Transcription
What's up, you guys? Thank you for joining me. Now, in this video, we are going to be covering admits form for chapter 1.1 functions. Right, so without further ado, let us get started.
So now, let's look at the first question. State whether each of the following relation is a function. So, for a relation to be a function, it has to fulfill one very important condition. And what's the condition? The condition is right here: every object can only have one image. Okay, I'm just writing short form: every object can only have one image. So, this is a very, very important rule. If it fulfills this rule, therefore, it is a function.
So, based on this, it's an arrow diagram, correct? So, let me just show you what is that, what is it about, which is the object, which is the image. So, this thing here on the left, let's say I put one, two, three, and this will be arrow diagram arrows, maybe two, four, six. Let's see. So, this circle here, on the oval shape on the left, is called the domain, and the one on the right is called the co-domain. The numbers in here, the numbers in here, they are the object, and the numbers in here, they are the image. Okay, so the question is asking, every object can only have one image. So, based on this example that I'm showing you, one is connected to two, so it only has one image. Two is connected to four, so it has one image. Three is connected to six, so it has one image. So, therefore, this example I'm giving you here, it is a function. Right?
So, now let's look at these examples. Question one: two is connected to one, four is connected to two, six connected to three, seven is left alone. Well, but it doesn't really matter why, because we just want to see the objects, correct? We just have to fulfill the condition: every object can only have one image. In this case, all three of them have an image, and they only have one image. So, therefore, this is a function. Right?
Now, let's look at B. So, question B: negative two and two is connected to four, and then you have three connected to nine. So, in this case, is it a function? The answer is yes, it is. It is a function. Why? Because every object there, negative two and two and three, all of them only have one image. Okay, so it does fulfill the condition.
Now, let's look at C. Is this a function? So, you can see that R here has two images: one is connected to eight, one is connected to ten. So, it does not fulfill the condition. Therefore, this is not a function. All right.
Next, determine whether each of the following graphs is a function by using the vertical line test. So, the question is asking, using the vertical line test. What is the vertical line test? So, the thing is, you have to remember the condition first. The condition is, one object can only have one image, correct? So, this is the condition. So, what is the purpose of this vertical line test? Now, the vertical line test is only used to solve when it is a graph. Like, you see the questions here, they are all graphs, correct? So, you can only use this to solve graph questions, okay, to determine whether or not it is a function.
So, the vertical line test, like this: when you cross, draw a vertical line on any part of the graph, and it only crosses one point, therefore, it is a function. Why? Because, let me show you an example. Let's say I draw a graph like this, got a straight line graph. Okay, so now, if I draw a vertical line, let's say I draw it here at this x-value, I only have one y-value, correct? Only one y-value. So, it means that x is the object. When my x, when I have an x-value, I only have one image. The image is the y-value here, okay? So, that's what it's trying to say. When you draw a vertical line and it crosses one point, it means it fulfills the condition of one object having one image. Okay.
But, however, let's say I draw a graph like this. Now, does this have only one image? Now, let's say I draw a vertical line here. You notice that when x equals, let's say this is x1, it has two y-values. Here is y1, which is a positive value, and here is probably y2, which is a negative value. So, in other words, this object here has two images, correct? So, therefore, it does not fulfill the condition, right, for it to be a function. So, this is not a function, and this is a function. Okay.
So, now let's look at the examples here. A: If I draw a vertical line anywhere here, doesn't really matter, I only cross one point, correct? I only cross one point. It means that every object only has one image. So, therefore, question A here is a function. Okay. What about question B? See, I can draw a vertical line anywhere. However, if I draw it here, I will have one image, here will be a second image, here, and here will be my third image. So, all together, I have three images. So, therefore, this is not a function. Okay.
Let's look at C. So, same thing. I can draw my vertical line anywhere. You notice that I only cross one at one point, correct? I only cross here. If I draw the line here, I cross here. If I draw a line here, I cross here. You see, all of them only have one image. So, because of that, I can say that question C is a function. Okay. Right.
