Verify if the function below is derivable on x1=0

F(x)

= -x^(2/3) if x<=0

= x^(2/3) if x>0

### 1 Answer | Add Yours

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.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes