 This is the study guide for the final exam for Math 1050. So if you've reached this point, congratulations, you're about to take the final and the course is about to end here. What I've done is to go through the material covered on each of the five exams plus the material since that last exam. And I've tried to pick out what I think are some of the most crucial points that you need to review. I wouldn't say that the list of topics that we're going to see here in a few minutes for each exam is completely exhaustive. In fact, you might want to refer to the website to get a more complete list, but I think this will have more than an hour's worth of material here to talk about that might help you to review for this exam. Now the final exam counts last for two hours. It will be a little longer than a regular test. I wouldn't say that it would necessarily take you the entire two hours to finish, but you certainly have two hours if you need it. Let's go to the first graphic here. We're going to begin by reviewing the five exams that have been given this semester, and I would say that one way to study for the final is for you to look through those exams and make sure you can work all of those problems. If you feel fairly comfortable with each exam, then you shouldn't have to review very much for this final. However, if there is one exam or if there's a portion of an exam that seems to be giving you trouble now, then you need to review that portion in particular. For the final exam, there will be no calculators allowed. Now I know that on exam two and exam five, you were allowed to use a calculator for some problems, and I'll try to show you as we go along how I would ask you a question that might have been interpreted as a calculator problem. You might only have to set it up or you might have to work it out longhand, but if there's a multiplication and division and so forth involved, I'll try to keep that to a minimum. Of course, you can't use notes or a text during the final exam. As usual, you'll want to turn in all scratch paper with your test, and you want to show your work for partial credit. Okay, let's go to the topics for exam one on the next graphic. Let's see. Now on exam one, one thing you needed to know, first of all, were the two equations for a circle. There's the center radius equation and there's the general equation of a circle. Let me just write those two down on the green board and I think you'll remember these. The center radius equation, the center radius equation looked like this. It was x minus h squared plus y minus k squared equals r squared. This is where h and k represent the coordinates of the center of the circle and r is the radius. But now if you multiply this out, you see you're going to get some x squared terms and some x terms and some y squared and y terms and what happens is you get what's referred to as the general equation or the standard equation of a circle and that equation ends up being of this form, x squared plus y squared plus some multiple of x plus some multiple of y plus a constant term equals zero. And one way these sorts of equations can be disguised, for example, in homework problems is the author might put a coefficient other than one on the two squares, maybe a four, four x squared plus four y squared and in that case what you do is immediately divide through by four or divide through by that common coefficient to get it into this form. And then you have to complete the square by grouping your x's and grouping your y's, complete the square and you make this equation go back into the previous form and you can find its center and its radius. Okay, let's go back to that same graphic. There were eight fundamental equations, eight fundamental functions and each had a graph and each one had target points that go with it. You should be able to graph each of those eight fundamental functions. I won't go over each one of those but I think they'll come up in our discussion during this episode. Now you should also be able to graph piecewise functions and let's do take an example of a piecewise function. Suppose I have a function such as f of x equals and let's say it's equal to x squared if x is less than or equal to one. And then suppose it's equal to x plus one if x is greater than one. So this is referred to as a piecewise function because it's defined in two separate pieces for x is less than or equal to one and for x is greater than one. How would I go about graphing it? Well let's see, let's put a graph right here in this space. I'm going to mark off my x-axis and my y-axis. Okay now here is one on the x-axis and I know that my graph is going to be divided into that portion to the left of one and to the right of one. To the left of one the definition is f of x equals x squared. So let's just think about that for a moment. f of x equals x squared is a parabola and it has three target points. The target points are at zero zero and at one one and at negative one one. So if I were going to draw this complete parabola, the parabola would just come down to the origin and go back up but you notice this parabola has to stop when I get to one. So when I draw the parabola I'm going to draw it so that it stops when I get to one right there. Now what happens after one? Well after one, after x equals one, I'll be graphing f of x equals x plus one. Now this is a straight line graph, its slope is equal to one and its y-intercept is equal to one and if I were going to graph it I would locate the point one on the