 Hello, everyone. So I wanted to upload this video where we discussed some of the solutions from our most recent quiz, quiz number seven from last Friday. And most of these questions were about things like injectivity, functions being one to one. We had one problem about interest rates and one problem about solving algorithmic over exponential equations. So without further ado, let's go ahead and jump in, talk about this a little bit. So this is, again, just quiz seven from last week. All right, so let's go ahead and start with the first question here, which asks us to state the definition of what it means for a function f from the real line to the real line to be injective. Just remember that this notation f from the real line to the real line, this is just me telling you what the domain and the range of this function are, really the domain and the codomain. And we discussed a little bit on the Thursday before the quiz, sort of two things. We have this horizontal line test, which will function as a mnemonic device for us. But we found that there was a slight problem with this where maybe if we get lucky, we have a function with a formula, or we're given the function in a form that's easy to graph, where the function is just one variable, but we've seen that there are some situations where maybe we don't know the formula for the function, maybe the function depends on multiple input variables. And so this idea of checking a horizontal line becomes more complicated. And coupled with that, we also found that even if we did have the graph, it would be a slight issue to check every horizontal line. Maybe we can eyeball it if we can see a sunspot where it fails, but if we want to actually prove injectivity about this function, we couldn't possibly go and check every horizontal line. There are infinitely many. It would just take forever. Totally impossible. So what we have to do is fall back on a definition, something we can check algebraically. And so this is just in the notes from that review session. F is injective. Injective is the keyword here. If and only if, which just means that you can take this as a definition. The logical implication goes both ways. The f of x1 being equal to f of x2 logically implies that x1 must be equal to x2. And what does this mean? This is really just going back to the idea we discussed, where we want to fix an output value. That's what f of x1 and f of x2 are corresponding to here. So we're in the range of this function. We're fixing one unique output value. And we're asking ourselves, what input values does it come from? And it could be the case that it comes from just one input value, which would be great. That would be saying that the function is injective. That's exactly the statement that x1 is equal to x2. Or it could be the case that it comes from two or more input values, both mapping to this same one output value. And this would be exactly the case of failing injectivity of the horizontal line test going wrong of the function not being injective. So that is the definition we'll use. And in this next part, it is asking us now to look at a number of specific functions, fi. And it's asking us to essentially investigate whether these functions are injective or not. If they are injective, then we need to provide a proof of this fact. We talked a little bit about the difference between proofs and heuristics. The horizontal line test will be a heuristic force that will help us make a conjecture. And for the proof, we can just fall back on this definition that we've just written above. And for each fi that is not injective, the question is asking us to actually exhibit to produce a pair of points x1, x2 where injectivity fails. So let's just dive right into it. The first question is asking us about this function here, f1, given to us by the formula alpha x plus beta, where alpha and beta, as is usual, for these quizzes, are just unknown real numbers. And we can assume that they are positive, so strictly bigger than 0. OK. And in the back of my mind, I'm thinking, OK, which way should I go? Do I suspect this function is injective or do I suspect it's not injective? And I think it is, just from heuristics. I'm looking at this and I'm thinking it looks kind of like a linear function. It looks like something of the form y equals mx plus b. And so this right is just, if I think about the graph of this thing, I just draw my axes here. And so that's the x-axis. That's the y-axis. And I'm suspecting that f1 of x is some kind of line of this form. And at least from my heuristics, I'm thinking, OK, a line won't be a problem. It'll pass the horizontal line test, except maybe in one case. If it's a vertical line, that'll be fine. If it's a line with any slope, that'll be fine. If f1 itself is a horizontal line, that will be a problem, right? Because then the horizontal line test will just be, you can pass a horizontal line through itself, and it's going to intersect everywhere. So that'll be a problem. And that'll be the