 This video is a continuation of another video we had done previously in this lecture series. And in the previous video, what we wanted to do is we wanted to, to first of all, determine the domain of the function, and then also look for the x-intercepts of the function. And the idea was towards trying to solve equations because if we can solve the equation f of x equals zero, then we can solve any equation, assuming f is allowed to take on any function whatsoever. And, but solving the equation f of x equals zero is equivalent to just finding the x-intercepts of the graph. Now, in these examples right here, I wanna introduce a technique that's extremely useful. And this is the technique that I'll call a U substitution. That is, we're gonna substitute out, we're gonna substitute out an expression for the new variable U. And we usually use the number, the variable U here for, well, for two reasons. One, U isn't a super common variable that we use in these algebra classes. You know, our go-to variables are usually x, y, and z. So U is often available for us to substitute out. So that's sort of the truth. And then also it leads to some awkward grammatical statements when I say things like U is, you know, it's gonna be fun. It'll be fun for the closed captioning auto-generated thing, U is, anyways. So what do U substitution mean? So U substitution, we're gonna do is, we're gonna take some function, you know, it's gonna be some function like f of x is some expression of f of x right here. We're just gonna replace a complicated form with a U. Basically what we're gonna be doing is we're gonna be recognizing there's two functions in play here. So we're gonna have like h of x is equal to some type of composition of functions. So like h, let's just say it this way, h is equal to some function like say g composed with f, right? And so we're just gonna be like, oh, let's just recognize we have this inner function U like here and use that to help us out here. Why could that be useful? Let's look at this situation right here. So you have f of x is equal to x to the 2 thirds minus three times x to the 1 third minus 10. If this was an equation, and this is a function we wanna find this x intercepts, we set this expression equal to zero. How would we go about solving this? Well, if you make the following observation, life becomes a whole lot easier for you. It's like, I see like x to like 1 third powers, right? You have like this x to the 1 third. If I try this idea of U substitution, I could actually replace that with a U. But what about x to the, what about x to the 2 thirds? Well, x to the 2 thirds is the same thing as x to the 1 third squared right here, right? But x to the 1 third is equal to U. So this thing is the same thing as U squared. And so if you make this U substitution, our function f, what we're saying here is f is equal to, well, let me say it this way, let's take our equation, we're gonna replace the x to the 2 thirds, it becomes a U squared, then the x to the 1 third becomes a U and you get negative 10 still equal to zero. So what this then became is a quadratic equation. So with the U substitution, we now have a quadratic equation which we can try to solve. I often refer to this as a quadratic like equation. It's like a quadratic equation if we do the appropriate U substitution. And so basically what we're saying is the following, we made the following recognition. If I take the function G of U to equal U squared minus three U minus 10, right? And I take the function H of x to equal x to the 1 third, what we've now recognized is that f of x is equal to G of h of x. That's what we've now recognized in this situation. Now you put the function x to the 1 third inside of a quadratic function. And so in order to solve for these x intercepts, I'm gonna first solve the outer function G of x, which is this quadratic function. And I can do that by factoring factoring we've done in previous settings or you could use the quadratic formula as well. And so we have to factor, let's see factors of negative 10 that add up to be negative three. We could take negative five and positive two, right? Negative five times two is negative 10 and negative five plus two is negative three. So that then gives us this one, U minus five times U plus two is equal to zero. By the zero product property, we get that U minus five equals zero or U plus two equals zero for which then we can solve for U. We get U equals five and we get U equals negative two. But what is U? U is, see how that happened there? U is x to the 1 third. So substituting back in the x, we get x to the 1 third equals five or negative two, like so. So how do we get rid of the 1 third power? We're gonna cube, right? So we cube both sides. In which case, then we get our solutions. x equals five cubed, which is 125 or x equals negative two cubed, which would equal negative eight, which then gives us the two solutions to this quadratic-like equation. So if we can recognize some type of U substitution, we can then solve many quadratic-like equations. Let's look at another example here. Let's take g of x this time to equal four x to the fourth plus seven x squared minus two. So I want us to recognize this has a quadratic-like structure, right? Because you have this x squared right here. You also have this x to the fourth. I could treat this like a polynomial of degree four, and I could try to look for solutions in factoring like we did previously in this lecture series. But the U substitution turns out really nice here. If you take U to be x