Common Core Algebra II.Unit 3.Lesson 6.Piecewise Linear Functions
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Learning the Common Core Algebra II.Unit 3. Lesson 6.Piecewise Linear Functions by eMathInstruction
Hello and welcome to another common core algebra two lesson. By E math instruction, my name is Kirk Weiler, and today we're going to be doing unit three lesson number 6 on piecewise linear functions. So piecewise functions, well, you saw those in common core algebra one quite a bit. We're going to see them a little bit on and off this year and you see them a lot in pre calculus. And the idea of piecewise functions, whether they're linear or otherwise is very easy.
We simply use different formulas over different parts of the function's domain to form, let's say, a more complicated rule that converts inputs to outputs. Yet, a function is a function, right? So for any value of X in its domain, there can be only one value of Y, keep that in mind. All right, let's jump into some piecewise linear problems. So exercise one, consider the piecewise linear function given by this formula. And I'm not going to read the formula off. But let's start talking about how to think about it. It says create a table of values below and graph the function. All right. Well, let's do this by hand. So when we have an input of negative three, what the piecewise function has to tell us is which one of these formulas to use? Well, it says that we should use the formula X minus three for every X value that's greater than or equal to negative three. And every X value less than zero. So if I put negative three into that function, well, negative three minus three is negative 6. If I put negative two in there, negative two minus three is negative 5. If I put negative one in there, negative one minus three is negative four. So I'm thinking I'm seeing a pattern here.
Now, if I put zero in there and I know that the formula doesn't apply at zero, notice zero is not there. I'm going to put a negative three here. I'm going to put a little dashed line for a second. Now, I realize that's not actually its value. But I'm going to have it there anyway. This is now what I'm going to transition over to the other formula. So when I put zero into this formula, one half times zero, plus four. Well, that's going to be four. When I put one in there, there's a little bit more of knowing, but I get four and a half. Okay? Fair enough. When I put two in there, I'll get 5. If I put three in there, I'll get 5.5. I think I see the pattern. It's just going up by one half every time. And I'll get 6. So let's plot this thing. Negative three, one, two, three. We have one, two, three, four, 5, 6. Negative two, one, two, three, four, 5. Negative one, one, two, three, four. Now, at zero, I'm going to have a three, but I'm going to draw it as an open circle, because that point is not really on there. So this is that part of the formula. Now really at zero, what I really have is a positive four. And then at one, I have a four and a half. Two, I have 5, three, I have 5 and a half and four, I have 6. So I have that. All right.
Now, let it be says state the range of F using interval notation, The Rain now the range is all the Y values. It would be tempting to just say it would be tempting to do something like this negative 6 to 6, right? Because that's the smallest Y value largest Y value. The problem is there's this whole stretch of Y values in here that we don't hit. So they're not part of the rage. We can't do that. In fact, the range starts at negative 6. It goes all the way up to negative three, but it never hits it. That's why I needed to calculate that thing. Union is what's called the union symbol. It joins two sets together into their union. Well, then we pick it up at four, and we go all the way up to 6. Notice square brackets on both of those things. Because we do hit four. So that's a much trickier range than what we're used to. We're used to just saying, here's the smallest Y value. Here's the largest Y value. And you go everywhere in between them. But this is what's known as a discontinuous function. Remember when we talked about continuous functions a little while ago, well, this is a discontinuous function.
There's a break in it, a gap. And that often happens at what I would call the cutoff point on a piecewise function, where we transition from one formula to the next one. All right, pause the video now if you need to. I'm going to clear this out. Okay. Let's keep working. Now, in the last problem, we took a piecewise linear function whose formulas given. And we changed it into a graph. Here we're going to go in the exact opposite direction. We're going to take a graph. And we're going to change it into the formula of a piecewise function. This is pretty challenging. Okay. Now let's just take a look at the graph. Let's get a lay of the land. Okay, there's really three distinct portions, three distinct portions. So ultimately, whatever I write a piecewise formula, F of X equals. We use these funny brackets. Now, we're going to have three formulas. One, two, three, and three intervals. Okay. That's what we're going to have. So let's tackle this first one, all right? That's a linear function. So section one, if you will. Y equals MX plus B so I really have to figure out M and I have to figure out B this one's going to be kind of a little weird. But let's play around with it.
