 One of the reasons that mathematicians are not satisfied with the I know it when I see it definition of continuity has to do with the useful property of continuous functions. This property is known as the intermediate value theorem. Continuous functions have the following property. Suppose a function is continuous over some interval between A and B, and let L be some value between F of A and F of B. Then there is some C in this interval for which F of C is equal to L. And this is quite a mouthful, but what it comes down to is the intermediate value theorem requires that a continuous function be equal to every value between F of A and F of B for some value in the interval between A and B. And to give you some idea of how fantastically useful this is, suppose F of X is continuous over the interval between 0 and 10. And suppose that besides being continuous, the only thing you know about F of X are its values at a couple of selected locations in that interval. Based on this information alone, we can tell how many solutions there are going to be to F of X equals 0 in this interval between 0 and 10. So since our function is continuous, we know we can apply the intermediate value theorem, so let's bring that in. Now one way of looking at the intermediate value theorem is through what you might call the Goldilocks principle. Suppose I'm trying to hit a certain function value. At some value of X, I might be too small. My function value might be less than what I want. And at some other value of X, my function value might be too big. It might be more than what I want. Then the intermediate value theorem guarantees that as long as the function is continuous between the two points, some place in between is going to be just right. So let's see if we can apply that. We want to find where F of X equals 0. So if I look at my function values, at X equals 0, my function value is 8, which is too big. At X equals 2, our function value is 1, which is too big. At X equals 4, our function value is minus 3, which is too small. And since my function is continuous, we can reason as follows. At X equals 2, we're too big. At X equals 4, we're too small. So some place in between will be just right. And so there is a solution to F of X equals 0, some place between 2 and 4. And by a similar argument, we see that there's at least one other place where we might have a solution. And so again, at X equals 6, we're too small, while at X equals 8, we're too big. So some place in between, we're just right. And so there's a solution between 6 and 8, and there's at least two solutions. One in the interval between 2 and 4, and one in the interval between 6 and 8. Now let's take a different tactic. We want to find an interval with the width of one unit in which we can find a solution to XQ plus 8X squared plus 50X plus 1621 equals 0. Now the first thing we want to establish is that the expression is a polynomial, which means that F of X is an algebraic function. And one of the things that's important to remember is that algebraic and transcendental functions are continuous everywhere they are defined. And that means we can apply the intermediate value theorem. So, since I have a target value of 0, if I can find a function value that is too large and a close by function value that is too small, then some place in between will be just right. So in order to do that, I might want to collect a couple of function values. So, for example, at X equals 0, our function value will be... And unless we're a politician, we can't tell anything from a single example. So we'll try X equals 1 and see what we get. And there's not a whole lot of insight we can get at this point, so let's try X equals 2 and see what happens. X equals 3 gives us... And if we were infinitely patient, we could try X equals 4, X equals 5, X equals 6, but let's stop a moment and smell the roses. While we could try X equals 4, X equals 5, X equals 6, and so on, we might observe that the values that we're getting are increasing, and so maybe we're going the wrong way. So let's see what happens at X equals negative 1. Well, again, unless we're a politician, we can't generalize from a single example. So at X equals negative 2 or negative 3, and here at least it seems we're going in the right direction. And if we have infinite patients, then eventually we're going to find out that we have at X equals negative 13, a value of 166, and at X equals negative 14, a value of negative 255. And now I can apply the intermediate value theorem, or the Goldilocks principle. At negative 14, we're too small. At negative 13, we're too big. So some place in between, we're going to be just right. So that means f of c is going to be 0, where c is someplace between negative 14 and negative 13. While this works, it's worth remembering that all human progress comes from somebody asking, isn't there a better way of doing this? And so one way we might do this is we might start with larger steps. So instead of evaluating f of negative 1, negative 2, negative 3, and so on, maybe I evaluate f of negative 10 and f of negative 20. And so my Goldilocks principle says negative 10 is too big, negative 20 is too small, so I know that I'll hit 0 for something in between negative 20 and negative 10. And this is great, except our problem actually asks us to find an interval with the width of 1 unit, and the interval between minus 20 and minus 10 is 10 units wide. So this interval is too wide, so let's see if we can narrow it down. So I know there's a solution someplace between negative 10 and negative 20, so maybe I'll get lucky and it's right at negative 15. Well, if you don't play, you can't win. So let's evaluate our function at negative 15. And we don't win. However, our Goldilocks principle says that negative 10 is too big, negative 15 is too small, so there is a solution someplace between negative 15 and negative 10. But our interval is still too wide, so let's narrow it down some more. So I try something between negative 10 and negative 15. How about negative 13? And it doesn't work. But we can apply our Goldilocks principle again. Negative 15 is too small, negative 13 is too big, and so there must be a solution someplace between negative 15 and negative 13. So now we try X equals negative 14. And again, we see that negative 14 is too small, negative 13 is too big, so someplace in between negative 14 and negative 13 will have just right. And notice that we do have an interval with a width of 1 unit, which is what we wanted.