<s>[INST] How to transform the graph $\sin^{-1}(3x+1)$? So there is this question in which you have to transform the graph of $\sin^{-1}(3x+1)$ and so is the following:
$\sin^{-1}(3(x+\frac{1}{3}))$ with $D: -1 \leq x \leq 1$
However, if one applies the transformation of moving the graph $\frac{1}{3}$ to the left, it results in $\frac{2}{3}$. Now, you apply the transformation of horizontally compressing it by a factor of $\frac{1}{3}$, to get $\frac{2}{9}$. However, using graphing technology, the graph domain is $-\frac{2}{3} \leq x \leq 0$ which isn't known how to get. 
Thanks. 
 [/INST] Here's why ...
$$\begin{align}
y &= \sin^{-1}\bigg(3(x-(-\frac{1}{3}))\bigg)\\
\sin y &= 3(x-(-\frac{1}{3}))\\
\frac{1}{3}\sin y &= x-(-\frac{1}{3})\\
\frac{1}{3}\sin y + (-\frac{1}{3}) &= x\\
\end{align}$$
so after algebraically flipping the graph $45^{\circ}$ we can easily see that we have a compression along the x-axis by a factor of three and a shift along the x-axis of magnitude $-\frac{1}{3}$. (I avoided the terms 'vertical' and 'horizontal' intentionally here to avoid confusion). 
So, if $\sin y$ ranges over $[-1, 1]$, then $\frac{1}{3}\sin y$ ranges over $[-\frac{1}{3}, \frac{1}{3}]$, and $\frac{1}{3}\sin y + (-\frac{1}{3})$ ranges across $[-\frac{2}{3},0]$, but this is just the domain of $x$.
</s>