Question

How do you find the domain and range of `f(x)=arcsin(3-2x)`?

Answer 1

Domain `[1/3,1]` Range `[-pi/2, pi/2]`

Explanation:

Domain of arcsin x is [-1,1] and range is `[-pi/2, pi/2]`

That means `-1<= 3x-2<=1`

Add 2 to the inequality to get `1<=3x<=3`

Now divide it by 3 to get `1/3<=x<=1`

Range would remain unchanged.