 All right, hi everybody In this video, we will go ahead and go through the solutions for the second exam So let me go ahead and bring that up All right, so I just have the second exam here. See there six questions So I'll try to add bookmarks on the video So you can just jump to the the question you want And just recall that we discussed a few of these in class, but I'll go ahead and work through them here Just so you have a recorded reference for them Okay, so in this first question, we have some unknown positive real numbers Capital lambda and capital omega. We know that they're real and we know that they're positive They're going to just be unknowns. We'll have to carry them throughout the problem And the name of the game for this problem is to solve a bunch of equations for this independent variable x Let me go ahead and zoom in on this first one So the first thing I will do with any of these problems is maybe just as my first step I'll write down a copy of the problem and just ensure that you know I'll double check to make sure I actually uh, wrote it down correctly Uh, especially if I'm doing my my scratch work on separate paper Okay, so I'm just going to write this down. We'll have log base 10 Of the quantity 3x Plus capital omega And this is supposed to be equal to capital lambda Okay, and before I make any other moves with this problem I will go ahead and double check that I actually wrote it down correctly and maybe triple check So this is a very common mistake. Um, you know just happened to write down the problem incorrectly And then it sort of throws off the rest of the problem So I think this is a good thing to build into your problem solving process of you know, just write down Record exactly what the problem said and then double triple quadruple check that you actually wrote it down correctly So my method is maybe I'll just check each side individually So on the the right hand side, I have a capital lambda And I do have that on the right hand side down here. So that that checks out Uh, maybe you want to check the base of the log Okay, these are both log base 10. So that matches up And then maybe let's just check this argument of the log the thing inside It's 3x plus capital omega And okay, that seems like it matches up. So now I feel pretty good about proceeding with the problem Um, so right, I want to in this kind of problem first, maybe uh, okay step zero, right? I guess was Write down the problem double check that it matches up step one is Maybe light up identify. What is the independent variable in this problem? And so I'll go ahead and take that x and just write it in purple Um, just to keep track of the fact that that's the thing we are trying to solve for and I can see It's trapped inside of a log. So maybe before you even write anything down for this problem Let me lay out what my strategy is going to be So I see something trapped inside of a log base 10 And I notice on the left hand side that that log base 10 is the only thing happening And I know I have this pair of inverse functions log base. Well anything say alpha And alpha so log base alpha of x is one of the functions And then alpha to the x is the other function And these are an inverse pair if I can pose them in either order these just give me x back Um, so maybe just to sketch out what I'm saying here is that I have this inverse pair We'll log base alpha of x And also alpha to the x So alpha can be any well, maybe a positive real number here, but it could be e it could be 10 It could be two Sort of pick your favorite one And I know that if I can pose these in either order. So what does that mean? I just plug one into the other So here I'm taking I'm composing them in one order. It's a long base alpha of alpha to the x That will just give me x back And similarly if I can pose them in the other order So I would need to take alpha and instead of x I need to plug in log base alpha of x That this too will just give me back x Okay And it really didn't matter that this thing was called x in this problem Or you know in this this inverse pair property this really could have just been anything. I'll call it question mark And what it's doing is freeing that question mark for me I have something up upstairs in the numerator. So alpha to the something no matter what it is I know that I can apply long base alpha to both sides And the effect is on one side the left hand side. It's going to move that question mark down Um, put it downstairs for me and the right hand side might be complicated because I still had to apply log to it But um, at least we freed something upstairs that we're trying to solve for similarly My this kind of property down here for the other function where it didn't really matter what the argument of that log was I I called it x, but it could be some long complicated expression In composing these in either order just gives me that whatever that argument of the log was it just gives that back Okay, so that's just an aside of you know, how are we going to approach these kinds of problems? What's the approach to take? and so the approach to take here is going to be well, I want to free that x so I definitely want to Undo this logarithm this log base 10 and so I'm going to raise both sides to the tenth power So I'm applying 10 to the x to both sides Okay, so that'll be my next logical step Or perhaps I'll just take exactly what I had And just do an operation to both sides And so I'm just going to raise both sides to the tenth power I'm just being a little bit careful here to make sure that I raise Both sides Well the entirety of both sides to the tenth power Hi, Camille is is visiting I'll have to bookmark this in the video She's trying to help Okay Back to the task at hand I'm very as both sides to the tenth power Um, and I've just been really careful to raise the entirety of both sides to the tenth power And what I'm seeing here is that well on the right hand side, I'm left with something complicated So let me go ahead and start into my next step here On the right hand side. I have this 10 to the capital lambda and that just is what it is It's a some some number lambda was some number that we don't know 10 is some numbers of 10 to the lambda is just some other number It's equal to something And the whole point of doing this raising to the 10 operation was so I could free the thing that I Free that argument that I see on the left hand side essentially everything Inside of the parentheses to log here the whole point of raising to the 10 was to bring that bring that out So this is going to come out as a 3x plus Omega and then maybe just to keep track. I'll light up that The x this independent variable that I'm really trying to solve for But okay, my my claim is that we've we've done the the hard part We've done everything from unit two now. We're just back to unit one Business and this is just some algebraic equation where I see the x and maybe I just need to rearrange a bunch of things to get The x on one side and then all of the other stuff on the other side So the first thing I'll do is just go ahead and subtract this capital omega from both sides So I'll get a 3x equal to 10 to the lambda except now that I have to Subtract this capital omega on the right hand side And again this thing that we're solving for this independent variable is x All right, and then I think we're just just about home free here Maybe I will multiply through this equation on both sides by one third from the left say so one third times this Stuff and then one third times all of this stuff is essentially the operation I'm doing next Just leave that out so I don't clutter up the equation too much But I get something it's Well x equals some stuff All right, because that one third I essentially multiplied by by that one third Exactly in order to get rid of that three And then what am I what am I left with paying the consequences of my actions on the right hand side? Well, I just have a one third times everything on the right hand side to 10 to the capital lambda minus capital omega And now I think I'm done I see something purple on the left hand side My independent variable x and I see some stuff on the right hand side, but the stuff on the right hand side is not purple I don't see any copies of my independent variable over there. So that's how I know I've solved four x I have x in terms of things that are not x um In this case, okay, maybe it's a little bit scary because we have some other Symbols floating around we have this capital lambda and we have this capital omega But that's okay. These are just numbers. We don't know what the numbers are, but they're just numbers So now I know that's how I know I'm done with the problem That I have x on one side and stuff not involving x on the other side. So I will go ahead and box it Put a period on it Maybe I will highlight in there something to indicate to the reader that this is my Final answer for the problem Okay, let's go ahead and move on to the next problem So we're given this capital lambda Now this this x minus one is appearing upstairs That whole quantity x minus one And this thing is equal to capital omega So I'm just going to do the exact same thing as I did for uh part a I'm just going to rewrite the equation first. So I have capital lambda to the x minus one equal to capital omega Which is quite difficult to draw Um Okay, so again, what is my my step zero here before I do anything? I just want to check that I've written down this equation correctly So I don't go through all of this this trouble of solving the equation and then find out that oh gosh I wrote down a two instead of a one or something and then have to redo all of my work So it's good again just to build this into your problem solving process literally every problem you solve That's sort of good mathematical hygiene Save yourself some headache down the line So again, maybe I'll just check the right hand side of the equality I had this capital omega originally and oh good. I have a capital omega showing up in the equation that I wrote down I have a lambda showing up on the left hand side and indeed the lambda is showing up there And maybe something that could go wrong here is that Here I had the x minus one all up in the exponent And I just want to double check that all of this is up in the exponent And what I wrote down as well. So it's very common to incorrectly You know have this one You know up here downstairs or something like that, but that's an entirely different problem Okay, so Sorry Camille is trying to help Put the lectures again, so Okay, so I I've written down the problem correctly So that's my step zero my step one is that I'd like to maybe independent Light up the the independent variable figure out what I'm solving for there it is I see it stuck upstairs. It's up in the exponent And so I have a tool to help me move things Upstairs downstairs right I have logs and really any log Will do the job here So all of these logs have the property that I can move exponents out of the log So maybe I will just go ahead and apply and apply Some log to both sides, but I won't I won't choose which one yet Maybe I'll choose one a little bit later to make my life somewhat easier Maybe we'll see what happens So what I will do is maybe just apply log base alpha to both sides So this is my next step. I'm just applying log base alpha both sides Being very careful to apply it to the entirety of the left hand side And then also to the entirety of the right hand side And then also just being careful that I kept all of that The stuff that was in the exponent before is still up in the exponent Maybe I don't want to skip too many steps here Because I might lose track of where that goes So I'll just light up again my independent variable. It's still stuck upstairs But I the whole reason of doing this log business was because it has this very nice property So let me make a little bit of room here with this off to the side What I see happening is that Inside of the log I see this this entire argument of the log All raised to some exponent So this is like my favorite situation to be in with a logarithm where the entire argument is raised to an exponent Because then I know I can always do this kind of maneuver where I take that entire exponent And just go ahead and move it out in front Okay, and I'm just going to add a set of parentheses here to be careful Um, you know in case there's so this is like a complicated You know, there are multiple terms happening here and maybe later on I'll need to multiply it Maybe distribute these terms out or something So I'm just going to put a set of parentheses around them to indicate that they're they're grouped together And they're all multiplying This term that's left just this logarithm Okay, but now I think, um Well, we could do a lot of things here Um We've done the hard part. We've done the the unit two stuff We've applied some logarithm to move