y-axis and the slope would be one so it would be a diagonal line coming through here. But I'm only going to graph it beginning at one. Now I'll put an open circle there because this is only for x is bigger than one and if I just extend that line on out imagining that it passed through one on the y-axis the graph continues like that. So this is a graph of the piecewise function f. It's part parabola and it's part straight line spliced together. By the way this is a dot here, not an arrow, that's a dot. Okay, let me try a different marker right there. There we go. Okay, so I may ask you to graph a piecewise function on the exam and you should be prepared to do so. Okay, let's go back to that list and look at some other topics from exam one. In item D there were three types of symmetry. There's a symmetry about the y-axis. Those functions are said to be even. There's symmetry about the origin, those functions are said to be odd and then there's symmetry around the x-axis and those graphs are not functions at all but they're still important types of symmetry. So you should be familiar with three types of symmetry. Let me give you an example of a symmetry problem that you could be asked or something like it. Suppose I told you that we had a graph that looked like this and suppose we were told that this function is an odd function, f is odd and so I've only drawn half of the function, there's another half over here on the other side and when I say it's odd what that means is it is symmetric about the origin and that's a clue as to how the rest of the graph is going to look because if I flip it over the y-axis and then I flip it over the x-axis and if I do this double flip over the y over the x then what happens is I get a graph that comes down like this and it turns and it goes back and it's just this graph flipped over and inverted so it's going to go up like so and go out the other side and that is a rough sketch of how the rest of the graph should look. Now you know when we say a graph is symmetric about the origin is if I pick a point like let's pick this maximum point right here if I go through the origin the same distance on the other side there should be another point just like it on the graph but opposite the origin, same distance on the origin. If I were to pick a point like right here if I go through the origin the same distance on the other side I should be on the graph again. Yeah, so that's what we mean by symmetry about the origin and this should all sound familiar to you and I wouldn't be surprised if you're thinking gosh it's been a long time since we talked about that that means you just better review that a little bit more and make sure you go over your notes, look through the book and of course be sure to look at exam number one. Now there were transformations of graphs you remember there are a number of ways let's go back to the graphic here in item E there were transformations of graphs you remember we can translate graphs up or down we can translate graphs left or right we can stretch them vertically we can compress them we can flip them over by changing the sign and we can also compress them and stretch them horizontally let's just take an example of a transformed fundamental function such as this let's take f of x equals negative 2 times x plus 3 cubed okay so let me let you think about this for a moment what are the changes I'll make to the fundamental function f of x equals x cubed now you know I'm thinking that the graph of x cubed I'll just sketch it over here in this little window there has three target points at the origin and at 1, 1 and at negative 1, negative 1 those are the three target points for the cubic function and the cubic function comes in sort of like a parabola but it's deeper a little flatter and then it turns and it goes down on the other side so we're going to be making some changes on that graph well the x plus 3 says the graph should be shifted to the left three units and the 2 says there's a stretch but it's a negative 2 so it's going to be inverted and stretched so I'll just say it's going to be inverted and then stretched so let's do those things to this graph right here let's see now you notice that this curve passed through the origin originally so if I move that over three units it's going to pass through negative 3 on the x-axis and normally I would go over 1 and up 1 to get to the next target point but I've inverted it and stretched it so when I go over 1 I'll go down 2 and then on the other side when I go to the left 1 I'll go up 2 so I've taken into account both the negative and the 2 here and then my cubic function comes in rather steeply it's going to look a little thinner but that's because it's actually a little taller and this is the graph of the function f that we see here now on the exam I think you can surely expect for me to ask you questions about transformed fundamental functions and ask you to draw the graphs or I may draw the graph of a transformed fundamental function and ask you to write the equation of that function but let's move on in our list of topics from exam number 1 let's go back to that item f remember that functions are used as models that's certainly a very important application of mathematics to other courses that you model real-life situations with functions we had a gardening problem we had the projectile thrown from the top of a building we had a number of situations that we modeled and then in that same episode we talked about variation there's direct