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# In mathematics, a function (or map) f from a set X to a set Y is a rule which assigns to each element x of X a unique element y of Y, the value of f at x, such that the following conditions are met: 1) For every x in X there is exactly one y in Y, the value of f at x; 2) If x and y are in X, then f(x) = y; 3) If x and y are in X, then f(x) = f(y) implies x = y; 4) For every x in X, there exists a y in Y such that f(x) = y.

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Okay, so here I haven't inequality that I'm trying to solve with an absolute value. And unfortunately, with these inequalities, you just have to think when you solve them, there's really not a great way. I mean, there's some some methods that you can sort of memorized, but the best way to solve them is just to think about it. Now. The first thing I'm going to do is isolate the inequality. So I'm going to subtract one from both sides, so I have absolute value of X minus two is less than three. Here's where I have to think. How can this be true? Well, X minus two could just be less than three. But at some point I'm gonna have a problem. And that's when if X minus two gets so small that it's actually less than negative three. Because if X minus two is, say, negative four when I take the absolute value, that's not going to be less than three. So I really have to things That must be true for this inequality toe hold. The first one is that X minus two must be less than three. But I also need so I'm putting and here to signify that this also needs to be true. X minus two must be greater. The negative three. And again, That's because what's inside the absolute value when I take the absolute value must be less than three. And if X minus two is less than negative three. When I take the absolute value, I'm going to get something bigger. Okay, so I need to solve both of these inequalities. For the first one, I get X is less than one. No, not one. I'm going to add two. So X is going to be less than five area and then here I'm gonna have X is bigger than negative one. So I get two intervals. I have X less than five. So that's negative. Infinity to five. And then I had X bigger than negative one. So that's going to be negative. One to infinity. And the key word is Aunt, I needed two things to be true and I got to distinct sets. So the and tells me I actually need to intersect them. So the solution is going to be the intersection of these to raise negative, intending to five intersect, negative one to infinity and What do I get? I want all the numbers less than five. Bigger than negative one. Well, that's just the interval. Negative one to find. So this is going to be my solution, SEC.

Georgia Southern University

#### Topics

Limits

Derivatives

Differentiation

Applications of the Derivative

Integrals