 I'm Zor. Welcome to Unizor Education. I will consider the second problem today in the graphs area. You learn the manipulations with graphs. If you add something to an argument, you multiply an argument for a function, add or multiply. Those were basically repetitions of the theoretical material. These will be real problems, which I will ask you to press the post button and try to solve yourself. My first problem is actually like three problems within the same area related to absolute value of the numbers. Now you know that absolute value is not strictly defined as this is the number without sign, which means basically for a positive, like 2, that's the 2 itself. And for the negative, you drop the sign. So for minus, let's say 25, the absolute value is 25. All right, this is a non-mathematical definition. What's the more rigid definition in this case is the following. Absolute value of x is equal to x for positive or 0x. That's what actually we were saying when the number without the sign, which means for the positive, that's the number itself. For the positive numbers, that's the number itself. For the negative numbers, let's say x is equal to minus 25. We know that absolute value in this case is equal to 25. What is 25 relative to minus 25? Well, that's minus x, right? So for negative x, absolute value of x is minus x. So that's the strict definition of absolute value. All right, so let's consider we know about this, keep it in mind. Now the problem related to graphs, we start with a very simple one. Well, let's draw the function graph of y is equal to absolute value of x. Here, there are many different ways of approaching this thing. One very simple way is the following. If you remember during the theoretical lecture, I was talking about odd and even functions where even where those functions which are not changing if you change the argument sign. Let's say if you change from 25 to minus 25, the value of the function will be exactly the same. Well, this is the case. This is exactly an even function. And as you remember, even functions have graphs which are symmetrical relative to the y axis, vertical axis. So how to draw this graph very simply? Well, let's do it this way. Let's draw it for positive x and then symmetrically reflect to the negative part, to the negative x part. So you know that the positive x module, absolute value of x is equal to x. So it's this function y is equal to x. But I have to restrict it only to the positive and the zero. Now, we have to symmetrically reflect it relative to the vertical y axis. So what will be the graph? Well, obviously this will be. So that's the graph of y equals to absolute value of x. So this angle type thing. Okay, from another perspective, we can do it based on the definition. Now, as I was saying, absolute value of x is equal to x for positive x or zero, which is this part. So from the function y is equal to x, let me return it back to the x. I have to cut only the piece where it's defined in this case for x greater or equal to zero. x greater or equal to zero. Now, for negative x, absolute value is minus x. So we have to draw a graph of function y equals to minus x. Now, why is it this straight line? Well, you remember that if you have a function y equals x and you would like to multiply it by minus one, y equals minus x, then multiplication of the function by minus one is actually reflecting relative to the x axis. So they're basically reflected this way. So the whole graph moves this way. What's interesting about the function y equals x, y equals minus x can be considered as the whole function multiplied by minus one, in which case it's reflection relative to the x axis. But at the same time, I'm multiplying argument by minus one. And in this case, if you remember, this is the reflection relative to the y axis. But what's again interesting is that this straight line, whether reflected relative to the x axis, you will get this. Or reflected to y axis, you still get exactly the same thing. It's the property of the y equals x, because that's argument and the function which are exactly the same. All right, so anyway, this is the function y equals to minus x. And I have to cut only those pieces which belong to negative x. So by combining these two graphs together, this piece on the positive and this piece on the negative, I come up with exactly the same graph as, sorry, as I did when I just used the property of the function y equals absolute value of x of being even. All right, so this is easy. Now, more difficult thing is y equals to x minus one plus x plus one. Here you can press the post button, think about this yourself. And I will start discussing this issue. So I hope you thought about this problem. And there are actually two ways, same thing as in this particular case, by the way, two ways of drawing this graph. One way is basically to have a graph of x minus one, absolute value, and x plus one, absolute value. And add them together. We know how to add graphs. It was part of the theoretical lecture which we had. All right, let's do it and we'll try to do it as simple as possible. Now, absolute value of x minus one, you know by now that