case where alpha equals 0. So we'll see how that shows up momentarily. But at the moment, at least, I am suspecting that this function is injective just from maybe investigating it. You could also plot this in Desmos or in your calculator. You could also try setting, say, alpha equals 2, b equals 7, just setting this to be some actual numbers and investigating the lines you get out of it. If you're working this in Desmos, you could set the alpha and beta to be parameters that you could move the slider with. And this might help you investigate the behavior of this function. So since I am assuming it is injective, or I'm supposing it's injective, this is my conjecture, I now want to sit down and prove it. So I can start with the assumption, if I just go back up to my definition here, I can start with the assumption always that the outputs are equal. And I need to logically conclude, after some number of steps, that the inputs, they must be equal. OK, that means that I can start with this equation right here. f1 of x1 is equal to f1 of x2. And to indicate where I'm going with this, I want to do some number of steps. And I want to logically conclude that x1 must be equal to x2. So that's what I'm trying to do with this problem. Maybe I'll just erase this for now because we need to fill in what these steps actually are. OK, so this first step I think what I'll do is, well, f1 has another name, or rather, it's given to us by this equation. So I can substitute in these arguments into these functions. And I will get alpha x1 plus beta is equal to alpha x2 plus beta. Again, just by substituting x1 and x2 directly into that equation, OK? From this expression here, well, I can subtract beta from both sides. That's no problem. Even if beta were 0, this wouldn't be an issue. So I can conclude that alpha x1 must be equal to alpha x2 by, again, just subtracting beta from both sides. And maybe just to indicate that, I'll put a red line there. But in the write-ups, we don't actually need that. OK, the next step I want to do is I want to divide through by alpha because I essentially want to cancel these guys. I want to conclude x1 equals x2. And to do that, I need to divide both sides by alpha. And so what I'll maybe do is write this. x1 equals x2 if alpha is not equal to 0. So there was some step there where there's a bit of subtlety. I tried to do a step that involved something where there's something being potentially undefined, dividing by 0 is one of the few things where we know things could be an issue. So we just need to watch out for this. But in this case, as long as we assume alpha is not equal to 0, then we have started at an f1 of x1 equals f1 of x2. And we have concluded x1 equals x2. OK, and now I need to actually answer what the original question was. The question was, is this function injective? And my answer is, yes. f1 is injective. And maybe I'll put provided alpha is not equal to 0. And that is my final answer, so I will box it. I will put a period on it because I have ended my mathematical sentence. And that is all for this problem. If you didn't catch this case of potentially dividing by alpha and you need to check that it was non-zero, that's totally OK. Sort of the most important part is noticing that you can run through this argument laid out here on the left. OK, so next up we have this function f2. And we would like to check to see if this is injective. Again, I'll step back and ask myself, do I have a conjecture about what this function might be? Do I think it's injective? Do I think it's not injective? Maybe if I didn't know, I would start with Desmos and I would let this variable phi be a parameter. And maybe let this parameter take on some numbers and I would investigate the behavior of this family of functions. And I think what I would probably include after doing that is, well, if I had something like phi equal to 0, so we're thinking about maybe like a parent function here. Let's see, so I have my y-axis, my x-axis. And this would be some kind of cubic function like g of x is equal to x cubed. Just move this over a little bit. OK, so this is kind of the parent function. And I guess if I have this other function, it's really just a horizontal translate. Let's see, so I'm subtracting phi from the argument. So this amounts to a horizontal translate to the right by phi units. So I've moved 0 to phi over here on the side. And other than that, the properties of the functions stay mostly the same. And this would be my f2 of x. OK, and if I kind of stare at this function for a minute, I think, well, it seems like it would pass the horizontal line test, or I might know from elsewhere in the class that we already did x cubed. So we know x cubed is an injective function. And we can see that, OK, if a function passed the horizontal line test to begin with, and then I just translated left to right, this isn't going to mess up any, you know, it's not going to add new intersection points or anything to any horizontal lines. It's just moving