In the math textbooks I’ve banged my head on, the absolute value as a function is defined thusly:
If x >= 0, then |x| = x. If x < 0, then |x| = -x
Which is all fine and dandy, except that you can shift the equal sign over to the x < 0 without much of a hitch. If you set f(x) = |x|, then:
f’= 1 when x > 0, f’ = -1 when x < 0 and at 0 f’ does not exist (DNE)
So hey, why not define |x| this way:
If x > 0, |x| = x
If x < 0, |x| = -x
If x = 0, |x| = 0
This gives us the same derivative EXCEPT for the case where x = 0. Now the derivative DOES exist and it equals zero.
I do realize that if you graph this function, there is a discontinuity at x=0 that would, at least on visual inspection, imply that f’(0) DNE. Still, given that most (all?) v-shaped discontinuities are due to having an absolute value present in the function. So why not define the derivative of this type of discontinuity as equal to zero? Given that there seems to be an arbitrary definition of 0! as equal to one, it doesn’t seem outside the realm of logical possibility.
DISCLAIMERS:
#1: This is on the Real number line; if there is a good reason that uses Complex numbers as proof, explain away. But since I am a statistician in training, do expect me to run from this thread screaming.
#2: If there is a good mathematical reason why 0! = 1 that doesn’t involve “well, it works” I’m all ears.