M1110 TB 2.3 Functions

so as we go on to topic free now this is called functions and I’m going to give you a little background here are the topics will be dealing with in this section but I think this might do very well in your notes as a general reference one of the words we’re going to talk about is relations and one of the best examples of a relation is an ordered pair and we’ve talked about that already now within the ordered pair we have the letter X and the X elements of any ordered pair make up what is called the domain and the Y element in any ordered pair makes up the range and then we have a special term in math which is called a function and that’s what this section is all about and we determine if something is a function if the X element in the relations is only used once and we’ll be seeing example of this a little later now one of the best examples of a function is f of X which is the symbol for function is equal to 2x plus 1 and you should also keep in mind that we said often we could substitute the f of X symbol with the letter Y so we’re saying y equals 2x plus 1 now you recall when we made a teach art and we said we could put various values in for X so if X is a negative 1 is 0 if X is a positive one so if X is a negative 1 then this becomes a negative 2 plus 1 gives me a result of a negative 1 now if x is 0 2 times 0 is 0 this gives me a positive one now if X is 1 2 times 1 is 2 plus 1 is 3 so these are skills you may already know but it’s showing how functions work and we have used these before so what we’re doing is we’re inputting a value for x and we get an output of a value of y so this value in this value out now what they’re talking about to in our text here is that there’s a dependent variable and an independent variable so we have the variable Y and we have the variable X well when we put a value for x here that is our independent variable and when we put that into the system and get this output this Y value is a dependent value or in this case variable ok so again some of the definitions and then as you read through this they’ll tell you that as we said before many items in our lives can be related as an ordered pair whether it’s the price you pay for 3 gallons of gas or any number of other things ok we’ll move on but again you should have this in your notes and should be studying and again they’re reviewing an independent variable was X the dependent variable was why so if you buy three gallons of gas this is what it’s going to cost if you buy eight gallons this is what it’s going to cost now when we see items in an ordered pair like this this is considered a relation because the x and y are tied together now there is a special relation that we call a function

and in a series of ordered pairs a function is such that the X element is only used once in the series of relations so let’s take a look at this an example one it’s to decide whether these are a function or not well let’s take a look at them so here we have a one we have a negative two and a negative three are the X elements repeated know so yes this is a function here we have a 1 111 repeated so this is not a function a negative 4 negative to a negative to the negative 2 is repeated it’s not a function now sometimes you see it shone like this where these are your X values these are your y-values now notice from the X values there’s only one line going from each value so this indicates that these items are only used once so this is a function now in this case we see from the negative 2 which is the x value two lines so that indicates that’s used twice so this is not a function now we’ll also see it in drafts from time to time as you look at that and you graph it notice the X is only used once so this is a graph of a function now a little later we’re going to show you what is called the vertical line test in fact i’ll show it to you right now now in the vertical line test we take a vertical line it’s an imaginary line that we move over the face of the graph now as we do this does this vertical line intersect our graph which are dots here in more than one place so as it goes over this one only one place it goes over this one only one place goes over this one only one place so yes this is a function now here again they’re relating a slope intercept formula to a function where your dependent variable results from putting some value for the X here as you put in some value for an X let’s say a 4 is what they’re showing then 2 times 4 is 8 and 8 is the output so we have this as the input with X the independent variable and then your Y comes out of the system now to tie this in to to graphing let’s say if we have a graph and we make a graph of this where our y-intercept is 0 we put a dot at 0 the slope is 2 over 1 so we go over 1 up to and then we make a line through that and in a sense this is the graph of this function we can change the wide f of X now does this line pass the vertical line test well as we move this line over the face of our graph does it intersect it at only one place as it moves across and the answer is yes so this is a function and that’s what they’re saying here as we go on they’re introducing domain and range which we’ve defined already the domain is the X element the range is the Y element so if you are asked to find the domain and range of a relation here are your coordinates I

