How do you find the domain and range of #f(x)=sqrt(9-x^2)#?

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Answer 1 · 2015-09-20T21:37:19

Refer to explanation

For the domain we have that

#9-x^2>=0=>(3-x)*(3+x)>=0=>-3<=x<=3# hence

#D(f)=[-3,3]#

and the range is

#R(f)=[0,3]#

The graph of the function is

graph{(9-x^2)^(1/2) [-10, 10, -5, 5]}

Actually the function represents the upper half of the circle with
equation

#x^2+y^2=3^2#