 Right as a next example Take the function f of x equals the natural log of x over x squared in terms of the domain What issues do we have to worry about well when it comes to the natural log of x? The input number x has to be greater than zero and in the denominator We can't have x squared equaling zero which would imply that x equals zero But as the natural already thinks that x should be greater than zero we get that our domain is going to be zero to infinity right there I'm going to circle it So we see that If we switch our attention to x intercepts and y intercepts I should say the y intercept doesn't exist because after all if x can't equal zero That's exactly where the y intercept is so there's no y intercept If we think of it terms of the x intercepts when does the natural log of x over x squared equals zero Times in both sides by x squared you get that when is the natural log of x equals zero And using the inverse function relationship here the natural log of x equals It implies x equals e to the zero ak one and so this gives us our y intercept Excuse me x intercept We have a single x intercept right there at one in terms of discontinuities The limit discontinuity we'd have to worry about would be when x goes to zero because that makes the genre go zero But that's already outside the domain So I guess there's an issue about what happens we approach zero that basically leads us automatically to the in behavior, right? What happens as x approaches? Say zero from the right the natural log of x over x squared and So if we investigate this one, what's the natural log of zero from the right over? zero plus Square Well, the denominator is pretty simple if you square a positive it stays positive So we're gonna be just a little bit to the right of zero zero plus here on the natural log as as Zero x approaches zero from the right the natural log wants to go towards a negative infinity And so we might be worried is this an indeterminate form? Well, if you divide by zero that's kind of like the same thing as multiplying by infinity So this really looks like negative infinity times positive infinity and that's actually not an indeterminate form whatsoever That would just be negative infinity and so this will represent a vertical asymptote on the graph That is as you get closer and closer to Zero you're gonna be approaching negative infinity We also should see what it does on the right-hand side as the limp take the limit as x approaches infinity the natural log of x Over x squared now in this situation if you just plug in x equals infinity here You're gonna the natural log of infinity over infinity squared. This one actually does have an indeterminate form infinity over infinity So to resolve this one, I would have recommend using Lopital's rule For which we can take the derivative of top which give one over x in the derivative of the bottom, which is 2x As x approaches infinity still Simplifying that fraction We're gonna end up with one over 2x squared as x goes towards infinity and in that situation We're gonna end up with one over infinity, which is zero. So this function has a horizontal asymptote at the x-axis All right, the remainder of what we need to talk about comes from the derivatives, so let's compute the derivatives of these things And so pretty feel pretty awful what I did earlier just give me through the derivative We should we should do the calculation. All right, we can do it Those steps I did skip before you can't find them in the lecture notes, but let's do the quotient rule this time So we get low D high Minus high D low Square the bottom here we go And so factor the top will simplify things first, I guess x squared times one over x is just an x we get minus 2x times the natural log of x over x to the fourth You can factor out the x that's in the top And that leaves behind one minus two natural log of x Over x to the fourth for which we can cancel The x on top with one of the x's on the bottom And so you have one minus two natural log of x over x cubed We want to remember this guy right here because we're gonna have to take the second derivative in just a moment But in the meanwhile, let's think of the critical numbers The denominator goes to zero When x is zero we already know that's the boundary of the domain. When does the numerator go to zero one minus two natural log of x? We solve this equation right here. We can add two times the natural log of x to both sides So we get one equals two ln of x Which would tell us the natural log of x equals one-half or x equals e to the one-half power Or if you prefer the square root of e And let's see the square root of e That's approximately I should have put this in my calculator beforehand. This is approximately 1.65 for future reference Let's compute the second derivative If we take the derivative of the function we had before Again, we can use the quotient rule to help us along the way So We get low D high the derivative of one minus two natural log of x would be negative two over x minus high D low Square the bottom here we go. So we get x to the sixth in the bottom right here And then trying to simplify this thing along the way x cube times negative two x will give us negative two x squared We get this negative three x squared times one minus two natural log of x All over x to the sixth So factor out the negative x squared that's common to the numerator Whoops, sorry about that and then that leaves behind what do we got a two plus three times one minus two Natural log of x And this sits above the denominator Which is x to the sixth So we can simplify the denominator a little bit. We get x squared cancel there leaving four there What's going to happen in the numerator? We're going to have three times one, which is three plus two, which is a five So we get negative five And then we're going to get a negative six natural log of x sitting above x to the fourth So the ppi's here again the denominator will go to zero at zero the numerator go to zero Well in that situation get five minus six natural log of x equals zero Very similar to what we did a moment ago, we're going to get the natural log of x is equal to five sixth And so our ppi I should say potential point of inflection We don't know it's an inflection point yet. It's going to be e to the five sixth power Which is approximately 2.3 So let's try to graph those points right here Now our first our critical number was 1.65 the square root of e so we get this point Well, we had an x intercept. Let's do that one first the x intercept was at one The critical number is going to be 1.65 approximately if you put it into the function You're going to get something super super teeny tiny If you put that into the function you get a point one eight So in terms of scale, let's actually make the y-coordinate be one point Just sorry just point one one tenth right there And so when you look at the y-coordinate, uh, this thing is just going to be a t little guy Right here. So we get this point 1.65 comma comma What did I say point one eight? Super small And then for our point of inflection, we got e to the five sixth, which was about 2.3 If you plug that into the original function f you're going to get point one six of the y-coordinate So What did I say 2.3? So we graph that it's going to be about right here 2.2 3 or 2.3, excuse me And then point one six that guy's not very big whatsoever And so we know we have a vertical asymptote at the x-axis at the y-axis We have a horizontal asymptote at the y-axis and if we fill out our sign chart here So we're looking at 1.63 And 2.3 If we look at the second derivative, which is off the screen right now back up Our second derivative was this thing right here The denominator is always going to be positive. There's a negative in front. This thing will be negative until it hits its It's x. I'm sorry. This this thing should be yeah, this thing will be Positive until it hits its x intercept and it'll switch to be negative But because of the negative sign actually switches things around So if we were to plug these into our function right here into the second derivative It's going to be negative negative and then positive And so what this tells us is that we have a point of inflection. It's going to be concave downward Like this going towards negative infinity And at this point it's going to switch to be concave upward as it approaches the horizontal asymptote Which is the x-axis so this gives us our graph right there looking at a computer image. We get this picture right here So notice that this point is an inflection point. It's concave down right there. It's concave up right there This point right here is a maximum. It was increasing then decreasing You see the asymptotic behavior as it approaches the x-axis. You see the asymptotic behavior as it approaches the y-axis there