anonymous
  • anonymous
Hopefully this will be my last question for the day...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Does anyone know how to get the domain and range from the following function?... \[y=-\sqrt{4-x^{2}}\]
phi
  • phi
do you know you are not allowed to take the square root of a negative number?
anonymous
  • anonymous
Yes, I do.

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phi
  • phi
any idea what x value will make a negative number inside the square root sign? or, more to the point, what x keeps the inside positive or zero? 4-x^2 >= 0
phi
  • phi
to "solve for x" add +x^2 to both sides 4 - x^2 + x^2 >= 0+x^2
phi
  • phi
4>= x^2 2 >= x and -2 <= x (we flip the sign if we pick the negative square root) so we need -2 <= x <= 2 that is the domain
anonymous
  • anonymous
"any idea what x value will make a negative number inside the square root sign? or, more to the point, what x keeps the inside positive or zero?" I'm sorry, but this statement is a bit confusing. I am not sure what exactly you are asking.
anonymous
  • anonymous
Also, how did you get 4-x^2 >= 0 from \[y=-\sqrt{4-x^{2}}\]
phi
  • phi
you want the "stuff" inside the square root to be 0 or bigger , right ?
anonymous
  • anonymous
I haven't done a problem like this in a while, so I don't really have an idea what I am supposed to be doing. Sorry.
phi
  • phi
you start with the idea that you don't want to take the square root of a negative number (because the answer will be undefined)
phi
  • phi
we are allowed to take the square root of zero or any positive number that means that the "stuff" inside the square root, that is 4-x^2 must be either 0 or bigger than 0 we write that as 4-x^2 >= 0 when we simplify that , we will get -2 <= x <-=2 x is from -2 to +2... if it is out side that domain, we will get a negative number when we do 4-x^2, and then we will be trying to take the square root of that negative number.. that is not allowed
anonymous
  • anonymous
Ah! I see now. Thank you :) So if that is how you find the domain, how do you find the range?
phi
  • phi
I would do sqrt(4-0) = sqr(4) = 2 and sqrt(4-2^2)= sqr(0)= 0 we will get numbers from 0 up to 2
phi
  • phi
in other words, I know x goes from -2 to +2 if we pick an "extreme number" like x=-2 and figure out y, we do \[ \sqrt{4 - (-2)^2} = \sqrt{4-4}=0\] I also notice that if x=+2 we will get the same answer for y then as we pick x's closer to 0, y will get bigger, until we get to x=0 at which point y = sqr(4-0)= 2 so the range will be 0 to 2
anonymous
  • anonymous
So you pick an "extreme number" to put in the x's place, and whatever answer you get for y, you then place it into the x's place to receive the next number? (my wording may be confusing)
phi
  • phi
No, not that we know the domain is x= -2 to +2 I also know (because we are doing x^2 in sqr(4-x^2) that the negative x's will give the same y value as the positive x's so the range will be the y values from x=0 to x=2 so I found the y value for x=0 and the y value for x=2 and I know the y values will be between those two numbers.
phi
  • phi
maybe this will be helpful https://www.khanacademy.org/math/algebra2/functions_and_graphs/domain_range/v/domain-of-a-function
anonymous
  • anonymous
Sorry, I had to step away for a bit. I do understand what you're saying now. I just needed to re-read it a few times.

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