The absolute value function

  • Thread starter Bipolarity
  • Start date
  • #1
775
1

Main Question or Discussion Point

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
828
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."
 
  • #3
pwsnafu
Science Advisor
1,080
85
  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
4,790
131
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).
 

Related Threads for: The absolute value function

Replies
12
Views
4K
Replies
2
Views
581
Replies
1
Views
2K
Top