variation, inverse variation and then there's a compound variation okay item g, the vertex of a parabola I think we need to mention that one here also if I have a quadratic function of the form f of x equals, let's just take an example let's say 2x squared minus 6x plus 5 and this is a parabola opening up because it's a quadratic function but a person might wonder where is the lowest point on the parabola if it opens up it has the lowest point and that's referred to as its vertex there's a formula for computing the vertex or you could complete the square I think the simplest way is to use the formula and that is that x equals negative b all over 2a now in this case a is 2 b is negative 6 and c is 5 so negative b will be positive 6 over 2a is 4 and that reduces to be 3 halves this is the x-coordinate of the vertex to find the y-coordinate you take f at that value which is 2 times 3 halves squared minus 6 times 3 halves plus 5 and that's 2 times 9 fourths minus, let's see if I cancel off a 2 3 times 3 are 9 plus 5 so I have 9 halves minus 9 plus 5 which is 9 halves minus 4 now 9 halves take away 8 halves is 1 half so what I've concluded here is the vertex of this parabola is that the ordered pair 3 halves 1 half and it opens up okay now if you look back at the episode where we talked about the vertex of a parabola there were applications related to that which we don't have time to work examples of today but you should be aware of how to find the vertex of a parabola and be able to apply it to a real life situation and the last thing that we covered on exam number one was the notion of inverse functions now a function has an inverse if and only if it's a 1 to 1 function and a 1 to 1 function is a function that passes the horizontal line test now inverse functions of invertible functions can be computed using an algebraic procedure which I don't think I'll do because I think we need to move on to other topics but you should be able to perform the algebraic procedure to find the inverse of a function you should be aware of how to draw the graph of an inverse function given the graph of an invertible function and that is to flip the graph across the line y equals x and you should finally know that the composition of a function and its inverse that is f composition f inverse of x is equal to x and in fact the composition in either order on x is equal to x okay let's go to exam number two and look at items from that exam first of all you should know how to graph fundamental polynomial functions and draw their graphs and variations of polynomial functions now let me give you an example of what I mean by that first of all you should be able to graph a function like this f of x equals x to the fourth power now this is a higher order polynomial but I would call it a fundamental function and it's going to look very much like f of x equals x squared it has target points and the target points would be at zero zero one one and negative one one and the difference between the graph of this function and a parabola is this function comes in even steeper and it flattens out even more in the middle so it flattens out even more there so now if you see me draw a parabola and you see me draw this draw them somewhat the same but at least I should be aware of the fact that this graph flattens out more in the middle than a parabola does and it rises more steeply now compare that with a different fourth degree function we'll call this one g of x and this one is given factored suppose the function is x minus two squared times x plus one times x minus three you see this is also a fourth degree function just like that was a fourth degree function if I multiply this out I'll get an x squared times an x times an x so the lead term here is going to be x to the fourth but there will be a cubic and a quadratic and a linear and a constant term as well so to graph this quickly what I do is locate the x intercepts now let me let you think about what are the x intercepts for this function before I say them out loud the x intercepts are x equals two x equals negative one and x equals three that's because x intercepts are the same thing as the zeros of the polynomial that is the numbers that make the polynomial zero and if you plug in a two or a negative one or a three this expression will be zero so my graph will be touching the x axis at two and at negative one and at three okay now the next thing I decide is as my graph comes in on the right toward three will it come in from above or will it come in from below and that depends on whether when I multiply this out I get a positive x to the fourth or a negative x to the fourth and this multiplies out to give me a positive x to the fourth so it comes in from above comes in from above now the next question I ask is when I get to this x intercept do I pass through or do I turn around and go back up and to decide that I look at the multiplicity for this factor and the multiplicity is one this factor appears only once it's to the first power so I pass right on through but I have to turn somewhere and come back to be at the x axis when I get to two so I'll turn right here and it turns smoothly it shouldn't be pointed and when I get to two I ask the question again should I pass through or should I turn around and come back to the multiplicity for the x intercept two and it's a square so it's going to look like a parabola in the vicinity of two so it's going to turn and come back down and not cross because the multiplicity was two now I have to I