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Use a graph to estimate the equations of all the vertical asymptotes of the curve $ y = \tan (2 \sin x) $ $ -\pi \le x \le \pi $Then find the exact equations of these asymptotes.

There appear to be vertical asymptotes of the curve $y=\tan (2 \sin x)$ at $x \approx \pm 0.90$and $x \approx \pm 2.24 .$ To find the exact equations of these asymptotes, we note that thegraph of the tangent function has vertical asymptotes at $x=\frac{5}{2}+\pi n$. Thus, wemust have $2 \sin x=\frac{\pi}{2}+\pi n,$ or equivalently, sin $x=\frac{\pi}{4}+\frac{\pi}{2} n,$ Since$-1 \leq \sin x \leq 1,$ we must have $\sin x=\pm \frac{\pi}{4}$ and $\operatorname{so~} x=\pm \sin ^{-1} \frac{\pi}{4}$ (correspondingto $x \approx \pm 0.90$ ). Just as $150^{\circ}$ is the reference angle for $30^{\circ}, \pi-\sin ^{-1} \frac{\pi}{4}$ is thereference angle for $\sin ^{-1} \frac{\pi}{4}, \operatorname{so~} x=\pm\left(\pi-\sin ^{-1} \frac{\pi}{4}\right)$ are also equations ofvertical asymptotes (corresponding to $x \approx \pm 2.24$ ).

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 2

The Limit of a Function

Limits

Derivatives

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this is a problem. Number 53 Stewart Calculus, eighth edition, section 2.2 Use a graph to estimate the equations of all the vertical as modes of the curve y equals tangent of the quantity two Synnex And we're only looking in the range of negative pi to pie. We're going to find the exact equations these sm toads after we estimate so. Taking a look at the graph first, you see that vertical as symptoms exist here, where there is, um, the function approaching negative, infinity and positive infinity. And it happens four times in this range. And we're going to estimate a certain value based on the markings of this craft. Here we see. Yeah, this ascent to occurs, it around 0.9 and then this one had around 2.2. So those are going to be our estimates 0.9 and 2.2, and we see that this these assets on the left hand side and the negative X axis occur at the at equal distances from the other ones at negative 0.9, a negative 2.2. So those are going to be our very rough estimations for this first part tax equals 0.9 2.2 negative 0.9 Negative. 2.2 We're going to find the exact equations by doing some calculation with dysfunction. First, we want to know where the vertical ascent oats exist. And for what reason? Detention function has ascent oats. When this input area the input to the tender tension function equals hi or two plus some multiple of pirate too multiple of pie added to pi over two and and represents an integer any positive or any positive negative integer or zero for example, five and a zero. Then tangent of pirate two equals undefined. And this is what we're looking for Vertical I sometimes have to do with function being undefined if n equals one hi plus power to is three power to And there is also another place where tangent is undefined. So we want to look for the X values that make this statement true. Two times the sine of X equals pirate two plus m pie. So we're going to proceed with solving for X divided by two on both sides. We get PIRA for when asked and hi or two. Okay. And we're going to try to figure out and narrow down the possible values of end by taking a look at the range of sign. Same varies from negative one two positive one. So if we take this into account, the largest value that this can be is positive. One so high before us and hi over to is less than or equal to one. And this gives us an end. Are you that is less than or equal to? And we do these calculations, we do one minus our reform. Multiply by two. Never had a bye bye and our end value here, our maximum and value is 0.1 three seven. And then we look for the minimum and value. Bye. Um, seen what the minimum and value would be nor afraid to reach negative one in here if we saw for. And you see that I am must be greater than or equal to negative one. I'm number four terms too. By the right time must be greater than negative 1.1 three in order initial assumption of labeling. What end is we said and would be any integer positive or negative or zero and the only end values here that fit within this range where it is less than 0.137 and greater than make it a 1.137 The only values that are applicable or zero and one. So we use these values and we get, um, two different instances of what sign could be of what X can be kind of X can either be, however, for when x zero nursery when and zero and Xanax can be negative power for because if and is one this is part two. So this is negative pi reform. And then we saw for the X values that are corresponding to this. In this case, we use a inverse sine function for both cases and for the first x value we to get 0.9 zero three, approximately 0.3 And then because tangent has a period of pie, the other X value will be hi minus 0.93 which gives us approximately two 0.24 And so our final answers are better. Assume toots are at plus and minus 0.93 approximately and also plus and minus 2.2 or

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