# How do you find the domain and range of arcsin(3x+1)?

Oct 29, 2017

Domain: $x \in \left[- \frac{2}{3} , 0\right]$
Range: $y \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

#### Explanation:

It is easy if you treat it as a transformation of your basic $\arcsin$ graph, knowing that the domain is $\left[- .1 , 1\right]$ and the range is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

Then taking your function $y = a f \left(b x + c\right) + d$, transpose it so that the actual function is on its own: $\frac{y - d}{a} = f \left(b x + c\right)$ and substitute each $y$ and $x$ transformation into the range/domain, and solve as an inequality.

Hence you get $- \frac{\pi}{2} \le y \le \frac{\pi}{2}$ and $- 1 \le 3 x + 1 \le 1$.

Of course the first one needs no solving; we already have the range.

Rearranging the second inequality we get the domain:
$- \frac{2}{3} \le x \le 0$