how do you know if f(x)=x^1/3 is differentiable at x=0?

Question

bahrom7893 · Answer

take the derivative and plug in x=0

anonymous · Answer

i dont think it is differentiable

bahrom7893 · Answer

okay here:
f'(x) = (1/3)x^(-2/3) = 1 / (3x^(2/3)) plug in x = 0, u end up with a 0 in the denominator, therefore.. no it's not differentiable, him is right. and that's how u test it.

anonymous · Answer

okay so if it is undefined for that point then it is not differentiable?

bahrom7893 · Answer

yes

anonymous · Answer

yes.  if the derivative is undefined at any point, then the function is not differentiable at that point.

anonymous · Answer

awesome...thanks guys!!

anonymous · Answer

if you look at the picture you will see why.  
$$y=\sqrt[3]{x}$$ has a nice corner at (0,0)

anonymous · Answer

called a cusp , however a zero in the denominator doesn't always suggest non differentiability.

anonymous · Answer

it is possible that the differernce quotient has different limits from the RHS and LHS

anonymous · Answer

true...but since the numerator is non-zero, it does :)

anonymous · Answer

cusp.  good word that...