# Absolute value inverse.

• Aug 23rd 2010, 03:21 PM
integral
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.

$\displaystyle \int$
• Aug 23rd 2010, 05:50 PM
Ackbeet
The absolute value function fails to be one-to-one 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?
• Aug 23rd 2010, 09:35 PM
integral
I would assume it would be the same as $\displaystyle x^2$ which is also non-injective.
The answer comes to be $\displaystyle \pm \sqrt{x}$ which is not a function, yet is graphed using only the real part.
• Aug 24th 2010, 02:19 AM
Ackbeet
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.