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anonymous

  • one year ago

Hopefully this will be my last question for the day...

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  1. anonymous
    • one year ago
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    Does anyone know how to get the domain and range from the following function?... \[y=-\sqrt{4-x^{2}}\]

  2. phi
    • one year ago
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    do you know you are not allowed to take the square root of a negative number?

  3. anonymous
    • one year ago
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    Yes, I do.

  4. phi
    • one year ago
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    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

  5. phi
    • one year ago
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    to "solve for x" add +x^2 to both sides 4 - x^2 + x^2 >= 0+x^2

  6. phi
    • one year ago
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    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

  7. anonymous
    • one year ago
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    "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.

  8. anonymous
    • one year ago
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    Also, how did you get 4-x^2 >= 0 from \[y=-\sqrt{4-x^{2}}\]

  9. phi
    • one year ago
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    you want the "stuff" inside the square root to be 0 or bigger , right ?

  10. anonymous
    • one year ago
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    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.

  11. phi
    • one year ago
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    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)

  12. phi
    • one year ago
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    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

  13. anonymous
    • one year ago
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    Ah! I see now. Thank you :) So if that is how you find the domain, how do you find the range?

  14. phi
    • one year ago
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    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

  15. phi
    • one year ago
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    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

  16. anonymous
    • one year ago
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    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)

  17. phi
    • one year ago
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    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.

  18. phi
    • one year ago
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    maybe this will be helpful https://www.khanacademy.org/math/algebra2/functions_and_graphs/domain_range/v/domain-of-a-function

  19. anonymous
    • one year ago
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    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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