Posted by Anonymous on .
Find the domain, range, and inverse of the function f(x)=arcsin(sqrt(x))

Calculus 
Reiny,
given:
y = arcsin √x
or
√x = sin y
domain: clearly 0 ≤ x ≤ 1
and the domain is : y any real number
for inverse:
interchange the x and y in
√x = sin y
to get
√y = sinx
y = sin^2 x
so f^1 (x) = sin^2 x or (sinx)^2
verification:
http://www.wolframalpha.com/input/?i=plot+√x+%3D+sin%28y%29