The absolute value function

  • Thread starter Bipolarity
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  • #1
Bipolarity
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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
 

Answers and Replies

  • #2
Robert1986
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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."
 
  • #3
pwsnafu
Science Advisor
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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.
 
  • #4
chiro
Science Advisor
4,815
134
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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