The absolute value function

[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).

 

[blue_raver22]

olarBear, to prove that the absolute value function has a global minimum at x=0, we can use the definition of a global minimum as a point where the function has the lowest value compared to all other points in its domain.

First, we can show that f(0) is a critical point of the function by taking the derivative of f(x) at x=0:

f'(0) = lim h->0 (f(0+h)-f(0))/h

Since f(0) = 0, we have:

f'(0) = lim h->0 (|0+h|-|0|)/h

= lim h->0 (|h|)/h

Since we are approaching from both positive and negative sides, the limit does not exist. This shows that x=0 is a critical point of f(x).

Next, we can show that f(0) is the only critical point of the function. This can be done by considering the cases when x>0 and x<0 separately. For x>0, we have f'(x) = 1, and for x<0, we have f'(x) = -1. This means that the function is increasing for x>0 and decreasing for x<0, and the only point where the function changes from increasing to decreasing is at x=0. Therefore, x=0 is the only critical point of the function.

Now, we need to show that f(0) is the global minimum of the function. This can be done by considering the definition of absolute value. For any real number x, |x| is always greater than or equal to 0. This means that f(0) is the lowest possible value that the function can take, and there is no point c in the domain where f(c) can be less than f(0). Therefore, f(0) is the global minimum of the function.

In conclusion, we have shown that f(0) is the only critical point of the absolute value function and it is also the global minimum of the function. This rigorous proof confirms that x=0 is indeed the point where the absolute value function has its lowest value compared to all other points in its domain.