would suggest all you do is start to list this is your domain item domain item domain item domain item and how would you list it well we call it roster notation I always suggest it as you keep notes you’d put addy for the domain put braces and I’ll just list your elements 34 now if the four is repeated you only write at once and six now since this is repeated is this relation a function and the answer is no now here you would do the same since these were your ex elements you would just list them in roster notation and then for the range again I would put an R and braces a negative 12 negative 1 comma 2 5 and a nun are repeated so that’s how you would write it now in in this case again these would be the same this by the way is a function because these elements are only used once here this is a function then you might say look at all these twos are repeated but those are the range elements it’s the domain elements that cannot be repeated it’s a relation but it wouldn’t be a function if it was repeated this is a relation but not a function because the fours are repeated ok and they give you the material in attack now in example 3 they’re asking you to look at these graphs and then pick out the domain and range and then give them I believe in interval notation so or whatever method they want you to use so let’s take a look at this first one here so again your domain elements would be a negative 1 0 a 1 and a four and the range would be usually we start with the lowest one here a negative for a negative 1 a1 and a2 and I believe you would put that in roster notation now for this one here we have a circle so the elements that are part of the domain again this would always be reflected onto the x-axis so it would start and it looks like four is included in the set so an interval notation for the domain you would put a bracket indicating that this negative 4 is in the set and it goes on to a positive 4 and the range and again you reflect the circle to the X elements I’m sorry the Y elements in this case starts at a negative 6 and goes to a positive 6 and the brackets indicate that those numbers are included in the set now here we’re looking for X elements and notice this line goes on to infinity in this direction so interval notation for the domain RX elements this would go from negative infinity to positive infinity and the range the same from negative infinity down here and eventually goes up to positive infinity I’ve had students when they wrote this instead of using a comma would use a period and then math lab counts that wrong but I would look at it and give them full credit now here’s a parabola notice that eventually this is going to go all the way out here to negative infinity for the domain and this would go out here to positive infinity but for the range it looks like it’s starting at three now it’s hard to tell whether 3 is included in the set the negative 3 so let’s

suppose it is then it would be a usually they would put like a dot there to indicate that so that’s a little tricky there you would say a negative 3 if it’s included it’s not included it would be a parentis e and this goes up to positive infinity up over there okay we’re going to look and see what the book tells us here in a moment too I’m just interr so we were good this is shown in roster notation this one is shown an interval notation and these numbers are included because of the brackets infinity and infinity that was good and then use of a bracket here 3 negative 3 is included okay now in this area where they’re asking us to look at agreement on domain we think of the number line and all the numbers that are on the number line are real numbers so how about the case where if a value such as let’s say 2x plus 3 is over 2x and we’re asked to determine the domain which would be the X values of this example well we have to keep in mind here and that’s what they’re referring to an agreement on domain that we can never have a zero for a denominator because that’s not in a real number system where we have a zero for a denominator we say anything with a zero for the denominator is undefined so are there any values of x here that would make this denominator a zero and the answer is you might say well how do i find that will just equal the denominator 20 and when we do that we get x equals 0 so if x were 0 that would make this denominator a 0 so we then have a restriction to what values we have for our domain in order for it to be a real number so we say that often in set-builder notation will say X is such that X is any real number but X cannot equal 0 so here you have a qualifier to your domain and this is set builder notation now you might say how would you put that in interval notation well again I’m looking ahead maybe hopefully it’s their interval notation that would come from negative infinity up to 0 but a parentis e that 0 is not included in the set and then we use a union symbol to indicate we’re linking it to the other side where it’s going to be the other side of 0 and positive infinity so an interval notation that would be the answer to this one so this is what they’re saying agreement on domain that X could be any real number but there are sometimes exclusions as to what X could be and we would find that always by taking our denominator equal to 0 solve the equation and that gives you the values that have to be excluded and I see here they’re introducing the vertical line test okay we’re just a little early and again imagine it’s a vertical line that you are going to pass over the face of the diagram over the graph and if it only intersects in one place then it is a function now here in the case of a circle it’s intersecting there and they’re so circle is not an exact of a function and they’ll give you many

