f(x)=arcsin(sqrt(x)) i think the domain is x>=1 inverse is sin(sqrt(x)) not sure about the range can someone help
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I think $\displaystyle x \in [0,1]$ and $\displaystyle f^{-1}= \sin^2{x}$ Now I have given you the domain, what do you think the range could be?
[0,pi/2]
Yep.
how did you come up with the domain
I asked myself 3 questions, what is the domain of $\displaystyle \sqrt{x}$ ? what is the domain of $\displaystyle \arcsin{x}$ and what is there intersection?
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