 In this video, we provide the solution for question number 25 for the practice final exam for math 12-10 in which case this is a curve sketching problem and we need to graph the function f of x equals e to the 1 over x. So the first thing about domain, the only concern with domain is going to be this fraction right here. If the denominator goes to zero, then the fraction would be undefined, the whole function would be undefined. So basically, we have to make sure that x is not equal to zero. If we write the domain in interval notation, we'll end up with negative infinity to zero union zero to infinity. Every number except for zero is included inside the domain. What about x and y intercepts? I'll do the y intercept first because that's always easiest. The y intercept is just looking at f of zero. That is we plug in zero into it, but it's like, whoops, that does not exist because that's outside the domain. So in fact, we'll just take a step back and we're like, oh, the y intercept does not exist. Even if the y intercept doesn't exist, you should comment that it doesn't exist. What about x intercepts? Well, the x intercepts occur when the function is equal to zero. So we have to take e to the 1 over x and set that equal to zero. If you proceed trying to solve that, I mean one, we could stop already and be like, oh, exponentials can never equal zero. But if you switch it to logarithmic form, you'd get the natural log of zero, which is basically like negative infinity, right? It does not exist. And so we're going to back that up as well and then make the comment here. e to the 1 over x doesn't equal zero, and therefore our x intercept does not exist. All right, so to make a comment about that as well. In terms of symmetry, we just need to look at f of negative x. That's how we always check for symmetry. So we get e to the 1 over negative x right there. Then that's going to turn into e to the negative 1 over x, for which there's really not anything we're going to be able to do to get rid of that, right? This is not the same thing as e to the 1 over x. And this is also not the same thing as negative e to the 1 over x. So we get no symmetry in the situation. What about discontinuities? Discontinuities is often connected to the issue about domain, right? Because we observed earlier that x could not equal zero. What's happening as we get closer and closer to x equals zero? So we want to investigate the limit as x approaches zero from above. And we'll do the left one first of e to the 1 over x. Because the natural exponential is a continuous function, we can actually bring the limit out. And we're considering the limit here as x approaches zero from the left of 1 over x, which that's going to give us as we approach zero from the left. So we get e to 1 over zero from the left here. If we approach zero from the left on the function y equals 1 over x, that's going to give us a negative infinity. That's equal to zero, of course. But we see something very different when we take the right-handed limit. Again, because of continuity, we can take the e out of the calculation. So we have to look at 1 over x as x approaches infinity from the right this time. This is going to give us e to the 1 over zero from the right, which gives us this time e to infinity, which is equal to infinity. And so what we discover here is there is a vertical asymptote at x equals zero, the y-axis. But it's a vertical asymptote. What's going to happen is that, again, as we approach zero from the left, we get a zero, but from the right, we get infinity. It's still a vertical asymptote, but the behavior is important to mention there. In terms of end behavior, as we go to the far end of the domain, what do we get here? So if you take the limit as x approaches infinity here of e to the 1 over x, again, by continuity, you can basically just plug in x equals infinity there. You're going to end up with e, take the limit as x approaches infinity here of 1 over x, which you might have just written that as 1 over infinity. That's going to give you a zero, so you get e to the zero, which is equal to 1. So you take the limit as x approaches negative infinity of e to the 1 over x. Same basic idea. By continuity, you can take the e out, and so you take the limit as x approaches negative infinity of 1 over x. Or many of you might just simplify this as saying, don't want to cover up the derivative on the next part. So we could simplify that just by 1 over negative infinity. That's still going to be e to the zero. For being precise, we're approaching zero from the left here. We're approaching from the right, but that doesn't make any difference in this case because as the limit exists, when we approach from the left or right, it's the same thing. We're going to end up with 1. And so we see that this function has a horizontal asymptote at y equals 1. This is all information that we're going to put into our graph, but we'll come back to that in just a moment. So let's investigate the first derivative here. It's given to us. We get that f prime of x is equal to negative e to the 1 over x over x squared. So our critical numbers, that is the numbers we need to be watching out for, the critical numbers. So what if you set x equal to zero? That makes the denominator go to zero. That is this denominator goes to zero, but it also makes this denominator go to zero. And so we do get a critical number at zero. If we try to make the numerator go to zero, e to the 1 over x. Well, this is the issue we had before. The e to the x can't, or no power of e can equal zero. So it turns out this is our only critical number. If we move on to the second derivative, what's going on here? What makes the second derivative go to zero make the second derivative undefined? Well, making x equals zero again makes the second derivative undefined. And so we're going to list this as a PPI. So x equals zero is of course a concern to us, but we also have this factor right here could go to zero. Two x plus one equals zero. That would give us a negative one half. So that's going to give us a potential point of inflection. And so with these PPI's and these critical numbers, we're going to build a sign chart. And I'm going