 Welcome back to our lecture series Math 1050, College Oddsworth for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In lecture 46, as we end up our series in chapter seven about precalculus, I want to review the idea of domain. Now, the way that one determines the domain of a function has a lot to do with how that function is represented. And we've seen many different ways of representing functions throughout this lecture series. The one you see on the screen right now is what if we just represent the function in some type of numerical format? What if we just know specific points like when x equals 10, f of x will equal negative 12. When x equals 14, f of x equals negative six. Or in other words, when you take 18, f of 18 is negative two. What if we have just some type of numerical calculation? Like we have a scatter plot. This is just sort of like a little baby scatter plot. She's so adorable. But if we just have this, right? And this is oftentimes how a function gets started. We collect data through some type of observation or survey or poll or what have you. We collect data. And then we try to determine an algebraic formula from that. It's very common to represent a function using this type of numerical expression. So what do we mean by the domain and range of a numerically given function like so? Well, recall that for a function f, its domain is the set of actual input values. And the range is the set of the actual output values. So if your function is given as a scatter plot or table or some type of numerical collection, then the domain is very easy to identify. You just look at what are all the x coordinates of the actual points inside of my function here. We get that the domain of our function f is going to simply be, well, okay, we have an x equals 10. We have an x equals 14. We have an x equals 18. We have an x equals 22. We have an x equals 26. And an x equals 30. That's all there is to it. And if you want to know what's the actual, you know, what's the range or the actual output values, well, we look at the y coordinates right here and we see that the range of our function f right here, well, what points do we get? We get negative 12. We get negative six. We get negative two. We get one. We get three. We get eight. And so this right here then gives us the domain and range. And so when you have that scatter plot or this table, a tabular representation of a function, finding the domain and range is really just coming down to reading it. Like what are the actual points used? What are the actual x coordinates? What are the actual y coordinates? Well, once you have a data set, you know, so maybe we have our scatter plot that looks to mean like something like this, maybe, again, just hypothetically speaking, maybe we have some type of scatter plot like this. Well, once you have your data, it might not fit a perfect function, but then we can use some type of statistical regression method to find a curve of best fit. And maybe we get something like the following. So then we move from this numerical representation of the function, perhaps to a geometric or a graphical representation of the function. So our function not given by a formula, but is given by here a graph. Okay, so if we have a graph, what can we find? How can we find the domain and range? Well, again, the domain and range are going to be the actual input values, the actual output values of the picture. We're following the usual convention, the horizontal, which we typically call x, will be our input, and the vertical variable, typically called y, will be our output. So as we look along this graph right here, the farthest left x value you can find is right here at one, two, negative three. And as we go through the graph, we get every point along the x-axis up until the point x equals three. And so then this indicates to us that the domain of our function f this time would be the interval negative three to three. So unlike the previous one, which we actually had a finite number of points for our function, this one, because of its continuous nature, we actually get an interval of points inside the domain. So we'll write this in interval notation negative three to three. Now we are going to use closed brackets because notice when we see this filled in dot, that means the point is included there. So there is a point whose x-coordinate is negative three and there is a point whose y-coordinate or whose x-coordinate is positive three, excuse me. To find the y-coordinates, that is to find the actual range here, this one gets a little bit more funky the way we draw it, but we basically want to ask ourselves, well, how far down the graph can we go? We can come all the way down to negative two for which there are points, right? You have this point right here, negative three negative two. This is a point that actually realizes negative two as its y-coordinate. We also have this point over here, two comma negative two. The fact that there's a multiplicity does not change the fact this is still a function. I mean it's not a one-to-one function because it does fail the horizontal line test, but that's not a requirement we need right now. So the smallest value we find in the range is going to be negative two and then what's the biggest value? We can go up up up up up up here to positive two. In fact, we have the point negative one comma two and so then the range will be negative two to two that we can see right there. And so when we, of course, have this graph, we can look at the graph to see what are the actual x-coordinates or the actual y-coordinates we have here. And so let's see the progression we had so far. We went from a data-driven that is numerical representation of the function, which has a limited domain and range. Then we can actually progress into a graphical one by statistical techniques of regression. But really our ultimate goal is going to be moving on to the algebraic representation of a function. So the most interesting consideration of course are these algebraic representations. That is we want to come up with a formula to represent the picture that we drew previously. Now according to the domain convention, the domain of a formulaic function f is going to be all real numbers x such that f of x is also a real number. So we need a real number coming in and a real number coming out. That's what our domain convention