in-example-4-we-considered-a-member-of-the-family-of-functions-fx-sinx-sin-cx-that-occur-in-fm-synth

Question

In Example 4 we considered a member of the family of functions $ f(x) = \sin(x + \sin cx) $ that occur in FM synthesis. Here we investigate the function with $ c = 3 $. Start by graphing $ f $ in the viewing rectangle $ [0, \pi] $ by $ [-1.2, 1.2] $. How many local maximum points do you see? The graph has more than are visible to the naked eye. To discover the hidden maximum and minimum points you will need to examine the graph of $ f' $ very carefully. In fact, it helps to look at the graph of $ f" $ at the same time. Find all the maximum and minimum values and inflection points. Then graph $ f $ in the viewing rectangle $ [-2\pi, 2\pi] $ by $ [-1.2, 1.2] $ and comment on symmetry.

Bobby Barnes · Accepted Answer

So, for example, for we&#x27;re going to consider the function efforts is eager to sign of X plus sign three X And what we&#x27;re gonna first do is Graff the function in the viewing window zero pie by negative 1.2 to 1.2. We want to find how many local maximum points it looks like and that they would know we could find. And then we want to use the first derivative and the second derivatives craft to see if there are any hidden maximum or minimum values. Um, and we also want to find what all these maximum minimum values and inflection points are going to be. Lastly, we&#x27;re gonna go ahead and graft f ine the viewing rectangle of negative two pi to two pi by negative one point to that. What point? And we want to comment on any symmetry you might see. So I went ahead and already graft aftereffects on zero pie, too. Her zero pie. By negative 1.2 to 1.2. I&#x27;m just kind of I get on this terrible Well, it looks like we have maybe a max here, so I invite Max, we have another Max right here way have to minimums here. So it looks like we have two maxes in two minutes, at least with this viewing rectangle. So I went ahead and raft the first and second derivative on the same graph. Uh, so the black line is a crime, and the red line is the second driven. It and I grafted on the same viewing the trouble of zero pie by negative 1.2 to 1.2. So all these blue dots here are going to be our inflection points, or at least what my graphing calculator said or inflection points. And all the green dots are going to be our possible critical batons, so looks pretty standard. But one thing that might seem a little bit off to you is over here around 0.5 and 0.7. Well, the red line, remember, is talking about our second derivative, and it says we&#x27;re gonna have two points of inflection here. Well, we have two points of inflection that would make you think, Why do we only have one minimum value or one extreme value in between them? It would make sense to say there should possibly be more. And if we were to actually zoom into that region, you would see that on 0.57 20.17 Far ex access and a longer lie axis negative 0.5 to 0.5 We actually have two Maur values that occur. And if we were just kind of look at this to the left of 0.591 the function is increasing into this point. Decreasing after that tells us that&#x27;s a maximum, uh on. That&#x27;s by using the first rule of the test this second value here. Well, it&#x27;s decreasing into it because it&#x27;s a negative, Soto left, and a positive soap to the right. So this is going to be a minimum and for other value, woods increasing into the point and then increasing after. So that would be a local max. And so if we were to go and look at our original function now and we look at the interval 0.55 to 0.71 ah, along the y axis 0.9999 to one point deserves their 01 You can see that so this we originally thought was supposed to be a maximum, but in reality ended up being a minimum about so it&#x27;s just gonna go back to the first draft will be quick. So this is that interval that we were looking at right here. And we looked at that and we said, Oh, that isn&#x27;t maximum. But in reality, it should have dipped down more like that. And so that point that we thought was a max is actually a minimum. So we&#x27;ve already found all of our maxes and men&#x27;s or at least X values for them along, um zero, the pie and all the inflection points. Now the last thing they wanted us to do was to go ahead and graft this on negative two pi to two pi by negative 1.2 to 1.2. And we need to comment on the symmetry. So you might notice one that from negative to pied a zero 0 to 2 pi, it repeats. So it&#x27;s periodic off a period of two pi, which makes sense because we&#x27;re working with signs and even more than that, it looks like it&#x27;s a metric about the origin, because if we were to come up here. Well, it kind of maxes out. I saved maxes out kind of in quotations because we know it should have a max two maxes and one men around there and on the other side, it looks like it becomes a minimum. So just for symmetry of this, we can say that this is a odd function with symmetry about origin. And so this is nice, because if we found all of our maxes and mittens on that interval zero the pie, we can go ahead and take those same ex fountains and just flip it if it&#x27;s a max or mint and then that would tell us from negative Pi tau pi were all of our Max&#x27;s mens are. And then we know that this is periodic from periodic to pie. And then we could just add to Pike 80 each of those exes to find the rest of our maxes end mids

Problem

Describe how the graph of $ f $ varies as $ c $ v…

Problem 27 Medium Difficulty

Answer

see solution

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