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Verify if the function below is derivable on x1=0 F(x) = -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)=`
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.
Posted by oldnick on May 12, 2013 at 1:01 AM (Answer #2)
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