has been all about. Now there are some stipulations why we might have to avoid this, right? Sometimes we have to specify the domain either implicitly or explicitly. Sometimes we might say that, oh, it doesn't make sense for x to be a negative number. So therefore we might restrict the domain to be non-negatives. This is very common in a story prompt situation. That's our fourth representation of a function, some type of verbal description, right? So there might be sort of real life stipulations that say, oh, f of x can't be that or x can't be this. Another stipulation we have to come up with sometimes is, you know, if we're doing like population growth, we can't have half of a person or we can't have three quarters of a bacterium or anything like that. So it might be that the domain, excuse me, the range actually has to be restricted to only whole numbers and rounding can compensate for those things. So there are sometimes real life stipulations that will require that we restrict beyond what the domain convention's already saying. But the domain convention's like, hey, we'll assume the domain to be as big as possible unless there's something else that tells us to restrict it. So we take our formula and we're going to say we'll accept any real number x so long as f of x is a well-defined real number. So we've studied many, many functions in this lecture series so far. We've talked about linear functions, quadratic functions, absolute value functions. We've talked about polynomial functions, rational functions, piecewise functions, radical functions, exponential functions, logarithmic functions. And that's of course just sort of the tip of the iceberg. These are some of these algebraic and transidential functions we've been studying. In a trigonometry course, such as Math 1060 at SUU, you can talk about some of the trigonometric functions like sine, cosine, tangent, arc tangent, arc cosine, arc, what did I just say? Well, just to list a few. We don't need to list every single one of them, right? And of course, we can also kind of combine these functions together, right? What if we take these functions, you know, power functions, exponential functions, they're inverses like radicals and logarithms. What if we start combining them together using the operations of addition, subtraction, multiplication, division, composition, so we could put functions inside of them. We could also use piecewise amalgamation. That is, we could take like one part right here and then break it up to another part over here. Those are options as well. And so we can build functions using these different families and these different techniques, right? If you throw in, if you throw in trigonometric functions into that mix, this gives us the family of so-called elementary functions. You take the trigonometric, the transcendental, the algebraic functions all together and combine them in all these different algebraic ways. You get what's called in calculus the elementary functions. Now in this lecture series, we won't be using any of the trigonometric functions, sine, cosine, tangent, etc. But we will stick with like log the algebraic and transcendental functions of exponentials and logarithms. If we have a function that's given in this algebraic sense, how do we determine what the domain is? Well, by the domain convention, we just look for the problems, right? What can cause certain types of discontinuities to the graph? And it turns out for all the functions we've studied and all of these ways we can create new functions from old functions, there's only three real problems that we have to look out for. So the first problem is division by zero. If you have a rational expression of any kind, it could be a rational function with a denominator, numerator, polynomials. But hey, the denominator could be a square root, it could be a logarithmic, it could be whatever. Whenever you divide by some variable quantity, we have to be concerned that a choice of x could make the denominator go to zero. And therefore, if you divide by zero, that will not be a well-defined real number. I mean, it's probably will give you a vertical asymptote because your function will probably be approaching infinity or negative infinity. That's not a guarantee because you do have removable discontinuities. But division by zero does cause problems with the domain. So whatever values of x make the denominator of anything go to zero, we have to throw it out of the domain. Okay? And I should also mention that when considering the domain of a function, we should always use the original expression, which is not simplified, or manipulated by factor and everything like that, because the original expression needs to be defined, not some simplified version. So things like the following of y equals x minus one over x minus one, the temptation is to simplify it and get, oh, this is just the constant function one. Well, that's almost true, except the original expression is not defined when x equals one, because when x equals one, you get zero over zero, which is not the number one. It actually is not a real number. It does not exist. So you have to make sure that when you consider domain, you always use the original unsimplified, unmanipulated form, because sometimes you get things wrong, otherwise, if you try to simplify it. So finding is a good thing, don't get me wrong, but you need to remember the exceptions to the domain that existed in the original form, because x minus one divided by x minus one does not always equal one. It equals one only when x doesn't itself equal one. All right? The second problem we have to look out for is taking the square root of a negative. And when I say square root, I of course mean any even root, so like a fourth root, a sixth root. And that's because if you take like the sixth root of negative one by exponent properties, you can actually factor this as the cube root of the square root of negative one. So whenever there's an even root in play, you really have like a square root in play. So we have to look out for square roots of negatives. And that's because, again, our rule is we have real numbers going in and we have real numbers coming out. Dividing by zero does not produce a real number. It actually produces, if anything, like an infinite number. That's not a real number. Infinity and negative infinity are not real