Question 503576
The square root function is a function which increases over its entire domain (ie going from left to right, the y values increase)



So the smallest possible value in the range will result when you plug in x = -3 (the smallest value in the domain)


So plug in x = -3, 


{{{f(x)=sqrt(9-x^2)}}}



{{{f(-3)=sqrt(9-(-3)^2)}}}



{{{f(-3)=sqrt(9-9)}}}



{{{f(-3)=sqrt(0)}}}



{{{f(-3)=0}}}



So when x = -3, y = 0. So the smallest value in the range is 0.


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Similarly, the largest value in the range will result from the largest value in the domain, which is x = 1


So plug in x = 1, 


{{{f(x)=sqrt(9-x^2)}}}



{{{f(1)=sqrt(9-(1)^2)}}}



{{{f(1)=sqrt(9-1)}}}



{{{f(1)=sqrt(8)}}}



{{{f(1)=2*sqrt(2)}}}



So when x = -3, {{{y = 2*sqrt(2)}}}. So the largest value in the range is {{{2*sqrt(2)}}}


So the range is {{{0<=f(x)<=2*sqrt(2)}}} or {{{0<=y<=2*sqrt(2)}}}



This means that the range in interval notation is 


<img src="http://latex.codecogs.com/gif.latex?\LARGE \dpi{150} \left[0,2\sqrt{2}\right]" title="\LARGE \dpi{150} \left[0,2\sqrt{2}\right]" />



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Thanks,


Jim