Differentiability of Inverse Functions

[The Student]

My notes say, "suppose f is continuous and one-to-one on an interval I and differentiable at a ∈ I. Let b = f(a) and denote the inverse function of f on I by g. If f'(a) = 0, then g is not differentiable at b". But, what about the case where f(x) = x^3? f(x)^(-1) = x^(1/3), f'(x)^(-1) = x^((-2/3))/3. f(x)^(-1)= 0^(1/3) = 0, so isn't f'(x)^(-1) = 0^((-2/3))/3 = negative infinity as x approaches both sides of zero?

 

[The Student]

I edited the OP to be much more clear.

 

[pka]

My notes say, "suppose f is continuous and one-to-one on an interval I and differentiable at a ∈ I. Let b = f(a) and denote the inverse function of f on I by g. If f'(a) = 0, then g is not differentiable at b". But, what about the case where f(x) = x^3? f(x)^(-1) = x^(1/3), f'(x)^(-1) = x^((-2/3))/3. f(x)^(-1)= 0^(1/3) = 0, so isn't f'(x)^(-1) = 0^((-2/3))/3 = negative infinity as x approaches both sides of zero?

I do not understand to point of your example.
\(\displaystyle {\displaystyle\lim _{x \to 0}}\dfrac{{{x^{-\frac{{ 1}}{3}}}}}{3}\text{~DNE}\)

\(\displaystyle - \infty \) is not a number.

 

[The Student]

I do not understand to point of your example.
\(\displaystyle {\displaystyle\lim _{x \to 0}}\dfrac{{{x^{-\frac{{ 1}}{3}}}}}{3}\text{~DNE}\)

\(\displaystyle - \infty \) is not a number.

Oh yeah, that was a silly mistake, thank-you.