Let's go next. Question: By using the following notation, express h in terms of x for each of the following arrow diagram. So, based on all these arrow diagrams here, you are required to form a function notation. What's a function notation? Function notation is basically a formula, or I would say an equation, to relate the x-value to the image here, so which is the hx-value. All right, so you want to find the relationship.
So, let me show you an example. Question A: two becomes one over two, three becomes one over three, five becomes one over five. So, what's the relationship here? So, hx, for you to get hx, you have to take one over x, right? So, if you're not sure, you can actually substitute. Let's say I get my object here, the x-value is two, so I get one over two, which is the image here, right? What if I substitute three? I get one over three, which is the image here. So, three becomes one over three. And if I substitute five, I get one over five. So, in other words, I'm actually following, I'm following this requirement here for number two to become one over two, and for three to become one over three, and for five to one over five, you have to go through this notation, okay? So, that is what it means by function notation.
Now, let's look at B. Negative five and five becomes five, negative four and four becomes four. So, what's the relationship here? You notice that all the negative numbers are becoming positive, and the positive numbers remain to be positive. So, what can you say here? We use this thing called modulus. So, hx equals to modulus x. What does this modulus mean? Modulus means whatever number that lies here, it becomes a positive number. Okay, whether if it's a negative number or a positive number, it still becomes a positive number. So, for example, if I have, um, if I substitute x as negative five, right, the object is five, I should get negative five and then modulus side by side, correct? So, this one, the answer becomes five because you're changing it to a positive number, right? Or, if let's say I substitute negative four with the modulus, I should get four. This is the answer, okay? So, that's the purpose of modulus. You try to make a negative number become a positive number, and the positive number will remain a positive number, right? So, this is the function notation for this arrow diagram.
Now, let's look at C. Negative three becomes negative twenty-seven, negative two becomes negative eight, one becomes one, four becomes sixty-four. From here, it's quite obvious that all of these numbers are being cubed, correct? So, you can say hx equals to x cubed, right? So, you can substitute all of these objects in here in the x-value, and you should get these answers, right?
Next, determine domain, co-domain, range, object, and image. So, before we start this, I want to explain to you what is the difference with all of this, right? So, let's say I have an arrow diagram. I think I showed you guys just now, right? So, if let's say I have one, two, three here, and I have, it's all connected. Well, let's say I have another four here. Here is connected to two, is going to four, is connected to six, and then let's say I have another seven here, let's say, okay? So, this is set x, this is set y. So, what is the domain? Domain represents everything here, right? The set x represents the domain. The co-domain is set y. Everything in here would be the co-domain, right?
So, what about range? So, range includes everything in the co-domain. However, it does not include any images that does not have an object, right? So, in this case, the range is only going to be two, four, and six, right? So, that's the range. Okay. What about the object? Well, object is actually the same as domain, but the difference is it only takes into account the objects that have an image. So, in this case, this, the object, this object four is not an object because it does not have an image, okay? There's no relationship. So, that is not an object, right? Same thing applies to image. It only takes into account, it's actually the same as co-domain, but it only takes into account, uh, the numbers that have objects. So, in this case, it's gonna be two, four, and six, right? So, these are the image. So, that's the difference between, you can differentiate what is domain, co-domain, object, image, and range.
So, now let us go through these questions, right? So, A: What is the domain? I'm just going to write short form, right? I just want to write D here so that it won't take so much space. So, domain is going to be, make sure you use your curly bracket, okay? Because why? Domain is the set, correct? You are representing it, um, domain is basically the set. So, because it's a set, the elements in there are written in a curly bracket, okay? So, in this case, what are the domain? It's going to be negative two, negative one, zero, two, and four, okay? So, you can see all these points here, when you extend to the x-axis, you'll get all your domains, right? What about the co-domain? The co-domain is, you same thing, take the points, connect it to the y-axis, and you will find all your co-domain. So, in this case, it's going to be one, two, oh, sorry, there's no two there, one, three, four, and five, okay? What about the range? Well, the range is going to be the same as the co-domain, why? Because all of these images here, they have an object, okay? So, therefore, it's going to be the same, so one, three, four, and five.