numbers. When it comes to square roots of negatives, right, you have a real number coming in like negative one, but the output is not a real number. You get an imaginary number in that situation. So for our functions, we can't be taking square root of negatives. And the same problem basically happens when you take the logarithm of a non-positive number. If you take the logarithm of a negative number, so you take like the natural log of negative one, this will actually produce an imaginary number like pi i, which we don't have to worry about what that is. But be aware that taking the natural log or any log of a negative produces imaginary numbers much in the same way taking the square root of a negative does. And then when you take like the, if you take like the natural log of zero, that actually produces negative infinity, so to speak. And that's because when you get closer and closer to zero on the natural log or any log without transformations, that's going to be a vertical asymptote. So it's kind of like dividing by zero. So logarithm, the logarithm of a zero and a logarithm of negative is kind of reproducing these problems right here. We don't want, we don't want vertical asymptotes. We don't want imaginary numbers inside of our graphs. So what I want to do is that I'm going to look at a list of functions and we want to analyze their domains based upon these three conditions right here. What makes a denominator go to zero? Do we ever take the square root of a negative and do we ever take the logarithm of a non-positive? Those are the only three problems we really are going to see when you are working with these algebraic functions because when you add together two functions, you, the only restrictions you have are the restrictions that were already there. Okay? If you subtract a function, two functions are, you multiply them as well. The only restrictions to the domains that the difference in the product will have will be those restrictions that already existed. Now if you take two functions and you divide them like f of x divided by g of x, you'll have the restrictions you already had but you could also introduce new restrictions based upon dividing by zero but hey, that's already in our list so we don't have to worry about that. When it comes to composition, when you put one function inside of it, so you take something like f composed with g, well if g has any restrictions then the composition will still have that because g had some problem with division by zero or square roots and logs. We put that inside. Those are still going to be there or we have to also make sure that what goes out of g fits inside of f but again that's because f either was divided by zero taking square roots or logs or whatever so you'll be able to see that very quickly. The last one when it comes to piecewise functions, piecewise functions could have restrictions to their domains that go beyond this list but by nature when it comes to a piecewise function you have these different pieces. You have to specify on this piece here's the domain, on this piece here's the domain, on this piece here's the domain so since the domain is actually explicit for a piecewise function that really doesn't cause us an obstruction whatsoever so it's just these three principles that one has to worry about and even still when you start introducing trigonometric functions like sine, cosine, tangent the only problems they ever have is actually division by zero because tangent is like sine over cosine so that's not a new problem so really then in trigonometry the only thing that they worry about is like arc sine arc cosine those inverse trig functions can't have restrictions on their domains which we would talk about that in a trigonometric setting which we're not going to do that here so let's look at our functions let's find the domain of the following algebraic functions assuming they're as big as possible what are the problems we see here well looking at the function we have a rational function f of x equals 49 minus x squared over 7 minus x all right it's a rational function there's no square roots there's no logarithms but there is division right so we have to see what makes the denominator go to zero so when we solve that equation 7 minus x equals zero we can add x to both sides and we see that x well 7 equals x so that's what makes the denominator go to zero so the domain of f is then going to be all real numbers x such that x does not equal 7 that's the only problem which oftentimes you're asked to put this interval notation so we're going to get the interval negative infinity to 7 union 7 to infinity for which we don't we put it we never put a bracket next to infinity or negative infinity because those are not real numbers so they're not included the domain we also don't put a bracket by the 7 because we don't include 7 7 is actually the exception there let's take g of x this time to be the square root of x minus 3 there's no division there's no logarithm so no worry that's there but we do have a square root right so the radicand the expression inside of the square root needs to be positive so we take we have to solve the inequality x minus 3 has been given equal to zero now the difficulty of this inequality comes down to the complexity of the radicand the more complicated the radicand is the more challenging this inequality could be this is just the linear inequality we could add three to both sides and get x is greater than equal to greater than equal to three and so therefore the domain of our function would then be we want all real numbers such that x is greater than equal to 3 or an interval notation this would look like bracket 3 to infinity so we want all numbers greater than 3 but we also can allow 3 itself because if you take g of 3 right here you get the square root of 3 minus 3 which is the square root of 0 this is a well-defined number that's just 0 itself now I do want to point out to you that if this problem was slightly changed if you took like the fourth root of x minus 3 this would make no difference on the domain the domain would still be 3 to infinity on the other hand if you took g of x to be like the cube root of x minus 3 in that situation the domain would actually be all real numbers all real numbers here and that's because odd radicals have no restrictions it's only the even ones we have to look out