# The absolute value function

1. Sep 24, 2012

### 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

2. Sep 24, 2012

### 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."

3. Sep 24, 2012

### 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.

4. Sep 24, 2012

### 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).