So, sometimes the question can also ask you, find the object of four. Okay, so when they say, find the object of four, so if they ask you to find the object, you know that four is the image. So, you have to look for four as the image. So, image is on the y-axis. So, when four, when image is four, it means you're looking at this point. What's the object? The object is also four. Okay. If the question asks you, find the image of, let's see, zero. So, you have to find the z. If they're asking to find for image, that means the number they give you is the object. So, object lies on the x-axis. So, when x is equal to zero, where is your point? Your point is here. So, the image is going to be three, right? So, that's how it works.
Next, let's look at question B. So, from here, what are the domain? Well, the domain here is going to be same thing: J, K, L, and M, everything in the set x, right? What about co-domain? Co-domain is going to be everything in set y, which is two, three, six, seven, and ten, right? What about the range? The range is the same as co-domain, except it does not take into account the numbers that does not have an object. So, in this case, the only range that you have here is three and seven. Why? Because two, six, and ten does not have an object, okay? So, that's how you solve domain, co-domain, and range. And the same thing can apply, right? If they ask, find the image of, um, K. So, the image of K would be three. Or the question can ask you, find the object of seven. So, object of seven, so this is the image, so object of seven would be M, right? So, that's how the object image work.
Let's solve question C now. If you notice that question A and question B, they are all discrete numbers, correct? They are all one by one, individual numbers, right? You see here, individual data. However, question C is a range of data, it's a continuous function. So, because of that, your answer will not be discrete numbers like this, it's going to be a range. So, how do you solve this? So, domain, you have to write in sentence, right? Domain of x is, sorry, not domain of x, domain of f x. My bad. Domain of f x, because it's a, this is an f x function, correct? So, domain of f x is, what's the range? So, you must see the x-value is between negative three here, right? Negative three all the way to five. So, you have to write negative three and x and five. So, this is your domain. What about your co-domain? So, co-domain of f x is, so where is your co-domain? It's starting from negative, from two all the way to six, correct? So, co-domain of f x is two up to six. But remember one thing, when you write here, make sure you're writing f x, right? Why? Because it is the y-axis, correct? Y-axis, they say y equals to f x, so you write here f x. If you write here x, then your answer is wrong, okay? Next, they ask for range. So, range of f x is, well, in this case, it's going to be the same as the co-domain, so it's two, f x, six, all right. Okay.
Now, next. Now, sketch the graph of each of the following function in the domain negative two x four, and state the corresponding range for the given domain. Now, when they want you to sketch, okay, the easiest way to sketch, of course, the easiest way is to do it freehand. However, you want to first know what are the data points, right? They give you the equation f x is modulus x plus one. So, you must first solve by, uh, doing the table. You want to know all your x and f x value, okay? So, let us do that together. You have, so the x-value, what are the x-value? That's what they give you, the domain, correct? So, it's from negative two, negative one, zero, one, two, three, four, okay? You just want to know roughly what is the, the range, okay? So, in this case, you got it right. So, this is your table, just to make it easier for you to plot. So, they don't make any errors. So, what you're going to do is, you're going to substitute negative two into this equation. So, when you have, for example, f negative two, and your object is negative two, you have negative two plus one, modulus, okay? So, you have negative one, in the end, you get one, okay? And then you can do for f negative one. So, you get negative one plus one, you should get modulus zero, answer is zero. So, you can do this for all of these x, all of this x-value, all the domain, and you should get one, zero, one, two, three, four, and five, okay? So, I'll just speed it up for you. You can do it one by one, like how I did here, okay?