more to look at too as we go on so here’s a whole bunch right here so this is when we had seen earlier and in fact I think I showed you this that again this is a function this is not a function this is a function think our line went up this way so the same and this is a function okay and the reason their functions because they pass the vertical line tests which states that can only go through in one point of intersection as it goes over the face of the graph if it goes over to it’s not a function or more ok let’s go on okay now we get into the meat and potatoes of using what we’ve built on so we have to decide whether each relation defines a function and then give the domain and range well one of the things you might do is if you have a graphing calculator is to put it in a graphing calculator and see what the graph looks like or if you remember your days from earlier courses where we made a graph of this y equals we put a dot at the y intercept which is for our slope is 1 so we go over 1 up 1 and this is the graph of that line now does this pass the vertical line test yes so this is a function so this does pass the vertical line test and therefore this is a function now what would be the domain of this well we had said already it’s going to be from negative infinity to positive infinity again you have to be careful in math lab you put the negative in front of the infinity sometimes students forget that they count it wrong I would usually give credit because it’s in the negative position and there’s a comma so but again put it right for best results now for letter B we had said that in order for this to be a real number under the radical sign this has to be 0 or greater it cannot be a negative number because then you would have imaginary numbers which we said we’re not using in this or at least at this time so how do you find the domain of this while you take away the radical sign and just equal our denominator or this radek add to 0 so we get 2x equals 1 then we divide and that would be greater or equal to 0 and then we get x is greater or equal to one-half so for our domain then you can’t have a number smaller than one-half because that would make your redic can hear a negative value so our domain then would be and this would be a bracket one half comma to positive infinity and how did we find it well we found it by saying you can’t have a negative value under here we equal it to greater or equal to 0 that would be our lowest number we could have and we saw what would make that and there’s our item right there and they’re going to tell us about that too so let’s go on okay you know for letter C we have here we solve for y by taking the the square root of both sides and we get y equals this and when we go to graph it the plus values go up here the negative values go

down here this gives us a sort of sideward parabola now is this a does it pass the vertical line test no and it’s not a function now the domain and range of this would be that it would start at zero for the domain and go to positive infinity and then for the range this eventually comes up from negative infinity to positive infinity ok let’s go on and again all the answers are down below so you could look at that too i’m just giving you a little interpretation of it so when you did a graph of this would there be any restrictions to the x and let’s look at what they say so as we go through these we had all of these correct and they are mentioning that the radicand cannot be less than zero so they’ve solve that the way we had and in letter C we did get the domain for the positive values and the negative values and got the graph of this it’s not a function and here they’re saying there is no restriction and it would be all real numbers for both the domain and range now here’s a case that I mentioned a little earlier that where we have something in the denominator and let me review that with you one more time if we said that we take our denominator equal it to 0 and we get x equals 1 well here if X where r 1 it would make your denominator is 0 so our value for our domain in a case like this is that it could be any real number but cannot be a positive one and let’s see how they’re showing that because i had mentioned away earlier and whether they’re going to use interval notation or not yes this is what they’re going to use so the domain would come from negative infinity up to 1 but not including one and then with Union on the other side starting at one just beyond it to positive infinity okay and the range applies the same way because in this case it’s 0 though let’s take a look at that now because this original denominator was a negative 1 then 1 is excluded so in a sense this is your vertical asymptote and then horizontally when we go looking for the other value it does it cannot be zero so our range is from negative infinity up to but does not include zero and then starting on the other side of zero going up so this one’s a little tricky so again study what they have here and you know that’s what you’ll be checked on in your practice and quiz me now keep in mind all of these fit within the definition of function and if I were you I would write them out in your class notes or your homework log or whatever so that you’ll have it as a reference for you okay we’re now going on to function notation many many of you have

