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

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Answer 1 · 2017-04-07T02:53:44

Domain $[\frac{1}{3}, 1]$ $[1/3,1]$ Range $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2, pi/2]$

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

That means $- 1 \leq 3 x - 2 \leq 1$ $-1<= 3x-2<=1$

Add 2 to the inequality to get $1 \leq 3 x \leq 3$ $1<=3x<=3$

Now divide it by 3 to get $\frac{1}{3} \leq x \leq 1$ $1/3<=x<=1$

Range would remain unchanged.

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