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I am given f(x) = sin^-1(3x) for -1/3 \(\leq\) x \(\leq\) 1/3.

I am supposed to find the inverse of the function above and state the domain and range of f^-1, but I do not know where to start..

 Apr 22, 2019
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First of all, to find the inverse, you have to switch the x and y values and solve for y.

So \(y=sin^{-1}(3x)\) becomes \(x=sin^{-1}(3y)\).

To begin solving for y, you must take the sine of both sides.

\(sin(x)=sin(sin^{-1}(3y))\)

The sin of sin-1 cancel.

\(sin(x)=3y\)

Then you divide both sides by 3.

\(\frac{sin(x)}{3}=\frac{3y}{3}\)

The 3's on the y side cancel.

\(y=\frac{sin(x)}{3}\)

So, the inverse of \(f(x)=sin^{-1}(3x)\) is \(f^{-1}(x)=\frac{sin(x)}{3}\).
 

By graphing, you can find the domain and range of f-1(x).

So, the domain of \(f^{-1}=\frac{sin(x)}{3}\) is \(-\infty < x < \infty\)  , and the range is  \(-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\).

 Apr 22, 2019
edited by MemeLord  Apr 22, 2019
edited by MemeLord  Apr 22, 2019

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