
Absolute value inverse.
$\displaystyle f(x)=\left  x \right $
I did this:
$\displaystyle \left  x \right = \left\{\begin{matrix}
x; x\in \mathbb{R}^+\\
x ;x\in \mathbb{R}^
\end{matrix}\right.$
$\displaystyle \therefore$
$\displaystyle f^{1}(x):= \left\{\begin{matrix}x;y\in \mathbb{R}^+\\
x; y\in \mathbb{R}^{} \end{matrix}\right.$
but, is there a more 'technical' way of computing this rather than using the definition of both absolute value and inverses such as in the following.
$\displaystyle y=2^x$
$\displaystyle \ln(y)=x\ln(2)$
$\displaystyle \ln(x)=y\ln(2)$
$\displaystyle y=\frac{\ln(x)}{\ln{2}}$
Thank you.
$\displaystyle \int$

The absolute value function fails to be onetoone on the real line (it fails the horizontal line test). Therefore, it has no inverse. Are there any artificial domain restrictions we should know about here?

I would assume it would be the same as $\displaystyle x^2$ which is also noninjective.
The answer comes to be $\displaystyle \pm \sqrt{x}$ which is not a function, yet is graphed using only the real part.

If you're only interested in an inverse relation, not a function, then I would agree that the situations are analogous. Incidentally, you should write the inverse as the following:
$\displaystyle \displaystyle{f^{1}(y)=\begin{cases}y &y\ge 0\\
y &y<0\end{cases}.}$
The graph of the inverse relation is merely the original graph flipped about the line $\displaystyle y=x$, as indeed, all inverse relations in 2 dimensions are produced in the same way.