So, now, once you've already obtained all your data points, now you can plot your graph, okay? So, your graph is going to look like so. Let's plot the x-axis first. You have negative one here, negative two, and negative three, then you have one, two, three, four, let's put one more, five, okay? This is your x. Here you have your f x. So, here you should have, um, one, two, three, four, and five, okay? So, now let's plot your points. When x is negative two, your f x is zero, so it should be, that's why it should be here. When x is negative two, x is one. X is a negative one, you get zero, should be here, then this one is next, here, here, here, here, and here, okay? So, when you draw your line, okay, when you plot them together, it should look something like this, okay? Of course, you use your own ruler to draw this, right? It's going to be a straight line. So, your graph should look something like this, okay? And then you can write here, um, fx equals to modulus x plus one, okay? So, this is how you sketch your graph.
Now, the second question is asking, hence, state the corresponding range for the given domain. So, they want to find the range for this domain. So, the range of fx is what is the range? [Music] So, it's basically for following the, um, whichever image that have an object, right? So, in this case, it's going to be from zero, right, on the y-axis, zero all the way to all the way through here, so it's going to be five. So, the range is zero, fx, five, right? So, this is your answer. All right.
Next, function g is defined by gx equals to three plus six over x minus one. Find the images of negative five, negative two, one over two. Okay, so in this case, they're asking you to find the image. When they ask you to find the image, they should give you the object. So, they gave you three objects here, so you should have three images, right? So, let's solve this question. A: Find the image. Okay, so the object is the first object is negative five. So, g negative five is equals to what's the equation here? Three plus six over the x-value is negative five, so negative five minus one. So, your answer should be three plus six over negative six. So, the answer is three minus one equals to two. Okay. Let's see, g negative two. Now, your new, you have a new object. Three plus six over negative two minus one, should get three plus six over negative three. So, you get negative two. So, three minus two is to one. Okay. Let's look at g one over two. The object is one over two. You should get three plus six over one over two minus x, sorry, minus one. So, you should get three plus six over negative one over two. So, you should get three minus twelve. So, the answer is negative nine. Right?
Now, let's look at B. Now, they are giving you any relationship here. They're saying, given the image of b, so b is the object, all right. Image of b is twenty-eight. Find possible values of b. So, they give you the object, right? The object is b. So, that means gb is equals to, they say image of b is two b. So, in this case, the image is two b. So, you deform your relationship, right? So, they're telling you if object is b, image is two b. So, now you just have to substitute into your formula. So, your formula for gx is three plus six over x minus one. So, in this case, since the object is b, you get b minus one equals to two b. Now, the first thing I'm going to do is, I'm going to take the whole thing here, I'm going to multiply by b minus one. Why am I doing that? Because I want to eliminate this so that it becomes, it doesn't have a fraction, right? Because fraction is much harder to solve. So, it's going to become like this: three b minus one plus six equals to two b b minus one. So, I get three b minus three plus six equals to two b squared minus two b. Right? So, I'm going to shift everything to the right because I want to remain my b squared is positive, right? So, I shift it, it should become zero equals to two b squared. So, negative two b minus three b, get negative five b. Negative three plus six, you get positive three. You bring over, you get negative three. So, this is your quadratic equation. So, what you have to do if you want to solve this, you have to factorize, okay? So, you should get two brackets. So, it's going to be two b plus one and b minus three. So, two b plus one equals to zero, b equals to negative one over two. The other b minus three equals to zero, b equals to three. So, these are the two possible values of b, right?
Next, function h is defined by hx equals to kx minus three over x minus one. Find the value of k such that h two equals to five. So, in this case, they've already given you, same thing, they've given you the object, which is two, the image is five. So, they want you to find the value of k, value of k in this equation. So, all you have to do is substitute, right? They form your equation. So, h two, right? So, it's going to be k times two minus three over two minus one, right? Substitute two into all your x-value, and you should get five. So, two k minus three equals to two minus one is one, so times five, you bring, you get five. So, what's the value of k? Value of k is going to be five plus three divided by two. So, eight divided by two is four. So, your answer is four, right?