Can we, can we figure out what M is just from the graph? Let's grab this point, and let's grab this point. What we do, we have to go over to, and we have to go up three. So that means M is three halves. Now the weird thing is it doesn't strike the Y axis, but if the line was extended forever, it would. And it would hit the Y axis at one, two, three, four, 5, 6, 7. So that's B so that first formula. Is three halves X plus 7. Now, what's its interval? Well, one, two, three, four, 5, 6, 7, 8. Well, it goes from negative 8. It's less than or equal to X is less than or equal to negative two. You might wonder, well, that didn't work out very well. It's negative 8 less than or equal to X less than or equal to negative two. Now it would be fair to kind of ask right now. Why did I put the equality there? Why did I leave it off? Well, notice this is, this is a filled in circle, right? It's not an open circle there. It's filled in that dot. So certainly in that formula does apply when X is negative two, you can even check it. Anyhow, I'm going to get rid of now I turn my eraser to being small. I'm going to get rid of all that writing. And I'm going to go back to my pen. Let's go with the second section. This one. Now that's actually pretty easy. And the reason that's easy is that is a constant, okay? That is an absolute constant. So it's a little different. It's just Y equals a number. And the question is, what's that number? Number is just four. So as weird as it may look as weird as this may look, the second part of this formula is just F of X equals four.
Now what's my interval? Well, interestingly enough, I could put the less than or equal to signs there. Because that formula does apply there. It's almost seems like this wouldn't be a function. But I'm going to leave it off just so there's no confusion. I'm going to say X is less than or equal to let me call it three. That's negative two. And that's three. And again, you can have the equals sign there and not there. Same deal there. All right, let's go for that last little section. So I'm going to get rid of this. Get rid of these two. Hopefully. Now I'm looking for that. One, two, three. Okay, well, let's try to figure out, again, it's a slanted line. So it's Y equals MX plus B well, one thing that's important to note right away is that M is negative, right? Because this thing is decreasing. Now it goes down to for every one that it moves to the right, so the slope is negative two. Now, what's its Y intercept? Well, let's see if I do this. Oh, I run out of graph paper. Now, a pretty sophisticated person, even if they run out of graph paper, can still figure out what it's going to be. But let's say for some reason it was harder than that.
Notice how the .3 comma four lies on this graph. At this point, three comma four. We could actually use that with the point slope form. In fact, we could even put the point slope form in for that. But I'm going to rearrange this into. Y equals MX plus B form, given that that's what all my other lines are in. So there is the correct formula. Negative two X plus ten. For three less than X, less than or equal to one, two, three, four, 5, 6. And there's our final answer. This is a big problem. Big problem. Getting through all of that. But very doable. We just have to find the equation of three lines, one of which is very easy Y equals four. Okay. Well, pause the video now, think about this. This is kind of a challenging one. All right. Now, piece one's functions can be very challenging, especially algebraically. All right. Because they have these different formulas for different parts of their intervals. So let's play around with another piecewise function. All right, here we have G of X equals that function. Letter a says determine the Y intercept of this function algebraically. Algebraically, then it goes on to say, why can a function have only one Y intercept? So anyway, think about this for a little bit, pause the video and consider both questions.