this is shifted to the right graph of y equals to absolute value of x. Why right? Because we're subtracting from an argument positive constant y. So it will be this way. Now, same thing here, but the shift will be to the left by one. So it's minus one. Obviously, it's zero. Both x minus one, absolute value and x plus one, absolute value are equal to one, both of them. So they cross this one. Now, how to add these two graphs together? Simple. Let's start with zero. You obviously get two as a sum. One and one graph and one and another graph. So you get two. Then, as you move to the right, one graph goes down and another graph goes up. Obviously, they are symmetrical. So whatever amount this diminishing, this one is increasing. This goes down, this goes up. So when you add them up together, you still get exactly the same value of two. So the function graph will be constant for a while up to this point and this point. Now, at these points, behavior is changing. This graph is increasing and this graph is increasing. So sum will be increasing and, by the way, it's increasing twice as fast. Right? Same thing here. This one is increasing in height and this one is increasing in height. So some of them will be increasing and faster. So faster here and faster there. So our graph will be this, this and this. But this is a qualitative analysis of the graph because I was talking about manipulation of the graph when you are adding, subtracting, multiplying graphs, etc. These are all qualitative analysis. Let's do it a little bit more precise, with precise numbers, etc. And that we can do based on the definitions which we have. We're doing problems. Okay. What's the definition? Well, definition of absolute value of x minus 1 depends on whether x is greater than 1 or less than 1. In this particular case, the definition of absolute value of x plus 1 depends on whether x plus 1 positive or negative, which means x is greater than minus 1 or less. So it's valid to consider the critical points which we have, which are minus 1 and plus 1. So this is a critical point for 1 and this is a critical point for another absolute value. So we will divide the whole x-axis into three segments, 1, 2 and 3. And we will consider the value of this function separately in these segments. Okay. If you do press the pause button, it's not too late because this is a nice clue and now we'll continue with this. Okay, so let's consider our function on this segment. So if x is less than minus 1, what happens with this expression? Well, simple. By itself, x minus 1, when x is less than minus 1 is negative. If this expression is negative, then absolute value is its reverse. So it's y is equal to x minus 1. How about this expression? Again, if x is less than minus 1, this is negative. So the absolute value is its reverse. I have to put minus x plus 1, which is equal to what? Oops. I forgot to put minus here. Sorry. Reverse of x minus 1 and reverse of x plus 1. So it's minus and minus. So what will be here? Minus x and minus x. It's minus 2x and plus 1 minus 1 nullifies each other. Great. Now, next interval is from minus 1 to 1. Minus 1 to 1. Okay. And y is equal to? Well, since x is less than 1, then this thing is still negative. So I still have to reverse. Now, but this thing now, since x is greater than minus 1, x plus 1 is positive. So the absolute value is itself. So I have to just retain what it is. And what is this? Minus x plus x nullifies 1 and 1, 2. It's constant. So during a changing of the x from minus 1 to 1, on this segment, it's constant. By the way, if you remember, our graph was like this, right? So this is basically this horizontal piece when you are researching how it looks just adding two graphs together. Okay. Let's swipe it out. And the third one, if x is greater than 1, then both expressions are positive. So we have to retain the sign. So it will be x minus 1 plus x plus 1 equal to x and x. It's 2x minus 1 plus 1 nullifies. So it's 2x. Okay. So basically, that's what we have. This expression is equal to this, this, and this on these three different segments. Just let's check it out. This function on a boundary when it's minus 1 is equal to 2 and this one is equal to 2. So they are coming together from this graph and from this graph. They're coming together into 1 and the same point. x is equal to minus 1 and y is equal to 2. Now, in this critical point, we also have to check if our graph on the left and our graph on the right are coming into the same point. Graph on the left, if x is equal to 1, is equal to 2. And graph on the right, if x is equal to 1, is also equal to 2. So they are coming into the same point. That's very important as a checking. So basically, you know the game right now. It's very simple. You have three graphs. You have minus 2x 2 and plus 2x. But from the minus 2x, we have to cut only x less than minus 1. So this is obviously minus 1 and this is plus 1. So we cut this piece from 2. We cut only the piece from minus 1 to minus 2. So we cut this guy and this guy. And finally, from y is equal to 2x, we have to consider only x greater than 1, which means this, which obviously leaves us with the same graph.