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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- Jul 28th 2010, 01:58 PMdat1611Find the domain range and inverse
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 - Jul 28th 2010, 02:13 PMpickslides
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? - Jul 28th 2010, 02:34 PMdat1611
[0,pi/2]

- Jul 28th 2010, 03:30 PMpickslides
Yep.

- Jul 28th 2010, 03:38 PMdat1611
how did you come up with the domain

- Jul 28th 2010, 03:47 PMpickslides
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?