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[-/1 Points]DETAILSSCALCET8M 10.2.527.XPFind Jn equalionLhe tungenltne curve wt the point correspondingtne given valueche jaramiclercos(0) sin(86) ,sin(e) cos(80);...

Question

[-/1 Points]DETAILSSCALCET8M 10.2.527.XPFind Jn equalionLhe tungenltne curve wt the point correspondingtne given valueche jaramiclercos(0) sin(86) ,sin(e) cos(80);

[-/1 Points] DETAILS SCALCET8M 10.2.527.XP Find Jn equalion Lhe tungenl tne curve wt the point corresponding tne given value che jaramicler cos(0) sin(86) , sin(e) cos(80);



Answers

$1-54$ Use the guidelines of this section to sketch the curve.
$$y=\frac{\sin x}{2+\cos x}$$

So for this graph, we know that, um, it can't be undefined. So the domain is all real numbers. Um, we do know that if sign is pi over a zero or pie or anything like that, because sign is in the numerator, that would give us our values of zero. So that would be our are ex intercepts. And then if we graph this, we see that the graph will always be increasing. There is never critical point. So no absolute men's No, no, um, no local men's air maxes or anything like that. We see that there are the a lot of inflection points. Um, negative two pi on. Then we have zero. And my assumption is that this is pie, so every pie integer would be an inflection point. We see that there are the X intercept security referred to, um, there's a vertical assim tow. It appears with this graph for their this graph to it's it's not defined at this This value we know that, um, if co sign the way we could make this equal to zero in the denominator would actually be by letting this equal pi so at pie or three pie or anything like that. We had end up getting zero in the denominator, so that would restrict the domain, actually, and it would also give the graph vertical awesome towards that point. So looking at the graph, we see we do have those vertical assam totes at pie and three pie. And we also see this these, um, ex intercepts like we expected. We see similar inflection points as we expected. Also, we see that there's con cave up. It's con cave up here con cave down here. And when we look at the graph, that is, in fact, what we do see, we also see that the graph is increasing the entire time, which is why are derivative graph was above the X axis the whole time. So just looking at those specific properties can help us really determine what our graph is gonna look like.

So the goal of this problem is to sketch the graph and analyze its properties. Um that kind of give the graph its shape that we know. So the graph in this case is sine of X over one Plus Cosine X. And there's all these properties that are wanting to analyze. So we see that the domain, first of all is going to be all real numbers except for these vertical ascent of values. And those are gonna occur at all the odd pi values if that makes sense. So this is this value right here. It's going to is just pie one pie that is and this is three pi five pi so on and so forth. So two N plus one times pi where N is an integer. That's going to be the domain. Those values do not exist on the ground but all other real values do why intercept? We see there is one of the origins zero. That's also going to be an X intercept. But then there's clearly a lot more X intercepts. That's all the even high values. So um that would be too and high. Um where N is an integer value. Again, those are going to be the X intercepts. Then we see that there is symmetry about the origin um based on this right here. So that means that based on that, we do have some symmetry about the origin, which is also odd symmetry. The vertical assam tote as we already discussed, is going to be at all those odd pi values. And then we want to look at the intervals of increase. So this graph is going to be increasing from here to here. Here. Here, here here. So we see is that the graph is always increasing. Um There's never a point at which it's decreasing. We see it's concave up. If we look, for example, it's derivative graph, we see that it is always increasing baits on this and then if we look at its double derivative, we see that it's common cavity is something that's changing. So it's concave up and then it goes concave down and then it keeps constantly changing. Um It's concave up between the positive pi or the even pi values and the occupy values. And then it's concave down from the occupy values to the even pi values. So keep switching back and forth. Um And then that will give us our inflection points as a result. Which then allows us to view our final graph which looks like this.

So the domain for this one is going to be ex. Cannot be Hi. So x cannot be pie. Um, for every period of two pi. So it's important for every period of to play. And because the reason for that is that ask himto exists at X equals pi for every period of two pi. Okay, so our period is two pi obviously intercepts. We have 100 We have one to pi zero. We have one at four pi zero, so you could see the pattern here. And it also goes in the negative direction as well. So that's where our intercepts or symmetry isn't it be odd? So now, usually we would try to find the first derivative second derivative, um, said equal to zero and do the 1st 1st round of tests second derivative test. But both of them are inconclusive and just very difficult. Um, once we try to do them. So, um, instead of doing that, what we can do is we can see the limit as it approaches the ass. Santos is the function. Approaches are asking toads. What is our? Let me get a look like from both the right and the left. So let's try that. The limit the limit as X approaches Kai from the right side of sine X over one Post co sign X. So it's gonna be negative. Let me negative infinity sorted. And then the limits as X approaches. Hi. From the left. Remember this You're asking too of sign X over one plus coastline acts were fun is going to give us a positive infinity. All right. And that's important. Keep that over here. So this is the information that we have to graft. Um, So go ahead and try that. So obviously, this is a periodic function, so it's gonna you're gonna see your repeat some I'm gonna do. Um, first of all, let's put in our let's say this is two plays doesn't readings, and this is negative to pie. And we said that we have Assam toads at pie. So if this is too pie, if this is too pie than pie would be Hey here. So that's gonna be your ass in two, then. Here is negative. Clay brother asked me to. And again, we're also gonna have asked him to us here. The cynics tie for every period of two pie. We're going to see this and it's gonna, you know, his periodic. So it'll continue. But we're just gonna do a couple here and now we have our intercepts. 00 is one. He said that to pi zero is one. And so negative two by zero is one Great. Um, And now let's look it. Our limit as X approaches are asked him to from the right. So coming from the right, it's going to go to negative infinity. So it's gonna look something like this, going too negative infinity. And when it approaches from the left is gonna go up to infinity. And it's simple is that it's just gonna be the same over here as it approaches from the left, Infinity from the right and negative Infinity and Sora graph is going to look like this

So keep in mind, this is a sign graph, so it will have a signing soil pattern. However, since we're multiple or we have Ah, one half X as X increases, we're going to be going to infinity, so we know that the domain is all real numbers. But now we also know that the range is gonna be all real numbers. It will be constantly increasing because as X gets larger, this value will get larger. Sign will always remain within negative one in one, so it won't prevent this part of the function from really growing eso. If we look here, we see that there are lots of critical points. Negative five power three negative power three, power three and so on. Um, we all see that the graph is mostly increasing. Um, however, there are small intervals of decrease small interval of decrease, esos air points to look out for and behavior to look out for him. We could let X equal zero on DWI would see that it would just be zero so that 00 is a Y intercept. That would also be an ex center set a Z see right here, then we can look at this graph when we see that we also have lots of inflection points located at zero pie two pi eso. When we look at the graph, we see, in fact, that the graph has these inflection point towards changing direction because it has that Sinus oil pattern. So just by looking at the properties of the graph were ableto figure out what our actual graph is gonna look like.


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