 So, in addition to giving us the slope of the line tangent to the graph, what else can we do with the derivative? One of the other things we can do is we can try to find inflection points on the graph of y equals f of x. Now, in order to proceed, we'll limit the problem to make it a little easier to solve. And we'll do that by making a couple of assumptions. First, we'll assume that f, f prime, and f double prime are continuous over some interval. And second, we'll assume that a and b are in the interval with a less than b. Since f of x is continuous over some interval, then we know that f of a must exist. Likewise, since f prime of x is continuous over the interval, f prime of a must also exist. The existence of f of a tells us that there is a point on the graph of y equals f of x where x equals a, and the existence of f prime of a means that we know the slope of the line tangent to the graph through that point. And if we put these two things together, we know we can always write the equation of the tangent line. Moreover, since we know f double prime of x is continuous over the interval, we know f double prime of a must exist, and so we can determine the concavity of the graph at any point in the interval. So here's an obligatory mathematical joke. How does a mathematician catch an elephant? Well, one strategy is that a mathematician figures out everything that isn't an elephant, and whatever's left over must be an elephant. Well, in this case, the elephant in the room is the inflection point. So remember that definitions are the whole of mathematics, and inflection point is a place where the concavity changes. So suppose the inflection point occurs of x equals c. Because we've assumed that our function first and second derivatives are continuous, we know f double prime of c must exist, so what can we say about it? If our second derivative is positive, then the graph will be concave up, and it cannot be changing from up to down at x equals c. And if f double prime of c is less than zero, then the graph will be concave down, and it cannot be changing from down to up at x equals c. And so that means our second derivative can't be positive and can't be negative. And since we've assumed that f double prime of x is continuous, it must have some value, and this means that the second derivative must be equal to zero. Now this follows because we've assumed that our second derivative is continuous and that our second derivative has a value at c. But there is always this other possibility if our second derivative is not continuous. If you can't be positive and can't be negative, you might be undefined. What's the geometry of an inflection point beyond the change in concavity? So suppose x equals c corresponds to an inflection point. Let's describe the location of the line through the inflection point with slope f prime of c. So again, since we've assumed f prime of x is continuous, then f prime of c exists, and so there is a line through the graph at x equals c with slope f prime of c. So remember, if you only learn one thing in calculus, you'll probably fail the course. But one of the important things that you should learn is that the derivative is the slope of the line tangent to the curve. And so we might describe this line as the tangent to the curve. But what does it look like? So we know the concavity is changing. So suppose the concavity changes at the inflection point from down to up. So for a little bit before the inflection point, all our tangent lines are going to be above the graph. Now remember, when you draw the tangent lines, you don't want to draw them too long because they'll clutter up the picture. So let's actually draw them short and in fact, let's draw them away from the inflection point so they look something like this. But if we're after the inflection point, all of our tangent lines are going to be below the graph and so they might look something like this. And what this suggests is that the backward part of the line through the inflection point with slope f prime of c is going to be above the graph while the forward part is going to be below the graph. And we might say that our line actually passes through the curve. And a similar argument can be made if the concavity changes from up to down. And so we get the following result. Let x equals c correspond to an inflection point on the graph of y equals f of x where y equals f prime of x and f double prime of x are continuous. The line through the inflection point with slope f prime of c will pass through the curve. So we know that if c is a point of inflection, then the second derivative at c is either 0 or doesn't exist. Does it work the other way around? Suppose our second derivative is 0. Does that give us an inflection point? And we might try to find a graph where the second derivative is 0 at some point but no inflection point exists. For example, suppose my second derivative is 0 at x equals 3. If I want this not to be a point of inflection, then I need to make sure that the concavity doesn't change. So maybe the second derivative before 0 is positive and the second derivative after 0 is still positive. And we'll randomly fill in what our first derivative looks like. So if such a function exists, then here's a case where my second derivative is 0 but we don't have an inflection point. So let's see if we can try and produce a graph with continuous f f prime and f double prime, which matches this description. So we'll start by setting down a few points and keep in mind that we may have to adjust the heights at the end. Since the first derivative is positive for all of these values, then we know that the tangent line's slope upward. But because the second derivative is also positive, we know that those slopes gradually get steeper. So our first tangent line might look like this, but the slope of the remaining lines would have to be even larger. So they would have to look like this. Next, since the second derivative is positive everywhere except at 3, then we know that all of these other points, the graph is concave up. So let's try and adjust the heights and fiddle with this to see if we can get a nice continuous graph. And it doesn't really seem possible to do so. Well, let's try at least one more time just to see if we can get this to work. Maybe we'll change our first derivative so that it's negative, then 0, then positive. And we can try and produce a graph with these characteristics. As before, we'll throw down some points with the i towards adjusting their heights later. Our first derivative starts out negative, so our first tangent line slopes downward. But since the second derivative is positive, we know that the slope is increasing, which means it goes from negative to less negative to 0 to positive to more positive. And since our second derivative is positive, we know that the graph will be concave up. And as before, we'll fiddle with the heights to try and make a continuous curve. And this time it seems to be possible to produce such a graph. And so even though the second derivative is 0 at a point, the graph itself does not have an inflection point. And so it's very important to remember, a second derivative of 0 does not guarantee an inflection point.