Step 2 of 2: Determine the x-coordinates of any inflection point(s) in the graphEnable Zoom/Pan10...

Find the inflection points on the graph of $y=\frac{1}{x^{2}+1}.$ (See Fig. 2.)

It just doesn't matter that inflection points. Uh Our first of the X. And Y. Value where you change con cavity. Now it doesn't even go from concave up to Con Cape down. But what it breaks down to is we need to have the second group of equal to zero. So as we examine the problem, uh Fx is equal to two E. To the negative X. Squared. We need to jump right into finding the first derivative. Now just a reminder of what you do is with an exponential. You leave it alone. They need to multiply by the derivative of the exponent, which would be negative two X. Now, before moving forward, I would rewrite this two times. Hang up to his negative four X. E. To the negative X. Square. So now I need the derivative of that which is going to be the product rule. So do the derivative of the left side. Which is negative for I'm doing the left side of the product. Leave the right side alone and then plus now lead the left side alone. And then we already did the group of the right side. But we had to do it again. Um And remember the chain rule where you have to do it by itself that you have to take the derivative of that. They have to works. So probably what I would do is factor out both of negative for uh e to the negative X. Squared. And then I'd be left with one. I took this out, I took this out. Um So I saw the minus two X squared equal to zero. Now the reason why this should be helpful is because this piece is always positive. Well it's actually always negative but you don't really care about that. The exponential is always positive, it's the negative that makes it negative. Um So as I'm considering the one minus two X squared equal to zero. I could add two X squared to the right side, divide by two. See and they like the X. Value as just living as plus or minus one over root two when you square root both sides. But you also need to find the wide corner, said the inflection points, the X and the Y coordinate. So we need to plug in both of those in right here. Which is kind of nice because if I square that one of the route to, I'm just back to being one half. So if I plug in one half in for X come on I figure out what the coefficient was to E. To the negative. It changes a sign of that term. And then the other one will be the negative one over root two. And you have to plug into the original so it becomes positive one half power. These should be to inflection points in the problem. Yeah. Yeah.

For the given problem, we want to plot F of X and indicate on the graph where the inflection points occur. Okay, so we have F F x equals two X minus x squared E V X. Yeah. Yeah. Yeah. And we want to indicate on the graph where the inflection points occur. So without using calculus, we see that this graph is concave down here and then it reaches a point right about here, are the graphs concave up, so we can say it's about a negative three and then it goes concave up and then it reaches another point where it goes, starts going concave down. We'll say that that occurs at 1/2 approx, maybe even less than that. And that goes concave down Using the second derivative, we get much more accurate representation, we see it that negative 2.732 and 0.732. So we were close but not fully correct. Um So it explains why the calculus is really necessary to know by taking the second derivative. We see where the graph is going from, changing from concave up to concave down, or changing from concave down to conchita. And the reason why is because that's when the slope of the tangent line, meaning the derivative graph is at its greatest point.

Two go over what the con cavity of this is and or to do that when you find the second relative. Once we get there and determined if there are any inflection points. So let's take a look first, taking the first route. If You have four x -5 taking the second derivative, we have four. Okay, well, so this is always greater than zero, which means that we've got talking about from negative infinity to infinity. Okay, So that means you shape upwards indefinitely. And the other thing here is that well, to be counted as an inflection point, there must be a change inside since it's always concave book, no matter what here, then you would say that there's no inflection point. That's right, sir.

We are going to find regions of con cavity as well as inflection points. So we won't need to get to the second derivative of this function for the first derivative of the one half is just a modification constant derivative of E. To the X. Is E. To the X. The derivative of E to the negative acts would be either the negative acts times negative one. And if we put the negative one in front, we can just change that negative sign the positive. The second derivative then would be 1/2 times E. To the X minus E. To the negative acts. We can rewrite this whole thing. We can make that 1/2 Times E. To the X -1 over E. To the X. And if we wanted to change the quantity to a single fraction, we would get the common denominator. E. To the X. Which would mean on the first quantity we multiply top and bottom by E. To the X and E. To the X. Times E. to the X would be to the two X. Mhm. Now this is never non differentiable because there's no X value that can make the denominator zero. So we're really interested in where that second derivative is equal to zero. So potential inflection points are going to come in, that fraction equals zero and fractions equal zero when the top is zero. So let's set e to the two x -1 equal to zero. We would add the one Giving us each of the two. x equals one. To solve that. We will then take the natural log of both sides. The reason we're going to do that is that the natural log of equal power is the power. So we now have two X equals natural log one. But natural log one is zero. And finally divided by two gives us the inflection point of zero. Now let's check around that point to see Arkan cavity. If we fill in -1, we would end up with on top E to the -2 -1. Uh that would be a positive number. And on bottom we would have a positive as well. So that's positive. If we fill in one we would have, let's back up there actually get a negative on the top because it would be on that first part negative, it would be one over e squared minus one. That is a negative. But if we fell in one we do get a positive. So it is concave down From Negative Infinity to zero. Concave up From 0 to Infinity. The X Value of zero is an inflection point. And we could find the y coordinate for that by filling in zero. But that would give me 1/2 times even the zero minus E to the zero, That's really 1/2 time zero because even the zeros one, So the inflection point is at the zero