 In this video I want to do some more computations of inverse functions, algebraically speaking. Consider the function f of x is given as 2 over x minus 3 and we add to that 4. The basic strategy is going to be the same, we start off with the relationship that gives the function f. We're going to replace the f of x with a y here, y equals 2 over x minus 3 plus 4 and then we proceed to swap the roles of y and x in that formula and this will give us a formula for f inverse. We're going to get x equals 2 over y minus 3 plus 4 and now we're going to proceed to do inverse operations until we can successfully solve for the y all by itself. So the first thing we do kind of reversing of order operations we're going to minus 4 from both sides that cancels the 4 on the left hand side and that would give us 2 over y minus 3 is equal to x minus 4. Now the left hand side is just a fraction right? We want to get rid of the y from the denominator. We have to kind of free from its prison, it's trapped inside of the fraction right there. One can accomplish that by a couple different techniques. In this situation I'm going to kind of treat the x minus 4 as a fraction x minus 4 over 1. That way I can cross multiply. You multiply the 1 by the 2 and you're going to multiply the y minus 3 by the x minus 4 and if we do that that will then give us 2 times 1 is equal to x minus 4 times y minus 3 like so and so 2 times 1 is pretty simple here. Now one has to be kind of careful on the on the next part right? What we don't want to do what we don't want to do is start spreading out the y's too far right? So it might be tempting to like foil out the the right hand side so you take x times y, x times negative 3, 4 times y, 4 times 3 there. The problem with that approach is you get an x y, you'll get a minus 3x, you'll get a minus 4y and then you get a plus 12. This is problematic for us because now there are two y's inside of the equation. We want one y right? We want it y equals so spreading out the y's is not going to help us. We want to keep them gathered in so instead of foiling the right hand side because our goal is to solve for this y right here instead look at the expression x minus 4 and pretend like that's the coefficient of this expression right here y minus 3. Divide both sides by x minus 4 and you have to do that to the other side as well x minus 4. Then you'll see that the the y you'll get a y minus 3 from the from the right hand side then you get 2 over x minus 4. Whoops minus 4 like so and now in this situation you'll notice we still have y all by it's I mean y is all connected together we still got to give rid of this stuff that's attached to the y that's pretty simple just add three to both sides and we end up with in the final form y equals 2 over x minus 4 plus 3 and so now that we solve for y we're going to get rid of the y and put back in the name of the function this is f inverse of x which we have this thing right here and you might be tempted like well should I add the fractions together like 2 over x minus 4 plus 3 you could but I mean we should do it for a reason we shouldn't just do something because we think we ought to there should be a reason motivating why we do it and so for example like we had that choice between foiling this side or dividing by x minus 4 and we actually took a moment and stopped to think about what will happen if we do that there are consequences to doing things to the equation foiling actually caused us to get farther away from solving for y as opposed to division so we don't do things just to do them we do them to make things better right to make to make things more to better but we'll just we'll just keep without look make things better so I'm actually perfectly happy with writing the function this form because in this form I can very well see the transformations you shifted up by three you've vertically stretched it by two and you shifted it to the right by four if I would start adding the fractions together the graph transformations would be concealed let's take a look at another algebraic function this these things right here you have a fraction of a polynomial of so you have a polynomial on top polynomial on bottom this is what we call a rational function this type of rational function where you have a linear a linear polynomial on top and a linear polynomial on bottom this is often what we call a linear fractional and it turns out that linear fractionals are one-to-one functions and we can solve for this thing algebraically and then there's kind of two ways one could do it one you could actually do a little bit of polynomial division like you could take this 2x plus one divided by sorry 2x plus one and you divide it by x minus one if you do that you'll actually get something that looks very much like this function right here in which case then you could find its inverse the way we did just a moment ago and that's a little bit extra flavor you know sometimes a little bit more spicy than we want it to be and we haven't really talked about division so we're not going to do that right now but instead let's let's just try the approach we've done so far replace the f of x with a y you get 2x plus one over x minus one and so then when you switch to the inverse function you're going to every y becomes an x and every x becomes a y and something different happens this time we end up with actually multiple y's in the equation there's a y in the numerator and a y in the denominator this is actually the situation we were trying to avoid in the previous example this might be one advantage of doing polynomial division we we actually would only have by doing the division you're actually only going to get one x to start with as opposed to two but like I said there's a price to paying division and the price isn't necessarily worth it considering what we have to do now so our goals we want to solve for y but first to solve for y we have to combine the y's we want to combine the y's together and what we're going to do is we have to free the y's from prison it's like we have two members of the family that are separated by war a wall has now kept one in one land and the other in the other so they have to break free in order to do that we need to liberate y from its prison which we call the denominator for which we're going to multiply the right hand side by y minus one because multiplying by y minus one will cancel the division by my y minus one but what's good for the goose is good for the gander we have to do the same thing to the