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For this problem, we have f of x and we have g of x. The operation they want us to do is division.
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So what we need to do is put one on the other to divide them. And when they would divide the house they are asking us to find what the domain is.
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So f of x divided by g of x, that is going to give me x squared plus 6 over the square root of 1 minus x.
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Now look at this problem and there is really nothing else I can simplify with this. So as far as dividing it, that is really about as far as I can go.
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So we have to find the domain. Remember the domain is all the x values that are defined for your function.
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So there is only two times that we have an x value where it is not defined. One is when we are dividing by our x value is 0 and we are dividing by it on the bottom.
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So x is not defined when we divide by 0 and x is also not defined when it is negative under a root.
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So what we need to do for this problem is, first of all when is this going to become 0? Well if I had x equal 1, what is it equal to 0?
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So when does this equal 0? Well if you square both sides, 0 equals 1 minus x, add the x on both sides, x is equal to 1. So therefore we know this function is not defined when x is equal to 1.
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Because when x equals 1, 1 minus 1 is 0, you can't divide by 0. So 1 doesn't work. What about when is it negative under the radical? Well let's test that. So it has to be, we know that the square root of 1 minus x for it to be defined for x has to be greater than or equal to 0.
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Again, let's solve for x. So he square both sides, 1 minus x, greater than or equal to 0. We add the x this time. So 1 is greater than or equal to x.
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Oh sorry, this x is less than or equal to 1. Now we said x canine equal 1. So this has to be all values that are less than or equal to 1.
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How are we set x canine equal 1. So therefore your domain is x is less than or equal to 1. This is not equal to because we said it can't be equal to.
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Got it? I don't think it's low sense. No, what don't you get? It says all this has to be greater than 0. So let's solve for what values have to be greater than or equal to 0.
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So when I solve for it that square, right? And then what happens is I just solve for that. So I move the x over here. So it's actually all values that are less than 1.
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The thing about this is all values that are less than 1, work. Let's put in 5 for there. 1 minus 5 is what? Negative 4, right? You can't take the root of negative 4. But let's say if it was negative 5.
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1 minus negative 5 gives you positive 6. So that's how you do the quotient. That's all you're really doing for those division cast.