 Welcome back to our lecture series Math 1050, College Oddsworth for students at Southern Italian University. As usual, I'll be your professor today, Dr. Andrew Missildine. In our last lecture, number 46, we discussed that length and notion of domain of a function, but not so much about range. In this lecture 47 and also in 48, I want to try to remedy that for algebraically represented functions. That is, how do you calculate the range of a function is given by a algebraic formula? Well, the issue says it's a whole lot harder to determine the range of a function than it is to find the domain of a function. That's because there's much more nuance to determining the range. Basically, we want to rely on the graph of the function to be able to determine its range, which requires us knowing a lot about the graphs of functions, their asymptotics and discontinuities, all of that type of stuff. But there is one simple example that I want to discuss in this video right here. One way to compute the range for a one-to-one function, that is, those functions which pass both the vertical and horizontal line test, is to compute the domain of the inverse function. Because the domain of a function becomes the range of its inverse function, and in the direction we want to go, the range of a function is actually equal to the domain of its inverse function, assuming it has an inverse function. If we know how to compute the domain of a function, then we can compute the range of it if it's invertible. Now, not every function will be invertible. It's got to be a one-to-one function, but in those cases where it is, we can actually find its range. So consider this one right here. Let's find the range of the one-to-one function, f of x equals the cube root of 2x cube plus 1 over 5x cube plus 1. So it might not be obvious from the get-go that this is a one-to-one function, but it in fact is. So we take this expression y equals the cube root of 2x cubed plus 1 over 5x cube plus 1. And when you draw your square roots here, do make sure that you extend it to the bottom of the fraction. So you see that the denominators also included in that. So this is the equation that determines f. If we want to switch over to the inverse function, f inverse, we have to swap the roles of x and y. So the y will become an x, and each of the x's will become a y. So we get 2y cubed plus 1 over 5y cube plus 1. And so this might seem intimidating. We have to solve for y here. We can do it. We have to first start freeing the y's from their captivity. So to get rid of the cube root on the right-hand side, I would want to take the cube, which what's good for the goose is good for the game. We have to do the both sides of the equation. So you're gonna get x cubed is equal to 2y cube plus 1 over 5y cube plus 1. So next we're gonna clear the denominators, multiply the right-hand side by 5y cube plus 1. But what's good for the goose is good for the game. And we have to do it to both sides the same. So it cancels on the left, and we end up with an x cubed times 5y cube plus 1. This is now equal to 2y cube plus 1. We're still trying to free the y. So we have to get y out of its prison. We got it outside of the radical. We got it outside of the fraction. And now we have to get it outside of these parentheses, which that's easy to do just distribute the x cubed. And notice you're gonna get a 5y cubed x cubed plus x cubed is equal to 2y cube plus 1. We wanna combine together the y cubed. So let's bring all the y cubes to the left-hand side and let's move everyone who's not a multiple of y to the right-hand side. So we end up with a 5x cube y cubed minus 2y cubed. This is equal to 1 minus x cubed like so. And in which case on the left-hand side, we can factor out the y cube. Cause notice that we have a multiple of y cubed right here. We have a multiple of y cubed right here. If we factor it out, we end up with y cubed times 5x cubed minus 2 equals negative x cubed plus 1 like so. And so in this setting, we can divide by the coefficient of y cube, which is 5x cubed minus 2. Divide both sides, 5x cubed minus 2. In which case then we end up with y cubed is equal to negative x cubed plus 1 over 5x cubed minus 2 like so. And therefore to solve for y, we're gonna take the cube root of both sides. Make sure you do it to both sides. And so in the end, we end up with the expression f inverse of x is equal to the cube root of negative x cubed plus 1 over 5x cubed minus 2. So we went through all this effort and now we have the inverse function. So the inverse function wasn't actually our target. Our target was to find the range of f. So the thing here that's critical is that the range of the original function f is gonna equal the domain of its inverse, f inverse. So what are the potential problems right here? Well, we have a cube root, right? There's no problems with the domain of a cube root, like taking the cube root of a negative is not a problem. So the only real concern here is going to be what makes the denominator go to zero. That's what I'm concerned about. So let's solve that equation. The denominator would go to zero when 5x cubed minus 2 equals zero. So we add two to both sides. So we get 5x cubed is equal to two divided by five to get x cubed equals two fifths. And then taking the cube root of both sides, we're gonna get x equals the cube root of two fifths. So that is the value that makes the denominator go to zero. In which case that's the only number in the domain of f inverse that would make, it's not inside the domain, excuse me. This is the only real number for which is