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How do you find domain and range for #y=sqrt((4 - x²))#?

1 Answer

Konstantinos Michailidis

Sep 25, 2015

Refer to explanation

Explanation:

Since its a square root we have that

#4-x^2>=0=>(2-x)*(2+x)>=0# which holds for #[-2,2]# hence

#D(f)=[-2,2]#

The range is

#y=sqrt(4-x^2)=>y^2=4-x^2=>x^2=4-y^2>=0=>(2-y)(2+y)>=0#

Because #y>=0# we have that #R(f)=(0,2)#

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