squared, then that means U squared will equal x to the squared squared. That is, it's x to the fourth. And so then our function here will look like g of U is equal to four U squared plus seven U minus two. We set this equal to zero, because after all, we're looking for the x intercepts here. When is this thing equal zero? We then can proceed to factor this thing. We need to find factors of four and negative two. So four times negative two, negative two is equal to negative eight. Can we find factors of negative eight that add up to be seven? Well, we could take eight times negative one. That'll do it. And so then we're gonna proceed to factor this thing by groups. So we get four U squared plus eight U minus U minus two. Equals zero. We're gonna take the first group and the second group. The first group can offer up a four U, leaving behind U plus two. The second group can offer up a negative one. So you get negative one leaving behind U plus two. Equals zero, which is good, because now we see that U plus two, U plus two is common to both. We factor out that common divisor of U plus two. You're gonna end up with four U minus one times U plus two. So once you factor this thing, you can substitute back in U at any moment. You could solve for U and go from there, or you could just plug in what U has X in right now if you prefer. It doesn't really matter when you do it. After we recognize the quadratic form, we can then stick back in the X squared. So that's gonna give us four X squared minus one times X squared plus two. This equals zero. In which case then we can set each of these factors equal to zero and solve. So we get four X squared minus one equals zero, or we have X squared plus two equals zero. In which case when you solve the first one, add one to both sides, you get four X squared equals one. Divide both sides by four. You get X squared equals one fourth. And then divide both sides by two. We're gonna end up with X equals plus or minus the square root of one fourth, which is plus or minus one half. All right, so you get those values right there. With the other one, when you try to solve that one, you're gonna subtract two from both sides. You get X squared equals negative two. If you take the square root of both sides, you'll get X equals plus or minus the square root of negative two, which is plus or minus I times the square root of two. So then you have to make a decision. Do I allow for non real solutions or not? Remember, our goal here is to find X intercepts of a function for which an X intercept must be a real number. So these imaginary numbers would be discarded. In which case then the only X intercepts would turn out to be plus or minus one half. So there's the two solutions. X equals one half and negative one half, like so. So this one, this one right here, you look at it, it's like H of X equals X squared minus one quantity squared plus X squared minus one minus 12. In this situation, many of you might be tempted to, I'm just gonna foil this thing out and combine like terms. I get a quadratic, then I can solve that quadratic. But it wouldn't be actually a quadratic. You're gonna get X squared times an X squared. Well, it's a degree four polynomial. I could try to do that. You could do that, but that'd be like taking a Mona Lisa thrown in a shredder and be like, we don't need this, we don't need this. I could probably draw a picture. You know, let's not destroy the beautiful architecture that's already there. Because notice here, you have an X squared minus one. Let's call that U. Then you have an X squared minus one squared. Let's call that U. And so it's actually would become a U squared. And so using the substitution, H of U would look like you're gonna get U squared plus U minus 12. This is a quadratic that's a much easier to factor, right? Oh, factors of negative 12 that have to be one. I could take U plus four and U minus three. Set this equal to zero. This would tell me that U equals negative four or positive three. Then we're gonna remove the substitution and go back to X. Cause U was equal to X squared minus one. We get X squared minus one equals negative four. And we get X squared minus one equals three. Add one to both sides. On the first one, we'll get X squared is equal to negative three. On the second one, when we add one, we're gonna get X squared equals four. Now again, so then we're gonna take the square root of both sides. We're gonna get X equals plus or minus I times the square root of three here. And then we're gonna get X equals plus or minus the square root of four. That is plus or minus two. We then have to make a decision again. Do we allow for non real solutions? Again, as our original goal was to find X intercepts, we have to discard anything that's not real. And therefore our X intercepts would be positive two and negative two. So this recognition of a U substitution can be very helpful, not just for doing these quadratic like equations, but for functions in general. But this video focuses on quadratic like equations. So this example here, capital F of X equals six times X to the two-fifths plus 11 times X to the one-fifth minus 10. Maybe you're starting to see the pattern here. Whenever you have something like X to the two over whatever P, this is just gonna be X to the one over P squared. So we see that we have two-fifths versus one-fifths power. So we're gonna set U equal to be X to the one-fifth and therefore U squared is the same thing as X to the two-fifths right there. So then our quadratic equation has the form six times U squared