The exercise goes like this: Let $f$ be an invertible function of class $C^k([a,b])$, prove that $f^{-1}$ is of the same class.

But wait a second: $f(x) = x^3$ is invertible and of class $C^{\infty}$ in any interval containing zero, but for the inverse $f^{-1}(x)= x^{1/3}$ the derivative is not defined in $0$. Am I missing something? Is there a hypothesis missing?


You have to assume that $f'(x) \neq 0$ for all $x$, I think. You have that $$ \left(f^{-1}\right)'(f(x)) = \frac{1}{f'(x)} \text{,} $$ and if that is to hold for all $x$, then neither $(f^{-1})'$ nor $f'$ can have any zeros.

  • $\begingroup$ So, as it stands, the claim of the exercise is not true. Thank you. I'll try to solve it assuming the derivative is $\neq 0$. $\endgroup$ – Gennaro Marco Devincenzis Apr 25 '14 at 14:29
  • 1
    $\begingroup$ @GennaroMarcoDevincenzis Another possibility would be to formulate this terms of tangents, not derivatives. $x^\frac{1}{3}$ does have a tangent at $x=0$, it's just that the tangent has infinite slope. But that problem goes away if instead of $f \,:\, \mathbb{R} \to \mathbb{R}$, you look at the function $F \,:\, \mathbb{R} \to \mathbb{R}^2 \,:\, t \mapsto (t, f(t))$, or in other words look at the graph of $f$. $\endgroup$ – fgp Apr 25 '14 at 14:35

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