to build it over here. So we have to watch out for zero. The monotonicity could switch at zero. The concavity could switch at zero. I'm also going to include the number negative one half because the concavity could switch at negative one half. We're going to consider the second derivative, the first derivative, and the actual function itself. So think about what's going on here. Some things to note that e to the 1 over x, this is always positive. So its sign is never going to change. X to the fourth is non-negative. It can only be zero at x equals zero. Otherwise, it's always going to be positive. So the reason, the reason I mentioned this is that these two circle parts of the second derivative are never going to change their signs. So the sign of the second derivative depends entirely upon two x plus one. And so as that's a line two x plus one, if you pick a number less than negative one half, you'll get something negative. If you pick something larger than negative one half, you're going to get something positive as well. You could also use test values like you could pick like one negative one fourth negative one. If you want to do test values as well, put those into the second derivative. All right. So what this says about our function is that we're going to be concave down. Then we're going to switch to be concave up and then stay concave up. So as we go from as we go past negative one half, we see that this is going to be a point of inflection because the concavity switched at that location. The second derivative switched. What can we also say about the derivative? Well, it turns out that by the, well, this function doesn't have any extreme. I should mention that yes, there was a single critical number at zero. But since the original function is undefined at zero, it has a vertical asymptote at that location that cannot be a local maximum. Because you might be tempted to say like, oh, the function it's concave up at a critical number. So that's going to be a minimum. But that's not exactly true because it's a vertical asymptote, not an extremum. But we still do care about monotonicity. If we look at the first derivative, which is up over here, notice what's going on right there. Just like we said before, e to the one over x, that's always positive. And what about x squared? Well, x squared just like x to the fourth is always non-negative. It's equal to zero when you're at x equals zero, but otherwise it's always positive. But then you have that negative sign. Don't forget the negative sign that's out in front. This tells us that the first derivative is always going to be negative with the exception at x equals zero. That's undefined. So it's negative, negative and negative. So what this means is our function is always decreasing with the exception at the vertical asymptote where it's undefined. But it kind of wraps around infinity and thus finishes the picture there. All right. So speaking of finishing the picture, let's come back to our graph. And what do we know about this graph? So we know there was a vertical asymptote at x equals zero. So I'm going to put just a red line over the y-axis to indicate to us we have this vertical asymptote. I'm also going to put a line here at y equals one just so we clear the tick marks here. Every two steps will be an integer. So this is four. And to make the graph look a little bit cleaner, we're going to do the same thing for the x scale. One, two, three, four, like that. So let's mention what else do we know? So we have these asymptotes. So we know that the function is undefined at x equals zero. When you are to the right of zero, you're going to be up towards infinity. So we see something like that. But also when you're approaching zero from the left, you get zero. That was the limit we found at this discontent. So we're going to put an open point right here on the graph. Okay. And so what else did we know? So as you approach zero from the right, you go off towards infinity. As you approach infinity, what's going to happen? Are we going to go something like this? Are we going to go something like this? Well, from the limit calculation we had before, where is it? There you are. So as we were doing something like this, we're approaching one, but we're going to be approaching one from above. This, on the other hand, will be approaching one from below. And that's a consequence of these numbers right here. So coming back to our graph, I want to draw this in one continuous swoop. So our function when you're above the horizontal acetote and to the right of the vertical acetote, we're going to get a graph that looks something like this. Okay. What else do we have going on? Well, another point we need to mention is that critical, not the critical point, excuse me, the point of inflection. So notice if we look at f of one, negative one half, that's what we need to do. We're going to take e to the one over negative one half. That's the same thing as e to the negative two, or in other words, one over e squared. And that's approximately 0.135 if we go to three decimal places, which also kind of justifies why I wanted to put such a big gap here, because if I'm going to do 1.35, that's going to be pretty small, pretty teeny. Again, a negative one half. So negative one half would be right here on my scale. And negative, or excuse me, 0.35 would be about right here. So this is my point of inflection. I'm going to label these like we have this remove point at 00. We have this point of inflection and negative one half. And then one over e squared is our point there. And so when you are, when you're between negative one half and zero, it should be concave up. That's what our graph told us earlier. It should be concave up here as well and decreasing concave up, decreasing in this interval right here. You're past negative one half going towards infinity. In that situation, your graph should be concave downward from what we saw earlier with the second derivative, but it's just to be decreasing. So we're going to get something like this. And we're going to approach a horizontal asymptote from below. So this is then the graph of our function f of x equals e to the one, or e to the one over x power.