Question

Someone help!!
find the domain and range of y=-(5-x^2)^(1/2)

Answer 1

if you graph the function you can find range and domain would be all real numbers i believe im not sure though

Answer 2

The domain is all the real numbers, since every number will be positive due to the square, and therefore fit on the square root.

Answer 3

The range is for -infinity to 0 inclusive.

Answer 4

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? :)

Answer 5

thanks guys!!! ^^

Answer 6

but what about the range??

Answer 7

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 \([0,\sqrt5]\).

Answer 8

Sorry, I mean \([-\sqrt5,0]\).

Answer 9

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 \([-\sqrt5,\sqrt5]\).

Answer 10

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

Answer 11

yup

Answer 12

http://www.wolframalpha.com/input/?i=-Sqrt%5B5-x%5E2%5D

Answer 13

Then follow her solution i took the negative sign inside the sqrt. HAHA :D

Answer 14

im confused.. please explain :(

Answer 15

so the domain is sqrt -5<=x=<sqrt 5 and the range is sqrt -5<=y<=0 is it?
Am I right?

Answer 16

Yes

Answer 17

thanks for the help!^^