things downstairs to now we're back to unit one stuff Which is just just solving an algebraic equation Um, so maybe I mean one thing I could do here is take this logarithm Maybe divide it over and then maybe I would need to add a one to both sides and then I would get x by itself Um, but maybe I can save myself a little bit of a headache here By noting that this alpha was something I just kind of I didn't specify. This was some symbol some arbitrary number. I I picked in the middle of the problem Um, or didn't pick, you know, but I introduced it. It wasn't part of the original problem So maybe now's a good time to go ahead and pick an alpha that'll Uh, somehow make a you know, maybe save us a few steps in this equation And as usual these kinds of problems, I have two two choices for alpha and both of them are pretty good Um, well, there are two natural choices. You could choose anything you could choose, uh You know alpha equals e to take the take the natural log and that would be totally fine You would just go through exactly what I said before of dividing stuff over and adding one Um, but maybe just to save myself a little bit of trouble I could set alpha equal to well this capital Omega is just some number and alpha is supposed to be some number So I could just set alpha to be capital omega and then this whole side would be one And that'd be pretty nice I can also Set alpha to be capital lambda lambda is the same. You know, it's just some unknown numbers So I could set my unknown alpha to this unknown lambda And maybe I'll go ahead and do that Because that saves me a step of dividing things And maybe just to be really explicit here, um, I'll write out exactly what's happening So this x minus one nothing's changed there And I want to simplify this term and this is a really really important property Um, what is this what is this equal to? Well, I asked myself what question does this answer This everything inside of this box What answers the question what power do I raise the pink to to get this blue term the argument of the log What power do I raise the base of the log to to get the argument of the log? Okay, and the argument here is the exact same thing as the base I can think of it as so they're both capital lambda I can think of capital lambdas capital lambda to the first power So what power do I have to raise lambda to to just get lambda to the first power? Well, it's definitely just a one So this whole thing is just going to be a one and that's kind of why I chose that It's because it made my life easier on the left hand side The right hand side just is what it is Long base lambda Of capital omega. It's just some number. We don't know what it is, but it is a number And maybe I'll go ahead and get rid of this one Hopefully it's clear clear to the reader once we go from this step to this step We only did one thing which is simplifying that log And it's just equal to one so I'm just leaving it like that And then maybe I can get rid of these parentheses too because now x minus one is the only thing happening on the left hand side So I don't have to worry about distributing distributing it out or anything like that And now the last thing I need to do is just take I guess add one to both sides So I get an x equal to a log base lambda of omega And then I plus one And I know that I'm done with the problem because I see purple on the left hand side and no purple on the right So I will go ahead and box this up Put a period on it Maybe highlight it for my reader just to indicate that this is the solution to the question they asked And again, the stuff on the right looks a little bit scary, but it doesn't involve x. That's the key point Everything else is just some unknown number Okay, so now we're at part c And uh, first thing I will do well the zero thing I will do my step zero all of the time is literally just write down the problem again So I have one over lambda to the x And then a plus one outside of that And that will be equal to one over capital omega Okay, and again step zero I've written down the problem Let me just really double and triple check that I actually wrote it down correctly So there's my one over capital omega sitting on the right hand side. No good. It looks like that matches up I have something in the denominator on the left hand side and just a one in the numerator And okay, this looks like something in the denominator and a one in the numerator I think I'm good so far What might be tricky here is that there's a lambda to the x happening there and a plus one is happening outside of the exponent So a common mistake here is to try and write this as lambda to the x plus one Which is a very different problem So I would just double check that I have this plus one Outside of what's happening to this lambda to the x totally separate And okay, it indeed looks like that's that's the case. So now I'll go ahead and proceed and start with the problem So I think the first thing I'd like to do is I see a bunch of denominators And gosh, you know, I really don't want to write things with multiple lines So maybe I can do some stuff to get it all on one single line and not have to worry about fractions anywhere So the name of the game is to clear denominators anytime I see fractions. This is a Usually a good idea To make forward progress in the problem And so I could do that one of two ways I mean one way to think about that is to multiply through by capital Omega so I just multiply this side by capital omega and the right hand side And I would get an omega upstairs on the left hand side And then after that step, I would multiply through by this lambda x plus one And that would cancel the one that's happening on this side and then put it upstairs on this side So that's exactly the same thing as the shortcut. We know of cross multiplication So I know that this will be equal to this. There is a process there. That's worth going through But we have a shortcut for it And I guess maybe if I'm being really careful, you know, it's the one appearing there times this omega appearing there Equal to this lambda x plus one from the left hand side multiplied by the numerator on the right hand side But okay, I'm being a little bit ridiculous. Like we don't need to keep track of the ones I'm just pointing that out because you know if that that numerator wasn't a one Then you would really want to keep track of you know, which side does it need to go on So, okay, maybe I'll just go ahead and delete those now that we have them there Okay, so cross multiply to have everything on one line Um, I see some stuff happening in the exponent, but also there's some other stuff This addition of one. So I think I just want to move everything. So again, what am I trying to do? I want to light up this independent variable this thing. I'm trying to find there it is. It's upstairs And it's still upstairs after I do this cross multiplication I'd like it so that Anything involving x is just the only thing happening on one side if I can arrange that situation I try to go for it So my so that was just a step we took The next step will be let's just subtract one from both sides So there's omega minus one on the left hand side And then my capital lambda to the x on the right hand side And again, the x is the thing I'm trying to solve for it's upstairs I know that well, I can apply a log any log my favorite log or even a log I don't even choose the base of Um, they all have this the same property that they'll help me move things from the exponent down So let me just do that So on the left hand side, I will have log base Little alpha. Maybe I'll pick it later of This omega minus that's not a great omega this uh omega minus one on the left hand side and Well long base alpha of capital lambda to the x on the right hand side And just to be explicit, maybe I'll put in one more step. Of course, you don't really need this in your work. You can You know do as much as you want on your scratch work As long as you're showing the reader the process I just see that the Argument of this log so everything inside of the parentheses is all raised to A power x. So I know that's my favorite situation to be in because I can just take that power of x And move it out in front And I'm just going to be careful to maybe put it in a parentheses here Um, yeah just in case there were more terms or something Okay, so now I'm in a situation where I have logs on both sides and gosh I guess I could take this thing on the right and divide it over and then I would have The answer I would have solved for x And that would be fine What I could also do And again, you could just you could have chosen log base 10. You could have chosen Alpha equals e you could have you know taken the natural log of both sides Um, and this would be totally valid here But I say let's let's save ourself a little bit of work Um, and let's go ahead and choose this alpha now. So this alpha was We just Applied some some arbitrary log and we didn't choose one, but let's choose one now And I think maybe to make my life easier. Let's just choose alpha equal to this unknown lambda I could also choose it to be equal to this omega minus one, but then I would still have to divide some stuff over Um, so I think this will be the easiest And why is that? Well, it's because I have this entire thing again And the whole point of this was that I can reduce this term here And I just asked myself what question does this answer? Well, it's what power do I raise the base to so pink? To get the argument and here that's that green And that thing is capital lambda to the one. So this whole thing is just going to be One it's all of that times one Uh, let me just light up this independent variable again That's the thing we're trying to solve for And it seems like we're done, right? So the one didn't matter. I can delete these parentheses And I see something purple on one side and no purple on the other side. So That tells me I've solved for x again in terms of some other unknown Numbers capital lamb down capital omega, but they're just numbers. So omega some number omega minus one is some different number log base lambda of some number is just some other number. So I have x and as some some number So we'll box it put a period on it highlighted to indicate to the reader that this is the solution to the question they asked, which was Again, just solve for x, right? Okay, so now let's go ahead and look at number two We're playing the exact same game as we were in number one We have some equations and we're solving for x The first uh, or the zero thing I will do in all of these problems is literally just write down Insume in a bit and literally just write down The problem that I had since the seven times two to the x is equal to eight times three to the x And I will double triple quadruple check just to make sure I wrote down the problem correctly So that matches up with that seems okay That matches up with that And that matches up with that and that matches up with that I just wouldn't want to do this entire problem and then find out that I'd written down something incorrectly at the start Okay, so I'm trying to solve for this independent variable x And I see it's upstairs So this is my my context indicator that You know, maybe a logarithm is going to help me out here logarithms help me move things from exponents downstairs So I guess I have some choices I could I could You know, maybe try to like divide this seven over and then apply logs to both sides that probably work Um, I could try to divide this eight over and apply logs to both sides that would still work Maybe I'll just go ahead and apply logs without doing anything because I see that Um, when I take the log of this stuff, I'm going to have a multiplication inside of the log And that's okay. I have a log rule that helps me move multiplications out of the log I would not want to do this if this was something like 7 plus 2 because then I would be stuck with An addition or a subtraction. So if this was like an 8 minus 3x I would not want to apply not want to apply logs to both sides here Because then I would have an addition or a subtraction stuck inside of a log And I don't have any log properties that'll help me You know separate those or move those out or anything So we are very lucky to be in the case where this is just multiplication So I know that applying logs will be a good idea So okay, I'll just apply a log base or something. I won't even choose it yet Say alpha to the left hand side And I'll just be sure to apply it to the entire left hand side And I'll also apply it to the right hand side Okay And what will I do now? Well, I have this this long property and that helps me move multiplication So I'll light it up in blue here Multiplication inside of the log will change into addition outside of the log So how does that go? I'm just rewriting so I'm using a log property here, but this will be log base alpha of 7 so