 I am Zor. Welcome to Unizor education. Today, I will continue problems related to graphing functions which include absolute values. Basically, problem number two was introduction to this type of problems, and this is a typical problem in the whole series, actually. I presented here for you just to take a look, press the post button, think about what you can do yourself. Obviously, the methodology which you can use is exactly the same as I was using in the previous problem, just basically determining the critical points and considering pieces of graph on each segment between the critical points. It's relatively straightforward, and the only problem is to do the accurate calculations. So try to do it yourself first, and I will continue with my solution of this particular problem. Okay. First of all, critical points. Critical points are those points where each of these expressions within the absolute value signs equal to zero. In this particular case, that's x equals to minus two. This one is x is equal to one critical point, and this one obviously is two. So I have one, two, three critical points, which means I have to divide my x-axis into three segments. This segment, this segment, this segment, and this segment. Four segments, actually. Three critical points, and they divide the whole x-axis into four different segments. Now, within each segment, I can determine the sign of the expression within the absolute values. It's always determined since it's in between these critical points, and that's why I can transform it without using the absolute values, right? So segment number one, minus two, sorry, x less, less than minus two. Okay. If x is less than minus two, then this expression is negative. So I should write minus two x plus four. This will stand in a way. So let's put it here. Minus two, one, and plus two. These are three values. Okay. So if it's less than minus two, then my function is equal to... This is negative, so I reverse the sign in front of the parentheses. This is negative as well, so I reverse the sign to make it positive. Now, two minus x, if x is negative less than minus two, is positive, so I can use it as is. So positive number in absolute value remains. Negative number, reverse the sign. Negative number, reverse the sign. Okay. And what is it equal to? Minus two x minus three x, that's minus five x, and minus x, that's minus six x. Minus four plus three is minus one, plus two is plus one. Seems to be like this. Case number two, from minus two to one, this becomes positive. So y is equal to two x plus four, that's positive. This, since x is still less than one, is negative, so absolute value I have to reverse the sign. Three x minus three. Now, if x is less than one, this is positive as it is, so it remains. Equals to two x minus three x minus one x, and minus x is minus two x. Plus four plus three is seven, plus two nine. Now, here we have to do a mandatory check. This function is for values which are less than minus two, and this function is from minus two to one. And when x is equal to minus two, they should come up with the same number. Let's check. If x is equal to minus two, this is twelve plus one, thirteen. If x is equal to minus two, times minus two is four plus nine, thirteen. Good. Check. So far, we didn't make any written to carers, which is a good sign. Now, three. From one to two. Two is our next critical value, right? This will be positive, so we can retain the sign. This will be positive, so we can retain the sign. And this will still be positive, so we can retain the sign. So, the overall result will be two x plus four plus three x minus three plus two minus x. So, all plus is here. Two x, three x, five x, minus x is four x, plus four minus three is one, plus two is three. Mandatory check. Mandatory, I mean mandatory. It really needs to be checked. It's very easy to make a mistake here. I'm doing mistakes very, very often. Okay. So, the common critical point for these two expressions is one. So, what is y if x is equal to one? In this case, it's minus two plus nine is seven. And this one, if x is equal to one, four plus three, seven. So, they needed the same point. Good verification is satisfied. And the last segment which we have to consider is when x is greater or equal to two. So, when it's very positive, this is positive, this is positive, and this is negative. So, we should retain the signs of these guys and change this one. So, y is equal to two x plus four plus three x plus, sorry, minus three. And here we have to reverse the sign. It's minus two minus x. Okay. Let's calculate two x, three x is five x, and x is six x. Four minus three is one, minus two is minus one. Looks like this. Check. The common critical point is two. This one, if x is equal to two, is eight, eleven. And this one is two times six, twelve minus one, eleven. So, we are good. Great. So, we basically have concluded our calculations. So, the only thing which remains to be done is to draw the graph which contains four different segments. And we have to draw it from these four different functions. Let me just rewrite that thing. I think it would be easier. So, we have x is not less than minus the first. x is less than minus two, then from minus two to one, and then from one to two, and greater than or equal to two. And the values of the function will be minus six x plus one, minus two x plus nine, four x plus three, and six x minus one. Okay. So, let's just draw these four functions. We'll probably do it quite approximately, as you understand. So, minus two, one, and two. Okay. Minus six x plus one. Again, let's calculate it on the critical point, minus two, which will be what? Thirteen. Right? So, if x is equal to minus two, y is equal to thirteen. This is where our e is supposed to end. Now, when it goes to minus infinity, minus six x plus one, x is like this, six x is like this, minus six x is like this, and plus one is like this. So, it's something like this will be the line which we are interested in. Next. Minus two x plus nine from minus two to one, so let's calculate it at one. At minus two, by the way, we already know that's thirteen, and at one it will be minus two plus nine, it's seven, so it's somewhere here. So, the graph will go on this segment as a straight line between these two lines, between these two points. Okay. From one to two, it's four x plus three. Let's calculate it at the critical point two, which is eleven. So, at two it will be eleven, something like this. This is seven. So, it will go like this. And finally, six x minus one, and again at two, it's equal to eleven, so it's the same number. And then it grows quite steeply, steeper. This is four x, so six x will be steeper than that. So, our graph contains four segments joined together, which looks basically like this graph. Well, again, if you have ten different absolute values, ten different expressions each containing absolute value, you just have to have ten different critical points and consider one of them after another sequentially. And within each segment, you know the sign of each component, so you can get rid of the absolute value. So, that's the general approach when you should take. Well, that concludes this problem number three. Thank you very much.