Well, to find a function's Y intercept algebraically, we always do the same thing. We plug X equals zero here. The question is which one of these two functions does X equal zero Lion? Well, it's got to be the one for this interval. X less than two, because zero is less than two. So 5 minus zero gives me 5. That's my Y intercept. Okay? And it can have only one because there can be only one output. For X equals zero. Now let her be find the X intercepts of each individual linear equation. X intercepts, not Y intercepts X intercepts. Well, let's do it. Like for the first equation to find X intercepts, remember we always set the function equal to zero. This is pretty easy. I'm not going to do much to solve this. I'm going to get X equals 5. The other one, a little bit more work, one half X plus two is equal to zero. That would give me one half X is equal to negative two. And X must be negative four. So those are the X intercepts of those individual formulas. Let her see, graph the piecewise linear function below. All right. Well, one thing I'm going to do, I'm going to put a little, I know this is going to sound strange. A little vertical line at X equals two, because that's where we switch over.
Now I'm not going to create a table of values personally. I'm just going to graph these two lines. Now, it might be a little bit easier for me to think about this line as negative X plus 5, right? Because one, two, three, four, 5. And then I go with a slope of negative one down one to the right one. Down one to the right one. Now, I stopped it right there because that's the cutoff point. And notice the equality isn't on that one. So I'm going to put I'm going to put an open circle there. And then I think I'm going to use this because it just makes my life a little bit easier or a little bit neater at least. All right. Likewise, one half X plus two. Let's do that one. Now, granted, it doesn't apply right here, but I'm going to still put maybe a dot right there. One half means I go to the right two and up one. To the right to enough one to the right two and up one. And that one's going to take off from there. Now notice this actually turned out to be a continuous function because we filled in that hole. In other words, both of these have the same value when X is equal to two. Well, look at D why does your graph contradict the answers you found in part B? Well, because. It has no X. Intercepts. I mean, look at it. It's got no X intercepts, and yet here they are.
Solved algebraically. So letter E then asks us, how can you resolve the contradiction from D? I mean, here's B, we found the X intercepts algebraically 5 and negative four. And yet when we graph the function, it has no X intercepts. Well, it's kind of cool. Taking a look at this function, right? We found that 5 minus X had an X intercept at X equals 5. But that formula only applies when X is less than two and 5 is not less than two. Likewise, one half X plus two will be equal to zero. Well, that was at X equals negative four. But that formula only applied when X was greater than or equal to two. So negative four doesn't count. So actually, we're all set. And that makes piecewise functions very tricky, because information that you get out of them, especially about X values, has to be checked compared to the domain intervals for which those formulas apply. That was kind of fancy. Anyway. Pause the video now. Think about what we did right down anything you have to before we go on. Okay, here we go. Exercise four. For the piecewise linear function, F of X equals blah.
Find all solutions to the equation F of X equals one algebraically. So in other words, I want to know all the X values where the output the Y value is one. So see if you can do that. Okay, let's do it. I have to set both of these formulas equal to one and see what happens. Negative two X plus ten, let's set that equal to one. That's going to get me negative two X is equal to negative 9. And then when I divide by two on both sides, I'm going to get X equals 4.5. Now this formula only applies though when X is less than or equal to zero. 4.5 is not less than or equal to zero. So that one's not correct. Let's try 5 X minus one. Equals one. Well, if I add one to both sides, I get 5 X is equal to two. Divide both sides by 5, and I get X is equal to two fifths. Now this is correct as long as X is greater than zero. Well, two fifths is greater than zero. So that's my winner. All right. Again, tricky. Because it seems like we're going to get two very valid answers. And yet only one of them checks only one of them works. Okay. Pause the video now if you need to. And then we're clear on this app. And I think finishing up. Okay, here we go. So today, we reviewed the idea of a piecewise function specifically piecewise functions where each piece is a linear function.
All right? These functions simply have different formulas for different parts of their domains. And that can be tricky, especially when you're using those formulas to find something algebraically. You always have to go back and check it versus the domain interval from which it came. I know that probably sounds kind of complicated. And we'll see piecewise functions throughout the course. So hopefully we'll get more practice on it. For now, let me thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking. I keep solving problems.