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The absolute value function

  1. Sep 24, 2012 #1
    The absolute value function [itex] f(x)=|x| [/itex] has a global minimum at x=0. How could we prove this rigorously? In other words, how could we prove that there is no point [itex] c \ \epsilon \ ℝ [/itex] such that [itex] f(c)<f(0) [/itex]

    (Obviously, the function is not differentiable at x=0 so we cannot apply Fermat's critical point theorem).

    BiP
     
  2. jcsd
  3. Sep 24, 2012 #2
    I think the rigorous proof just goes something like "by definition, the absolute value of a real number is non-negative, therefore no such c exists."
     
  4. Sep 24, 2012 #3

    pwsnafu

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    1. If ##x>0## then ##f(x) = x## so ##f(x) > 0##,
    2. If ##x=0## then ##f(x) = x = 0##,
    3. If ##x < 0## then ##f(x) = -x## and so ##f(x) > 0##,
    and that's it.
     
  5. Sep 24, 2012 #4

    chiro

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    You can also use the derivatives for each analytic portion of the function (you have to break it up into two analytic functions each with their own domain).
     
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