where they are. And so I would guess that this function is indeed injective. And so now I would set out to try to prove it. So again, let's think about what I'm allowed to assume. I'm allowed to assume that f2 of x1 is equal to f2 of x2. And I want to dot, dot, dot, dot, dot do some work and conclude that x1 must be equal to x2. OK, so I'll erase that because we need to fill in the steps. And it's really just going to go the same way. I have another name for f2 of x1, namely x1 minus pi cubed. And I have another name for f2 of x2. It's x2 minus pi cubed, again, by just plugging the arguments of these functions into the formula that we're given. OK, and what I want to do in this step is now take the cube root of both sides. OK, I guess this should be an x1 on the left-hand side and an x2 on the right-hand side. And I guess we should be a little bit careful here because it could be the case, like we saw, that taking a square and then a square root, there was something funny going on there. But we saw in class or in some of the worksheets that taking the cube, taking the cube root of it, just gives us that thing back. These are an inverse pair as long as we're restricting, I guess, to an appropriate domain. And so we can actually conclude from this without too much issue that x1 minus pi is equal to x2 minus pi. And now by subtracting or rather adding pi to both sides, we can conclude that x1 is equal to x2. And this is telling us that f2 is injective. So let's take a look now at c, which is this function here. f3 of x is x minus pi to the n. And the point of this question is that we want to sort of generalize. We've done the case of n equals 3 beforehand. And we want to now investigate what happens for n equals 4, 5, 6, and so on. So what I would do, again, if I didn't know much about this function, I would go and investigate it. I would put it in Desmos. I would now make phi and n a parameter. And I would try to vary n to see kind of what happens as you change this in. And if you do this sort of analysis, I can just tell you what will end up happening. Let me make two copies of, or rather, let's see what two functions we sort of get. So what happens is that there will be this value phi down here on the x-axis somewhere. And if you have a cubic function, well, OK, if you have an odd function, essentially what will happen, so if n is 3, 5, 7, and so on, you will get something that looks a bit like this. So it really just resembles the cubic function, but it'll be sort of skewed out a little bit as you vary the parameter n and make it bigger. And if you have an even, so sorry, let me write this. This is the n odd case. And then if you have an even value, it'll come down. And it will intersect the axis, something like that. So it'll still have a 0 at phi. This is the n even case. And if you did a little bit of this analysis, you might notice that, OK, the blue one, the n odd case, looks like it's passing the horizontal line test every time. So I would guess that that one is injective. The n even case, if I investigated this, I would see that it's always, as I go up to positive infinity, it's going to positive infinity. As I go up to negative infinity, it's also going out to positive infinity in y values. And so you always get a shape like what's in orange. And you might expect that if you run a horizontal line through that, you're definitely going to intersect at two points. So we'd expect that the n odd case is injective and the n even case is not injective. So for this problem, even if you didn't get it completely right, if you showed that you were just doing some analysis on these functions and kind of noticing the difference between the odd and even case, then you likely got most points. But I'll just walk through one way to just show this. So we can start with the assumption, let's assume n is odd. We can start with f3 of x1 is equal to f3 of x2. That will logically imply that x1 minus 5 to the n equals x2 minus 5 to the n. And while I need to be a little bit careful, x1 minus 5 to the n to the 1 over nth power be equal to x2 minus 5 to the nth, to the 1 over nth power. And what I'd like to conclude from this is that x1 minus 5 is equal to x2 minus 5. And this is kind of a question mark, right? We don't really know if this is something that works. But as long as you are in the case where n is odd, this will actually be something that's true. And so as long as you showed that you were doing some analysis of even an odd case and you went through this algebra, for example, that's totally fine for this class. Let me just shrink this down a little bit so I have slightly more room. Well, if I'm at this last step and I had x1 minus 5 equals x2 minus 5, then I can add 5 to both sides and conclude x1 equals x2. OK, in which case this f3 is injective. That's only half of the argument here. So I need to say something about what happens when n is even. And again, even if you didn't know how to show this, as long as you showed that you were putting