 Let's do another example. This time we have a rational function. f of x equals x plus one over x. So what are issues we have to do with here? Well, because of the denominator, there could be issues that make this thing undefined. And so if we think about the domain of this thing, the domain is gonna be everything, so that the denominator doesn't go to zero. So basically we have to throw out x equals zero. That's not inside of our domain. Now, I often like to do domain with the discontinuity category together, because problems with the domain often are associated to discontinuities. I mean, after all, to be continuous, your function has to be defined. The limit has to exist in the limit in the function after a degree. Well, if you're undefined, that's already an indication of a discontinuity. So what's going on there? Well, as we approach x equals zero from the right, if we take the limit here of our function f of x, well, as you approach zero from the right, you're gonna end up with this one over zero plus, which is positive infinity. And as you take the limit as x approaches zero from the left of f of x, you're gonna end up with one over zero minus, which looks like a negative infinity. And so this is indication that we have some type of vertical asymptote at the y-axis. We'll put this on the screen a little bit later. In terms of intercepts, right? Well, because x equals zero is outside the domain of the function, turns out there is no y-intercept, right? The y-intercept does not exist. In terms of x-intercepts, well, if we looked at the equation f of x equals zero, this would imply we look at x plus one over x equals zero. This gives us that x equals negative one over x. Give myself a little bit more space there. Times both sides by x, you get that x squared equals negative one. I'm gonna stop you right there. x squared can't equal negatives. It has to be non-negative. If we were to take the square root, we're getting some imaginary numbers at that point. So this tells us that x-intercepts also do not exist. So this graph has no x-intercepts nor y-intercepts. That might seem like bad news, but this is actually great news because if you have no x-intercepts or y-intercepts, this tells us that we can never touch the x-axis or the y-axis. That's gonna help us draw our picture actually quite well. In terms of symmetry, this graph does actually have some symmetry. I normally don't care about it, but if you look at f of negative x, you end up with negative x minus one over x, which if you factor that out, x plus one over x, that is factor out the negative one, you end up with negative f of x. So this indicates our function's odd. And so if we were graphing it, it should be symmetric with respect to the origin. All right, let's see, in behavior, this one's a good one to know as well. If we take the limit as x goes to infinity of the function x over one over x, well, what's gonna happen? As x goes to infinity, the one over x is just gonna go to zero, it's gonna vanish. And so this thing will just look like the limit as x goes to infinity of x. And as this is a polynomial, it'll just look like infinity. And similar things will happen as we take the left side, take the limit as x goes to negative infinity of x plus one over x. Same thing's happening as before. The one over x will go to zero and this thing will look like the limit as x approaches negative infinity of here x. This will go to a negative infinity. And in fact, this graph has what commonly called an oblique asymptote. This is some diagonal line where our function f of x will be approximately the same thing as x as you go to the extreme. So we're gonna actually add that oblique asymptote to our graph in just a second. Let's look at the derivative first. Actually, now that I've moved down a little bit, I guess we are kind of in the right spot to start graphing this thing. So what did we say earlier we had? So we have this oblique asymptote, which is the diagonal line y equals x. You get something like that. And we get something like that. And then we also had a vertical asymptote at the y-axis. There was no x-intercepts, no y-intercepts, but we do have these asymptotes right here. So we're gonna find some critical numbers by looking at the first derivative, y prime is gonna equal one minus one over x squared. If we set this equal to zero, we end up with one equals one over x squared, which gives us x squared equals one, which gives us x equals plus or minus one. So we get some critical numbers at plus or minus one. Critical numbers we also should be worried about where does the derivative not exist, which there's the same domain issues we had before. If you divide, if x is zero, you're gonna divide by zero. But don't worry about that necessarily because the original function was undefined at zero, the derivative will also be undefined at zero. And that's not a critical number, that's just a discontinuity from before. And so let's add these critical numbers to our graph here. If we take x to equal one, plug one into the original function, you're gonna end up with one plus one over one, which is two. And so we're gonna plot that point there. We're gonna get x is one, y is two. So we get a point right here, one comma two. I wanna figure out what happens at negative one, which we could plug into the function as well, but we can also use symmetry since we know it's odd. We know there's gonna be a point right here at negative one comma negative two. These are our critical numbers. For the second derivative, we take the derivative of this bad boy right here. And so that's gonna give us positive two over x cubed, in which case that actually never equals zero. If you have a fraction equal to zero, it's because the numerator was equal to zero, which here the numerator is two. There are problems at x equals zero, but again, that's a discontinuity. That's not a point of inflection. So what this tells us is that the derivative could change its inflection at zero. So we do care about it. I mean, don't be ignoring it. The first and second derivative are undefined at zero. And so that's gonna come into our sign chart that we're gonna build right over here. So the numbers we wanna care about are negative one, zero, and positive one. And so what does the second derivative do at these locations? Well, looking at the function two over x cubed, two is always positive. x cubed will have the same sign that x does. And so when x is positive, the second derivative will be positive. And when x is negative, the second derivative will be negative there as well. So we have a minimum at negative one. We have a maximum at x equals one. And because you went from a negative to a positive, this indicates a change of inflection, right? When you're less than zero, your graph will be concave down. When you are greater than zero, you're gonna be concave upward. And so putting this all together, let's take our point one, x equals one here, which is gonna be a local minimum. The graph has to approach infinity when you're to the right of zero. And it has to approach this asymptote as well. And so we get this concave up picture right here. It went from decreasing to increasing because it's a local minimum. And then by symmetry, we're gonna get the same type of picture down here. When you do these curve sketches, you often get more information that's actually necessary to draw the entire graph. Those type of redundancies are okay. We like them because it helps us get a better understanding of the graph right here. And so this is our final picture. Let's see how we did the next page. We can look at a computer generated image right here and voila, we have exactly that picture. We see the two little cup shapes. One is concave up right here. The other one's concave down right here. It was increasing then decreasing. It was decreasing then increasing. The oblique asymptote is illustrated. I did not include the vertical asymptote here just because it's the y-axis and so you can't really see it very well if I were to draw it there. So we did pretty good.