How do you find the domain and range of f(x)=arcsin(3-2x)?

Apr 7, 2017

Domain $\left[\frac{1}{3} , 1\right]$ Range $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

Explanation:

Domain of arcsin x is [-1,1] and range is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

That means $- 1 \le 3 x - 2 \le 1$

Add 2 to the inequality to get $1 \le 3 x \le 3$

Now divide it by 3 to get $\frac{1}{3} \le x \le 1$

Range would remain unchanged.