for so let's consider the function this time h of x equals the absolute value of 4x plus 2 so this is an absolute value function which if you want to you could think of it as a piecewise function right where this looks like 4x plus 5 when you're greater when you know when you're positive right or it's like negative you could try to break it up in this piecewise function but really it doesn't you don't need to do that absolute value is a very special piecewise function that again you can't benefit from think about the piecewise function but you really have to do that often basically here think of this in terms of composition right this function is going to look like the absolute value of u composed with the linear function 4x plus 5 so that as we put the linear function inside of the absolute value now linear functions have no restriction on their domain so their domain here so for the first one it's going to be all real numbers right and as for absolute value absolute value also has as its domain all real numbers and so as both functions in the composition here have domains as real all real numbers there's no problem right the operations in play are addition multiplication and absolute value none of those are restricted so the domain of h here is going to be all real numbers like so so oftentimes when you look at a function there are no restrictions they get the word about there's no division by zero there's no square roots there's no logarithms we can move on and be happy that the domain is all real numbers so how about this one right here f of x equals x squared times two to the x plus the natural log of x minus one so searching through this function we have a power function that has no restriction on its domain we have an exponential function two to the x it has no restriction on domain um we have a logarithm aha so notice here you have actually two functions in play you have a function plus a function like we mentioned earlier when you add two functions together you just look at the restriction of each of the terms in the sum the first one has no restriction so no big deal we look at the natural log of x but x minus one excuse me which much like the square root in order for this thing to be well defined real a well-defined real number we need that x minus one has to be positive so we saw the inequality x minus one is greater than zero this is of course is the critical difference between square roots and logs here the square root while you could take the square root of zero the square root of the natural log of zero is not a real number so we actually need a strict inequality but then you solve it you add one to both sides you get x is greater than one and so we see the domain here of f is going to be all real numbers x such that x is greater than one or an interval notation we get one to infinity so we only look at the pieces that have restrictions the ones that don't we can kind of move on from them we don't have to worry about it all right let's look at example e right here so this one's got a couple issues right so we notice the numerator has the sixth root of x squared minus one because this is a sixth root this isn't even root so this suggests to us a potential problem that we could be taking the square root of a negative so we need to solve the inequality x squared minus one is greater than equal to zero equal to zero is okay here another concern we have to deal with is the denominator right we have this one plus two x e to the x right here if there was a choice of x that makes this go to zero we need to know it so we need to solve the equation one plus two e to the x equals zero so we solved both of these here let's start with the inequality the inequality it's a quadratic inequality which we can the right hand side's already zero that's great about these ones factor the left hand side as a difference of squares you get x minus one times x plus one is great equal to zero if we think about this graphically right x squared minus one looks like a parabola uh that's been that moved down by negative one so you're gonna have two two markers here uh you have plus one and negative one where is it above the x axis the it's gonna be above the x axis when you're to the left and to the right of these values so you take the wings of the bird here so this suggests to us that the domain restriction from the numerator is that we have negative infinity to negative one bracket union bracket one to infinity that's what the sixth root says that the sixth root will help you find when you're between negative one and one uh what about the division by zero what could the denominator go to zero well as you start solving this one here let's let's minus one from both sides of the equation right that then gives us two e to the x is equal to negative one divide both sides by two we end up with e to the x is equal to negative one half now some of us might have already detected the issue right here an exponential function can't equal a negative that's outside of its range for which that would then be enough to tell us that there's no solution to this which if we didn't see that we could keep on going the next thing to do would be to take the natural log of both sides in which case you then get x equals the natural log of negative one half which that should then cause us some alarm especially if you throw in our calculator because this is not a real number this is not a real number which we should have that very much in our minds right now because we're looking for domains we know the natural of a negative is not going to give us a real number so this tells us there's no solution so this equation one plus two to e to the x equals zero has no solution there is no value of x that'll make the denominator go to zero so the denominator actually provided no restriction it could have we didn't know that but it turns out there's no restriction and therefore the domain of g is going to be negative infinity to negative one union one to infinity that the sixth root provided an obstruction to the to the domain but the denominator which potentially did it didn't know and that's the thing is we have to investigate that we have to look at all allegations of domain irregularities but sometimes those allegations turn out to be false the denominator really didn't go to zero so be vigilant in your search there and then let's