outside the domain of f inverse. So our domain here is gonna be all numbers from negative infinity up to the cube root of two fifths. And then you union that with the cube root of two fifths to infinity. So the domain of f inverse is defined for everything except for the cube root of two fifths. And therefore that's gonna be the range of the original function. You get every y coordinate except for the cube root of two fifths. Let's look at another example of this. Let's take this time the function g of x equals two e to the x plus one over three e to the x plus seven. And so if we think of the function f right here, it's given by the equation y equals two e to the x plus one over three e to the x plus seven. And so as we switch to the inverse function f inverse right here, we're gonna swap the roles of x and y. So we get x equals two e to the y plus one over three e to the y plus seven. And so this functions to solve for the inverse functions can be very similar to what we did last time. We need to solve for y. So I'm gonna begin by multiplying both sides by the denominator. That is I wanna clear the denominator. So they cancel here, but what's good for the goose is good for the gander. We have to do it to both sides the same. So we end up with x times three e to the y plus seven. And that's equal to two e to the y plus one, for which then we can distribute the x, distribute it there. So we end up with a three x e to the y plus seven x is equal to two e to the y plus one. So again, combining like terms, let's move all the multiples of e to the y to the left hand side. And everyone who's not a multiple of e to the y will move to the right. So we get three x e to the y minus two e to the y is equal to negative seven x plus one, like so. On the left hand side, as everyone's divisible by e to the y, we can factor it out, we factor it out. In which case then we get e to the y times three x minus two and this is equal to negative seven x plus one, for which then we'll divide by the coefficient three x minus two on both sides, like so. We end up with e to the y is equal to negative seven x plus one over three x minus two. And so how do we solve for y in this situation? We have to get rid of the base e. So we're gonna take the natural log of both sides, the natural log, the natural log. This then gives us y is equal to the natural log of negative seven x plus one over three x minus two, like so, for which now we wanna determine what is the domain of f inverse, right? So this right here is our inverse function, f inverse of x. So remember the range of f is gonna equal the domain of its inverse function, like so. In which case then the next thing to do is to see what problems does this have with its domain. We have a natural log, right? So we have to be concerned with negative seven x plus one over three x minus two, this needs to be greater than zero. We also can't let three x minus two go to zero, but that will be taken care of that when we think of the vertical asymptote of this rational function. If we were to graph this real quick, right? There is a vertical asymptote at two thirds, positive two thirds, so that's a problem. There's gonna be an x intercept at one seventh. So let's mark that, one seventh is smaller than two thirds. The y intercept, if we plug in x equals zero, we end up with negative one half. So y intercept would be something like this. And this is a balanced rational function in which case as x approaches infinity or negative infinity, this thing would approach negative seven thirds. Negative seven thirds of course is further down than the negative one half we saw right there. So connecting the dots, we would anticipate a picture looks something like the following, something like this. So where are we gonna be greater than seven? I was not greater than seven, excuse me, where are we gonna be greater than zero? That looks like it happens only in this sector right here as we go from one seventh, one seventh to two thirds. That's the only place where that thing is gonna be positive, one seventh to two thirds. So that gives us then the domain of, I guess I called it f inverse, I started this problem with a g, g or f, it doesn't really matter too much the name of the function. Sorry about that. So we get the domain of g inverse is gonna be one seventh to two thirds. Now the domain of the inverse function gives you the range of the original function like so. So what I wanna do here, when you look at these examples, they looked really intimidating, right? So these examples A and B right here looks really intimidating, but I promise you that it turns out that this is sort of like a cheap little trick that makes it look super hard but turns out it wasn't as bad as it might seem to be. So first of all, when you have a function like, f of x equals like A x plus B over C x plus D, right? These are called, this function's called a linear fractional function for which it's a linear function divided by a linear function. Makes sense. And these guys are always gonna be one-to-one functions. And I wanted to kind of show you that we can kind of shortcut the calculation of this inverse function very, very quickly by the following principle, right? If you take y equals A x plus B over C x plus D, if you look for the inverse function, we're gonna get x equals A y plus B over C y plus D. Kind of proceeding forward here, the first thing to do is clear the denominator. So you get x times C y plus D is equal to A y plus B. You're gonna distribute the x, you get C x y plus D x is equal to A y plus B. Combining like terms, you're