Question

How do you find the domain and range of `arcsin(1-x^2)`?

Answer 1

Domain: `abs x <= sqrt 2`
Range: `[ - pi/2, pi/2 ]`

Explanation:

`y = arcsin ( 1 - x^2 )` is constrained to be in `( -pi/2, pi/2 )` and,

as `( 1 - x^2 )` is a sine value, `- 1 <= 1 - x^2 <= 1`

`rArr -1 <= 1 - x^2 rArr x^2 <= 2 rArr -sqrt2 <= x <= sqrt 2`.

See graph, depicting domain and range.
graph{(y-arcsin(1-x^2))(y-pi/2 +0y)(y+pi/2+0y)=0}

Had it been the piecewise wholesome sine inverse

`(sin)^(-1)( 1 - x^2 ) = kpi + (-1)^k arcsin( 1 - x^2 )`, you can use

the common-to-both inverse `1 - x^2 = sin y`, to get the

wholesome unconstrained graph, for range unlimited.
graph{1-x^2-sin y = 0[ -20 20 -10 10] }