just this first term there And then the multiplication comes out as plus And then I get another log base alpha of 2 to the x And that's the the left hand side the right hand side. I'll just play the same game It's log base alpha of 8 And this multiplication inside the log turns into addition outside of the log So I get a log base alpha of 3 to the x I'll just light up to remind myself That the x this independent variable x was the thing I was trying to solve for Gosh, this is getting pretty complicated Um I think maybe now is a good time to go ahead and choose this alpha because I think we can make our lives a little bit easier Let's just make a copy of this Because there are so many terms floating around Um Actually, let's let's take one more step before we choose the log So just remember that these were all pluses And let me just stretch this out a little bit to give ourselves some room What I want to do now is I see In this step up here I had some log of something And the argument of that log while the entire argument was raised to some power x So I know that's my favorite situation to be in I can pull that Exponent again because it's applied to the entire argument Um, that's that's what lets me pull it out So I get this x out in front There we go And I'm just left with this log base alpha of 2 on the left hand side I'm just going to play the exact same game on the right hand side I see that this Uh argument here is all raised to some power so I can take that power and move it out And I'm left with log base alpha of 3 and it comes out as multiplication Okay, but now let me look at this this step here. This is maybe where I want to Choose this alpha. Maybe just to simplify things and make my life a little bit easier So I don't have to carry around as many terms So I could choose log base 10. That's fine alpha equals 10 Um, I would still have four terms hanging around. I would just have to juggle them to solve for x I could use, you know alpha equals e it could be taking the natural log Sort of the same situation. It's not going to simplify any of these terms They're all just going to be natural logs and I'll still have to juggle a bunch of stuff Um, so maybe I could choose alpha. Well, okay, I could choose it to be seven In which case this thing would become a one which would be pretty nice Um, I could choose alpha to be equal to three in which case this thing would become a one which would be pretty nice But I also noticed that I could choose We'll say alpha equals two and that might make my life, uh, the most easiest here because Let's just go ahead and do it and see what happens So let's just do two Everywhere And the observation I'm going to to use here is that Well, okay, so this one over here on the left-hand side this this thing here is just going to be a one But I also for free get that this one is going to simplify a little bit too because the inside Is a power of two so eight is just two to the third power and So what I'm what I'm going to see happen here is that this thing is equal to one That's what'll happen in my next step And what'll happen here is well, again, I just asked myself what question does this this thing answer Well, it's what power do I raise pink to? to get the blue So what power do I raise the base of the log to to get the argument of the log? Well, the argument here is two cubed and so the power I need to raise two Two in order to get two cubed. It's definitely just a three So I'm going to get some actual numbers out of this which will be pretty nice So what I get is well log base two of seven Who knows what it is? I might plug it into my calculator later. I might not Let's just keep it around for now I get an x times a one because this thing was just log base two of two And that's all that's on the left-hand side on the right-hand side. I'm seeing that this thing was a three And again, I'm left with whatever this was. This was like a plus x times log base two of three Okay Gosh, we are really in the weeds. What are we trying to solve for again? It was this x This independent variable And maybe let me just delete this one. We don't have to keep track of multiplying by one Okay, so I mean the name of the game always is where we're down to like a unit one situation now We've done all of the hard unit two stuff. So now it's just juggle everything around and get x by itself on one side so what I'll do is I'll keep this x on this side There it is And I'll take this and move it to that side as well. So I get a minus x log base two of three And I just want all that to be on the left-hand side because then I see my my purple x is there And let me just move everything else off to the other side Well, I see a three that I left over on this right-hand side And then there was this thing which was on the left. I need to move it to the right So it'll come in with a negative sign if you want What I just did here is while I subtracted log base two of seven from both sides And then I also added x log base two of three to both sides Just kind of did both at once. I'm just rearranging some of some of these equations Okay, we're running out of room, but fortunately we are just about done. I think so let me go up here and just copy down what we have Okay, so that that was what we had before What I want to do now is uh Yeah, I really want to factor out this x. So I see an x being multiplied to two terms here So what I want to see is x by itself just one copy of x Well, there's my one copy of x And I just need to figure out what goes in this set of parentheses And this will just be equal to I won't even do anything to the right-hand side Okay, and this is a little bit tricky um Yeah, just you have to be a little bit careful Um, when you pull an x out of this you're left with a one here So there's the one When you pull an x out of this you're left with that log base two of three And this minus sign of course stays around Um And this is a step where I would actually try so I've run it forward, uh, but really I kind of Re reasoned about it starting from this and going backwards So I thought about this thing and then I realized that if I multiply This out and distribute it in I get exactly this thing I saw in the previous step So that's kind of how I knew what to do for the next step So anytime you do an operation like this where you're maybe not sure if you did it quite right So we've tried running it forward by pulling out an x just make sure you can run it backward So try and multiply this x in to those two terms and make sure you get the thing you had before So if you want this is like an arrow that goes in two directions I mean they all are but Um, but okay, so now that I have this I'm just about done So I'm just going to take this thing Put a Well, I'm just going to divide this thing out, right? So I wanted to go over to the other side And I just have to be careful to divide the entire right hand side by all of this stuff And the only thing I'm left with on the left hand side is x So I have purple on one side purple x my independent variable that I'm trying to solve for And everything else on the other side is not purple. So it does not involve x So that's how I know that I am done Okay, and I will just box it put a period on it highlight it It's telling my reader that that's the the solution. Uh, they were looking for Okay, so let's look at to be Uh or not to be But yes, definitely to be Um step zero as usual Uh, well pick the right stylus. I guess is step negative one step zero is uh I'll just write down the question again. This is a log of x squared Plus one minus a log Of x plus one being said equal to two And let's just remember the name of the game. We want to solve for this independent variable x then I'm lighting up And I should be careful. I just skipped a step Right, that was my step one was finding this independent variable. My step zero is just double checking Did I write down the question correctly? Okay, the two's match up the x plus ones match up The x squared plus one matches up And the logs match up I guess I should be a little bit careful So it's it's a log that's just written there with no base Um, and so we you know when I'm writing it in class, I don't do this But this is a convention out there in the world. Um, you know, if people write a log with no base Depending on what context you're in it's it's usually a log base 10. So I will just go ahead and add that in when it's not specified Say on a calculator or something Um, it's usually a log base 10, but it really depends on what field you're in for if you're doing computer science or something A log might always be log base two because you're counting it in binary, you know, zero and one If you're doing, you know, sort of mathematical stuff, you might log might always mean a log base e the natural log So as a good practice just always put, you know, a base on whatever log you write Okay, so now that I've identified the problem, I know what I need to solve And I guess there are a number of things I could do here I'm seeing this log base 10 and I know that Well, I have an inverse function pair for this where I raise You know 10 to the something power and I know that I can undo the log base 10 by doing this operation Um, but maybe I don't want to raise both sides to the 10th power right away because if I do that I know 10 squared on the right and then 10 to all of this stuff on the left I can't exactly cancel immediately But I really need to see is 10 to the log base 10 of just this stuff You know some longer than applied to some stuff and that'll free that stuff for me But the problem is is that I have all of this other stuff kind of in the way So I can't directly apply my my inverse function theorem Of course, I could like split out the exponents here that would that would work fine. I could use my exponent properties Uh, totally equivalent equivalently I could use my log properties first to combine them and then do a long base 10 later So I'll go ahead and do that just because it's a little bit less writing, but both ways actually end up working So the first thing I'll do Is use the fact that a subtraction Outside of a log Well, maybe just to be careful. Let me do it this way I'm gonna copy this down And okay, I'm gonna think of this in sort of a funny way Um When I think of this as adding on well negative one Times that log I saw before Why am I doing that? Well, it just helps me keep track of this negative one which might get lost otherwise um What I can do in the next step Is I have something out in front of a log and I know that what I can always do is just take something in front of a log And put it up in the exponent. So let me just do that Um, what I need to do is take the entire argument of the log when I saw before And then I can apply the exponent to all of that stuff Okay, so that's x plus one to the the negative first power So that's the result of moving this upstairs This whole thing is equal to two And just the point of this maneuver is that um Gosh, yeah, I don't remember all of these log rules But I definitely remember that a a plus outside of a log gets changed to a multiplication inside of a log And it turns out that you know minus gets turned into a division You can just go directly to that but maybe I just only want to remember this this plus rule But now I'm in totally fine shape because I just have this plus I've Cooked up a situation where that's the only thing that appears between them And so now I can just do This log combination rule again addition outside gets turned into multiplication inside And so this x squared plus two well, it just comes along for the ride. There it is And everything in the other log also comes along for the ride There it is And then they get combined with well this plus on the outside It's changed to a Well not a plus on the inside but rather a multiplication And all of that happens inside the argument to this this log And that whole thing is still just equal to two Okay, maybe I'll just do one quick step here If I zoom in a little bit So I'm being a little bit silly. This is something raised to the negative first power That's the same thing as one over that thing So this is really Log base 10 Of I'll just go ahead and combine them. It's you know x squared plus one over x plus one Okay, because this Multiplied by one over this I can just go ahead and multiply those those fractions together And this whole thing is equal to uh two Okay, so let me just make a copy of this we can modify it elsewhere Okay, so that's what we had before and What I will do in the next step is now I'm in a good situation where I just see log base 10 applied to some stuff That's the only thing happening on one side. So that's a great situation to Take the log of both sides or sorry apply Raise both sides Okay, apply 10 to the something with power to both sides. There we go 10 to that power Is equal to 10 to the power of the other side Okay, but