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Question y=3x 1.what is the x intercept2.state the domain in interval notation...

Question

Question y=3x 1.what is the x intercept2.state the domain in interval notation

question y=3x 1.what is the x intercept2.state the domain in interval notation



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Find the domain of the function and express the domain in interval notation. $$y=\log (3 x-1)$$

We're being asked to find the domain of the given function. Well, I see we have a fraction here, which means our denominator can never be equal to zero. So we have to figure out Are there any values of X that will make it zero? So we'll three x ever be equal to zero. Well, this will happen if x zero, because three times zero is zero. So we know X cannot be equal to zero in our domain. However, X can be any other value, and we will have a defined value for our function. So what we will say for our domain is that we will have all really numbers. Except, like we said, X cannot be equal to zero. So our domain is all real numbers except

Let's find the domain of this function P of X first. Before we do anything else, I want to rewrite this when you see the negative 3/5 power. Well, that's pretty confusing. Negative powers and fraction powers aren't easy to work with, so we're going to rewrite this in nicer terms. That is, the X minus one squared is fine on its own, so we can rewrite that. But then, well, this is a negative power, so that means it goes into the denominator of the fraction is we know of X minus one squared over X squared minus nine raised the positive 3/5 power and the positive 3/5 power I'm also going to rewrite. That could be rewritten as the fifth root and cubed that is X squared minus nine. Take the fifth fruit and then cube that entire thing. So that is how this power would be rewritten. And now that it's in this form, let's talk about what it means to take the domain of a function that is, you need to look for restrictions, these air things that would cause the function to not exist, like the square root of a negative or dividing by zero. Let's look for restrictions in this function. So for square root of a negative well on the bottom, we have this fifth root. So what happens if that's negative? Well, there's just the problem that this is 1/5 root, not a square root with 1/5 fruit or with any odd route. No matter what number you plug in, it will exist. That is, the fifth root of negative one is just negative one. You can take the fifth root of negatives just fine. Any odd routes are perfectly tolerant to negative numbers. It's even roots, which stop existing with negative numbers. So there is no way that in this problem we would end up with squared of a negative. That's good. That means we can focus solely on the other type of restriction dividing by zero, and this we can have. So if you notice here on the bottom, we have X squared minus nine, surrounded by a whole bunch of stuff. And if export minus nine equals zero, the whole denominator equal zero and we divide by zero. So let's figure out what makes this happen is X squared minus nine, equal to zero a simple addition of nine gives that X squared equals nine. And in the square root gives X equals not just three but positive or negative three as taking square gets positive and negative. So if X is either positive or negative, three dysfunctional cease to exist. But every other number is acceptable. There's no other numbers that would cause this function to stop existing. Therefore, we can write our Domaine de as follows any number you could think of from negative infinity all the way up to negative three is fine, but way use around in parentheses. Here around bracket. You may have heard us to show that the three or the negative three is not included. Everything up to is included, but not the negative street. Then we use this you symbol meaning union connecting us to another interval from negative three to positive three once again with rounded brackets. This is showing that numbers like negative two or zero are allowed just not the threes. And then we joined it with a final interval from positive three to positive infinity. This is showing every single rial number besides positive and negative three, and that is the domain of our function

Why is equal to look off two x plus one toe The best three We have to find the domain to find their domain. We have to know that the argument two X plus one must be greater than zero. It implies that X is greater than minus one. Divided by two. It means X belongs to open interval minus one divided by toe to infinite, it means their domain off roboto X plus one to the base study is open interval minus one, divided by two to infinity.

Okay, so we want to find the domain of the function. So it's no that here we're undefined when we have division by zero. So that's when I denominated. Her is equal to zero, so that's two is equal to X. So here X cannot be equal to or we have division by zero. So our domain is from the good of infinity. Up to two union with two to infinity.


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