 Hi there and welcome to the screencast where we're going to find all the inflection points on the graph of f of x equals 5x to the seventh minus 14x to the sixth. And unlike a previous video where we did this entirely graphically, here we're going to do it entirely without graphs just by using the formula and what we know about the first and especially second derivative here. Let's review the main concept that's going to underlie all the calculations we do here. And that is, if you have an inflection point on the graph of a function, it means it's a place where the concavity of the function changes. And something we know about concavity is that it's measured by the second derivative. Namely that when a function's second derivative is positive, it means the original function is concave up. And likewise, where the second derivative is negative, it means the original function is concave down. And so therefore, an inflection point happens at a place where the concavity changes, which is equivalent to saying it's where the sine of f double prime changes. So this process is going to look a lot like finding local extreme values that we did a few screencasts ago by using a sine chart, except now we're going to be interested not in the sine of the first derivative, but rather in the sine of the second derivative. We're going to be interested in finding where that second derivative changes sine because that's where the inflection points are. So first order of business before we do anything is to actually get that second derivative. Very easy to do since f is a polynomial, f prime, first derivative of x is equal to 35x to the sixth minus 84x to the fifth. And therefore, the second derivative, the main object of study here is going to be 210x to the fifth minus 420x to the fourth. And there's quite a bit of factoring we can do here. We'll factor out the greatest common factor, which is 210x to the fourth, and we're simply left with x minus two as a result. So we're going to use this fully factored form of f double prime of x to find the inflection points. What we're interested in finding here is where f double prime is positive and where f double prime is negative. We're especially interested in places where that sine changes from positive to negative or negative to positive. To find those areas of positive or negative behavior, it's going to be helpful first to find where f double prime is zero, because everywhere else f double prime would have to be positive or negative. So where is f double prime equal to zero? Well, that's fairly easy to see if you look at the factored form of f double prime up here. If this thing were equal to zero, then it would mean that either x to the fourth is zero, which means that x itself would be zero, or x minus two would have to be zero, which means that x would equal positive two. So those are the two places where f double prime is equal to zero. Now, I want to pause here and give you a very important warning. Just because the second derivative is zero, these two points, it does not automatically mean that f has inflection points at those two points. It is possible for the second derivative to be equal to zero, and yet there is no inflection point there. An inflection point is not a place where the second derivative equals zero necessarily. It's a place where the concavity changes. And until we know whether or not the concavity of f changes at these two points, we are not allowed to conclude that there are inflection points. We have a lot more work to do, in other words, before we can say anything about inflection points. But let's use the information we just got to set up a little sign chart, just like we did for f prime when finding the local extreme values. I'm going to draw a number line and plot my two zeros on here, x equals zero and x equals two. And as you can see, this splits the number line into three distinct intervals. On each of those three intervals, f double prime is going to be either all positive or all negative. We will not change sign on those intervals because if it did, we would have picked up another point where the second derivative is either zero or undefined, and we've already got all of those points. So I'm going to use the chart here to record two pieces of information. I want to find the sign of f double prime on each of those intervals, and then I'm going to record the behavior of f in terms of concave up or concave down. As we said before, if f double prime is negative, then the f is concave down, and if f double prime is positive, f is concave up. So really what I'm finding here is the sign of f double prime on each of these three intervals. The way I'm going to do that is to look at the factored form of f double prime up here. f double prime is a product of three things, 210, x to the fourth, and x minus two. The 210 obviously doesn't contribute to the sign of f double prime, it's always a positive number. So I'm going to focus on whether x to the fourth is positive on each of these three intervals, and also whether x minus two is positive or negative on each of these three intervals. So let's walk through each of the three intervals and do that. On the first interval here on the far left, we have numbers that are less than zero. Now if I evaluate a number that is less than zero into x to the fourth only, I'm obviously going to get a positive answer, because x to the fourth is something raised to the fourth power, that makes it as always positive unless x is equal to zero. On the other hand, if I put a number from this interval into x minus two, I'm going to get a negative result. For example, x equals negative five. Negative five minus two is negative. So f double prime is a product of plus 210 times a positive quantity, times a negative quantity. That makes f double prime a net negative on that entire interval, therefore f is concave down on that entire interval. Moving on to the second interval between zero and two, on that interval 210 is obviously positive, x to the fourth is positive, and x minus two is negative again. For example, if x were equal to negative one, then x minus two would be a negative result. And since f double prime is the product of 210 times a positive quantity, that makes f double prime negative on that interval. So I'm still concave down on this interval as well. Finally, on the last interval from two to infinity, x to the fourth is positive, and x minus two is also positive. For example, if I chose x equals ten, which is on this interval, that would make ten minus two, of course, is eight. That's a positive number. That means f double prime is positive, therefore f is concave up. Now, let's take stock of what we know now and decide where the inflection points are. The inflection points happen wherever the concavity of f actually changes, or you can think of it as the sign of f double prime actually changes. That's only in one spot. That's only right here at x equals two. Notice again that even though x equals zero made f double prime equal to zero, there was no inflection point there. Here's an example again where just because the second derivative is equal to zero, it doesn't mean there's an inflection point there. For example, there's no inflection point here at x equals zero. There is only one inflection point on this graph, and that is at x equals two. So I have an inflection point, one, and only one inflection point at x equals two. And the sign chart helps me keep track of all the bookkeeping so I will know in a little more structured and organized way where those inflection points happen. Thanks for watching.