plus 11 U minus 10. This is equal to zero. We could factor this, right? We could also use the quadratic formula, whichever we prefer. This one does actually factor, right? So taking your coefficients together six times negative 10, that's gonna give you negative 60. Can I find factors of negative 60 that add up to be 11? And that's gonna be 15 times negative four. And there's a little bit of guesswork that happens there, absolutely. But the quadratic formula can be used if you don't feel comfortable with this guesswork that happens. So you're gonna get six U squared plus 15 U. That's gonna be our first group. Then the second group will consist of, let's see, negative four U minus 10. This is still equal to zero. The first group can offer up a three U leaving behind two U plus five. The second group can offer up a negative two. We're just back down to the GCD there. And that gives us a two U plus five again. You'll notice, of course, that the two U plus five is the same. So we can factor it out. That then gives us the factorization of three U minus two times two U plus five equals zero. At this moment, I usually like to substitute back in my value. U equals X to the one fifth. So we get three X to the one fifth minus two. And then we're gonna get two X to the one fifth plus five is equal to zero. The zero product property gives us two equations. So we have three X to the one fifth minus two equals zero. And we have two X to the one fifth plus five equals zero. Be very cautious on what operations you're doing here on the first one. We're gonna add two to both sides. This gives us three X to the one fifth is equal to two. Divide both sides by three. So we end up with X to the one third. Is equal to two fifth or two thirds, excuse me. And then we have to take the fifth power of both sides. And so we end up with as our value there, X is equal to two to the fifth over three to the fifth. Let me scooch it up just a little bit more. For which two to the fifth is 32 and three to the fifth is gonna be 243. So that's one solution. The other one, if we go solving for that one, we're gonna subtract five from both sides. This gives us two X to the one fifth is equal to negative five. Divide both sides by two. Whoops. Divide by two. This is gonna give us X to the one fifth is equal to negative five halves. Then the next step is we're gonna take the fifth power. For which, since we are taking the fifth, we have X to the fifth root of X, excuse me on the left hand side. There's no reason why I can't be negative. So we don't have a domain problem here because the right hand side is negative. We end up then with X is equal to negative five to the fifth over two to the fifth. And that should equal negative 3,125 over 32 again. And so these will be our two solutions to this quadratic like equation. So we get one over here and then we get one over here. All right, I wanna do one more example in this video. This time let's do one involving logarithms, so you can perhaps see what it is by now. What's the U substitution? We have the natural log of X quantity squared minus the natural log of X minus six. So we're gonna set U equal to the natural log of X right here. And so then our function G of U becomes U squared minus U minus six, which that's a nice one to factor there. You can factor that thing as U minus three times U plus two. So notice we took factors of negative six, negative three times two that have to be negative one, negative three plus two. And so then getting rid of the U, we go back to the natural log. We're gonna have the natural log of X minus three times the natural log of X plus two. These things equals zero. We're looking for the X intercepts here. Setting both factors equal to zero, you get the natural log of X minus three is equal to zero and we get the natural log of X plus two equals zero. Pay attention to problems with domain and range for which the natural log potentially could have, right? If you add three on the first one, both to both sides, you get the natural log of three, natural log of X equals three, excuse me. Then you exponentiate, you're gonna get X is equal to E cubed, right? The natural log doesn't have any problems with the range. It's the domain that's an issue, which we're okay here. And then if you subtract two from both sides on the second one, you'll get the natural log of X equals negative two. Now this is what I'm saying here. You can actually exponentiate both sides even if the right hand side's a negative. You're gonna get X equals E to the negative two or if you prefer one over E squared. It doesn't really matter too much here. So you're gonna get these two X intercepts. You're gonna get E cubed and then one over E to the square, right? So now unlike all the other examples here, we were looking, so remember, our original goal was to find X intercepts but also to find discontinuities. Where are our locations where the function could potentially be undefined? Well, so the domain of the function is of concern to us here, right? Notice that because of our U right here, this is the natural log. The quadratic doesn't have any problem with domains but the natural log does. So the domain here of G is going to be zero to infinity, which none of our X intercepts fell into that range so we didn't have to discard them but we do wanna pay attention to where is this function undefined? In particular, this function will have a vertical asymptote. There'll be a vertical asymptote at X equals zero because the natural log has that vertical asymptote at X equals zero.