some analysis into it, I tried to give people points for that. And I want to explain kind of what happens here. You go through the same first three steps, no big deal. But what happens here is that you get something that looks like, well, really something that's like absolute value of x1 minus 5 equals absolute value of x2 minus 5. Something roughly of this nature, sort of like what happens in the n equals 2 case of squaring something and then taking the square root. This gives you the absolute value back. You can run through some mathematics and essentially get a result like this. You can see it spelled out on the written solutions on the website, how exactly to show this. But if you get this, then all you really know is that x1 minus 5 is equal to plus or minus x2 minus 5. And from here, you can't really say too much. In fact, I think it's pretty hard to come up with a pair x1, x2 that specifically show that these are different. But as long as you did some analysis like this and you showed that it breaks up into an even and odd case and the odd case is fine and the even case looks like it has some issues, then you're totally fine for this question. OK, so let me zoom out a little bit. Shrink some things to make some room. OK, so next up we have this f4, which is x minus 5 to the n and this x minus psi. And I think this is a little bit tricky. But hopefully, we remember from maybe one or two quizzes ago there's this notion of roots. And so one easy way to know that a function is not injective is if it has multiple roots, multiple distinct roots, because the easiest line to take is the line y equals 0. So intersecting the graph with y equals 0 is just giving you places where this function is equal to 0, i.e. a root of this function. And if there are multiple distinct roots, then you've already failed the whole horizontal line test. So let me just kind of draw a picture of what's happening here. Again, you could also put this into Desmos. You could make phi n and psi parameters. And you could move the sliders and investigate the behavior. OK, we can't really say much about what this function does. Maybe if this n is odd, then this whole function will be like an x to the n plus 1 out in front. So it'll be an even degree function. So it'll be something like this going off to infinity. At some point, it will cross phi, maybe psi is out here. So it might do this. Doesn't have any other roots. So it will circle back up at some point and cross through there. Again, not the most accurate picture in the world. But it's just giving us an idea of the fact that there are two roots and they're two distinct roots. And so the horizontal line that exhibits the failure of injectivity is just taking really just the x-axis. And it's intersecting the graph at two places. There's one and there's the other. And OK, to show that a function is not injective, we need to find an x1 not equal to x2 where f4 of x1 is equal to f4 of x2. Or maybe I'll say but here. So these are two distinct input values mapping to the same output value. And OK, the output value I'll take is 0. And I can check that I'll take x1 equal to phi and x2 equal to psi. And let me just plug in what both of those are. So I have f4 of x1 is equal to, well, let me just do this, f4 of psi. OK, and then if I swap that into the equation, I get, well, psi minus psi to the nth power, psi minus phi minus psi. And the important thing to notice here is that this is 0. And so this whole thing is just equal to 0. So we verified one of these. Let me just rearrange some stuff to make some more room. Move this over here. We can play the same game now with f4 of x2. This is equal to f4 of psi. And OK, this is equal to just plugging it into the equation, psi minus phi to the nth power. And then in the second term, we have psi minus psi. And what's important to notice here is that that second term is just equal to 0. And so this whole thing is equal to 0. And so I can conclude that f4 is not injective when, well, as long as phi and psi are distinct numbers. If you didn't include that, that's OK. As long as you did some kind of analysis like this, where you're checking maybe the roots of the function, these are kind of the easiest places to find failures of injectivity. And if you notice this key property that this function has two roots, namely just plugging in x equals phi there and x equals psi there, you get a 0 output from potentially two different inputs. OK, so this last one here, I'll talk about. So this is f5 of x is equal to, well, at some combination of things from f1. So maybe I'll just put a reminder here of what f1. This was alpha x plus beta. So first order of business is to figure out what exactly f5 is. And I guess I should just do now some kind of function composition. So this is equal to phi. And now let's see. I need to take f1 up here to the right. f1 of psi x would be equal to alpha psi x plus beta. And this is saying I need to put that whole thing in for this slot. So that's alpha psi x plus