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How do you find the domain and range of #arcsin(x^2+y^2-2)#?

Answer 1

Nolan Alexander

Domain of #(x,y)# is the ring between the circles #x^2+y^2=1# and #x^2+y^2=3# and range is from #-pi/2# to #pi/2#.

As the range for #sinx# is #-1 <= x <= 1#

#-1 <= x^2+y^2-2 <= 1#

i.e. #x^2+y^2 >=1# and #x^2+y^2 <=3#

Hence #x# and #y# can take values between #1# and #3#

in other words domain of #(x,y)# is the ring between the circles #x^2+y^2=1# and #x^2+y^2=3#

graph{(x^2+y^2-1)(x^2+y^2-3)=0 [-5, 5, -2.5, 2.5]}

and range for #arcsin(x^2+y^2-2)# is as that of any #arcsinx#, that is between #-pi/2# and #pi/2#

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Answer 2

Caroline Adams

The domain of arcsin(x^2 + y^2 - 2) is -√2 ≤ x^2 + y^2 - 2 ≤ √2, and the range is -π/2 ≤ arcsin(x^2 + y^2 - 2) ≤ π/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.