now I'm seeing my my cancelling pair. I'm seeing 10 to the log base 10 Maybe I'll just go ahead and take off these parentheses now since we know that we're Just raising putting both sides up in the exponent the entirety of both sides Um, and I see that there's nothing in the way if there was some kind of number here I'm out in front of this log that would kind of be in the way and I wouldn't be able to Um, just cancel things Um, but here I just see these two functions composed And so I know that that Allows me to free this whole argument here So I just get that by itself It's x squared plus one over x plus one On the right hand side. Well, it just stays. What it is 10 squared. Let's go ahead and write that as 100 um Okay So again, we're in the weeds. What was I trying to do? I was trying to solve for This independent variable x. So I'm just going to light it up I'm not quite done because I don't see x equals something I do see a lot of purple on the left hand side and no purple on the right hand side But I'd like to just see x equals something to solve for x So I see stuff in the denominator though and I know that uh, my favorite trick is to move everything out of denominators So to clear denominators And how will I do that? Well, really just by multiplying both sides by x plus one here So I'll just be really careful to take 100 and then multiply it by x plus one The quantity x plus one just to remember that I have to distribute that so I'm going to keep the parentheses around And okay now I have no denominators, but I have purple on both sides But that's fine. Let's go ahead and fix that It's all I'm going to do is well, I want to move This term onto that side So that's fine. I'll just do that But it comes in with the negative sign And then the right hand side is equal to zero All right, and now I'm seeing something. It's you know, it's a little bit better. All the the purple x's are on one side I don't see x equals something yet um So maybe what I want to do is Notice that I I have something that looks like a quadratic. So I see a bunch of powers of x Well, really just this x to the two and there's an x to the one But the highest power of x is a two so that's telling me this is a quadratic in x So let me just try to write it in the sort of standard form so maybe There should be an x squared on front and that's just picking up that term And then there should be like a plus x times something And what should that something be? Well, there's a negative 100 x coming from there And then I should have a plus some well some constant term So I just have to figure out. What did I what did I not take care of? Ah, yeah Well, there's a plus one that I didn't have Well, let me just put a a one there leave the plus out And there by distributing this in I get a minus 100 So there's my minus 100 Okay, but I'll simplify this immediately one minus 100 is well negative 100 minus one So negative 99 And that whole thing is equal to zero Ah good, but now I see I'm in a situation where I just have a formula that does the work for me Um, I guess this is telling me that I can think of you know a equals one And then b equals negative 100 And c equals negative 99 And I can do the quadratic formula. All right, so I can just get uh, let me just Mark that we had a bunch of x's here quadratic formula, let's just go to x equals Stuff and we can just write out what that stuff is And gosh, uh, I'm always hard pressed to remember exactly what the quadratic formula is. Let's see. How does it go? B squared our Let's see. It should it should be negative b, right? So plus 100 Plus or minus some stuff and the stuff will involve b squared. So negative 100 squared Which is the same as just 100 squared. Just simplify that right away Minus four times a which was one. It's only bit out Times c which was negative 99 And let me go ahead and move this negative sign out. It becomes a plus there I need to raise all this stuff to that one half power same thing. It's the square root if you like And then I take all of this over two times a a was just equal to one so I mean there it is And you know what honestly, I would probably just leave it like this. Maybe Put all this under a square root if you like and There you go Now I have x equal to some stuff and the stuff just involves numbers In this case, all the numbers are known And I would just leave this as the answer because if you plug it into your calculator I imagine that you will get Some decimal and you would have to take some kind of approximation But as usual, you know, we'd rather just keep the exact answer around in case, you know Yeah, if you if you want to approximate it with your 10 digit digit calculator, that's fine. But maybe you know if you're doing a NASA You know trying to land something on another planet. Maybe you need many many more digits than that So you can just give them this exact form and they'll compute it on their super computer for As accurately as they'd like Okay, so that's some exact form. I will just go ahead and leave it like that I will box it period put a period on it and highlight it so the reader knows that that's the answer they should go looking for Okay And I guess we Have one more problem here. This part c First thing I will do or the zero thing. It's just write it again e to the x squared plus one Is equal to e to the 2x And I will double and triple check that I actually wrote it down correctly The things to notice here that the 2x is all up in the exponent on the right hand side and okay, I got that right. Everything's upstairs And over here on the left hand side this x squared plus one again that whole quantity is in the exponent And that's great because that's exactly what's happening in what I've written down Okay, so I have a bunch of x's upstairs. I have this independent variable. I'm going to light up Then I want to solve for Everything's upstairs. I know logs are going to help me move things downstairs In this case, maybe I'll just go directly instead of doing this log base alpha business I see that my base is e so I might as well just do a natural log So what's going to happen is that I'll take a natural log of well e to the x squared plus one Really just a natural log of both sides e to the 2x And what I can do now Is my favorite game. I see the entire arguments of these logs applied to some power And I know that well any log would give me this property, but in particular the natural log does It lets me move things out So it goes from upstairs to downstairs I'm going to put a set of parentheses around it and then note that it comes out with a multiplication I'm just left with this natural log of e And then just the same game on the right hand side. This is 2x Again, just double checking that this entire argument is raised to the 2x power. That's what lets me do this So I can move it out And okay now I'm just being a little bit ridiculous. I have right the natural log of e This is asking me what power So natural log is a log base e. What power do I raise e to to get e? Well, that's definitely just a one And I have the same thing over here Okay, so I can get rid of these parentheses and kind of clean things up a little bit So I've x squared plus one is equal to 2x The thing I was trying to solve for was x and I have purple on both sides That's fine. Let's just go ahead and fix it in our next step So what I'll do is well, I want to move this 2x to the same side as everything else and it's to come in with a negative sign because I did that And on the right hand side I'm just left with the zero. So that's my next step And okay, maybe in just one more step. I'll just write it out in a little bit nicer form. It's x squared minus 2x plus one And that's equal to zero And I guess you could maybe just do the quadratic formula here. I mean that always works I I have a quadratic so it's an x raised to the second power It's the highest term appearing I think this factor is though let me see Just double check. I want to say it's x minus one times x minus one And why does this work? Well, this thing is supposed to be the the product of the two Numbers appearing. So this thing times this thing should be equal to that And that's true negative one times negative one is one And then the sum of these two guys Is supposed to be equal to this thing in the middle Oh, that's also true negative one plus a negative one is a negative two So I know that this is the the factorization And okay, I'm being a little bit silly. I've written it twice, but it's really just that thing squared equal to zero And now I'm just going to use my favorite property of the real numbers that if uh Well, okay, maybe I'll I'll do this If anything squared is equal to zero then well say x minus one is equal to zero Okay, it's just taking the square root of both sides and taking the square root of zero is totally fine And So I think I want to say now that x is just equal to one Actually, I think I do want to do this slightly different way That maybe uses something we've seen before x minus one times x minus one is equal to zero So I just have two things that are equal to zero So I know that I'm in the real number So if a product of things equals zero then I can conclude that one of them is equal to zero So either the first one is equal to zero Or that's from that one The second one equals zero And it's from that one and they just happen to be the same And so from this I can conclude that x equals one Maybe a little bit cleaner. You don't have to worry about Uh, when you take the square root of something squared. Oh, gosh, something goes You know, you have to put in a plus or minus. Maybe you have to check some things So maybe this is a little bit nicer way to do it Okay, and That x was the thing I was trying to solve for all along and now I have x On one side and a number on the other side And in particular Purple on one side and no purple on the other side. That's I know I've solved for x So I'll box it, highlight it, put a period on it and that is the the solution Okay So let's move on to to question three so The name of the game here is that we're given some situations And we want to uh, you know come up with a come up with a formula or build a function To model these situations Okay, so in this first one we're given Some some kind of exponential growth Uh, we're given two data points for this exponential growth Um, so I guess I should say yeah, what are we what are we looking at here? Um, we're looking at some exponential growth function w of t And I'll say what that we have like a general equation for that and I'll say what it is in a second Um, and then I'm giving these two data points. There is w of zero is equal to four I'm just writing down the information from the problem w of Zero It's equal to four And then also w of two Is equal to six So two data points and then we have some function and we want to to build that function from these data points And I know that if I so the in bold here, I have that this is exponential growth and from class I know that we have a formula that that models these these types of situations We have a general form for it. It's well Uh, w of t is equal to w not times e to the r t And I'll just be careful here because T is just the name of my independent variable Didn't really matter what I called it. It could have been t or x or l. I mean, it's just a name Uh, but I also have these two unknown numbers or these parameters in the problem r and w not And if I want to write a function uh to model something I want to plug things into that function so I can go and evaluate it, right? If I want to predict something out in the future or learn something about some time in the past Or in general just learn things about because I only measured these these two data points But I want to learn something about other data points that I didn't measure Um, then I want to actually get numbers For this this r and this w not so I can plug in a time and then get it on get out an actual number Okay, so we saw a little bit of how to do this in class But this is kind of a nice situation. It's what I'll do is let's let's work on this one first It's w of zero is equal to four So I'm just going to write that over here w of zero is equal to four So that's one name for w of zero But I have another name for it Well, here's w of well anything So let me go ahead and oh we lost a Should be an r there So let me just copy this over here So this was w at t but here I'm plugging in this this purple t equals zero So this should be a Time zero happening up there And then well this thing here I'm raising something to the zero-eth power and anything to the zero-eth power is just going to be one The only tricky situation is if the the base is also zero, but here it's not. Sorry. What is zero to the zero? Okay, you have to decide Um, but this is just some some positive number to the zero-eth power. So that's just a one So this thing is just equal to w