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2 - I What is the domain of the function Express your answer using interval notation I + 1...

Question

2 - I What is the domain of the function Express your answer using interval notation I + 1

2 - I What is the domain of the function Express your answer using interval notation I + 1



Answers

Find the domain of the given function. Express the domain in interval notation. $$R(x)=\frac{1}{x^{2}-1}$$

What is the domain of this function? Big F of X equals one over X squared, plus one. To find out we need to look for restrictions now. Restrictions with regards to domain are anything that would cause this graph to not exist. It is the graph of this function. So taking the square root of a negative number, for example, or dividing by zero those things are restrictions because they would cause this to not exist. So are there any restrictions here? Well, there's no square roots, so we don't have to worry about that. But this denominator is a bit troubling because what if x squared plus one were equal to zero, then we would be dividing by zero. So let's figure out which exes would cause this to happen so we can avoid them. Well, solving for X, we would get X squared equals negative one or X equals the square root of negative one. Well, the squared of negative one. That's I, but we're only dealing with real numbers, not complex ones. So there are no real Number X is that would cause this to not exist because we're not allowing I So no matter which exceed plug in. Since we're only using real numbers, this is always going to exist. Meaning that the domain of this function is every possible ex. You can't pick a bad extra this So the domain for this function is negative. Infinity all the way up to positive infinity.

So F index is equal to arc co tangent of the square root of two x minus one. Now, typically we would talk about the arc tangent and think we can plug anything we want in. But basically, because of that square root, We need that to X -1 to be some value that is greater than or equal to zero so that will determine what our domain is, So adding one to both sides and then dividing by two. We find out that X must be some value that can be as low as 1/2 and up to positive infinity. That is our domain. What?

Yes. This problem gives us a function F of s is equal to one over as plus one, and it wants us to find the domain of dysfunction and interval notation. Okay, so, um, tradition is bound in all real numbers that we have no issue there about in division, um, you cannot divide by zero. So in this case s, uh, if it equals negative one, we will have a problem. Because division is not found in all real numbers on So if s can't equal negative one, that's really the only issue we have. So, in writing our domain, we're gonna start it. Negative, independent e we're gonna stop at negative one already use parentheses because it can't actually equal negative one. We're in a union that with the interval that starts at negative one and goes to positive infinity, can you notice that you soft parentheses around the negative ones? Because that means it cannot equal negative for Okay, thank you very much.

Let's find the domain of our function. R of X. First off, before we do anything else, I want to rewrite this function into an easier to look at form. That is, the X squared is fine, but this negative 1/2 power is very hard to interpret. Well, the negative means denominator, and the 1/2 means square root, so it's a square root of three minus two X in the denominator. Now that that's established, let's talk about finding domain. In order to find the domain, you need to look at restrictions. These are things that caused the function to not exist at a point, such as taking the square root of a negative or dividing by zero. So let's examine what would cause these things toe happen. First off, dividing by zero, we would only ever have zero in the denominator if the stuff inside the square root were equal to zero. That is, if three minus two X, we're equal to zero so we can solve this and figure out which X would cause this to happen, adding two extra. Both sides gives three equals two x, and dividing by two gives X equals three halves for 1.5 if you prefer decibels, so we have to make sure that we avoid having X equal to 1.5. That's pretty good. Now let's move on to our other restriction square root of a negative. So that is what would make three minus two X being negative or less than zero. Once again, we're going to add two x two. Both sides Giving us three is less than two x, and dividing by two shows that X must be greater than 1.5 or three halves if you prefer fractional form. So not only do we have that X can't equal 1/2 but now ex can't be greater than 1/2 either, because if X is greater than 1/2 than three minus two, X will be less than zero. Ah, and this function will cease to exist. So we could include any number as long as that number is not greater than 1.5. Therefore, the domain of our function. Well, we can go as low as we want. We can go to all the way to negative infinity with low numbers, but we can't go any higher than three halves. And in fact we can't even hit. Three has itself because of right here that would cause the function to equal something over zero. So we use a rounded Brackett showing exclusivity, not including three halves. Thus, our domain is from negative infinity to three halves Exclusive.


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