 One method we have of analyzing optimization problems is known as the second derivative test and the most important thing to remember about the second derivative test is that it's mostly useless. So a quick review here, the critical points occur where the derivative is zero or undefined and we should determine whether the critical points correspond to a local maximum or a local minimum by analyzing the sign changes of the derivative around the critical point. But if the derivative is actually zero as opposed to undefined, we can look at the second derivative. If the second derivative is positive, then we know the graph is concave up, so the tangent line is below the curve, and if we think about what this looks like on our graph, that means our critical point is going to correspond to a local minimum value. Similarly, if the second derivative is negative, then the graph is concave down and the tangent line is above the curve, and so the critical point will correspond to a local maximum. And so this leads to the following mostly useless method. Suppose the derivative is actually zero at some point. If the second derivative is positive, then x equals a corresponds to a local minimum value. And if our second derivative is negative, then x equals a corresponds to a local maximum value. And here's the most important thing to remember. If our second derivative is zero or does not exist, the second derivative tells us absolutely nothing. And since finding the second derivative is usually tedious and time-consuming, the second derivative test is usually a waste of time. Well, let's say at least one case where the second derivative test is somewhat useful. So here's a bunch of information about a function and its derivatives, and we might want to keep some other information in mind, namely that f of x is continuous, f prime of x is continuous except that x equals 2, and the table shows all of the critical points. So first off, we might want to identify that the critical points are going to be at x equals 2, x equals 6, and x equals 10. So let's take a look at our second derivative. At x equals 2, our first derivative does not exist, so our second derivative does not exist. And so the second derivative test tells us nothing for x equals 2. How about at x equals 6? Well, here our second derivative is zero, and again, the second derivative tells us nothing for x equals 6. Our third critical point at x equals 10, our second derivative is negative 1, so y equals f of x is concave down at that point, and that means that x equals 10 is going to correspond to a local maximum value. So we do have some information from the second derivative, but we still have to characterize the other critical points. So we notice that f prime of zero is positive, and we know that f prime of x is continuous except at x equals 2. And so that means that f prime of x can only become negative at a critical point. And in fact, this gives us a more general observation. Since f prime of x is continuous except at x equals 2, the sine of f prime of x can only change at a critical point. And so this means that since f prime of x is positive at x equals zero, it has to stay positive until x equals 2. And since the derivative is positive, then I know f of x is increasing until we get to x equals 2. Similarly, since f prime of 4 is negative 1, and f prime of x is continuous except at x equals 2, we know that we can't change signs unless f prime of x is zero or undefined. So that means that f prime of x must be negative after x equals 2, which tells us that f of x is decreasing after x equals 2. So we know that f of x is increasing until x equals 2 and decreasing after, so that tells us x equals 2 is a local maximum value. We also know that f of x is decreasing until x equals 6 and increasing after, so that means that x equals 6 is a local minimum value. And since there are no other critical points, there are no other extreme values. That's about the only type of problem where the second derivative test is actually worth doing. In almost every other case, it is a waste of time. So the best thing you could do now is skip the rest of this video and go on to something that's actually useful. Since you're still watching, you want to see the second derivative test in use, so let's consider this problem. Now, we've already solved this using the first derivative test. So again, the best thing you can do is just skip this analysis using the second derivative test, because it is largely a waste of time. But if you insist on using the second derivative test, it's going to go something like this. So first we find the critical points. We differentiate, and we have to put this in factored form in order to find the location of the critical points. If we insist on using the second derivative test, we'll take the second derivative, which gives us this rather nightmarish form, but we think that it might actually be useful. So let's evaluate this at our three critical points. At x equals 11 thirds, our second derivative is zero, so the second derivative test tells us absolutely nothing. Which means we have to go to our first derivative test, which tells us that x equals 11 thirds is neither a maximum nor a minimum. How about our second critical point? At x equals seven halves, the second derivative at seven halves is zero. So once again, the second derivative test tells us absolutely nothing. And we have to go to the first derivative test. If we analyze the sign of the first derivative, we find that seven halves is a local maximum value. And finally, for our third critical point at x equals 151 over 42, we substitute that into our expression and we find the second derivative is positive. And so we find there is a local minimum at x equals 151 over 42.