Question

How do you find the domain and range of `arcsin(3x+1)`?

Answer 1

Domain: `x in [-2/3,0]`
Range: `y in [-pi/2, pi/2]`

Explanation:

It is easy if you treat it as a transformation of your basic `arcsin` graph, knowing that the domain is `[-.1,1]` and the range is `[-pi/2,pi/2]`.

Then taking your function `y = a f(bx+c) + d`, transpose it so that the actual function is on its own: `(y-d)/a = f(bx+c)` and substitute each `y` and `x` transformation into the range/domain, and solve as an inequality.

Hence you get `-pi/2<=y<=pi/2` and `-1<=3x+1<=1`.

Of course the first one needs no solving; we already have the range.

Rearranging the second inequality we get the domain:
`-2/3<=x<=0`