can wander away but I have to come back and cross and touch the x axis at negative one but I figure I can probably wander away more when I come back because I have a wider span of values that I can go through and when I come back when I come back to negative one what's going to happen well at negative one the multiplicity is one so I pass right on through let's try this one more time here so I pass right on through at negative one so this is a rough sketch of the graph now if you compare that with the graph of f over here you notice these are both fourth degree polynomials and they both go up on the n's in this case now if there had been a negative in front this graph would have been inverted and both n's would go down but generally a polynomial of even degree like a fourth degree both n's the n behavior is the same on both sides they both go up or they both go down okay now you would know accuracy intended for telling you how low this graph goes this could go lower, this could go much lower or maybe not quite this low but this is just a rough sketch of the general shape of the graph and that's what we're after let's go back to our graphic for exam number two and look at some other items you need to be familiar with now you remember long division you've covered long division and in fact synthetic division I think in intermediate algebra we've been using long and synthetic division what are factors of polynomials later in this material for exam two so of course you should be familiar with each type of division for the final exam then there are remainder and factor theorems and the rational root theorem and without going into details on each one of those I think their exam number two covered each of those theorems and you might want to take a look at your exam to review that now those theorems help us in item E to factor and then graph polynomial functions and by once I can factor the polynomial function I can graph it much like I did in the example here just a few minutes ago okay let's go to item F and G fundamental rational functions and then more complex rational functions let's spend a little bit of time talking about these two things there were two fundamental rational functions that we've considered they are F of x equals one over x one over x and then another one G of x equals one over x squared and to remind you how these look how these graphs look let's plot them using their target points for one over x the target points were at one one and negative one negative one but not zero zero you notice this function is not defined at zero so there's no point to be plotted on the y-axis once we plot the target points we have to know the shape of the graph and the graph appears in the first and the third quadrants you know now that we're at the end of our course it makes more sense I think now if I tell you that this is actually a hyperbola rotated 45 degrees and covered on the fifth exam we talked about the conic sections and you can have a hyperbola with branches that open left and right or branches that open up and down well if you imagine opens a left or right and then you rotate it 45 degrees you can get a graph like this this is a rotated hyperbola but we didn't know about hyperbolas at the time we talked about these fundamental rational functions now the graph for G of x it also has two target points and they are at one one and negative one one in other words this graph will be in the first and second quadrants now it comes down a little bit steeper and it flattens out a little bit faster relatively speaking than one over x did over here and on the other side it comes down a little bit steeper and it flattens out a little bit faster and as was the case on exam number two I'm not going to be greeting you very carefully on how rapidly you make this approach the horizontal asymptote but these are just facts that you should be aware of okay so there's a horizontal asymptote which is y equals zero or the x-axis there's a vertical asymptote which is the y-axis or x equals zero the same is true for the function one over x okay now let's try graphing a variation of these fundamental functions such as this function let's say we have f of x equals one over x minus three plus two well let's see this looks to me like it's a variation of one over x what changes have been made well I subtracted three from the x that's going to cause it to shift three units to the right and I added on a two on the outside that's going to raise it up two units otherwise it will be just like this graph so let's graph it right here okay we just saw the graph of one over x and it was centered at the origin so that origin is going to move over three and up two now if you remember one over x has two asymptotes it has a horizontal asymptote which I'll draw in right here and it has a vertical asymptote which passes through that same center so I draw in these lines clearly as a tool to help me sketch the graph this is not officially part of the graph but it's there to help me graph it and you should show these asymptotes as well now from the new origin I'm going to go over one and up one and from the new origin I'll go back one and down one and my graph looks like this and this is the graph of the function f that you see on the left-hand side now while we have the graph up here let me ask you a few questions about it does this graph have an x-intercept? yes it does right here it looks like it has an x-intercept somewhere between it looks to me like between two and three how could I figure out where the x-intercept is? well the way you find x-intercepts is you let y equal zero now you might say but Dennis there's no y in this equation well y is expressed as f of x so I'm going to let f of x equal zero so if I put zero equals one over x minus three plus two then I'm going to solve this equation for x to do that I think I'll subtract two from both sides and now I think what I'll do is flip both sides over I'm going to flip this over and call it x minus three I'm going to flip this over and call it minus one half now you might have said you would multiply both sides by x minus three but that would be another way to solve this but continuing along my approach if I add three to both sides I get two and a half or if you prefer five halves so there is an x-intercept it's at five halves so the x-intercept would be at five halves zero by our calculation now on this graph is there a y-intercept? yes there is a y-intercept right about here and would like to know now how to find the y-intercept well to get the y-intercept you let x equal zero and if I let x equal zero I'll just plug in a zero into this function and f at zero is one over negative three plus two that's two minus a third which is five thirds so it crosses this graph crosses the y-axis at five thirds and that makes the y-intercept zero five thirds so we have that y-intercept and we have this x-intercept okay now let's take an example of a bit more complicated rational function if you remember we spent two episodes talking about rational functions first the fundamental functions we transformed but then we looked at more complex functions that might have slant asymptotes or holes in the graph so let's take an example of a function with a slant asymptote suppose we have f of x equals let's say x squared plus one over x now immediately I can see this graph is going to have a slant asymptote and of course if you weren't aware of this you may be wondering how do I know it has a slant asymptote it's because the degree of the numerator second degree is bigger than the denominator in fact it's one larger degree two over degree one so I'm going to divide x into x squared minus x squared plus one x goes into x squared x times and when I subtract I have a one left over x plus one over x so this is the same thing as one over x plus x you notice I'm putting the rational expression first and the x second now if this had been a constant this would represent a shift up and it would be a horizontal asymptote y equals whatever that constant is but instead the slant asymptote will be y equals x constant so when I draw my graph I'll put that slant asymptote in it so let's see y equals x is a 45 degree line so I'll just draw that 45 degree line through there there's also a vertical asymptote and where does this function have a vertical asymptote well the vertical asymptote is when I might be dividing by zero you see this x won't cancel out and therefore I'll have a vertical asymptote at x equals zero which is the y-axis right along here okay now you see what happens is my graph is going to be caught between the asymptotes either between this vertical asymptote and the slant or between the slant and this vertical asymptote down here that is my graph is either going to be drawn approaching the slant asymptote and going down or above the slant asymptote which one will it be? well if my graph comes down I have to have an x-intercept on the right-hand side does this function have any x-intercepts? well to find the x-intercepts I would let y equals zero which would say zero equals x squared plus one over x and this is zero only if x squared plus one is zero and that never happens there are no x-intercepts so my graph can't come down and approach the negative y-axis it has to stay above so it's going to turn here and go back up now of course the question is where do you make this turn should we come down further and go up or maybe we shouldn't have come down so far well this is only a rough sketch so I'll just indicate that there is a turn here somewhere on the other side where I have to cross the y-axis the x-axis so instead I'll have a turn and it goes back down so this is the graph of x squared plus one over x and you see it's quite important to know where you don't have x-intercepts as well as where you do have x-intercepts okay this is just one of a variety of more complex rational functions that we've discussed and you should be able to graph things like this let's go to the graph, a study list for exam 3 now on exam 3 we looked at fundamental exponential functions and then we looked later on in this exam at graphs of logarithmic functions and they are related let's look at fundamental exponential functions first of all if I were going to graph a function such as f of x equals 3 to the x then there were three target points for this function starting from the origin I should go up 1 and if I go to the right one I should go up 3 because that's the base 3 and if I go to the left one I should go up 1 third and then on that basis I should know the general shape of the graph and that is that it comes down and it turns and it levels off along the negative x-axis this function has a horizontal asymptote but it does not have a vertical asymptote now you have to be careful when you have functions with a negative in the exponent for example suppose I have e to the let's say e to the negative x this is the same thing as 1 over e raised to the x now now that I put this in what you might call