not Well times one if you want but we don't have to keep track of that if it's multiplication But okay, I have two names for the same thing One of them is four and one of them is w not and w not is one of the things I needed to find a number for so that's one piece of the equation Or one piece of what we one piece of information that we need Um that we have now I'm going back to the original equation. So there was one of the pieces That's the other piece that we need this r And how are we going to get that? Well, I've already used this piece of data So I guess I need to use now this piece of data Right, it wouldn't quite make sense if I had um You know, I have this equation and I have two unknown numbers that I need to pin down And one piece of data is not enough information to pin that down with one piece of data If I'm lucky I can pin down like one of these parameters maybe but if I have two unknowns I kind of need two equations So let's use This other one now w of two is equal to six So I have a name for w of two Well, it's six. It's given to us by the problem and I have another name for w two Let me just go up here and grab it Uh, let's copy that So there it is Oh gosh, but that is a t in it. No problem. I am plugging in two for t in this equation. So I get a times two there instead And actually I can do one better. I actually know a name for w not as well if I go back over here We found it. That was uh one piece of this problem right there So we have w not is actually equal to four and that's going to be the key piece of information that lets us solve this problem Okay, so now I have two names for the same thing so I can set them equal to each other these two guys So four e to the two r coming from this It's just equal to this which is six So that's a step. I need to solve for r. This is the parameter. I want to pin down So I'm just going to take a step by dividing through by four and I get e to the two r is equal to three halves after I take six and divide by four And now I'm going to apply a natural log to both sides. Maybe I'll skip a step Say that that is two r which is equal to the natural log of three halves Again, I've just skipped that I've taken the natural log of the left hand side I've moved this power down after doing that or move it out in front of the log because that's what logs do for me And then I've used the fact that the natural log of e is one In fact, that's why I chose the natural log was so this this thing would cancel and just become a one Let's just keep track of what we were trying to solve for was this r And I need to see r equals something So I'll just do that in my last step that r equals one half log of Three halves the natural log of three halves And that's another key piece of this equation that I need was what was this value of r? Okay, now I just have to sort of summarize what I've done put it all together I have this equation here This is the thing I oops Uh, so I've deleted it, but there we go So this is the thing that will be the answer right because they were asking for a function w of t So I need to tell them what that function is So there's my w of t And I had this w not equals four. So let me just sub that in And over here I have this r equals one half log of three halves So this was one half Natural log of three halves And then there's still this times t on the end And now I actually know I'm good to go because I have a formula Um in terms of numbers and there there are no unknowns You might be a little bit worried about the t but here I mean t is just the independent variable It's telling you how to map inputs to outputs Um, so that can stay in the problem And I'll just mention here. So if you wanted this would be totally fine as an answer sorry Camille is Coming up to help again So this would be totally fine as an answer so you could box this and Put that as your answer if you want. I'll just point out that there is a really nice simplification here Just notice that this this right hand side is just equal to four times e to the So everything up here is in the exponent But I have something in front of the log so I can just move it up as an exponent to the argument of the log so I get e to the Well, three halves to the one half power So gosh, I guess that's square root of three halves And then I have a times t on the outside And I can actually cancel a little bit here. This whole thing is just equal to four times root three halves times e to the t So I guess I've I've used a step here where I've written this as four e to the log root three halves Times e to the t or uh, oh, yeah, sorry. I've made a mistake here. Let me roll this back a little bit Yeah, it's okay moral of the story here is don't skip too many steps So I guess I should be a little bit careful So this thing is equal to four times e to the log of root three halves And then this whole thing raised to the tth power Right, because that's how multiplication upstairs happens it happens when you exponentiate something Um, if this was a plus then I could do what I was doing before which was separating it out into two different bases So just yeah, I guess that's a mistake to to watch out for even I make it Um, but okay, so now what happens here is I can simplify this this middle thing in the same way and this is four times Well square root of three halves To the tth power So, I mean, yeah, I don't know. It's it's a little bit nicer I guess than having all of this I see a log a natural log upstairs and I see an e in the base and I think oh Well, you know, maybe we can combine these and simplify some stuff Um, you can get a little bit of a simpler formula, but yeah, don't worry about it too much Okay, so let's look at this next question, which was part b Um, and this one is a little bit of uh, you know, free points as long as you studied and we're keeping up with the class notes Uh, this is something that's just written down directly and uh, I think it was like the friday's lecture before this exam And so what it's asking here is uh, well We want to find a functional inverse of this function f of x equals log base alpha of x And alpha is some unknown positive real number So this could be like alpha equals e so like what's the inverse the functional inverse Of the natural log of x This could be log base 10. What's the functional inverse of log base 10 of x? So the first thing But you can see that this is bolded you have to kind of know what the What this this vocab word functional inverse means So what I need for this to be a functional inverse Is that well f of what we need two things we need f of f composed with g Evaluated at x should just be x And then we also need the g composed with f Evaluated at x should also be equal to x And I've just written these in this this funny compose notation. This is exactly the same thing as f evaluated at g evaluated at x and that needs to be equal to x and So I'm just these are the same thing if you want And the second one is just saying g evaluated that f evaluated at x Should be equal to x So that's just what it means to be To have a pair of functions that are functional inverses They both compose to the identity function the function that just takes your input variable Your independent variable and just gives it back to you. It doesn't do anything to it. It's the identity And then if you compose it in the other order it has the exact same effect you send in an input It just gives you back your output okay, and well For better or worse, this is just something it's it's sort of a fact from class that what is the the inverse of So we have f of x is equal to log base alpha Of x This is something we've used a bunch of times at this g of x If you want you can think of this as like f inverse of x Is just equal to alpha to the x And why is that? Well, maybe you just want to do a quick check Uh, essentially this this check here compose them in both orders So what is f evaluated at g of x? This would be log base alpha Of now I need to replace this x with all of this function. So it should be alpha to the x And okay, that's x Times log base alpha of alpha. I've just taken this exponent. I see it's on the entire argument. So I could just move it out And then this thing is just one. So this is just equal to x So I know that works And then you just have to check that in the other order. This is something that we just have as a fact from class alpha All right, so let me let me just say I'm getting this and we're composing G compose f of x So this says take g of x. There it is alpha to the x Except everywhere I see an x plug in all of this stuff So that x becomes a log base alpha of x And this it's just a fact from class that that's this is what we've been using in all of the previous problems that Really when you send anything in here You get that thing back And then in particular it works if that thing is just x This was the uh the x everywhere Okay, and that's all you really have to do to to show that these these are functional inverses Um, I guess the question didn't even really ask you to show that. Uh, it just said, you know, say what the functional inverse is Um, and hopefully if you got the first couple of problems you were using this fact Um, so hopefully you just remembered that you know, if you have log base alpha Just needs to be alpha to the x and that this this works for more than just like the The natural log in e or like 10 and log base 10 It's really log base alpha any alpha pick your favorite alpha and then alpha to the x Maybe don't pick your favorite alpha if your favorite is negative or something um, but okay uh, okay, so this next question is about um Injectivity and it's highlighted in bold here because this was sort of a Technical vocab word from class and we had a definition of it Um, and here we're we're going to be using one of these these tests that we had to check Injectivity So remember that this is exactly the same as being one to one and then the word Injective is just to remind you that there's some some technical thing you have to check There was this algebraic equation about how inputs and outputs are related So we're given the graph so we have some function again Remembering that functions can be all kinds of things but this function happens to take real numbers as inputs And output real numbers so we can graph it Um, just here on the Cartesian plane we've graphed out this function It's something that sort of looks like part of a sine wave Um, and it looks like it's only on some restricted domain and some has some restricted range So it isn't taking in all real numbers. Just maybe only the real numbers that are You know in this this sort of area here All right, and so we need to decide in this first question I guess this is asking here is this function Injective Um, and then we need to justify sort of why or why not in one sentence And the hint here is a pretty good hint. I think uh, you may appeal to a line test Um, but if that's the case if it like say fails injectivity Then you really need to like show the reader what line that is it needs to fail one particular line test um So I mean, let's just go ahead and do it. Let's find one line On this this thing that fails injectivity that shows, you know, it fails injectivity And the difficult question the hard thing to remember for this kind of problem is Which line test you do right there are two you can do the vertical line test and the horizontal line test So, I mean, maybe if you haven't memorized you might remember that the vertical line test Is something it's telling you about whether something is or is not a function Right, this is telling you you fix one input And you want this function to have this like deterministic property to where there's only one output you could send it to That's what it means to pass the vertical line test because you're I mean, it's exactly what I said in words. You're fixing one input this x here And you're looking at all of the the outputs there And in particular, so these are all the different, you know, y values Anywhere this intersects the graph that's telling you that function takes on that y value there And if that function takes on that y value Twice then that function you send in an x and then the function has to make some decision right like what what why do I choose? And so I know maybe functions aren't smart enough to make choices most of the time So we say that something is a function right if you send in one input And there's only one unique output that it goes to so that's exactly the same thing as the vertical line test Being able to go back and forth between these two interpretations is really important Um, but okay, we're looking forward here. So if we know that that one's the one the test for functions Then it's definitely the other one the tests for one-to-one this injectivity. So it's the horizontal line test And here I see that well I could just sort of pick randomly right if I just pick