had this introduced in an earlier course that we can always substitute the letter Y for a function notation symbol this is referred to as f of X now this parenthesis not mean multiplication as it usually does but it’s just part of the function notation so we can change from slope-intercept and its standard form to function notation and throughout this you’re going to be given values for this X here and that would be your independent variable that you then put into your expression and then work it through to get the output which would be a y value or the value of the f of X in that particular situation so let’s take a look at some of these and often i mentioned to my students to make a little teach art so even though they’re not showing you that technique you’d put your ex there and then just put f of X here keeping in mind that would be a Y and then they want you to deal with the tomb and then they may want you to deal with other numbers here in this case we’re just starting with a two so you just put the two there this gives you three times two is six so this would be a one I like this because it helps to organize your data and I think very useful on your blue sheets when you’re doing a test and the teacher can see specifically what you’re doing so they started us off easy and now they’re giving us some more challenging items part of it is to understand what they’re asking here they’re saying let f of X equal this function but then they’re also showing G of X is this function so here for the F function we would put the two where the X is in there and the later we’re going to put a cue wherever the X is and they show this substitution down here and then for our G function we’re going to put a plus one where the X is and then just work that through as you see here ok and then again in this case maybe a t-chart might not be appropriate but you’re going to do them as different examples so here this was the easy one this one came out as three and this one will then come out as an expression where you’re just substituting the cue for the X and then here you’re putting in where the X is the a plus one the a plus one and then just multiplying through and ending up with an answer that we can’t quite see there let’s take a look at it there we go so by the time you group like terms and everything you end up with this and you’ll be getting some practice ok now for something like this for each function fine f where the x value would be three so some of this is straightforward and we can use what they show we’re just going to take the three put it where the X is nine nine minus seven is two ok and then in the case of B we’re looking for where X is 3 and we have to look where X is 3 which is this one so why is going to be a one ah so something a little different and then here where F is three right there it’s

going to be five and then on this graph where F is three well we look let me make a little picture of it for you here there’s where it’s three we go up here to right there and what is our ordered pair there well it’s three and that’s going to be up for so our Y value is 4 so a number of interesting ways in which we use function notation where they give us part of it and we have to determine the other remember they’re giving us the independent variable we have to find the dependent variable from that information whether it’s an equation series of ordered pairs sort of like a Venn diagram and then a graph ok let’s go on see what they show us we did the first one and there the others there’s our five and there’s our four okay let’s take a look at this well here they’re just asking us to find an expression in our funk using our function symbol so the key is you solve the equation for y and then replace the y with the function symbol so this one’s pretty easy because the Y is already there you’re just going to replace that with the function symbol in letter B we have to actually solve this equation for y which we did in earlier lessons in other courses put it in slope intercept form and then substitute the Y for our function symbol ok I always get for the end of this lesson we’re seeing we want to determine whether something is increasing decreasing and whether it’s a constant function well again in our language we go from left to right generally so as we go here and this blue line it’s increasing as we go in this reddish line is constant and as we go down it’s decreasing so really good example of what’s going on here so illustrate these ideas here is increasing here it’s decreasing and here is constant now as you look at this parabola remember going from left to right it is decreasing here then is constant for a moment and then starts increasing and let’s read what they say here the concept of increasing and decreasing functions apply to intervals of the domain not to individual points all right that’s interesting and now we’re taking a look at something here determine intervals over which a function is increasing decreasing or a constant well I’ll let you look at this in this first one as we go from left to right this is coming from negative infinity so that’s decreasing up to 1 but notice that one has a circle on it so we use a parentheses and then it’s increasing from one which is a bracket to indicate one is included in the set up to notice we’re reflecting on the x-axis three and then from three again bracket because three is included in that point up to infinity which is a constant alright and on this word problem I’m

going to let you read it because the tape is running 40 minutes wow hopefully it’s helpful for you give you some background and I know you could read your textbook yourself but this might be useful all right we’ll wrap it up