look at one last example here to illustrate what's going on here so we have h of x equals the square root of x plus one divided by x minus three minus one over three minus the natural log or the log base two of 16 minus x oh boy there's a lot of stuff going on here there are two pieces there's this square root piece and there's this division by a logarithm piece because we have a subtraction of two functions we have to look at their domain separately so let's look at the first one right here if we pull that out what problems do we see well inside of the square root well let's just stop there for a second the square root hey do we have to take the square of x plus one over x minus three so x plus one over x minus three that needs to be gradient equal to zero because if we take the square of a negative that's not going to work but we also have the concern right that we have a denominator this x minus three this x minus three here needs to be non-zero x minus three we need to figure out when that's equal to zero so that's that that equation is easy to see here right x equals three is the problem so we need to make x not equal to three right but then when we come over here back to our inequality we can think of this rational function if we were to kind of graph it really quick and we don't need a very good graph it's a very quick one will work so we have a vertical asymptote at three that's something to remember the numerator is going to go to zero at negative one so it's like an x intercept because it's a balanced rational function we see that there'll be a horizontal asymptote at y equals one the ratio of their leading coefficients like so and then finally if we think of like the y-axis right here we can think of the y-intercept which you plug in x equals zero you'd get one over negative three so negative one third and so connecting the dots we get a picture that would look something like the following like so and then like this right here for which then looking at the picture we see that it's positive when you're to the left of negative one and when it's positive when you're to the right of three if we put these two pieces together we see the domain is going to be negative infinity up until negative one negative one will be included in there because negative one makes the ratio go to zero which zero can be greater than equal to zero right then we're going to be at union we're going to go from three to infinity in that situation now three will not be included this time because notice dividing our plug in an x equals three divides by zero let's move on to the second part right here let's look at this this rational expression well I see some two issues going on here so we have a logarithm so the the operand of the logarithm needs to be positive so we need 16 minus x to be greater than zero this tells us that 16 should be greater than x or x is less than 16 okay but we also need that three minus the next or the log base two of 16 minus x this should not equal zero so when is it equal to zero if I move the log to the right hand side of the equation we're going to get three is equal to the log base two of 16 minus x I want to move the base two to the other side it switches from a logarithm to an exponential so we're going to get 16 minus x is equal to two cubed which is in fact equal to eight subtracting 16 from both sides we get negative x is equal to negative eight therefore x equals eight so x equals eight of course is the value that we shouldn't have when x equals eight that makes the denominator go to zero so we don't want eight but we have to also be less than 16 so the second function the second part of the function suggests to us that we're going to go from we have the less than 16 so we're going to go from negative infinity up to eight jump over eight play a little leapfrog right there up to 16 we're 16 is not included and so we have to put these things together like where are these things overlap and if it helps you can draw a picture of such a thing think of something like this let's think of these important values so we should be thinking of the number negative one we should be thinking of the value three in which case if we color that in we have this portion right here and this portion right here but when we think of the blue one we have eight which is like over here this is not drawn to scale of course in 16 so we want to be less than eight and we want to be between eight and 16 we're going to like a open circle right here this suggests we don't have that so where are these things double overlap right so they're going to overlap on this portion when you're less than negative one as you then go from above three up to eight and then from eight to 16 so if we look at the intersection here this tells us that the domain of our function h will be given as negative infinity up to negative one negative one was included into that we then take the union where we're going to go from three three was not included because it divides by zero up to eight eight's not included because that also led to division by zero then you're going to jump over eight and go up to 16 so that one is a little bit more complicated there's a lot of moving pieces going on there but we see we have the skill set we need to determine the domain of any elementary function again i'm not talking about trigonometric functions here but those those extra trigonometric functions don't really make the domain much more difficult to do than we've already done in this example here so i hope this review of computing domains of functions was helpful we learned how to find the domain of numerical functions graphical functions and we spent the most time on learning algebraic functions the fourth representation of a function of verbal one like a story problem uh that one's i i really can't give you too many general tips at this moment because it depends on the problem can we accept uh x or y to be a decimal or a fraction or does it have to be a whole number can they be negative uh again that that depends on the context and so those would be things you have to treat on a case by case situation if you feel like you learned something about domains today uh please feel free to hit the like button also feel free to subscribe so you can see more videos math videos like this in the future if you're interested thank you everyone see you later