# UNR Math Center Review Session Math 126 Bagheri Exam 1

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### UNR Math Center Review Session Math 126 Bagheri Exam 1

because these aren’t actually labeled by section unfortunately i’m wondering i think that should be this page piecewise functions is is that section 2.4 yes okay piecewise functions yeah okay that’s reasonable to go on like graphing them um finding their domain and range let’s see if i can find a question that has a piecewise graph no no finding range i don’t understand the different types of range okay hmm okay having trouble with the isolated at the non-isolated values hmm i’m not finding any piecewise things on here it’s kind of close oh here we go perfect this this is definitely piecewise right i wonder if there’s if there’s another one yeah there’s a couple here okay perfect yeah so let’s start with here um so we have this graph f and it’s defined uh right they tell us it’s defined as follows um so it’s four plus four x if x is less than zero and it’s x squared if x is greater than or equal to zero um so right we wanna do all of these things we want to do all these things and below below this part is is going to be the answer so maybe i’ll try to cover that up for now until we can get an answer ourselves okay so we want to find the domain of the function um well if you um how would i find the domain of this function does anybody have an idea remember this this function is piecewise the infinity position yeah yeah right because um we don’t really have any domain restrictions in either of these two functions right um i’d have to be looking for like a square root or a like a denominator right for both these functions but both of them i don’t have either right i don’t have a square root or a denominator so and and it’s defined from like like to the left of zero and to the right of zero right so um the domain in this case would be in fact um how can i put how can i write this maybe i’ll put like equals negative infinity to positive infinity yeah there you go there’s that first one and now we want to locate the intercepts how can i locate the intercepts it’s all x and y yeah right um you’re basically solving for x and y when um when like the other one is zero right so if i’m looking for if i’m looking for like let’s say the x-intercept right in order to do that i would set y equal to zero right and similarly if i want to find the y-intercept i would set x equal to zero right so that would that would allow me to find the intercepts um in this case i feel like it might be easier to find the y intercepts so let’s do that one let’s do that one first right if we set x equal to zero

zero yeah um i would say one important thing to check is you want to make sure that this value um correspond or here let’s do this and write again you want to make sure that that the answer here right because because this answer came from this equation which comes from over here right so you want to make sure that that that this x equals negative one um fits with what is what is stated right here right is is if x is negative one is that less than zero right yeah right because so because of that i could definitely say that my one of my y-intercepts is going to be negative 1 0 right and then same thing on this side right this was my answer for this equation right so i just want to i want to make sure that it it it fits with what is stated here right is is 0 greater than or equal to 0 well yeah it is right so then we can say that another one of my x-intercepts is zero zero right and there’s that thing that i i said right we’re going to get um the origin we’re gonna get twice because it’s both an x-intercept and a y-intercept and that’s what we should expect once we see this once we should ex we should expect to see it again over here okay does that make sense does anyone have any questions on on this part all right so it looks like we have zero zero i’m just going to write that once and negative one zero so yeah they say you say the the intercepts are negative one zero and zero zero perfect all right hopefully that was okay um yeah let’s move on oh and now we want to graph the function yeah so this is probably a an area where we might be having some trouble so let’s graph this um there’s a couple different ways we could approach this um what do you guys like to do what does what does your professor like to do is anybody does anybody remember what your professor likes to do like how do they like to start this right i mean we know we know a couple points that are supposed to be on here right we know the the intercepts so let’s let’s let’s start by plotting those um i think oh yeah let’s do this in red so i know that there’s one point at the origin zero zero um and i know that there’s one point at negative one zero i’m not sure how big this is going to be so try and just make it um reasonable that’s at negative 1 right so that means that that point is at like negative 1 zero this point is zero zero so i know that those two points are on there what other points are how can i get like more points what am i gonna have to do where’s a good starting point for this guy anyone have any ideas we want to graph this right okay um yeah so there’s there’s a couple different ways we could start this um i think it’s reasonable to just like pick you know one of these to start with and just start by graphing it you know

