting g of x into f of x we're substituting f of x into g of x so the same thing but the opposite now we look this is our mean one now because wherever we see x we're going to substitute the entire f of x into here and that's what we did so wherever we saw x we put 6x squared minus 1 because that's the function f of x and from there we simplified to get our result. And that's exactly how we do composite functions. And note, yes, we can compose a function with itself. That means you can substitute f of x into f of x and g of x into g of x.
Okay, so when I say you can compose a function with itself, you can substitute a function into itself.