standard form sort of like the graph over here I'm going to plot three target points again and at 0 I go up 1 if I go over 1 I go up the base and the base now is 1 over e that's about 0.36 and if I go to the left one I should go up the reciprocal of the base which is e so this number right here is intended to be e so my graph now looks like looks like this and this is the graph of g over here we have the graph of f those were two fundamental exponential functions let's go back to our list of topics for exam 3 you should be familiar with two compound interest formulas one of them is for discrete compounding such as compounding monthly or daily but the second compounding formula is one for continuous compounding and that involves an exponential base e I'll let you look those up in your text or look in previous episodes but now we come to logarithms in item c you should be familiar with what a logarithmic expression means and maybe we should just review that for a moment you remember a logarithm is an exponent so if I were to write an expression such as log base 5 equals 2 this means the same thing as an exponential statement 5 to the second power equals 25 so here I've given the expression in exponential form and here I've given it in logarithmic form now let me just ask you suppose we had the log base b of of let's say the square root of 10 is equal to one half so the question is what does b equal in this logarithmic expression right here well to figure out what is the value of b I'm going to change this to its exponential form sort of like I just changed this logarithm to its exponential form up above and writing this in exponential form it says base b raised to the one half power equals the square root of 10 but this is equivalent to saying the square root of b is equal to the square root of 10 and so b must equal 10 so b is equal to 10 so I've been able to solve for a variable in this logarithmic equation here by first converting it in the exponential form and solving it by more traditional methods ok now there are four laws of logarithms that you need to be familiar with and if we look on item on the graphic for exam 3 if you look at item d the laws of logarithms let me state those four laws of logarithms and then I'm going to add to it the change of base formula which is not one of the original four laws so the laws go like this law number one says that if you have the log base b of m plus the log base b of n when you add those together you get the log base b of mn ok now you probably remember the other three based on seeing this first one here law number two says the log base b of m minus the log base b of n is equal to the log base b of m over n law number three says the log base b of the quantity m to the c power is c times the log base b of m in other words an exponent inside a logarithm comes out as a coefficient outside a logarithm or vice versa if you're given a coefficient in front of a logarithm you can bring it in as an exponent inside ok law number four this is I think the most fundamental of the of the laws although it comes at the end of our list if you have b and you raise it to the log base b of m power the answer is m that is a log is an exponent so I place the log where an exponent belongs in the superscript and if the base in the log is the same as the base expression then m will be the answer will be a simplified answer to this expression now I'm going to add to this an item number five we didn't call it law number five in our previous episodes but we called it the change of base formula and it goes like this if you have the log base a of a number m let's say this is log base two like seven and on your calculator you don't have base two you can change this to any other base let's say log base b of m however you then have to divide by the log base b of the old base a so I take a ratio of these two logarithms and this is called the change of base formula ok now how would I go about graphing a logarithmic function suppose I wanted to graph the function f of x equals the log base four of x well the graph the graph of this function depends on being familiar with the graph of its inverse function which is the exponential function four to the x power and if I can graph this one I can graph the inverse of it which is the log base four of x and to summarize how you do it I'm going to plot three target points starting from the origin I'm going to go to the right one to graph the log base four of x if I go up one I should go over four and if I go down one I should go over one fourth and now I'm graphing a logarithmic function and this is the graph of f of x equals the log base four of x okay so if I'm familiar with the graph of four to the x when I flip it over a 45 degree line I get the graph of its inverse function which is the log base four of x okay let's go back to that graphic and let's go to item f exponential and logarithmic equations now I don't think we have time to work examples of both exponential and logarithmic but I think we can take one of those for example what if I wanted to solve the equation five to the x plus one power equals let's say twenty-eight let's come back to the green screen to solve this problem five to the x plus one power equals twenty-eight now this is an exponential equation but to solve it I have to use logarithms because I have to get my variable out of the exponent and I'm on both sides and I could take a log using any base but I would like to take one in which I could use a calculator so I would