this horizontal line Then certainly it's it's running into the function. Uh, just all over the place there and there There and there And so what is this? What is this telling me? So I fixed by doing a horizontal thing I fixed one output. I fixed one y value and I'm scanning to the left and right and asking myself Uh, where does the function take on this y value? Are there multiple inputs that get mapped to this one y value? So this is like subtle but slightly different than being a function a function is about fixing one input And asking is there only one output it goes to? And then this horizontal line test is now about fixing one output and asking What inputs did it come from? And if it's only one input then we'll say it's one-to-one or injective But in this case, we're seeing that it's coming from a whole lot of inputs. It's intersecting the graph in multiple places Which means that for this output, which is I don't know 3.1 or something, whatever it is Well, here's an input it comes from It's like a negative 1.5 or something Here's another input it comes from. This is like a negative 0.25. I mean, whatever it is Here's another input it comes from 2.5 ish And there's another one and 3.75 or something like that So these are a bunch of different inputs all getting back to the same output That's telling you that the function is not Injective It's exactly the same same thing as this horizontal line test and in the same way that going back and forth between this Line test and this input output perspective is extremely useful and extremely important Okay, so, uh, what have we concluded from our analysis here? Uh, no It fails the horizontal line test And well, it says that we can appeal to a line test, but we should show a specific line So maybe I'll just say at the line L or something or maybe capital L And then maybe I'll just go up here and label this line that we drew capital L That's my line Okay, so that seems seems fine for that Next question asks us to determine. What is the domain and range of this function? And we should write it in interval notation as is standard Let me erase all of this stuff And let's just look at see what we're working with here So the domain right is the set of all possible Uh, or all of the inputs to this function that makes sense all the inputs for which there is an output So if I look at something like negative four, that's not in the domain because I don't know What function value to assign to negative four the graph doesn't exist there. I don't have that information Uh, but I do see that if I'm at like negative one, for example Then I can look up up and down above negative one and oh, I see a point on the graph above or somewhere on that line x equals negative one And that's telling me that there's a y value for it. So I know what the function is there That's telling me that this point is in the domain And so if you want to I mean, yeah, you can just continue that kind of process Um, you can also do this sort of projection process we talked about Which just like look at the sort of shadow when you shine a light from above and you shine a light from below And just see how much of the the axis you hit Also, if you kind of zoom in here So this point if you shine a light from above Means you're getting a little square bracket here Right, I include that point when I'm doing this interval because if I look above this, uh, so this is negative two, I guess Then I have a function value there. That's a closed circle. So this is telling me that say negative two, uh One is on the graph So negative two is in the domain of this function because I know the output and negative two And then I see a bunch of stuff all here under the graph from shining this light from above And then I get here and okay now I'm shining a light from below and I'm getting all of this stuff And of course, I'm including You know where these intersections are these are definitely in the domain Then I switch back to shining a light from above I'm doing getting all of these x values And then just here I do the same thing. This is a closed dot So this is something like x equals four and I guess this is y equals one Is a is a point on the graph So that tells me I can include four in the domain because I know What the output of the function is there. It's just y equals one So that means it's in the domain the function knows how to handle it And now I just see what I'm left with well, it's something like negative two It's positive four where I include both endpoints Okay, so I'll go ahead and write that just record my findings here the domain of f Is what was it negative two to four And I'm just double check that I've included the endpoints Now I want the range of f Okay, so let's go back up here and I'll get rid of all this stuff so we can Keep the clutter down So the range again is all of the attainable y values for the function Um, if I just kind of look up here or maybe down here y equals negative four is not attainable because I'm I'm fixing this y equals negative four here And looking out to the the right into the left and seeing that this doesn't hit the graph of the function anywhere So the function is not taking on y equals negative four anywhere This is not in the set of it's not in the range But I see essentially everything in here will probably be so let me just be a little bit more precise about that Um, I think what I'll probably use here is this this idea of for the range I just kind of want to shine a light from the left and shine a light from the right And see what values on the y axis I hit So doing that here. I see that You know starting here if I shine this light I get Down to here on the axis so I can include that point again This closed dot is telling me that this y value is achieved so that'll be included I should be a little bit careful up here. I'm shining a light from the left This point is on the graph and that's like the highest point. So that's getting mapped over here And I can include this point this y equals uh Looks like y equals five I guess Because there's a a point on the graph where y equals five is attained Okay, and then I'm getting all of this stuff here in the middle Coming from the shadow of all of this stuff and all of this stuff And I can keep going through in this point I can include Um, and now I need to shine a light from the the right hand side now And it's the exact same game. This this point is on the graph So when I cast it over here, I include this this, uh, this is negative three And the shadow of all this stuff will show up as this part of the interval And it'll go up to here and I include that point in the middle two And now if I'm thinking about you know casting a light from over here Well, I get all of this stuff, but oh good. I've already accounted for all of those y values So now I just need to record my findings here. This went from y equals positive five Or really y equals negative three to y equals positive five So let me just record that it was from negative three where I included that End point to positive five And again, I if I were doing domain and range problems the thing I would always check at the very end is Um, did I get the end points right? Did I include them or exclude them appropriately? And so I would just double check did I need to include negative three? Yes, I do because there is a point on the graph that achieves y equals negative three So this is something like one negative three And that's an actual point on the graph And so negative three is an achievable y value for this function Which means that it's in the range Okay, so I have the domain and range and I guess maybe I'll highlight this for the reader Just in case and then maybe I'll also highlight for that previous problem that the answer was no Okay, and the next part's a little bit tricky. It's asking us To do something a little bit new determine some new smaller restricted domain for which this function Is injective so this function isn't injective, right? We saw that in part one. We proved it with that line test. We showed it Um And we want to fix it somehow like not being injective is not good Right because we saw that injective functions are great because they let us like apply things to both sides of an equation And then keep going in that uh equation solving process. So that was really nice Um, and it helped us define inverse functions so If we can restrict the domain to make this function injective then we can have like a nicer function Where maybe we can apply it to both sides of an equation and help us solve more equations So that's the point of this Um, so we need to somehow find a smaller domain for which this this injectivity works and then we need to explain Why exactly that's that's true So let's just go back after this graph and delete All the clutter we have from the previous one And so I need to restrict the domain. So I need to Um throw out some x values Um, and I don't know so what's what's preventing this function from being injective, right? It was really just that that problem we ran into at first Where it was just failing the horizontal line test in some places But so all I have to do I guess to make it injective is just make it not fail the horizontal line test So I can just pick some some little region of the graph where it's clear that you know, maybe every horizontal line only intersects one Intersects the graph once So there are a lot of of good choices here plenty of good choices But notice that there are some bad choices. For example, I can't take uh Let me see just this chunk of the graph for example would not work If I tried to do all of this I would be uh, not in a great situation because I would still be failing the horizontal line test in a few places There are some intersections. That wouldn't be great Um kind of the same thing like anytime I crest over One of these include both sides of it or something. I'm failing the horizontal line test This is telling me that maybe I don't want to uh to go farther than this this maximum or something if I just take this half That's fine But okay, let's just keep going. I mean if I just Continue it a little bit farther. That's fine. I can still keep passing all of these horizontal line tests So just keep going see how far you can go while I get down to this and I've taken a lot Passing the horizontal line test everywhere But I think as soon as I go past this uh This min here and start coming up again. I'm going to run into problems Um, so I'm going to start failing that horizontal line test again So again as soon as I cross that at a minimum I can go kind of arbitrarily Close to that minimum no matter what as long as I'm taking anything past it I know I can zoom in zoom in really far and I'm going to fail the horizontal line test because every little region Uh around that this like point here if I kind of put a magnifying glass around it it's going to look like this kind of thing So I need to like leave out all of all of this stuff on the left hand side if I want to make this work It's okay. Just leave it out. There we go And I guess I need to be a little bit careful. We're just going to take everything From here Up to everything here Take all of this part of the graph And that'll be fine. All of these horizontal lines will only intersect once And now I just need to figure out what does that domain correspond to? I'll just Play this projection game shine a light from below And this will go down to uh looks like one and I can include that one. That's fine uh right because if I If I include anything off to this side of one then I'm in trouble because if I include this point Then there's going to be a corresponding point over here and horizontal line going between those But it's fine if I just include this this point at the mid because there's there's no matching pair for it to Have a line with And over here I can play the same game just project down and I'll get this point three And for the exact same reason I can go ahead and and include three And then just looking at the shadow under this I get all of this portion of the x-axis here looking at the shadow of all this stuff And so I think this gives me one to three is a perfectly fine region where this stuff will work Or where this function will be injective So let's just record that So new domain And whatever we had was uh, so it was one to three Box it highlight it let the reader know that that's what they the new domain they should take And okay, I'll just check that this matches up with the hint the hint says we may appeal to a line testing in So I guess we need to justify and explain why this is true. Why is f-injective on this new domain? We just look at this new domain you realize that It'll pass the horizontal line test everywhere. That's kind of how we we cooked it up in the first place Was by picking some really big domain or just kind of picking a little bit And seeing that it passed the line