inverse of a sin function

[btrfly24]

f(x)=2-sin(pix-pi) Find the inverse and state domain for 1/2<x>3/2

I know to switch x's with y's but do you factor out the pi first? This is what I have so far:

x=2-sin pi(y-1)

First subtract the two to get x-2=-sin pi(y-1)
then divide by the -1 to get 2-x=sin pi(y-1)
divide by pi to get (2-x)/pi =sin(y-1)
Now where?? Did I do it right so far?

 

[mark]

you can't take out pi because its in the sin function

f(x) = 2 - sin(pix - pi)

x = 2 - sin(piy- pi)

x-2 = -sin(piy-pi)

2 - x = sin(piy - pi)

arcsin(2-x) = piy - pi

arcsin(2-x) = pi(y - 1)

arcsin(2-x)/pi = y-1

arcsin(2-x)/pi +1 = y

 

[btrfly24]

I get it now!! Thanks a million for your help Mark!!

 

[btrfly24]

how do you figure out the domain?

 

[mark]

the domain of y = arcsin(2-x)/pi + 1 is from 1 to 3

you can find this by looking at the graph

or

because the range of sin(x) is -1 to 1 and the domain of arcsin(x) is from -1 to 1
the range of f(x) becomes the domain of f(x) inverse

so if you plug in and 1/2 and 3/2 into f(x) = 2 - sin(pix - pi) you get 1 and 3 for the range, so 1 and 3 become the domain of f(x) = arcsin(2-x)/pi +1