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f(x) = sqrt 5/x-2

Mathematics
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Is it\[f(x)=\sqrt{\frac{5}{x}-2}\]or\[f(x)=\sqrt{{5}{x-2}}\]?
lol i wish i could write codes but naw :(

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umm no let me draw it out
Blarg, second one was supposed to be\[f(x)=\sqrt{\frac{5}{x-2}}\]
yes that is correct
please go slowly :D
Ok. (I forgot to put the \frac in the first time.) First, we'll start with assuming that the domain is all real numbers, and then restrict it as we go. Now, we cannot have any values of x that would make 5/(x-2) be negative. The only way that could be negative though, would be if x was less than 2. So we know that x has to be greater than or equal to 2. BUT when we look at the denominator (x-2) we know that we can't let it equal 0. So we can't let x equal 2, or we would end up dividing by 0. So this leaves us with x is greater than 2. So the domain is\[2 < x < \infty\] To get the range, we look at what happens to f(x) as we vary x throughout the domain. The closer and closer we get to 2, the smaller (x-2) becomes. And when we're dividing by an extremely small number we end up getting a very large number. Because there is no number that we can point to as being the next largest number from 2 (basically this means that if I were to try and claim that some number a was the next largest number from 2, you would always be able to point to a number that was between 2 and a). So as x goes to 2, f(x) goes to infinity. A similar argument holds for when x goes to infinity. As we divide by ever larger numbers, we get very small numbers out. However, there's never going to be a way to divide 5 by a number and get exactly 0 out (we can only get close to it), so we can say that y will be greater than 0. So the range is\[0 < y < \infty\]
hold on...getting there ..
so x cannot be negative or cannot be 0 ?
Well, anything inside of a square root sign cannot be negative. Also anything on the bottom of a fraction can't be zero. There are different restrictions for different functions. \[f(x)=x^{2}\]has no such restrictions for example, while\[f(x)=\frac{5}{x^{2}-7}\]has the restriction that x^2-7 cannot be 0.
oh i see
so when it's under sqrt it's restricted
Right.
Ok, thanks for your detailed words :D I must practice more to understand Will get more helped if i need, thank you !!
You're welcome. I have to head back to my apartment now, but I'll be back online in about 20 minutes if you need more help then.
umm i think that's it for the night! i will do some science homework now lol ! thanks so much!! sorry for wasted your time :(
You didn't waste my time. I enjoy helping people learn (that's why I'm here! :P)

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