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Math Help - Inverse of absolute value

  1. #1
    Member Greengoblin's Avatar
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    Inverse of absolute value

    If f(x)=|x| then what is f^{-1}(x)?

    It is undefined isn't it? since a particular object maps to two images?

    Can someone confirm this? thanks
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  2. #2
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    Hello,
    Quote Originally Posted by Greengoblin View Post
    If f(x)=|x| then what is f^{-1}(x)?

    It is undefined isn't it? since a particular object maps to two images?

    Can someone confirm this? thanks
    Define f as following :

    f(x)=\left\{\begin{array}{ll} x \quad \text{if x $\ge$ 0} \\ -x \quad \text{if x $<$ 0} \end{array} \right.

    Then define an inverse function for x > 0 and for x < 0. It should be okay from here
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  3. #3
    Member Greengoblin's Avatar
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    so:

    f^{-1}(x)= \left\{\begin{array}{ll} x & \text{if x $\ge$ 0} \\ -x & \text{if x $<$ 0}\end{array} \right.

    So the inverse of |x| is |x|?? (I got that by solving f(f^{-1}(x))=x for both intervals.)

    I thought it might be undefined, since I was thinking of the reflection of the graph in y=x, where f:\{x\in\mathbb{R}:x>0\}\to\mathbb{R} is undefined due to any x having multiple value of f.

    Cheers

    P.S. Why isn't my code working for the first bit? It's almost exactly the same as your array.
    Last edited by Greengoblin; October 8th 2008 at 12:31 AM.
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  4. #4
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    Technically speaking, the inverse of |x| is not a function. However, the inverse of y = |x| for x>0 is y = x, and the inverse of y = |x| for x < 0 is y = -x. However, both inverses are only defined for x > 0. The inverse of |x| is not |x|.
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  5. #5
    Member Greengoblin's Avatar
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    Quote Originally Posted by icemanfan View Post
    Technically speaking, the inverse of |x| is not a function. However, the inverse of y = |x| for x>0 is y = x, and the inverse of y = |x| for x < 0 is y = -x.
    I'm confused now. if:

    f(x)=|x|=\left\{\begin{array}{ll} x & \text{if x $\ge$ 0} \\ -x & \text{if x $<$ 0} \end{array} \right.<br />

    And we solve f\circ f^{-1}(x)=x for x\ge 0:

    f\circ f^{-1}(x)=x \implies<br />
f^{-1}(x)=x

    And for x<0:

    <br />
f\circ f^{-1}(x)=x \implies<br />
-f^{-1}(x)=x \implies f^{-1}(x)=-x

    so we have the inverse of an absoulte value is:

    <br />
 f^{-1}(x)= \left\{\begin{array}{ll} x & \text{if x $\ge$ 0} \\ -x & \text{if x $<$ 0} \end{array} \right. = |x|

    So this is my reasoning for thinking the inverse of |x|=|x|. I can't see anything wrong with the way I've done this.

    However, both inverses are only defined for x > 0. The inverse of |x| is not |x|.
    When I think about the inverse graphically, it's not the graph of a function at all. Look at the attached image |x| in red, and it's inverse in blue (reflection in y=x) which is undefined for x<0 since it doesn't take x values in this range, and also for x>0, since any x here, maps to multiple values of y.

    Both methods seem to be correct, but giving me different answers.
    Attached Thumbnails Attached Thumbnails Inverse of absolute value-inverse-absolute-value.jpg  
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  6. #6
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    Actually, only bijective functions have an inverse one

    f has inverse <=> f is bijective

    f is bijective <=> f is injective and surjective

    Now, if we define:
    f:R ->R
    f(x)=|x|
    it has no inverse, since its not even injective

    but if we define:
    f:R+ ->R+
    f(x)=|x|
    we obtain a bijective function

    so it depends on how you define the domain and the codomain
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  7. #7
    Member Greengoblin's Avatar
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    ah yes I see now The function |x| is not invertible over the domain (-\infty , 0)\cup[0,\infty) since it is not bijective, but each 'piece' of the piecewise function is invertible induividually.

    so -\infty ,0)\to(0,\infty ) " alt="f-\infty ,0)\to(0,\infty ) " /> defined by  f(x)= -x has 0,\infty )\to(-\infty ,0) " alt="f^{-1}0,\infty )\to(-\infty ,0) " /> defined by  f^{-1}(x)=-x

    and f:[0,\infty )\to [0,\infty ) defined by  f(x)=x has  f^{-1}:[0,\infty )\to [0,\infty ) defined by  f^{-1}(x)=x

    So unless the domian is specified as otherwise, the function f=|x| is not invertible.

    Thanks

    ugh I always have problems with LaTeX. Anyone care to explain what is wrong this time?
    Last edited by Greengoblin; October 9th 2008 at 09:32 AM. Reason: latex problems
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  8. #8
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    right
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