um that’s that’s what i would that’s what i would do um let’s try this first one if we graph this first one right that’s that’s the equation of a line let me write it like this right this is like like mx plus b right so i know that my um i know that my y-intercept is at four right so if i go up 1 2 3 4. i know i’ll have a point there right and let’s see let’s not let’s not worry about this condition quite yet let’s just try to graph this like like if if we only had this what would this graph look like let’s let’s try and uh figure that out all right so um i start out i start out the y-intercept of four and then i use the slope from there right so i can go up four more units so i can go up one two three four more units and then over one unit right so i would have another point like up here i know that’s kind of out of the graph unfortunately maybe i can extend this right um and then i could also use the the for the slope going downwards right so i could go down one two three four units and then left one and then i have another point right here right and then i’d have another point right here which is why we’re getting this point right and then i and then i would connect those two or three that’s not good let’s do something like that oh i have a good idea there we go that looks good okay um right so this would be the equation that we get just from that first one just from this first guy right here now let’s think about the condition that x has to be less than zero right and and what i’m going to be doing is i’m just going to be erasing everything that’s not fitting that condition right so i know that x has to be less than zero so what that tells me is that um i i need to be on like the left side of of you know like the y-axis i need to be on the left of this so pretty much everything on this side is going to work everything that’s not here is not going to work so i need to erase everything else right so i’m going to erase this part of the line i’m going to erase that dot there um and if we if we kind of examine this closely i’m also going to erase this point right here i’m going to instead of making this a closed point i’m going to make this a whole because if we look at this condition x is not allowed to be equal to zero right x is not allowed to be equal to zero so what i need to do is take this erase the point and make it like a like a hole there is that okay right so the the graph doesn’t exist at that point we haven’t we have an open like singularity right there or something like that right so there there um there’s that top equation filled in right so um the way that i i approached it is i just started by graphing this and then i i erased everything that was not um fitting this condition basically so that’s that’s definitely one way you could do this you could also just if you don’t like doing like the erasing method you could just start by graphing this whole thing you know taking both into account um if you’re if you’re skilled at that then yeah you can do that too um both those are definitely great ways of doing it does anybody have any questions for this first part of the graph

this side and then a straight line going downward on this side where there’s a hole on the y-axis right so that’s that is in fact the same graph okay perfect so now we want to based on the graph we want to find the range right um i will i would think that you guys um have dealt with domain range already um if you recall domain has to do with like all the x values that you could put into a graph and get a answer out of it right so that means that range would correspond to um all of the y values right so now um instead of thinking about the domain which is if you think about it it’s kind of going this way right for that’s for domain domain right um but range is is just the opposite right don’t range i’m thinking about what are my values going up and down right that’s for range so with that with that and the graph what would you guys say the ranges of this graph does anyone have a guess no is range the difficult one or something um we could so um right so this these graphs like extend infinitely right so i mean this graph keeps going oh i mean that kind of looks weird this graph keeps going on like it goes it goes up forever it goes upward forever and it goes to the right forever right so because it goes to the right forever that’s why um that’s why we have our the upper bound on our on our domain right because because it goes to the right forever that’s where we’re getting this upper bound on our domain right um so and it’s it’s the same reason for our range right this graph goes up forever right so on our range the upper bound that we should expect is going to be infinity right because it goes up forever and then similarly for the for the other side um this graph goes right it just goes this way forever on the left which means uh it’s it’s kind of like the same thing it goes left forever and it goes down forever right um right and it the because it goes left forever because it goes left forever that’s why we have negative infinity as our lower bound on our domain right and it’s going to be the same reason because it goes down forever that’s why we’re going to have negative infinity as the lower bound on our domain on our range sorry is that okay so that’s how we get the range we just look at the the y directions of the graph is that okay with everybody so this would be our range okay so we have a question here i understand that part of range but how do we tell if there’s an isolated value okay yeah yeah yeah good question yeah that was the other part of your question um let me give you an example let me i think that that might be the best way to look at it um like an isolated value um i think i know what you’re talking about and let me let me see if we can think of something um okay so this question is done let’s let’s make sure that that’s our range that is indeed our range let’s check the answer real quick um d um okay so d uh they say that the range does not have any isolated values it is just this negative infinity to positive infinity yeah perfect so that that is correct um so yeah so now let’s let’s take a let’s think about like a different example here let me let

