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# Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain. State the domain. $B(x) = \dfrac{\tan x}{\sqrt{4 - x^2}}$

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##### Kristen K.

University of Michigan - Ann Arbor

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So in this problem we are given this function Vfx is tangent of X divided by the square root of form honest. X squared were asked to use the rooms 457 and nine to show that B of X is continuous. And the state the domain for B. Fx. So first of all, bye there. I'm seven attention X is continuous over its domain. Well, what is the domain of this? Well, first of all, the domain for tangent X. Is all the face of X. Such that X is not equal to two n Plus one. Um Spy over to where in is a natural number. Okay. Mhm. Then also by there I'm seven. The skirt of four months x squared. Yeah. Is continuous over its domain. Mm What's the domain of this? Well, let's just see the main of this is what? Well, if I put in two for X I get zero. I can do the screw to zero. Let's put something greater than two. And their let's say three, three squared is nine. So I have four minus nine. That's negative five. I can't do the screw to -5. Oh, that's interesting. So what happens if I put in negative two in here? Well, negative two squared is positive for four months. 40. Okay, so this domain then turns out to be all the real numbers between negative two and two. Inclusively in here. Okay, now then when we look at our theorems, my serum for and nine we have then that vfx is continuous over its domain. Well, what's the domain of the fx mm. Well, first of all for the denominator to not be zero or afford to be defined at all. Okay, then, I have values between minus two and two, don't I? And I can't let the denominator be zero. All right. So this means this is all the values of X. Such that X is in -2-2 without -2 and two being in there. Because those may make the denominator zero. And I can't have this denominator can't be zero. Okay. But I also can't have this right has to be valid over this domain. And what do I notice? I notice that minus pi over two and pi over two are both in here. So I need to add on to this. And also X is not equal two pi over two or minus pi over two. Because both of those show up in that ah in this interval, don't they? Okay, so there is my domain B of X. And by the theorems, provided we know that this is continuous over this domain

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Oklahoma State University

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