beta. And then there's this plus delta on the end. OK, let's just kind of collect up terms because there are so many symbols floating around. Looks to me like there's a phi alpha psi multiplying the x. And then there's plus a beta delta in the constant term. And at this point, I mean, you could go through sort of the same argument. But you could just think of this as let's call it something like alpha not. Think of this as something like beta not. And so just by the way we've written it, this is alpha not x plus beta not. And all we've really done here is just multiply a bunch of numbers together and give them a different name. But you'll see that this is essentially the same situation as part A. And this is just coming from the fact that we're just multiplying together a bunch of real numbers. We get some arbitrary real number out at the end of the day. But this is what's kind of nice is that in f1, we showed that functions of this form, f1 and alpha x plus beta, f1 of x equals alpha x plus beta, all functions of this form provided alpha is not 0. This whole family of functions, they're all injective. So we've kind of already done the work here. And we just found out that f5 of x is a function of this form. It's just now, OK, we've given a different name alpha versus alpha not and beta versus beta not. But these are just numbers. As long as alpha not is non-zero, then we're in business. So the answer to this would just be f5 is injective. And I guess we maybe need to note as long as 5 times alpha times psi is not equal to 0. That's kind of the punchline to this problem. But of course, if you didn't notice this, that's totally fine. You can just go through and let's see, starting with this, that I'm boxing in orange, you could now assume this is injective, set two of these things equal to each other, and just play the same game of, all right, subtract this term, circling in red, subtract that off of both sides. And then you would be dividing by this from both sides. And you would get x1 equals x2. And that's a perfectly valid way to go about it. I would just mention that, yeah, it's worth noticing here that somehow we did some kind of transformation of f1. So we scaled it horizontally, and we scaled it vertically, and we shifted it vertically, and none of these things change the injectivity of this kind of function. OK, so I want to now, I think I'll just say something about 2. And then I will wrap it up there. I don't think there's too much to say about 3. But let's see what this is asking. This is question 2. So we have a bunch of Greek letters. They're all positive real numbers. And I am defining a function for you, g, which takes real numbers as an input, and it gives you real numbers as an output. And I'm specifying this function g by a formula, involving a lot of unknown numbers. But you can see that the argument, x, is appearing as the exponent of theta. OK, and it's asking us to solve something. And it's the thing that we're asking, being asked to solve. The thing that we're being asked to solve is g of x equals capital lambda. OK, and I won't say too much about the hint. I will say that the first thing that I would do here is start with the thing that we're given, g of x equals lambda. And we just need to parse, what equation do we actually need to solve? And what I would notice here is that I have another name for g of x, namely all of this stuff. So that will logically imply that if I just change my name of g of x to alpha beta x plus gamma, then I haven't really changed anything about the problem. I've really just swapped out what g of x means in terms of its formula. So this is now the equation we have to solve. And this is another place where math goggles are kind of a nice thing. Put this on there, and we'll highlight it in red. x is kind of the only important thing in this equation. There are a lot of symbols floating around, but the Greek letters are all just unknown numbers, just some parameters in the function. And so what we want to do here is somehow isolate the x. So that's the end goal, is like somehow, I want to get to x equals something. So that's how I know which direction to go when I'm doing this algebra. Well, I will subtract gamma from both sides. So I'm left with alpha beta to the x is equal to capital lambda minus gamma. I will now divide by alpha. So there's beta to the x is equal to capital lambda minus gamma, this whole quantity over alpha. And okay, we have some options here. Anytime I see something in an exponent that I want to extract, I know I'm going to have to take some kind of logarithm. Here, I would say maybe the nicest thing to do is take the log base beta of both sides. So this is important to note. Log base beta of beta to the x is equal to log base beta of capital lambda minus gamma over alpha, that whole thing. And maybe what I can do at this point is I can use this exponent rule for logs to pull the x out. I'm left with log base beta of beta again. That's equal to log base beta of capital