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Hey guys,

 

How do I find the range of f(x,y)=sqrt(9-x^2-y^2)?

 

Thanks for the help!

 Aug 4, 2017
 #1
avatar+101769 
+1

How do I find the range of f(x,y)=sqrt(9-x^2-y^2)

 

 

\(\sqrt{9-x^2-y^2}\\ =\sqrt{9-(x^2+y^2)}\\ \text{So for real solutions}\\ x^2+y^2\le9\\ -3\le x \le3\qquad and \qquad -3\le y\le3\\ \text{The range is } -3\le y\le 3\)

.
 Aug 4, 2017
 #2
avatar+101405 
+2

 

f (x,y)  = √ [ 9 - x ^2 - y^2 ]  = √ [ 9 - ( x ^2 + y^2) ]

 

We have a 3D surface here

 

Let  f(x,y)  = z

 

In real terms, the minimum for z, 0, will be achieved when x^2 + y^2   = 9

 

And the maximum for z, 3, will be achieved when   x , y   = 0 

 

And the range of z will be      0 ≤ z ≤ 3

 

 

cool cool cool

 Aug 4, 2017
edited by CPhill  Aug 4, 2017
 #3
avatar+101769 
0

Thanks Chris,

I guess it was the the restriction on z scores that were wanted.   :/

Melody  Aug 7, 2017

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