take either common log or natural log let's say we take a common log log log of five to the x plus one power equals the log of twenty-eight now if I bring the x plus one out in front as a coefficient this is using law number three the equation looks like this so that says that x plus one is equal to the log of twenty-eight divided by the log of five and that says that x is equal to this ratio the log of twenty-eight over the log of five minus one now we would say that this answer is the exact answer because I haven't had to round off using a calculator yet because on the final exam you don't have a calculator to use you're not allowed to use a calculator if I gave you a problem like this I would say give the answer in exact form and this is the answer in exact form but you wouldn't be able to approximate this answer without a calculator so I couldn't ask you to do more than this but it doesn't mean I couldn't ask you a question like solve this exponential equation you just will have to stop a little bit early okay and then finally you need to know on exam number three how to solve certain applications such as population growth radioactive decay or Newton's law of cooling you had at least one of these applications on your exam number three and we talked about each one of these in class quite extensively so you should be able to solve an application such as this but let's go on to exam number four okay now we're getting to material recent so I haven't left as much time to review exams four and five and what goes beyond exam five but let's go through this a little bit faster in items A and B here we have our two matrix methods for solving systems of equations systems of linear equations there's a Gaussian elimination with back substitution and then there's a Gauss-Jordan elimination now if you remember in both cases what we do is we set up the augmented matrix using the coefficients and constants from the system of linear equations and then we reduce it the augmented matrix by either getting zeros below a diagonal of ones or getting zeros below and above a diagonal of ones that's the primary distinction between Gaussian elimination and Gauss-Jordan elimination now you should be able to recognize systems of equations that have no solution and systems of equations that lead to infinitely many solutions and you should be able to represent infinitely many solutions then there are applications of systems of equations and the ones that I would ask you about would either be partial fractions I think we've done several partial fraction problems on TV and they're in your book and also traffic flow problems we had one at least one example of a traffic flow problem in a previous episode then we come to matrix arithmetic of course you should know how to add and subtract matrices how to multiply them by a scalar and how to multiply matrices together now let's just go to an example of matrix multiplication let me remind you of a subtlety for this multiplication suppose I wanted to multiply two matrices such as matrix A and let's say this matrix is one, two, four three, zero negative two now this matrix is said to be a two by three matrix it has two rows and three columns and then matrix B for matrix B let's say our matrix is five one negative four two, this is a two by two matrix okay now let's say we were asked to multiply A times B well let's say now A is two by three and B is two by two and you notice these two numbers do not match up the number of columns in A three is different than the number of rows in B therefore this product is undefined and this multiplication cannot be done now what if I reverse the order B times A well now B is two by two and A is two by three these two numbers match up the two and the two therefore this multiplication is defined and if I look at the other two numbers two by three that will be the dimension those will be the dimensions of my answer my answer will be two by three so it's going to have two rows and three columns let's see before I fill that in maybe I better write the two matrices down so I can actually multiply them let's copy down B and A B is five one negative four negative two and A is one two four three zero negative two okay now I'm expecting my answer to be two by three so two rows three columns and to get the number in the first row first column I multiply the first row times the first column and I get five plus three is eight and to get the number that goes here first row second column I multiply the first row times the second column and that will be ten plus zero is ten now to save time let me go ahead and fill in the other numbers in this matrix and you can verify these as I go or you can verify them afterwards the number in the first row third column will be eighteen the number in the second row first column will be negative ten what do you think is the number in the second row second column you should get negative eight and the number in the second row third column is twenty I believe okay so you notice that the multiplication in one order is not defined in the other order it is defined so this is a clear example of the fact that multiplication is not commutative A times B and B times A are not necessarily the same thing okay if we go back to our graphic for exam four item E no item F to find the inverse of a matrix now you remember the way we find the inverse of a matrix is we set that matrix within a larger matrix suppose I have a square matrix A suppose it's for example three by three A has an