 In this video, we'll provide the solution to question number 14 for practice exam number two for math 1210. In this video, we're asked to prove that the equation cosine of x equals x has a solution. And as this is a proof, there's going to be some theorem results we need to use. And so we need to name specifically which major theorem we're going to use. Now the basis of this proof is going to be based upon the intermediate value theorem. So what we have to first do is establish that the assumptions of the intermediate value theorem are satisfied. So the first thing I would do when you're given this equation is to set one side equal to zero. So take for example cosine of x minus x is equal to zero. So if we solve this equation, then we solve this equation. So we can make this shift to the right hand side equals zero. So we'll say something like let f of x equal the nonzero side of this equation. Let f of x equals zero. Let that be cosine of x minus x, which we need to mention, which is continuous. It's continuous for all real numbers. Starting in three, two, one. All right. So continuity is an essential ingredient in this one. The intermediate value theorem only applies to continuous function. So if we don't ever mention the functions continuous, then we don't actually know if we can use the intermediate value theorem. So we need to make sure that is mentioned. The other thing we have to mention, because what our goal is, is we're trying to identify that the function is somewhere positive, somewhere negative. And so since it's continuous, the intermediate value theorem implies that there will be an x intercept somewhere. Notice that this equation, cosine of x minus x equals zero is equivalent to saying f of x equals zero. We're trying to prove that f of x has an x intercept somewhere. And so in that regard, we're going to say something like the following note, we need to find some value for which it's going to be negative, interesting or positive, we can go out of the way. So start with something easy, like if we take f of zero, right, this would give us cosine of zero minus zero. Cos cosine of zero is equal to one. So we get one minus zero, which is equal to one, that's greater than zero. So that gives us that we have some point above the x axis. That's great. We should get something else. What can we, what can we do to guarantee it would be negative? We could do something like say, maybe pi, right, f of pi is equal to cosine of pi minus pi, for which we're going to get cosine of pi is negative one. You subtract pi from that, you're going to end up with, well, negative one plus pi, which is certainly negative. So we got something negative, which means that there's going to be some value below the x axis. And these numbers are not unique. I'm just trying to pick values that are going to be very convenient. We could have also done something like f of pi halves. That's a pretty good choice because cosine of pi halves is equal to zero. And if you subtract from a pi halves, that's definitely a negative value. So we don't have to find the tightest bounds. We just have to find some number, which the function becomes positive. And for some other number, which makes the function become negative. And so then at that moment, we didn't say something like therefore, so thus, by the intermediate value theorem, intermediate value theorem, there exist some number, some number. And what can we say about it? We can call that number C, right? And it's going to be between, I'm going to get rid of this picture. The picture is not actually part of the proof. It's just illustrated. There's going to exist some number C between the two numbers we used above between zero and pi for which we could have actually done pi halves, but we'll just leave it at pi because that's the number we had there. There's going to exist some number C between zero and pi, such that I will just say this way, which solves actually, I mean, I guess we could just simplify even better. There's going to be some solution. C between zero and pi. I'm just leaving it at that. Because after all, finding the next intercept of F, like we said, solves this equation, which then solves the original equation as well. And so let's summarize what we had going on right here. We actually don't need this portion. Again, that was just to kind of illustrate what's going on here. What are the essential ingredients to this type of proof? One, you should be working with a specific function. Tell me what that function is. Two, you need to establish that function is a continuous function for which the intermediate value thing doesn't apply. I wouldn't apply if it's not continuous. You need to find some place where the function is negative. You need to find one place where that's actually positive. Excuse me, find a place where the function's positive. Find a place where the function's negative and then invoke the intermediate value theorem to find the solution in question. If you have those ingredients, then you're going to have a good proof and you can expect full marks on a proof similar to the style you see right here.