Next, function f is defined by fx equals to modulus four x minus three. Calculate. Okay, so a, really gave you the objects, correct? They want you to solve it. So, in this case, just substitute negative two into the x-value. So, you get four negative two minus three. So, you get minus eight minus three. And so is negative eleven. So, since you have modulus, it becomes a positive, right? What about f negative one over two? So, in this case, you should get four times negative one over two minus three, modulus. So, you can divide, so negative two minus three, answer is negative five. So, answer is five.
Now, let's look at question B. The value of x such that fx equals to one. So, they already gave you the image, the image is one. So, in this case, you're going to be, you're going to write here fx equals to one. And what's the fx value? Fx value is four x minus three, right? Modulus. Now, when you shift this to the opposite side, it's going to become four x minus three equals to plus minus one. Why am I doing this? Why is that plus minus? The reason is because, you see, when I put modulus negative two, I will get two, correct? When I put modulus two, I also, sorry, when I get modulus two, I also get two, right? So, regardless of negative or positive, I still get the same answer. So, in other words, when there is modulus, there are two possibilities that it can be a negative value, it can also be a positive value, all right? So, that is why in this question here, when we shift the nega, modulus, modulus, when we shift to the opposite side, it becomes plus minus one. So, in this case, you should have two answers because two, first one is plus one, the other one is four x minus three equals to negative one, right? You have two answers. So, the first one, you get x equals to one plus three over four. So, you should get one. The second one, you get x equals to negative one plus three over four. So, you should get plus two over four, one over two. So, these are your two answers, right?
Now, let's look at question C. Calculate the domain of fx less than one. So, fx less than one. So, what is fx? Fx modulus four x minus three, modulus less than one. You see, I want you to pay attention. See the difference between question B and question C, right? There's a, there's a difference here. Here, question B, they are using equal sign, correct? They're using equal sign. But question C is an inequality, right? It is less than, less than. So, when you do question B, right, when you shift the modulus to the other side, you get plus minus, correct? However, when you do question, four x, I mean, question C, when four x minus three, when you shift the modulus to the opposite side, when it is an inequality, you will have same thing, plus minus, but the sign will change, right? So, let me show you. It becomes like this: four x, right? So, it becomes four x minus three. So, when it is less than one, you're going to do this. You're going to write here less than one, and here you're going to write less than negative one, right? One is negative, one is positive. Why am I doing this? The reason is because, look here, four x minus three, I told you there are two possibilities, right? The first one is going to be positive one, the second one is going to be four x minus three negative one. But when you write negative, the sign must change. So, you have two ranges here, right? So, if you notice, when I draw a number line, let's see, here is less than one. So, if let's say here, one is here, less than one means this side. And then this is negative one. Let's say negative one is here, then this is more than. So, you notice that where does four x minus three lies? Where does it lie? It lies between negative one and one. That is why I'm telling you, when you see less than one here, when you shift the modulus, it becomes like this, okay? One and negative one is in between.
All right, so coming back to this question, how do you solve this? Straightforward, shift one by one. So, you get negative one plus three, you get four x here, one plus three, so you get two, four x, four. Then you shift the four, so you get two, there by four, less than x, less than four over four. So, you should get one over two, x, one. So, this should be your answer.
Now, let's see the last question. D: Calculate the domain of fx more than five. Now, this question, you might think that it is similar to question C. Well, it is similar to question C. However, the difference is, one is less than, this one is less than, this is more than. Okay? So, what's the difference? The difference is when it is less than, like what I showed you here, you should get this. But if it is more than, and you shift the modulus, it becomes the two values that I showed you just now. You should get four x minus three more than five, and the other value is four x minus three less than negative five, right? Remember, when you change it into negative, the sign must also change, okay? The sign must also change. So, these are the two values. Now, the question is, why can, why do I not make it like this? The reason is because if you draw a number line, you'll notice that it doesn't fall in between, right? So, let's say I have a number line here. Here is more than five, correct? So, if five is here, I should get more than. And negative five is here, I should have less than. So, you notice that four x minus three lies on two sides, one is this side, four x minus three, and here is the other side, right? So, there are two possible values, it's not in between. So, that is why you can't combine the inequality.
