Verify if the function below is derivable on x1=0 F(x)   = -x^(2/3) if x<=0   = x^(2/3) if x>0    



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Condition of derivability of a function in a point `x_0`  is there exist both left  and right derivatives and     `D^(-)f(x)|_(x_0) =D^(+)f(x)|_((x_0))`           (1)

Now the problem is that  `f(x)`  is not defined in `x=0`  even tought exists   `lim_(x->0) f(x)=0`  

`D^(-) f(x)|_(x_0)=` `lim_(h->0)( f(x-h)- f(0))/(-h)=` `=lim_(h->0) -h^(-2/3-1)=`

`=lim_(h->0) h^(-5/3)=0`

On the other side, we can see that   `D^(+) f(x)|_(x_0)=0`

Note that if `x_0 != 0` function cannot be derivable there, i.e  `x_0=1`

`lim_(x->1^-) f(x)=-1`  while:  `lim_(x->1^+) f(x)=1`

And function wouldnt contiunion neither derivable. 


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