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# Trig Question

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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