How Can We Prove the Global Minimum of the Absolute Value Function at x=0?

[Bipolarity]

The absolute value function f(x)=|x| 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 c \ \epsilon \ ℝ such that f(c)<f(0)

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

BiP

 

[Robert1986]

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."

 

[pwsnafu]

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.

 

[chiro]

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).