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Math Help - Absolute value inverse.

  1. #1
    Member integral's Avatar
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    Absolute value inverse.

    f(x)=\left | x \right |

    I did this:

    \left | x \right |= \left\{\begin{matrix}<br />
  x; x\in \mathbb{R}^+\\<br />
-x ;x\in \mathbb{R}^-<br />
  \end{matrix}\right.
    \therefore

    f^{-1}(x):= \left\{\begin{matrix}x;y\in \mathbb{R}^+\\<br />
-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.

    y=2^x

    \ln(y)=x\ln(2)
    \ln(x)=y\ln(2)

    y=\frac{\ln(x)}{\ln{2}}

    Thank you.

    \int
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  2. #2
    A Plied Mathematician
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    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?
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  3. #3
    Member integral's Avatar
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    I would assume it would be the same as x^2 which is also non-injective.
    The answer comes to be \pm \sqrt{x} which is not a function, yet is graphed using only the real part.
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  4. #4
    A Plied Mathematician
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    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{f^{-1}(y)=\begin{cases}y &y\ge 0\\<br />
-y &y<0\end{cases}.}

    The graph of the inverse relation is merely the original graph flipped about the line y=x, as indeed, all inverse relations in 2 dimensions are produced in the same way.
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