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no, the derivative is undefined at x = 0

the derivative as x approaches zero from the right is 1 , the left hand slope is -1

What algorithm?

This one: \[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]

the derivative must equal the right hand limit and the left hand limit if it is to exist

I don't think you're really paying attention to what I'm saying.

We have an accepted definition of derivative. I don't know why you want to violate it.

Your picture shows there isn't a unique tangent line at x=0.

^

local max/min can occur when f'=0 or when f' dne

because of the definitions already here
we can make a new definition maybe

and I should say if g'(x) doesn't approach one of the infinities

g star=1 if x>0 and -1 if x<0 and 0 if x=0
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@Michele_Laino Would you like to comment on this problem?