 Today we're going to talk about the intermediate value theorem, one of the most fundamental theorems in the calculus. What the theorem states is that if a function f is continuous on a closed interval from a to b, then for all k between f of a and f of b, there exists at least one value of c, that is an element of the interval from a to b, such that f of c equals k. And we're going to look at this graphically in just a second. This is an example of an existence theorem, meaning that it simply tells us something exists. It does not really tell us how to go about finding what we want to find. Notice that there can be more than one such number c. It is not unique. This is also the basis for the location principle, which you probably learned about in algebra 2 or pre-calculus, for locating zeros of a function. It's actually how your graphing calculator calculates the zeros of a function. So let's look at this graphically. So here we have a continuous function f of x, and we're talking about from a to b. What the intermediate value theorem states is that if there is a k value in between f of a and f of b, so let's take a look at this. Here's the x value of a, trace that up. Here you'll see f of a right here, and if we do the same thing over at b, trace that up on the curve, and over f of b is right here. And here's the k value right in between f of a and f of b. So what the intermediate value theorem states is that provided k is in between f of b and f of a, and it doesn't matter which is bigger or which is smaller. That doesn't matter as long as k is in between. It guarantees that there is going to be a corresponding c value, the x value that corresponds to the y value of k, that then lies in between a and b. So hopefully it kind of makes a little bit of sense. If you think about it from a graphical point of view. So let's look at an example problem. Suppose we have the function 4 plus 3x minus x squared on the interval from 2 to 5. We want to verify that the intermediate value theorem applies and find the value of c that is guaranteed by the theorem, so that f of c is equal to 1. Now this 1 is our k value, but we are going to have to show somewhere that's in between a and b, in this case 2 is our a and 5 is our b. So somewhere along the line we will have to verify that. So let's think about what the intermediate value theorem states. It states that we must have a continuous function. Well, f of x is a polynomial function and therefore it is going to be continuous for all x's, so that satisfies that one criteria. Now we also need to show that the k value of 1 is in between f of a and f of b. So let's go ahead and find f of 2 and we'll need to find f of 5. And simply substitute in to find those. We find that f of 2 is 6, f of 5 is negative 6, and obviously 1 is in between. And again, it doesn't matter which is bigger or smaller, f of a or f of b. So what we want to find is the c value, and that's going to have to lie in between 2 and 5, such that f of c is going to equal 1. So if we were to substitute c in place of our x's, we would have 4 plus 3c minus c squared, and that needs to equal 1. Well, that's simply a quadratic that we can solve. We can rearrange it, and that would give us a negative c squared plus 3c plus 3 equals 0. It is not factorable, however, you can use the quadratic formula on your graphing calculator. And you do get 2c values. You get a c value of negative 0.791. And a c value of 3.791. Now remember that c needs to lie in the interval in between a and b. So the other only answer that is really good here is that one. Now let's take a look at a graph of this, and we can see that it actually does work out this way. So here's a snapshot from my graphing calculator of that curve. Right here is 2. I scaled this, so here's 2, 3, 4, and 5. And remember our k value was 1, so here's 1 right here. So what this is telling us is that this would be f of 2, and here's the y value for 5 right here. So our k value of 1 does lie in between f of 2 and f of 5. So we trace this over to the curve. Right here is your c value. That's what we found to be the 3.791, and notice it is in between 3 and 4. So most problems involving the intermediate value theorem can be solved this way.