Suppose I have some function $f$: $\mathbb{R}_+\rightarrow\mathbb{R}_+$ that is bijective, continuous and strictly increasing. Moreover, it is (at least) twice continuously differentiable everywhere. Since $f$ is bijective, its inverse $f^{-1}$ exists. What conditions need to be satisfied so that $f^{-1}$ is also (at least) twice continuously differentiable everywhere?

From wikipedia's Inverse FT article: when $f$ is a continuously differentiable function with nonzero derivative at the point $a$, then $f^{-1}$ is continuously differentiable. Given that $f$ in my case is strictly increasing $\implies$ $f'>0$ this condition seems satisfied for all $a \in \mathbb{R}_+$.

Counterexamples in comments assume that $f'(x)=0$ at some $a$, but this is ruled out by the assumption that $f'(x)>0$.

  • 2
    $\begingroup$ You might want to look at $f(x) = x + c \sin x$ for $0 < c < 1$ to get an idea of what might go wrong $\endgroup$ Dec 16, 2017 at 22:25
  • 3
    $\begingroup$ That function is not strictly increasing and is not bijective. $\endgroup$ Dec 16, 2017 at 22:32
  • $\begingroup$ every injection $\mathbb R \to \mathbb R$ is already strictly monotone. $\endgroup$ Dec 16, 2017 at 22:38
  • $\begingroup$ the derivative of $f^{-1}$ is not defined at every point where $f'(x)=0$. $\endgroup$
    – Masacroso
    Dec 16, 2017 at 22:39
  • $\begingroup$ @MarkFischler: Did you notice that I restricted to the case where $0 < c < 1$? That guarantees positive derivative. $\endgroup$ Dec 16, 2017 at 22:40

1 Answer 1


The condition is exactly that $f'(x)$ is nonzero (or equivalently, positive) for all $x$. Indeed, if $f^{-1}$ is differentiable everywhere, then differentiating the identity $f^{-1}(f(x))=x$ gives $(f^{-1})'(f(x))f'(x)=1$ so $f'(x)$ can never be zero.

Conversely, if $y=f(x)$ and $f'(x)\neq 0$, then $f^{-1}$ is differentiable at $y$ with $(f^{-1})'(y)=1/f'(x)$ (indeed, you can prove this directly from the definition, and the difference quotients to compute $(f^{-1})'(y)$ are the reciprocals of the difference quotients for $f'(x)$). So if $f'$ is always nonzero, then $f^{-1}$ is differentiable everywhere with $(f^{-1})'(y)=1/f'(f^{-1}(y))$. If $f$ is twice continuously differentiable, we can then differentiate $(f^{-1})'$ by the chain rule to find that $f^{-1}$ is twice continuously differentiable.

Note, however, that merely assuming $f$ is differentiable and strictly increasing does not imply $f'$ is positive everywhere. For instance, consider $f(x)=x^3$, which is strictly increasing and infinitely differentiable but $f'(0)=0$ and $f^{-1}(x)=\sqrt[3]{x}$ is not differentiable at $0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .