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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.

$ A(t) = \arcsin(1 + 2t) $

[-1,0]

02:17

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

University of Michigan - Ann Arbor

University of Nottingham

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Boston College

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So in this problem were given the function A F T equals arc sine of one plus to T. And were asked to use our theorems to prove that this function is continuous over its domain continuous over its domain. And to state what that domain is. All right. So let's let's start with this first of all, if you remember the graph of sign, right graph of sign goes like this doesn't it keeps going okay. And oops, graph of sign goes like, yes, sorry, Where this is Pi over two And this is -9/2. Okay. And remember the definition of arc sine mark signed by definition Y equals the arc sine of X. Yeah. And only if X equals the sign of y. Small words. This is a transpose that's happening. So if I draw the line Y equals X then look what happens right then than the arc sine here coming up to there and going up right like that, something like that. Okay, where This is -1 and this is a positive one. Okay. And so the domain, let's start from their domain Is from -1 to 1 inclusively. And then window by Farum that our Trigana metric functions are continuous over their domain. All right now. So, one thing we need to look at is that ours is arc sine of one plus two. T. Okay, this is the domain of Y equals arc sine X ours is a little bit different. So we need to simply sol minus one is less than a good one plus two. T less than equal to one. And subtract one everywhere. That's -2 is less than or equal to two. T. Let's ankle 20 divided by two So -1 is less than or equal T. is less than or equal to zero. So this means that our domain is from minus one 20. So there's the domain. And using our theorems that is why this function is continuous over its domain.

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