anonymous
  • anonymous
Someone help!! find the domain and range of y=-(5-x^2)^(1/2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
if you graph the function you can find range and domain would be all real numbers i believe im not sure though
ivancsc1996
  • ivancsc1996
The domain is all the real numbers, since every number will be positive due to the square, and therefore fit on the square root.
ivancsc1996
  • ivancsc1996
The range is for -infinity to 0 inclusive.

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AkashdeepDeb
  • AkashdeepDeb
We need to find the domain here and that is all the values of x that are allowed in the RHS. Now as it is under square root. x2-5 is greater than or equal to 0 because if it is lesser than 0 then the RHS becomes imaginary. So x2-5 >= 0 So domain is [sqrt 5, infiniti) Union (-infiniti, - sqrt 5] Understood? :)
anonymous
  • anonymous
thanks guys!!! ^^
anonymous
  • anonymous
but what about the range??
anonymous
  • anonymous
Note that \(f(x)=-\sqrt{5-x^2}\) is the lower half of a circle with radius \(\sqrt5\) centered at the origin. This means the range is \([0,\sqrt5]\).
anonymous
  • anonymous
Sorry, I mean \([-\sqrt5,0]\).
anonymous
  • anonymous
Also, @AkashdeepDeb's domain doesn't seem right. For the function to be defined, you have that \(5-x^2\ge0\), or \(5\ge x^2\), which gives a solution set of \(|x|\le\sqrt5\), or \([-\sqrt5,\sqrt5]\).
AkashdeepDeb
  • AkashdeepDeb
ISNT THE WHOLE THING UNDER THE SQRT?? IF NOT THEN Yeah the domain must be [-sqrt5,sqrt5] BUt is the negative sign outside the sqrt? @Mandy_Nakamoto
anonymous
  • anonymous
yup
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=-Sqrt%5B5-x%5E2%5D
AkashdeepDeb
  • AkashdeepDeb
Then follow her solution i took the negative sign inside the sqrt. HAHA :D
anonymous
  • anonymous
im confused.. please explain :(
anonymous
  • anonymous
so the domain is sqrt -5<=x=
anonymous
  • anonymous
Yes
anonymous
  • anonymous
thanks for the help!^^

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