 take the function f of x equals 3x squared over x squared plus five. So I always like to start with the domain because it's a rational function. I care about what makes the denominator go to zero x squared plus five equals zero. That would imply that x squared equals negative five and jk. x squared can never equal negative five, right? That's some imaginary number. This tells us that our domain is going to be all real numbers, negative infinity to infinity. Okay? Now, if we jump to discontinuities, for a rational function, the only way you'll have a discontinuity is if your denominator goes to zero. So since this denominator never goes to zero, this function has no discontinuities. That is to say it's always continuous, always continuous. All right? And let's see, moving on to the next piece. In terms of intercepts, I do care about that, intercepts. Looking at the y-intercept, you get f of zero. This will equal zero over five, which is zero. In terms of x-intercepts, we want to set the function equal to zero, 3x squared over x squared plus five. Now, like I said just a moment ago, if a fraction equals zero, it's because the numerator equals zero. You could multiply both sides of the equation by the denominator, x squared plus five, and do the other side as well, x squared plus five. Well, of course, if you times anything by zero, that'll just be zero, and on the right-hand side, the cancel. So we end up with just zero equals 3x squared, which will imply you can divide by three, take the square root, that x equals zero itself. So this is our x-intercept. It was also y-intercept. So this function is going to go through the origin. If we looked at symmetry, f of negative x, this will equal three times negative x squared over negative x squared plus five. If you take a square of a negative number, it becomes positive. So you get 3x squared over x squared plus five, that's equal to f of x again. This function is even. It's symmetric with respect to the y-axis. If you skip this step, it's not a horrible thing, but it is an even function. Let's use that to our advantage. In terms of in behavior, take the limit as x approaches infinity of our function 3x squared over x squared plus five. When you have a rational function to determine the in behavior, you want to think of who's the dominant term, like who's the fastest growing function on the top and bottom? Well, the numerator is only 3x squared, so that's what it's going to be there. And on the bottom, you have an x squared as well. So this is what we would call a balanced ratio. And so in terms of the in behavior, you only have to look at the leading terms. And so this will look like 3x squared. Sorry about that. 3x squared over x squared as x goes to infinity. The x squared basically cancels out and this will just look like three. That is y equals three. We have a horizontal asymptote. Now for rational functions, the in behavior on the right, if it's a horizontal asymptote, we'll match that on the left. So if we did the same thing here, 3x squared over x squared plus five. In terms of in behavior, only the dominant terms matter. So you get 3x squared over x squared. The x squared cancels out again. You still end up with three. And so this is going to be the same on the left and on the right. All right. Now with this rational function, the derivatives are going to be a little bit more challenging to calculate. So be patient with me here. The quotient rule is going to be necessary as we look for y prime. So by the quotient rule, sing the song if you have to. So we're going to take low x square plus five d high 3x squared prime minus high d low square the bottom. And here we go x square plus five squared, right? And so the derivative of the 3x squared, that's going to be a 6x times x squared plus five. And then the derivative of the x squared will be a 2x. So that gives us 3x squared times just 2x. This sits above the denominator still. We're going to leave the denominator alone. Don't multiply it out. It never does you any benefit to do that. Leave denominators factored. They like that way. In which case, if we keep on going, let's see, where were we? If we keep on going, I'm noticing there's a common factor of 6x on the top and bottom. 3 times, sorry, on the top, you have a 6x right here. And then you have 3 times 2x. If you take out the 6x, because we do want this thing to be factored, that'll be behind x squared plus five. And then we get minus just an x squared for which we see that the x squareds are going to cancel. That's kind of convenient. And this still sits above the denominator, x squared plus five squared. Simplifying that a little bit more, you get 6 times 5, which is 30. So you get 30 over x, 30x times x squared plus five squared. And so in terms of critical numbers, you're going to end up with just zero. Because if zero is what makes the numerator go to zero and the denominator never goes to zero in the situation. The second derivative is going to be a little bit more involved here. Let me actually kind of, well, let's see how much space we have right here. It's a little bit crowded, so I'm going to bring it down here. It's really helpful to simplify your derivative as much as possible before you go on to the next step. So if we were to carry this on down here, that's kind of too much. We want to bring it up over here. So for the second derivative, again, using the quotient rule, we're going to get low d high. Take the derivative at the top, you get 30 minus high d low. So by the chain, we'll get 2x squared plus five times 2x. This all sits above your denominator, which is x squared plus five to the fourth, because we just squared it. Let's try to simplify that. There is a common factor of x squared minus plus five here and here. So factor that thing out. Can we factor anything else out of the fraction? Just give it a quick inspection there. I guess you take a 30 out. That's also acceptable. So we get 30x squared plus five. What leaves behind x squared plus five? And then we have a 4x squared. I think, no, that should be negative, my bad, from the negative sign that happened earlier. So that'll be a negative sign. And this sits above our denominator, x squared plus five to the fourth. Now, because we took out an x squared plus five on the top, we can't lower this power by a little bit. We should combine these x squareds to give you a negative 3x squared. And then from that, I think we have a pretty good second derivative. Let's find a little bit more space here, for which case we're going to get 30 times 5 minus 3x squared all over x squared plus 5 cubed. And so our potential points of inflection, this time around, will look like... Well, we have to set this guy equal to zero. And if you do that, you're going to end up with plus or minus the square root of five-thirds, like so. And so we want to build our sign chart from this. So let's take our PPIs. So we have this negative square root of five-thirds. We had zero as a critical number. And then we had positive square root of five-thirds. What is our second derivative doing right here? So if we pick something less than... If we pick something, like I said, zero right, at zero, this thing ends up being positive. So we get positive and positive. If we pick something greater than the square root of five-thirds, we'll get something negative. And then over here, we also get something negative. We plug some values in there. So this tells us that we're going to have some type of maximum value, I'm sorry, minimum value at zero. It's concave down, then concave up. It's still up and then concave down. And so let's try to put all this information back onto the graph that we had. So if we plug in x equals zero into the original function, we're going to get zero. We had it so it was an x-intercept and a y-intercept. Square root of five-thirds, that's approximately 1.29. And if you put that into the function f, you're going to get like three-quarters approximately. So it's not a really high point. You're going to get something like right here, if you do that point of inflection, and then you get the other point of inflection basically over here. We had, remember, a horizontal asymptote at three. So we get some examples like that. And if we use the notions of concavity and monotonicity, we had a local minimum at zero, and it was concave upward at that location. And so then at these other points, it's points of inflection, the concavity will change. It should be concave down as it approaches this horizontal asymptote, like so. And so this gives us our graph. Let's see what the final result looks like. Computer image gives us something like this. Hey, that's pretty good, pretty good. And so looking at the time, that's about it we have for right now. You can see that these kerf sketch examples can take a while to do. We got through three examples in about 50 minutes. I want to do some more examples of that, and we'll do that next time in lecture 37. See you then. In the meanwhile, if you have any questions or comments, please post them in the comments below. If you'd like a script of what we talked about today, you can find that in the description under the link to my Google Drive. And again, I'll see you next time. Bye.