So, coming back to this question, you have to solve one by one. So, you get four x more than five plus three, so you should get x equals is more than eight over four. So, x is more than two, right? Now, so you get four x less than negative five plus three, so x is less than negative two over four. So, x is less than negative one over two, right? Clear, right?
Let's see the next question. All right, given gx equals to six x, six minus two x, modulus here. Find the values of x if gx equals x. So, they gave you the relationship here, so gx equals to x. So, what is gx? Gx modulus six minus two x equals to x. So, same thing, when you shift the modulus, you should have two possible answers. The first answer is the same one, the second answer would be negative x, right? So, there are two possible values. So, here you should get six equals to three x. So, x is equals to six over three, you get two. Right? I already like this easier. The second one would be six equals to, bring it over, positive two x, so becomes x. So, x is equals to six. Right? Clear?
Next. All right, this is the final question. Function f is defined by fx equals to mx plus c. Given f two equals to seven and f four equals negative one. Find the value of m and of c. So, in this case, question A, they gave you two unknowns, right? M and c, both are in the same equation. So, whenever they ask you to find two separate, two different unknowns, they should give you at least two of this, right? Because why? You need to form your equation. You should have two equations to solve two unknowns. If you have three unknowns, you need three equations. So, in this case, let us solve the first one. So, f two equal to seven. So, mx, what? M times two plus c equals to seven. So, two m plus c equals to seven. So, this is your first equation. What about your second equation? So, f four equals to negative one. So, f of x is m times four plus c equals to negative one. So, four m plus c equals to negative one. So, this is your second equation.
Okay, so now what are you going to do? What I'm going to do is, I'm going to take two equation two minus equation one. Okay? So, I should get four x plus c equals to negative one, minus two m plus c equals to seven. Right? I'm using elimination method to solve this. So, four m minus two m, we get two m. C minus c, you get zero. Negative one minus seven, you get negative eight. So, what's your m value? M should be negative eight over two. So, in other words, it is negative four. Okay. What about your c value? So, I'm going to use one, one of the equation here. So, I know that two m plus c equals seven, correct? So, two times negative four, sorry, plus c equals to seven. So, what is c? C equals to seven plus eight. So, you get fifteen. So, now you should know that your equation is what? F x equals to m is negative four x plus fifteen. Fifteen is c. So, this is your equation, right?
Let's look at B. Find the image of two under f. So, they are telling you to find the image. So, what is this? Two is the object. So, in other words, it is f two equals to one. So, f two equals to negative four x value is two plus fifteen. So, you get negative eight plus fifteen, you get seven, correct? Yeah.
All right, lastly, let's look at C. Value of x that is unchanged under the mapping of f. So, what does that mean? It means that the value of x remains unchanged. That means if you draw an arrow diagram, the x-value, the object remains unchanged. That means the image is also x. So, when you form your equation, it's going to be fx equals to x. So, fx is what? Negative four x plus fifteen equals to x. So, fifteen equals to five x. Therefore, f is x is equals to three, right?
So, to summarize, these are the three things that we have learned in this video. The first thing is graphical representation and notation. So, you remember, we learned how to use, um, vertical line tests, right, to determine whether or not it's a function. We also learned, there are other ways to determine whether it's a function or not, as long as it fulfills the condition of one object has one image, correct? At the same time, we also learned how to form a notation, right, an equation based on the relationship between the x and the y, the object and the image, right?
Next, we also learned about domain, co-domain, range, object, and image. And lastly, we learned how to determine image, or we determine the object. How they give you, for example, f two, you have to find the image, or they give you f say, you don't know what's the x-value, and they tell you f x, or maybe this is, they say f x equals to five, so you have to find the x-value, right? So, these are the three main things you need to know for this subtopic. So, I hope you learned something from this video, and I will catch you on the next video. Take care.