 In this video, we provide the solution to question number nine for the practice final exam for math 1050. We're given a function, this one looks like f of x is equal to 1 over x times 1 over x minus the natural log of x over x squared. In this, with this function f, we have to identify all of the x intercepts and all of the discontinuities here. The x intercepts are what's going to make the rational expression go to zero, the numerator go to zero, I should say, because the rational expression will go to zero when it's numerator is zero. And then the discontinuities are going to come from perhaps things that make the denominator go to zero. You also have a natural log in there, so there's some things to watch out for. So let's put this thing together. Notice that 1 times x, 1 over x times 1 over x is going to be 1 over x squared. That gives us common denominator of x squared. I can rewrite this thing with an x squared in the denominator, and then you end up with a 1 minus the natural log of x right here on the top here. So what can make the numerator go to zero? So that's the first thing to look at there. If you get 1 minus the natural log of x equal to zero, because again, the fraction can only go to zero if the numerator goes to zero. So if 1 minus the natural log of x equals zero, that would suggest that the natural log of x is equal to 1. So you just move the natural log to the other side. And then switching to the exponential form, we get that x equals e to the one that is e, right? So this gives us the x intercept for this graph. That's something to pay attention to. We also care about what makes the denominator go to zero. So if x squared goes to zero, we can take the square root of both sides. We end up with x equals zero, like so. So we have an x intercept at e. We're going to have a vertical asymptote at zero. The other thing to pay attention to, of course, is with the natural log. We do have a natural log in play here. When is, what's the extremal values for the natural log? Basically, as x approaches zero from the right, notice the natural log of x will approach a negative infinity. So there is also going to be a vertical asymptote at zero, but that's actually an agreement with what we already discovered. So it turns out the natural log did not add any new discontinuities. There's a vertical asymptote at zero, an x intercept at e. So if we list the intercepts, the x intercepts, and the discontinuities, we would then get choice h, zero and e are the x values, these critical numbers we were looking for. Now out of curiosity, if you're wondering why was the function written like this, one over x times one over x minus the natural log over x squared. This formula right here is exactly the derivative of a function and when you calculate the derivative, it would look exactly like this. The next step in a calculus problem would then ask with this derivative, what are the x intercepts of the derivative? What are the discontinuities of the derivative? These are known as the critical numbers of the original function. So I presented very much in this calculus context.