 So, this is lecture 37 in our course entitled Curve Sketching II. This lecture is intended to be a continuation of what we talked about last time in lecture part 36 about curve sketching. The examples for this lecture will be taken from James Stewart's calculus textbook, particularly section 4.5 entitled Summary of Curve Sketching. And so, in the last lecture we talked about the basics of curve sketching and we worked through three examples and that took a good honking 50 minutes to get through them all. One curve sketching example can take a lot of time. So, I want to do some more examples, three more this time. It shouldn't take as long because we're not learning the basics of it this time, but I want to do some more examples. Last time we had focused entirely on functions that were polynomials or rational functions that is quotients of polynomials. I wanted to look at some which were a little bit more exotic today. So, for example, let's take this function f of x equals x squared over the square root of x plus one. That is there's a square root involved in the calculation this time. How does that affect things? Well, if we start off looking at the domain of this thing, the issues with the domain are going to come from the denominator because we don't want to divide by zero and we also don't want to take the square root negative. So, if we think about the square root issue first, we need that x plus one is greater than or equal to zero, which would suggest that x needs to be greater than or equal to negative one. But likewise, we also don't want to divide by zero in which case the square root of x plus one, we have to figure out what makes that equal to zero. Well, squaring both sides that implies that x plus one equals zero, in other words, x would equal negative one there. So, earlier we had mentioned that x should be greater than or equal to negative one, but here because of division by zero, we actually don't want to equal negative one. So, the domain of this function is going to equal negative one to infinity, where negative one's not included in that. I also want to mention that right off the bat that if its domain is negative one to infinity, we can already say that in terms of symmetry, no, there's not going to be any symmetry here. You can't be even or odd if your domain isn't symmetric. Therefore, I'm not even going to bother calculating it because I know it's not going to work. In terms of intercepts, right, if we were to look at f of zero, we would get zero squared over the square root of zero plus one. That gives us zero over one, which is zero. It's our y-intercept. And then if we were to try the x-intercepts, we have to look at x squared over the square root of x plus one equals zero. Whenever a fraction equals zero, you can really just multiply both sides by that denominator, cancels out. It's only the numerator that matters in this situation, x squared equals zero, and then we get that x equals zero. So, what we've discovered here is that our x-intercept is also our y-intercept. It's going to be zero zero, and the domain is going to be negative one to infinity. So, let me highlight sort of the important things we found here. So, we have an x-intercept and a y-intercept, and our domain was given. Great. Let's consider the in behavior of the function. So, if we take the limit as x approaches infinity here of our function x squared over the square root of x plus one. As x goes to infinity, let's again pay attention to the dominant terms going on right here. In the denominator, the dominant term is going to be the square root of x. In the numerator, it's going to be an x squared. And so, this thing is going to look essentially just like x squared over x to the one half power as x goes to infinity. And so, this would simplify as a fraction just to be x to the three halves as x goes to infinity. And so, as a power function, as x goes to infinity, some positive power of x will also go to infinity. So, this thing points up on the right hand side. But on the left side, things are a little bit different. We can't ask the question what happens as x goes to negative infinity because this graph terminates at negative one. So, we can actually ask what happens as you approach negative one from the right as we get x squared over the square root of x plus one. And so, what happens as x approaches negative one right here? Well, on the top, you're going to get negative one plus squared over the square root of negative one plus plus one. And so, when you square something, it's always going to be positive. And so, the negative one is just going to look like a one on top. Now, we have to be careful on the bottom because if you're approaching negative one from the right and you add one to it, is that going to be a positive number or a negative number? And it's going to be positive because we're a little bit shy of negative one from the right. When you add one to that, that'll give you zero plus on the right hand side. We take the square root of that. Well, that'll simplify to be one over zero plus, which looks like plus infinity. So, it turns out that this is going to give us a vertical asymptote on the graph. And we're going to point up on the right hand side as well. So, let's record what we have so far on the graph. So, we know that this function will go through the origin. It's its x and y intercept. We know that on the right hand side, it wants to point up. Maybe I do a different color for that because we'll have to come back to this a little bit. We know it points up on the right hand side. We also know how to vertical asymptote at x equals negative one. Let's draw that. And so, we also can see that this thing is going to be pointing up on the right hand side. Back to those a little bit later. Let me erase them for now. All right. So, let's think about the derivative for a little bit. How is the derivative going to affect things? Well, the original function, let me write it back on the screen so we can see it here. The original function y equals x squared over the square root of x plus one. And so, as we look at the first derivative, we're going to have to apply the quotient rule again. And I'm not going to go through all the details of calculating the derivative this time just so you're aware. But if we take the derivative, we're going to end up with three x squared plus four x over two times x plus one to the three halves, like so. All right. And so, there's some details to be there. And I apologize for not doing all the details there, but that's going to be the usual quotient rule type of thing. And so, what we want to be looking for is when does this thing go to zero? That'll happen when the numerator goes to zero. So, we're interested in three x squared plus four equals zero. If you factor that thing, you can take out a factor of x. So, you get three x plus four, when does that equal zero? And we can see that this will happen when x equals zero. And negative four thirds. We also have to worry about when the denominator goes to zero. And the denominator is going to go to zero. The two doesn't make much of a difference. The denominator will go to zero when x plus one equals zero, which happens here at negative one. Now, negative one's outside the domain of this thing. So, nothing's really to worry about right there. We need the second derivative, which again, you're going to excuse me. I'm just going to jump immediately to the derivative. Use the quotient rule again on this thing above. And when you take the second derivative, you'll end up with a three x squared, right? Three x squared plus eight x plus eight. And this sits above four times x plus one raised to the five halves power. And so again, the denominator will go to zero when x equals negative one. That's the boundary of the domain. When does the numerator go to zero? Well, the denominator won't make much of a difference there. We have to look at the denominator there, the numerator there, excuse me. So, we have three x squared plus eight x plus eight. When does that equal zero? And when you try to solve this quadratic polynomial, you could try to factor. It doesn't really work. You could use the quadratic formula, but it turns out that when you try to solve for this, it turns out it never actually equals zero. This thing is strictly positive. And so this tells us that the graph has no PPIs. It'll be either always concave up or always concave down back up here. And so if we build our sign chart, the numbers that we should be interested on from what we looked at before, the second derivative never equal to zero, we had a single critical number at zero. And actually, I forgot to mention this earlier, because didn't we have two critical numbers, right? We had zero and negative four thirds. But remember the domain, right? The domain was negative one to infinity, negative four thirds. It's outside of that domain. So we only have the one critical number at zero. So that's all we have to contend with when we look at the second derivative here. Now the second derivative, remember what it was, is the three x square plus eight x plus eight was always positive, the four is always positive, and x plus one will be positive when you're greater than negative one, and it'll be zero when you're less than negative one. And so what we see is that the second derivative is always positive right here. And so this graph is going to always be concave up, concave up, concave up. And so we actually get a local minimum right here on the graph. So x equals zero represents this minimum value. If we approach our asymptote, we're going to go up towards infinity. And then we're also going to approach infinity on the right hand side as well. And so the graph's going to look something like this. Let's take a look at a computer generated image. And we see voila, we had a pretty good picture there.