Find the domain and the range :

y = sqrt^(9-x^2)

Question

Find the domain and the range : 

y = sqrt^(9-x^2)

anonymous · Answer

Better one : $$y=\sqrt {9-x^2}$$

hba · Answer

9 - x² ≥ 0

anonymous · Answer

I worked it out, and found out the domain -- [-3,3] .. but i'm stuck with the range..

hba · Answer

If x = 3 or -3, then the y value is 0. If you put in any value between -3 and 3, it makes 9 - x² some positive number smaller than 9 which will have a square root less than 3. This continues until when x = 0, y works out to be 3. But then there is a negative sign out front. This makes the y values of the function go down into the negatives so instead of going from 0 to 3, it goes from 0 to -3. The range is [-3,0].

anonymous · Answer

Oh wait! Y's value would never be less than 0 -- we have a sqrt there!

shubhamsrg · Answer

its [0,3] //ofcorse..

anonymous · Answer

I'm not really sure with my answer, but it could be [0,3] :/

hba · Answer

haha.my bad :(

shubhamsrg · Answer

you may either use calculus to find the maxima..
or you may simply understand that maxima will occur at 0..

anonymous · Answer

oh no no ! :P No calculus please :P I mean , i am not supposed to put that maxima-minima formulae here to find out these :P

shubhamsrg · Answer

nevermind,,[0,3] seems correct..