left hand side so that equality is preserved this then will give us two y plus one equals x times y minus one now what a slightly different predicament than we were a moment ago in this situation we have a y imprisoned right inside of these parentheses here uh the the guard keeper of course is the multiplication by x previously i told you not to multiply it out because that would spread out the y's on the other hand though since the y's are already spread out in order to combine them together we need to liberate them and that's not only going to happen if we distribute the x it almost seems like i'm giving you different advice in one situation i told you not to multiply it out but in a different situation i told you to multiply it out why this the important difference well again the point is on how we do things not because we're told to because there's a reason for it we didn't do it earlier because it's spread out the y's in this situation since the y's are already spread out in order to gather them together we have to free all of them so i would actually advise we multiply out the x you get x y minus x is equal to two y plus one you'll now notice that with all of the walls gone our y's are free to be reunited they can go back to their their missing family member right and so we have a x y on the right hand side what if we subtract two y from both sides that'll put all of the y's back together and what if we add x to both sides that way we put all of the multiples of y on the right hand side and everyone who's not a multiple of y will be on the left hand side uh the two y's will cancel the x's will cancel on the right hand side this then gives x plus one on the left hand side and this will give x y minus two y on the right hand side now if i was giving you something like five x minus x what would you do with something like that well you'd be like i'll combine like terms in which case you would get four x but why is it four x um why isn't it five x five times one why do you or why isn't it uh you know why isn't it seven you know why why'd you get a four right there and when you're probably like well okay you have a five right here you have a negative one right here you add together the coefficients and let me give you some explanation of why that's what we do you'll notice with five x and with x there's a common divisor we call it the gcd the greatest common divisor there's a common divisor of x and so we can factor out that common x which would then leave behind a five minus one times x notice if i redistribute the x through um you'll get five x minus x factoring is just the opposite of distribution you just pull out the thing that was multiplied beforehand and then you can see that once you have five minus one times x that'll then become a four x why am i mentioning that right now well because we're actually in the same situation when you look at this right here there's a y there's a y if you have x y minus two y how do you combine those together add together their coefficients which is the same thing as just factoring out the y uh factoring out the y you're going to end up with an x minus two times y now look at your equation the left hand side is currently x plus one the right hand side is currently x minus two times y you'll notice that there's only one y in the equation now hooray we have now solved for why we combined the y's together that's our first objective uh the next part is then once you have the y's together now you're going to isolate the y that is you want to get everything away from the y now the thing attached to the y is is an x minus two and it's attached to y via multiplication so we perform the inverse operation which is division here so the x minus twos will cancel and then you're going to get y equals x plus one over x minus two and this would be the inverse the inverse function of this linear fraction right here that might seem quite intense right it seems like there might be a lot going on there but it turns out um it turns out that we have all of all of the skills in front of us to do these type of calculations uh like if we go on to if we look at the next example here we have another function f of x equals four x minus one over two x plus three i can promise you that this calculation is actually quite methodical you do all the same steps each and every time although you know the numbers change a little bit so compare this example to the previous one the relationship for y is going to equal y equals four x minus one over two x plus three switch the rules of x and y as you transition to the inverse relationship you get x equals four y minus one over two y plus three multiply both sides of the equation by two y whoops two y plus three what's good for the goose is good for the gander so that this cancels on the left hand side you then get x times two y plus three is equal to four y minus one so step one switch x and y step two clear the denominator step three distribute the x uh that is on the left hand side there you get two x y plus three x equals four y minus one now you're going to gather all of the y's on the left hand side you're going to put everyone who's not a multiple y on the right hand side you end up with two two x y minus four y on the left and on the right you have a negative one minus three x on the left hand side factor out the y since you put every multiple of y on the left hand side there will be a common factor of y factor it out you get y times two x minus four this equals the right hand side negative one minus three x and whatever was left over when you factor out the y's divide by it two x minus four divided by two x minus four that will cancel and now you will have the inverse function f inverse of x is equal to uh negative one minus three x over two x minus four see we got we got all of the things we needed to do just right then and there um the steps are the same every single time for these linear fractals they seem a little involved but if we're intentional if we focus on the algebraic steps and more importantly why we are doing the algebraic steps we can change the numbers and follow the same process this is sort of the beauty of algebra if we can focus on what's essential and we can ignore what's non-essential what's irrelevant then we can repeat the process over and over and over again and that's really what algebra is all about focusing on the important things and ignoring the irrelevant things and that's exactly how we can solve for an inverse function for one of these linear fractionals