inverse what I do is I put A on the left-hand side and I put the identity on the right-hand side and then I perform elementary row operations so that I end up with the identity on the left-hand side and what appears on the right-hand side will be A inverse now if you're not able to get the identity matrix on the left-hand side it merely means that A has no inverse so it will not allow you to get the identity the procedure will not so that you cannot come up with a matrix over here on the right so if you can get the identity then there is an inverse matrix okay back to the graphic one more time determinants and Kramer's rule I will surely be asking you something about determinants on this exam you remember it's quite easy to evaluate a two by two determinant you just take these cross products and subtract them but if it's larger than two by two you have to expand by cofactors now which row or column would you normally choose to expand along and I think in most cases you'd want to expand along the row or column that has the most zeros in it so that you have to compute fewer cofactors and then finally Kramer's rule Kramer's rule allows you to solve a consistent system of linear equations by taking a ratio of two determinants okay so I think enough said for that let's go to exam number five okay now here we come to our most recent material so I think we can go this fairly quickly items A, B and C are the conic sections you should know fundamental equations for parabolas, ellipses and hyperbolas you should be able to determine their foci, their intercepts in the case of a parabola you should be able to find the directrix and in the case of hyperbolas you should be able to find their asymptotes I may draw the graph of a conic section in label, key points and ask you to tell me the equation of it I may give you the equation of a conic section and ask you to tell me is this a parabola or an ellipse or a hyperbola and then sketch its graph and then solve for other information about it items D and E and F I think go together this is on sequences and series you should certainly be able to distinguish between a sequence and a series and you should be able to distinguish between arithmetic and geometric sequences and series now there are two formulas that go with arithmetic sequences rather for arithmetic series and that is for adding up consecutive terms of a finite arithmetic series in the case of a geometric series there are two formulas one of them is for summing a finite geometric series the other one is for summing an infinite geometric series now let me ask you a question here what is the restriction on using the formula for finding an infinite geometric series and applying that formula and the answer is that R B of R has to be between negative one and one if you're going to use that formula otherwise the geometric series doesn't converge to any finite sum but it goes on to infinity there may be one or two exceptions for that but generally it's not going to converge to a finite sum okay and then in item G you should know the formulas for annuities the formula for computing the present value and the formula for computing regular payments for installment purchases now when we gave an exam on chapter 5 rather exam number 5 you were allowed to use a calculator to use these formulas on this final exam you won't have a calculator to use but I still may be asking you questions about the formulas I may ask you to write the formula down I may ask you to set up and solve for the annuity for the present value or the regular loan payment value and so you would have to set it up but you wouldn't be able to actually calculate it since you don't have your calculator available to you but it should not be taken that I wouldn't ask you about these these formulas now you know we've covered some information since exam 5 let's look at the last graphic okay first of all we talked about mathematical induction if you remember that is a procedure for proving infinitely many cases of a theorem we don't have time to go through a proof by mathematical induction but you might be asked a question about that also we've talked about the binomial theorem and related to that Pascal's triangle you should be able to expand a binomial raised to a power larger than a second power but not so large that it would take you too much time to ask you to cube raised to the fourth power or maybe fifth power let's say possibly a bit higher a binomial which could be a sum or a difference of terms and Pascal's triangle will help you to expand that and you should be able to compute individual binomial coefficients then we concluded with the fundamental accounting principle permutations and combinations let me just remind you of the formulas for permutations and combinations if you're computing if you go to the green screen if you're computing the number of permutations on n objects taken r at a time the formula is n factorial over n minus n minus r factorial permutations are problems in which the order of selection makes a difference now the number of combinations of n objects taken r at a time is factorial over r factorial times n minus r factorial know these formulas for the exam because you will be asked some questions about permutations and combinations and then finally we talked about problems dealing with probability and you should be able to solve some rather basic probability problems best wishes on the final exam it's been a pleasure teaching you this semester and we'll see you again soon