1. Given: f(x)=arcsin(x/3)+5, and -3<=x<=3
Unknown: f(x)^-1=?
I got y= sin^-1(x-5)*5
Any suggestions?
Mmk. So to do this problem, just switch x and y.
Original: $\displaystyle y=\sin^{-1} \left(\frac{x}{3} \right) +5$
Switch x's and y's. $\displaystyle x=\sin^{-1} \left(\frac{y}{3} \right) +5$
$\displaystyle \sin^{-1} \left(\frac{y}{3} \right) = x-5$
Take the sine of both sides. Can you finish it?