 In this video, we're going to prove that the equation x cubed plus x minus one equals zero has exactly one root. There's one solution to this polynomial equation. Now, this is similar to a problem that we've approached previously in this lecture series. We showed using the intermediate value theorem that's an equation like this one would have. We're going to show in this video that it has exactly one solution. Exactly one. The way we're going to show that is it has at least one or repeat that argument, but we also the show has at most one and that's what's going to be new in this video. So let's remind the viewer how do you show it has at least one solution. So this part right here, we're going to show it has at least at least one root. So we're working with an equation, but we really want to turn this into a question about a function. So set one of side of the equation equal to zero, which we've already done, x cubed plus x minus one. Take that non-zero side and we're going to make that into a function. So take the function f of x, which is equal to x cubed plus x minus one. Note that this function is continuous. Continuity of f is going to be important for the forthcoming intermediate value theorem that we're going to see in just a second. Well, this function, it's a polynomial function. Note that the in behavior of this polynomial functions determined by its leading term x cubed. And so since this is x cubed, notice that as x goes to infinity, f of x also has to go to infinity. Why is that significant? Well, if our function goes off towards infinity on the right hand side, that means eventually the function has to be positive. You can't get close to positive infinity without some time become. So this tells us that for some values x equals t, right? So for some t value in the domain, which has a polynomial function, the domain is all real numbers. There's some number t in the domain, so t is positive. It's somewhere greater than zero. Now, if we want to be precise, we can actually find numbers. It's not like they're hard to find. For example, if you take f of one, notice that if you plug one into our function, you get one cube, which is one plus one minus one. That's equal to one, which one is positive. So that's a specific number if you care for, but I'm perfectly happy with just knowing that eventually it's positive, right? It's also at some point negative because, again, looking at the leading term, it looks like x cubed, right? So the in behavior means it points down on the left and points up on the right hand side. And who knows what happens in the middle, okay? On the left hand side, the limit as x approaches negative infinity of f of x will likewise be negative infinity. So that means for some value x equals s inside the domain, it must be that f of s is negative. So at some point, the function must be positive, specifically f of one is positive. And at some point, the function might, right? If we want specifics, we don't actually need them here, but specifically at f of zero, you'll get zero plus zero minus one, which is negative one, right? So we see that our function is somewhere positive. It's somewhere negative because it's a continuous function that's somewhere positive and somewhere negative. That means by the intermediate value theorem that somewhere we must cross the x-axis, which if we cross the x-axis, we find an x-intercept of our function f of x, that would be a solution to the original equation. So there's some number r so that f of r equals zero, thus giving us a solution. This is an argument we've seen before, and this shows that it has at least one solution to this equation. What I want to now do is show you why there's at most one solution. So this part right here, we're going to show there's at most one solution because you have at least one at most one. The only way to resolve that is to say that there's exactly one solution. So what we're going to do here is use a logical technique called proof by contradiction. It's a very nice proof technique, which means that if you want to prove something is true, you can assume the opposite and get a contradiction. The only way you can get a contradiction is because if we want to show there's at most one solution, what we're going to do is we're going to assume there's at least two solutions. Let's call them a and b, which if a and b are solutions, that means f of a equals zero and f of b equals zero. Why do we want to do that? Well, since we have a polynomial function, it's continuous on the intervals a to b, which we don't actually know what a and b are, right? We just know there's some numbers in the domain of our function f, but f has a polynomial, its domain is all real numbers, right? So a and b could be any real numbers. But since it's a polynomial everywhere, so be continuous on the closed interval a to b, right? You'll likewise be differentiable on the interval a to b, whatever those numbers turn out to be. So if you have a continuous function on the closed interval a to b, which is also differentiable, then the mean value theorem applies to this situation. And what does the mean value theorem say? The mean value theorem says that there's going to be some number c that's between a and b such that the derivative of f at c is equal to zero. Well, how do you get that? Well, the mean value theorem says that f prime of c is going to equal the average rate of change. It's going to equal f of b minus f of a. For the sake of argument, I'm going to assume that b is the bigger of the two. It doesn't really matter though. f of b minus f of a over b minus a, which f of a and f of b are both zero because they're both x intercepts. We get zero minus zero over b minus a. You're going to get zero over b minus a, which is equal to zero. So the mean value theorem guarantees that the derivative is zero at the number c, some number. But the derivative we know what our function is, right? I mean, it's not on the screen anymore, but our function is still going to be x cubed minus or x cubed plus x minus one, right? I'll just write this on the screen again. f of x equals x cubed plus x minus one. If you take the derivative by the usual power rule, you're going to end up with the derivative. f prime of x is 3x squared plus one, which x squared is at smallest zero. If you take a non-zero number, its square will be positive, but if you take zero, it'll be zero. So 3x squared worst case scenario is z1 to that. 3x squared plus one is at least one, probably bigger. In particular though, the derivative is always positive because it's at least one. This is our contradiction, right? So the mean value theorem guarantees there's a point where the derivative is equal to zero. But in actuality, the derivative can never equal zero, thus giving us a contradiction. And therefore, if we have a contradiction, what assumption was wrong? That there's at least two solutions to the equation. So using the mean value, we can then guarantee there's at most one solution. Therefore, since we've used the intermediate value theorem to show that it has at least one solution, when you put those together, at most one solution, at least one solution, we get the solution to this equation. We don't actually know what that solution is, but we can guarantee that there is a unique x-intercept to the function xq plus x minus one. And I do want to point out to you that this argument actually applies to any odd degree polynomial. Well, at least about having at least one solution. The fact that it has at most one solution, that is dependent on xq plus x minus one. And that's because this is an always-increasing function. If your function's always increasing by the mean value theorem, you can have at most one x-intercept, which we were guaranteed by the intermediate value theorem.