test there and then just keep extending until we fail the line test So that's what I would write down if you're you know trying to justify it And then does this uh, do we have this this thing where it says the new domain should be some interval contained in the interval We wrote above for this domain Well, here is this domain negative two to four. Let's just draw that really quickly on a number line So there's zero There's negative two There's four And this was my first domain and now my new domain is this one two three And oh good. That's an interval contained in the the one we had before And so that's just a copy of the real line And okay, so I know that this solution checks out And again, there could be many many solutions here You could take sort of any other part of the graph where the horizontal line test has passed everywhere So you could also take You know this little branch of the curve Or you could take like this branch over here everything up to that max point Um, you could just take this branch over here Sort of a lot of different options You could also take something something kind of weird like you could take like just this stuff and Then some of this stuff down here if you wanted and you would still pass the horizontal line test if you were careful about it Okay, so this next part of the the problem is asking us to show now using the the definition of injectivity That this function is injective and let's just remember that there was Injectivity we use this word because there was a technical meaning There was this algebraic equation associated with it that we needed to remember And what is that definition? so the definition was that F is injective if and only if So right this i f f to mean that which just means this is a definition the implication goes both ways if and only if Whenever you have f of a equal to f of b You can logically conclude That a is equal to b So what does that mean in english? This is saying that if you have two outputs that are equal If your function is to be injective then those two um the two corresponding inputs should have been equal to Um, all right. Hopefully you have to think through this a little bit. What does it mean to not be injective? Well, it means if you had uh, you had one output Sorry, I think I phrased this incorrectly. Let me let me back up. Um, what does it mean to be injective? This means if you if you start with this one output, so this is one these two things are equal Um, but it's the output of the function where it was coming at potentially two different points a and b that were mapping to it So pretend so this one output potentially coming from two inputs But you can logically conclude that the two inputs had to be equal. There was really only one input to begin with So that's what that's what it means to be injective What does it mean to not be injective? Well, it means you had one output Which looked like they were from two inputs and then oh, yeah, they were they were from two inputs or potentially more So one output coming from many inputs is not injective one output coming from exactly one input is injective But okay, you just have to remember this algebraic equation, you know, we have a formula for it this kind of thing So let's just do it. We want to start with this thing So they've kind of spoiled it for us in the question, right? They've said to prove that it's injective Which means that it probably is injective. So we just have to start with this thing Do some steps and conclude this thing So what does it mean for f of a to be equal to f of b? So we have f of a equal to f of b Well, now I'm just going to plug. So I have a name for f of x, right? It's f of x equals one over one plus x Hopefully that's what it was. Yeah Oh, I guess there's a little bit of interpretation here Yeah, I should I should mention this So we have this kind of this funky notation for the function And it's worth, you know parsing out. What do what exactly does this mean? How does this match up with things we've been seeing in class? So just remember that this is the the name of the function. I could call it anything I could call it f. I could give it some long word of a name like sine or cosine or something Um, and then I've told you kind of the universe where the inputs live and the universe where the outputs live And then I've told you how the inputs map to the outputs. In other words, I've given you a formula for how it works Um, and just remember that functions are very abstract. Not all of them have formulas But in this case we have one and then let's just remember that this is exactly the same thing as saying f of x is equal to one plus x And just remember the problem was is that this didn't tell you all of this this other information Like what is the name of the independent variable? What kind of inputs and outputs does it take? And even like what is the the name of the function? It's all kind of there, but it's all kind of mixed up together Okay, so f of a is equal to f of b. What does that mean? Well, I have a name for how f maps inputs to outputs and it maps the input a to one over one plus a It maps the input b to one over one plus b So again, remember that I get to start with this assumption And I just need to do a bunch of steps and conclude a equals b. All right. That's what I need to get to A equals b at the end of the day So that's the goal. How do we get there? well, um All right, I see denominators again, so I'm going to play the game of uh clear everything from the denominators so I get One plus b Is equal to one plus a if you want just by cross multiplying these guys Or if you want really just by multiplying through by one of one plus b And then this side becomes a one this becomes a one plus b over one plus a And now multiply through by one plus a to clear that denominator and then you get a one plus a on this side It's okay. Whatever floats your boat What I can do now is subtract one from both sides and get b equals a and that's exactly what I wanted to show right, this was the the whole the whole definition of injectivity was that I just had to Start with this assumption do some steps conclude a equals b So I started with that assumption I did some steps and I concluded a is equal to b. So I've shown that this is injective Okay, so I think this one was really just an exercise and You know making sure you're kind of uh, you know You went back and looked at the lecture notes to see what this definition of an injectivity was We stress really the the importance of it has this very special definition And then also just being familiar with this notation that we've been using in class Um about how to you know write a function and map inputs to outputs and how to sort of Get the usual, you know notion you're seeing in the pre classes and worksheets back out of that Okay, uh, so let's look at the next question Next up we have something about uh investing money into a fund with a certain interest rate and it's compounded once every month So see we have uh, let's try to scan out the keywords here We want to write some mathematical equation and probably solve it So what information do we need to to glean from this? Well, there's a number 1.1% Here's some important info. It's compounded once per month. So that sounds kind of like a number once This is probably gonna be an important part of our equation We have an initial investment initial right is like time t equals zero or something And there's some initial amount some number And how much money will be in the account after three years is what we're trying to find out So gosh, I have all of these words I would really like to to write down a mathematical equation do some translation process Solve this mathematical equation And come up with an answer for this So the first thing I need to realize is that This is asking about compound interest and we have some formulas for compound interest And it's important to recognize that this is once per month. So this is not continuously compounded Which would be like not just once every second or every microsecond or every nanosecond Continuously every possible instance in time But this isn't that this is just once every 30 days once every month or something So the first thing I need is the equation for compound interest compounded discreetly So we had this this notation for it in the notes, which was a d of t And this was equal to well, there was just a general form for it There was a 1 plus r over n To the end t This was like the important part of it and then there was also this a not out in front Okay, so I want to before I do much with this I want to light up. What is the independent variable t? It was just a name I gave for how the inputs map to outputs And this helps me find out the other things that I need to identify in this problem If I want to to answer this question I might need to identify a not r and in So let me just try to write those out. I need to identify a not It's equal to question mark r is equal to question mark And in is equal to question mark So what are these things? Let's see if we can if did the problem just give them to us maybe Let's try to find this out So the interest rate is 1.1 and that's that's actually corresponding exactly to my r my r stands for rate So I should just be a little bit careful here because right this was 1.1 percent And the problem is is that percentages aren't numbers their ratios and so percent right is mean this means per So okay 1.1 per cent means 1.1 Per is like a division Cent is like 100 percent divided by 100 It's okay if you want this is equal to zero point Uh, oh gosh, I don't even think I want to do that. I just want to leave it as let's do 11 over 1000 just leave it as some some nice ratio of integers Um, okay, so we have the r We need the n. I think we we can probably find that this was once every month Just remember that this in in this equation was how many times you could pound uh per year Plus this equation was set up to be in years And so if we're once per month then we're 12 times per year so that this in is 12 And now we need to find out what this a not was So let's just take a look at this um equation Uh So one one way to see what that is is if you plug in t equals zero In this equation this doesn't always happen, but in this case You plug in t equals zero and this whole term while you're raising something to the zero with power That something is a a positive number. It's a number bigger than one even So it's some positive number to the zero power, which is just one Okay, so this whole term just becomes a one Which means that this thing evaluated at zero is just equal to this a not And okay, I guess that's what this is telling us here that this thing evaluated at zero Was equal to this number 250 000 So there's my a not 250 000 Okay, so I think I have my equation now. It's a d of t a sub d of t Well, it's equal to 250 000 Whatever that number is times one plus So this was r 11 over a thousand divided by something that was this in 12 And then I needed to raise it to the n t power. This is 12 times t I just want to check am I actually done did I get a full equation? Well, there's my independent variable. That's fine It can still be a letter Everything else is a number. So I know I actually have a model now. So I have a everything's a number I can plug Times into this and I can get things out of it like real predictions. So I've eliminated all of the unknown variables So that's great. Uh, is this what the question was asking? Let's go back and figure out what the original problem was There's the question mark at the very end Let's read backwards. Uh, they wanted to know how many or how much what is the balance in the account after three years? Well, the whole point of this equation was that we could plug in time values t and this would just tell you the balance at that point Again, this was like we're gaining interest on money in an account. So ad of t is like how much money is in your account After t years remembering that this whole equation was set up to work in years to begin with So that just means that I need to plug in ad Of three And I won't even rewrite it. It's literally just all of this stuff Except for this purple t became a purple three And at this point, I will leave it up to you And at this point, I will leave it up to you So you plug it into your calculator And get some number and then just be careful to indicate to your reader that you're taking some some approximation here Um, right like you can imagine like if it's you looking at your savings account, maybe just uh Estimating it to the cent or whatever like to two digits in decimal places is fine But you can imagine if you're like a high frequency trading firm or something Maybe you want to know it all the way down to like the millionth of a cent because that You know adds up over 10 million transactions to some significant amount of money or something So indicate to the reader that you've taken some approximation Um And then you know, that's your answer if you've plugged in t equals three to get the balance of three three years So I won't I won't bother writing that out here. I'll just leave it up to you to plug whatever this is into your calculator In fact, if you if you wanted to you don't even have to plug If you wanted to just leave This is your answer that would be totally fine for this kind of class Because that's it's an exact it's a precise answer. Anybody else can plug it in if they want Okay, so I think this is either the last question or the second to last Um, and this is probably the most difficult one, but this is one that's very very similar to what we did in class And there's a little bit of a trick to it So we have this this same situation as class where we have a petri dish filled with bacteria And we're supposing that it's it's modeled by an exponential equation Or rather the the weight of this dish is modeled by an exponential equation And we're supposing it's exponential decay of some sort So we don't have a bacteria colony growing over time. It's it's the weight of it is going down. It's shrinking And it's an exponentially shrinking And so we see we see it's given by w of t some function that models exponential decay And we have like a general form for that equation So that'll be like the first thing we find And then we're given some data points. So after t equals 30 days We go and weigh it and we find that w of 30. Okay, so what does that mean? It's this function w of t Evaluated at 30 days. So this is the weight at day 30 And that's 200 grams And then something goes wrong We leave the lab for vacation or something for 20 days We come back and we measure Gosh, our our bacteria colony is dying at 50 days. We only have 150 grams of it left. So What we'd like to do is build a model for this And use it to predict some data point that we didn't collect So here we want to we only collected data at t equals 30 And t equals 50 and we want to make some prediction about what did it look like at t equals zero Days, what was its initial weight? How much has died over time and that that 50 days like total So let's let's see if we can figure that out So the first thing we need to do is we need a Like a general form For this kind of equation if we want to try to build this model So we have this in class that anytime we have an exponential sort of growth or decay situation We can model it by this this function here w of t Is equal to w not e to the r t So remember that t can stick around because t is our independent variable And then we have some unknown numbers r and w not these parameters And if we want this to be a model if we want to Make actual predictions and get numbers out then we need to Determine what these unknowns are we need to make them known and find actual numbers for them Okay, so let's just record some other data that we have so this was just some general form that we had from class And you can actually see that it's the exact same equation as continuously compounded interest Which is a form of exponential growth in that case And the only difference here is that if it's decay, maybe this r might be negative or something But it's this is kind of one of the miraculous things about math is that one equation can model Sort of a lot of very different seeming seemingly different things So let's record some other things that we have about this problem One piece of data we have is that w of 30 is just equal to 200 So there's w of 30 equal to 200 And then we also had some information about w of 50. This was equal to 150 grams Okay, so this is at least giving me hope that I can solve this equation. I have uh two unknowns appearing in this equation And I also have two pieces of data So I have two equations and two unknowns. I'm I'm hoping that I can probably solve this If I just had one equation, I would not be so hopeful like one equation and two unknowns Uh chances are I won't be able to pin it down precisely Okay, so just following the exact same procedure we we did in class. Let's work on This first piece of data. So we have w of 30 Is equal to 200 This is just one name for the output of the function when I plug in 30 But I have another name for it Here it is make a copy of it And uh gosh, so this is really I've just written down w of t, but I need to plug in This purple t equals 30 So that needs to become a 30 And I get some equation out of this So these are just two different names for the same thing. So that means they're equal So I get something that's like 200 Is equal to w not e to the 30 R Okay, that's as far as I can go with this But I want to now do the exact same thing I've I've used up my first piece of data somehow use that information already now. I want to use this piece of data And I'm just going to play the same game w of 50 I have a name for it 150 It's that name there I have another name for it Well, it's the original equation This thing here But oh gosh, that thing was w of t And here I'm taking purple t equals 50 So let's plug in 50 there And now I have two names for the same thing. So I can set them equal to each other So this tells me that uh, let's set it up to be similar to the first one 150 is equal to Well, w not e to the 50 Times r Okay, so we just did that process twice on our two pieces of data Uh, and now we need to do something kind of clever We need to solve for uh, w not an r in this and the clever thing to do Is to take these two and divide them And so we saw that if you divide you can essentially yeah, let me You can take one equation, uh, like inequality and divide it by another equation And you just do this by dividing both of the two sides So what I can get from this is that I have 200 over 150 Is equal to so I've just taken this and divided it by that And that has to be equal to taking this and dividing it by that Okay, so that is w not e to the 30 r Over w not e to the 50 r But okay, some nice things happen here. The the reason why this was clever is because Um, I've cancelled out this w not that's kind of the clever bit So that's that's the uh, the trick to sort of solving this And what that means is that I've gotten something that looks like well, let me just can uh, condense the left hand side so this is something like Uh, g it's 20 over 15 which is like Oh, let's see. It's divided by five. So four over three So I get four thirds on the left hand side On the right hand side i'm left with e to the 30 r over e to the 50 r But okay, now I can use my exponent rules. This is e to the 30 r times e to the negative 50 r And I just use this fact that anytime I have Something raised to a power and it's in a denominator I can move it up to the numerator as long as I flip the sign on it and just remember it's not Um, always just changing it to a negative if it was already a negative You would have to flip the sign and change it to a positive So anytime you take something in the denominator Move it upstairs to the numerator. Just flip the sign and turn it Turn this division into multiplication And now I'm going to use another exponent rule that if I have so I have two, um Two things raises some powers and they they have like bases. So that means that I can combine them into a single base so e to the 30 r And this multiplication downstairs becomes addition upstairs And it's plus a negative 50 r And okay, I'm being a little bit daft about it. I 30 plus negative 50 But it's just to be clear that you only have to memorize the one rule about how multiplication Uh an addition work and then you get the ones for subtraction and division for free Um, so 30 minus 50. I guess this is a e to the negative 20 r And okay, just summarizing what I have I have that kind of equality there. So let me just write that down here So four thirds Is equal to e to the negative 20 r And right now I can just take the natural log on both sides Maybe before I do that just to make things a little bit cleaner um I will Take reciprocals of both sides. So I'm just going to get rid of the the negative sign So if you want I multiply both sides by e to the 20 r Let me just do that like an e to the 20 r times four thirds equals e to the 20 r times e to the negative 20 r And I'm just going to use so the reason I did this is because I see this multiplication downstairs Which I can turn into addition upstairs And this whole thing becomes zero Let's e to the zero equals one So that's the whole reason I did that And then I'll just multiply through by three quarters and that'll cancel this and I'll get a three quarters on that side It's just to get rid of the negative sign just a silly trick here You could just go ahead and apply logs, but you could have another negative sign floating around to keep track of Um, but another way to see this is I've literally just taken the reciprocal of both sides The reciprocal of four thirds is three quarters That's showing up there And then one over e to the negative 20 r is e to the positive 20 r because I can just move it upstairs and flip the sign Okay, but now I can just take logs so I get 20 r Is the natural log of three quarters And lo and behold, I've solved four r and it's just in terms of numbers So it's one 20th times the log The natural log of three quarters So that's not my final answer, but I'll box it because I need it. It's part of part of the answer I need Uh, so that's r equals one 20th of natural log of three quarters What else do we need? We need this, uh, this w naught, right? So we have this r and we need now this w naught And you can really just pick, um I guess we can go back to the yeast equations these Either of these will give it to us. So let's just pick this one Totally arbitrarily to work on now And so I just have that this this is inequality that I have But I know this r now So I I can write this as 200 equals w naught e to the 30 r but whatever r was r was 120th natural log of three quarters all right and okay, I want to uh Yeah, I really want to solve for this w naught. So maybe I'm just gonna all of this is just some number But maybe I'll simplify it really quickly So this is w naught times e to the 30 over 20 times the natural log of three quarters And I really want to cancel this to these two But there's this pesky number out in front in the way. So I need to do something with that So there's w naught e to the natural log while there's three quarters And I just remember that I can move this stuff up into the exponent if it's just multiplying out in front So that's a 30 over 20 up in the exponent And okay now I can finally cancel these guys I get a w naught three quarters to the 30 over 20 And let's just remember that this was equal to 200 You know good now. I just have w naught equal to 200 divided by three quarters To the 30 over 20 or I guess we can just call this uh simplify. This is three halves Okay, that's an important part of my solution too. I'll need that w naught And so now I can actually get a full full on equation for this uh, so I have This here was my equation And the whole goal was to find the two unknown parameters But there we go. I have w naught So let me just substitute that in Here we go. That's our whole w naught And then I had an r. So let me just substitute that in So here here it is There I guess maybe shrink it so we can all fit in the exponent Okay, and then I get uh, gosh this this huge awful equation. I guess you could simplify it a little bit um But I guess that just is the equation All right, everything I'm seeing now. I'm just seeing that one purple independent variable appearing in my formula And oh oops That r we substituted in was all of this stuff And the w naught we substituted in was all of this stuff And so everything appearing everything else other than the t is a number All right, and that just means that now we have a model we can plug things into it. We can extract information What was the question asking? It was asking how much did it weigh at t equals zero days So all we have to do is take our model I make a copy of it here So there's our model And if I just plug in t equals zero So I just plug it a zero there And I get a zero there And I'm just raising e to the zero power. So that just becomes a one And then the only thing that's left is just whatever that is And I guess if you plug that into your calculator, you'll get some uh, some weight So I'll leave it up to you to figure out what that is But maybe if you're Recording that on a you know for telling your reader about it if it's not some exact number be sure to tell them that it's an approximation Okay, so I think that is I mean double check. I think that's it for this exam Hopefully that is helpful to you guys If you guys have any questions about any anything on the exam, I'm happy to go over things in more detail Although that's already quite a lot of detail But yeah, feel free to contact me via email At any point. All right. Thank you very much