lambda minus gamma over alpha. And here's where I notice something nice that the log base beta of beta is just equal to one. And this is something a very general phenomenon if you're taking the log base something and that something is also in the argument. The question is asking what power do you raise the base to to get the thing in the argument or the power you raise beta to to get beta is just one. So that's exactly what that is. And you can conclude from this that x is just equal to log base beta of everything on the right hand side. And that's all you have to do for this one. So really, this is more of a mess of notation than anything. The actual application of the log rule isn't too bad. I do want to point out that this step going from this to this, you could also take the natural logarithm. And so you'll still be able to use this property of pulling the exponent out. It's just that you won't be taking the log base beta of beta, you'll be taking the natural log of beta. So there will be one more term floating around. Maybe that's worth actually just writing out really quick. So if we did this, you get log beta of the x is equal to natural log of all of this. And what you would get here is x natural log of beta now equals all of this stuff again. Let me rearrange some stuff here to create some room. Almost worked. And then all that you would get here is that x is equal to, so you'd have all of this stuff up here again. And then you would have something like this in the denominator. So that's the only difference. These two expressions are totally equal. So either one would work. It's just in one case taking the log with beta in the base makes things a little bit simpler. Okay, and I won't say too much about what's going on in three. Three was really just, okay, the formulas for the functions for discrete compounded interest, so compiled or compounded it at some discrete number of events like maybe once per month would be like 12 times in one year. Once per year would be one time in one year, but at some discrete event. So this was really just saying, what is the general formula? Can you write it out and explain what the parameters mean? That's all coming from the notes. And then this A C of T was again just what happens when the interest is compounded continuously. And we saw that this was just another formula again in the notes. And what you had to do for three was really just sort of set up two equations and compare them. So really you have, maybe it's worth saying here, A D of T, well it's equal to P naught times one plus R over N to the N T. Okay, and in this case the P naught was, well, something that we don't know. So we can just keep P naught as an unknown. Okay, we do know the rate. The rate is 1.05%. So let's remember that percent means divide by 100. So we need to have 0.0105 divided by now N. So let's see, this is compounded 12 times per year. So that's a 12. And this is to the 12 T power. And if we take the second option, continuous interest one, well the general formula is P naught E to the R T. Again, we don't necessarily know what P naught is. We don't know how much we're going to take out in this 10 year alone, but we can plug in all the rest of the things. So 0.91% becomes 0.0091. And there's just a T in the exponent there. And I've kind of written it suggestively. So we don't know how much we want to put in, but we can compare by ratios. And if I just put this here and then put this in the denominator, then something magical happens. Okay, what happens here is that somehow the dependence on this P naught cancels out. I can just divide through by T. And now I have, well, okay. So I'm actually also plugging in a T equals 10. So I can plug in times 10 here and times 10 there. Okay, and the thing on the top is just the number. The thing on the bottom is just the number. I can plug these into my calculator. I can take the ratio, just divide one by another in a calculator, what you'll find is that this is approximately equal to 1.01 or so. But the punch line is that it's bigger than one. And okay, if a ratio is bigger than one, this means that the thing in the numerator was bigger than the thing in the denominator. And so the thing in the numerator was this AD of T. And this was, we have to go back to the original problem. Which loan is preferable and why? Well, AD of T is computing how much we have to pay back on the loan. So this is saying that we have to pay back more if we take out this loan, AD of T. And so in this case, AC of T is preferable because it's the smaller number. In this case, it's not like the interest in your savings account or something like that. It's how much you owe at the end of the loan. So as long as there was some analysis of these two equations, you plugged everything in right and you maybe compared ratios as described in the hint, then you probably got most of the points on this problem. And with that, I think I will go ahead and end the video. Please feel free to email me if you have any questions.