made another point like right here let’s say that my piecewise graph was something like this i think that’s what what’s meant by the isolated value right this one is just a point right it’s just a point so we would want to include that in our range right so if i add this point this range is no longer completely satisfying this entire graph right i need to include something in this range so i would have to um i’m not sure how to denote it um it might be like we union it with like six or something like that might be like um let’s move this might be like like this i’m not 100 um but i think that’s what what um you’re talking about when you say um isolated value right because this is like an isolated point right we want to make sure to include that in our range but nothing else we don’t want to include any other point is that cleared up yeah okay perfect yeah and actually if i do if i put the point there this is no longer a function um because it doesn’t pass the vertical line test if you remember right i have to draw like i have to be able to draw like a vertical line and only intersect one point so instead of putting the point there um let’s let’s move the parabola let’s move it to the right a little bit let’s still keep it at eight um but let’s move it to the right i don’t know like two uh let’s move it like three units so i’ll have a point here probably one around here probably one up here or something and then let’s redraw the parabola right so just so that i can move this point like like i don’t know over here right this is like two or something two and six right so this would still be a function this whole thing would still be a function because it still passes the vertical line test but yeah this is kind of clearly an isolated value now though and actually you might even see like the point like on the uh on the y-axis that would still be an isolated value still be an isolated value it would still be some kind of case like this yeah the range would actually stay the same right um moving the moving this part of the graph to the right is going to change my domain though right so you want to make sure that that you also have your domain correct right um let’s actually analyze that real quick if this is our graph what would our domain be so anyone i mean we’d approach it in the same way right we’d start with you know this portion of the graph we’d say we’re looking left and right now right so what’s the what’s the least it can be what’s the lowest what’s the lower bound right well it goes left forever so the lower bound would be um negative infinity right right and then when does it stop when it goes when it comes back to the right when does when does this part of the graph stop it stops at i’m not looking at four anymore i’m looking at the x direction so it stops um when it touches the y axis right and that’s at zero oh yeah someone’s got an answer infinity is zero union eight infinity yeah exactly um yeah i think that’s perfect zero right it stops at zero and then we union it with the other part of the graph h to infinity right because this part of the graph starts at eight or actually it starts at three in this case yeah yeah you you caught it you caught it it starts at three because i’m looking i’m looking at the x direction now right and it goes it goes right forever yeah so the only difference with those is you’re looking at the just different directions and um and one more thing is one last thing is um since i have a point here um i i actually would want to include zero in my um domain

so instead of being a soft bracket i think i’m gonna want a hard bracket here even though they they um the point kind of jumps it still exists as our domain it’s still like if i plugged in zero into this function right this graph i would still get a y value out of it right the only time where where you might um not do that is if i were to move that point right let’s say i move that point kind of drew it in a weird spot let’s say i move that point um that’s not what you want let’s say i move this point um back over here then it has that isolated value sort of thing again right so we would make sure to put this as a soft bracket again because we don’t have any any value on the y axis and then we want to make sure to like um union the domain with like like 2 or something like that i think that might be how you denote it um i’m wondering if this professor has any of those types of questions with isolated values let’s see um that one doesn’t look like it has that might be it for um the piecewise on this review but i i think i think we covered most of most of it honestly so i think it’s i think it’s okay does that make sense with everybody does anybody have any questions on you know any other questions on piecewise graphs i know we kind of went over a lot there so um yeah what are your questions okay okay so if there’s no questions um i think we’re good to move on um does anybody have like a specific like uh another like section or topic you want to go over right now um i know piecewise piecewise functions that’s a good one um are there any other ones that you guys are having trouble with okay so let’s clear that um if we don’t have any other ones we could let’s see um i think i’d like one of these questions down here um like the you have you guys dealt with um shifting compressing stretching reflecting oh do we have one can we go over something like translation yeah yeah yeah that’s exactly what i was about to uh go over yeah because this this one is another area of like error for students with the like the transformations right so yeah let’s try this one uh actually this one might be a little bit easy just because it only has one transformation i wonder if there’s a harder one we could try um i mean we could make up a harder one probably yeah i think we’ll make one up okay so for this one um okay let’s just let’s just go over the the general idea of 46 here um right so they give you the graph of y equals square root x right so that’s this graph here if you can’t see it maybe i’ll draw just a little bit larger it looks something like this that’s square root x right and the only thing that that has changed in this they’re asking us to graph um negative square root x right does anybody remember what the negative does to our graph negative out in front reflects right it reflects over the someone has it here over the y um let’s see so when i when i think about it um what i think about is like the original graph was was this right the original graph was this and