Thread: 1 Trig Problem with inverse

1. 1 Trig Problem with inverse

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?

2. I'm confused on your notation. What does f(x)^1 mean? Is that supposed to be the inverse, or $\displaystyle f^{-1}(x)$?

3. Re:

yes I am sorry

4. 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?

5. Re:

No I got to thant point, I can finish it.

6. Re:

What happens when you take a sin^-1 back across the equal's sign. Does it change back to just plain sin ?

7. $\displaystyle \sin^{-1} \left(\frac{y}{3} \right) = x-5$

Take the sine of both sides.

$\displaystyle \frac{y}{3}=\sin(x-5)$

$\displaystyle f^{-1}(x)=3\sin(x-5)$

8. Re:

Thanks Jameson, you are the Master of Math!!! And God's Gift to us all!!!