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?

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- Nov 6th 2006, 10:20 AMqbkr211 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? - Nov 6th 2006, 11:20 AMJameson
I'm confused on your notation. What does f(x)^1 mean? Is that supposed to be the inverse, or $\displaystyle f^{-1}(x)$?

- Nov 6th 2006, 12:19 PMqbkr21Re:
yes I am sorry

- Nov 6th 2006, 12:44 PMJameson
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? - Nov 6th 2006, 12:47 PMqbkr21Re:
No I got to thant point, I can finish it.

- Nov 6th 2006, 12:49 PMqbkr21Re:
What happens when you take a sin^-1 back across the equal's sign. Does it change back to just plain sin ?

- Nov 6th 2006, 12:56 PMJameson
$\displaystyle \sin^{-1} \left(\frac{y}{3} \right) = x-5$

Take the sine of both sides.

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

$\displaystyle f^{-1}(x)=3\sin(x-5)$ - Nov 6th 2006, 12:59 PMqbkr21Re:
Thanks Jameson, you are the Master of Math!!! And God's Gift to us all!!!