gonna get C x y minus A y is equal to negative D x plus B. Factor in the left-hand side, you get y times C x minus A is equal to negative D x plus B. And then dividing both sides by C x minus A, you're gonna get y, which is actually the same thing here now as f inverse of x. This is equal to negative D x plus B over C x minus A. So notice here, if you compare the original formula with the inverse formula, what happened? You'll notice you have these coefficients A, B, C, D. There's some movement around here. Notice that the A and the D swap places and you switch their signs. So now it's a negative D, negative A, okay. And then the B and C actually don't move around. So if you see something like that, it's like, hey, I can notice a pattern here. This makes the computation of their inverses a whole lot easier. Like for example, if we come back up to this example right here, I'm gonna quickly make the substitution u equals e to the x. Then this thing would look like two u plus one over three u plus seven. The inverse of this thing, the inverse of this function would then look like negative seven u plus one over three u minus two. And I feel like we saw that somewhere, right? Oh no, negative seven plus one, three and negative two. So the coefficients we are able to predict but what if the natural log, right? So when you look at this function right here, this function g of x, I'm gonna erase it for a second. The thing that we wrote right here. This function g of x can actually be decomposed in terms of composition. That is it looks like, it looks like negatives, excuse me, negative positive two u plus one over three u plus seven. And then you've composed it with the function e to the x. That as you put e to the x in for you, you have this composition of function. Okay, and I compute the inverses of these things individually. So the inverse of this linear fractional I can calculate by the formula. And then who's the inverse of e to the x? We've seen it somewhere before. Aha, it's the natural log. See that connected? That's natural right there. So we can actually do these things individually. And so when you have a function decomposition like this, so we have something like, so f of x is equal to g of h of x like so. Then if you wanna calculate f inverse of x, what you get is the so-called shoe sock principle. That is if f is equal to g composed with h, then the inverse function is gonna swap those roles around. And we call this the shoe sock principle for the following reason. In the morning, if you put your socks on, then your shoes, when you take them off, you first take off your shoes, then you take off your socks. When you invert it, you have to switch the order of these things. And so with this example in mind, our function g of x, which was given as, what was it again? Two e to the x plus one over three e to the x plus seven. We can factor this thing as two u plus one over three u plus seven composed with e to the x right here. Well, what I know here is that by the shoe sock principle, the inverse of g is gonna look like switching the order. I first take the inverse of e to the x, which is the natural log of x like so. I'm actually gonna call it u. I always call the second one u here. And then you're gonna take the inverse of this linear fractional, which we saw was negative seven x plus one over three x minus two. And so then when you compose these things together, this function goes in for you. You end up with the natural log of negative seven x plus one over three x minus two. So when you have sort of like those, when you can see the function decomposition, you know how to calculate the inverse of the linear fractional. That problem of finding the inverse becomes so much easier. What about the previous example, right? What if we were to do f of x, which on the previous slide, I'm just gonna write it here on the screen. We have the cube root of two x cube plus one over five x cube plus one. Well, in this example, let me illustrate to you that this has a natural decomposition. This looks like the function, the cube root of u composed with, we're gonna say two v plus one over five v plus one composed with x cubed. All right, so we can see three functions here. I'm gonna put x cubed inside of the fraction and the fraction then goes inside the cube root. So therefore, when we compute the inverse function, we're gonna end up with the shoe sock pencil reversing the order. We're gonna take the inverse of x cube, which is the cube root of u. We're then gonna take the inverse of the middle function, which as that's a linear fractional, you're gonna get negative v plus one over five v minus two and then you have to take the inverse of the cube root, which is the cube like so. So because these guys right here are inverses of each other, it doesn't look like we went in the right order, but we did, we in fact did. And then when you put these things back together, we're gonna put a cube root inside of this and then this goes inside of, we're gonna put a cube inside of the fraction, the fraction goes inside the cube root. We end up with the cube root of negative x cubed plus one over five x cube minus two for which you can then see that was, you can double check earlier in the video here. I guess I'll just bring it up over. Let's see, let's go back. Where was our good friend? Aha, here it is. We actually were able to find the inverse. We can find inverses very, very efficiently, but remember the reason we brought this up in the first place is that if you have a one-to-one function, its inverse can be used to find the domain.