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

1 Answer
Oct 29, 2017

Answer:

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#