I've been set this problem recently and I'm having a lot of trouble with it. Any help would be much appreciated!

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function with continuous derivatives of all orders and suppose that, for some $x\in \mathbb{R}$ the derivative $f'(x)$ is non-zero. Thus there exists an interval D containing x such that $f'(y)\neq 0$ for all $y\in D$. We can also show that f is Lipschitz with Lipschitz constant less than 1.

Show there is a unique function $g:f(D)\rightarrow D$ such that $f\circ g$ and $g\circ f$ are the identity maps on $D$ and $f(D)$ respectively. Also show that g is continuous and differentiable.


$f'$ is nonzero on $D$ so $f$ is strictly monotonic. Thus it is one-to-one. This means that $f:D\to f[D]$ has an inverse $g:f[D]\to D$. $g$ is diffentiable with $g'(x)=\frac{1}{f'(g(x))}$.

  • $\begingroup$ How do we know that $g'(x)=\frac{1}{f'(g(x))}$ ? And that it is unique? $\endgroup$ – jackwo Aug 20 '15 at 14:32
  • $\begingroup$ see this. $\endgroup$ – user795571 Aug 20 '15 at 14:42
  • $\begingroup$ user chain rule in $f\circ g =id$. $\endgroup$ – Alcides de carvalho jr Oct 25 '15 at 0:27

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