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..
+81 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 \(-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\).
