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I try to understand the inverse function theorem by trying learn what will happen if one of conditions isn't satisfy. I try to find if it's possible to find an example such that

a function $f:[a,b] \to [a,b]$ is bijective,

$f$ is infinitely differentiable,

$f$ has some continuous inverse function $g$,

but $g$ is not differentiable at some point of $[a,b]$

I don't think there is an example satisfy all of above, but I'm not so sure about that.

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    $\begingroup$ $x^2$ on the interval $[0,1]$. $\endgroup$ – user61527 Mar 23 '14 at 18:02
  • $\begingroup$ may I have a question ? from your example $g(x)= \sqrt {x}$ which is defined on $[0,1]$, but how is it not differentiable on that interval? Maybe this is a stupid question, but I forget alot about calculus and really want to regain my knowledge. $\endgroup$ – Diane Vanderwaif Mar 23 '14 at 18:35
  • $\begingroup$ It's not differentiable at zero, since it has a vertical tangent line there. $\endgroup$ – user61527 Mar 23 '14 at 18:36
  • $\begingroup$ oh, I see , thank you very much. $\endgroup$ – Diane Vanderwaif Mar 23 '14 at 18:37
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Whenever the derivative of $f$ has a zero, the inverse function $g$ will not be differentiable. Thus, a function such as $x^2$ on $[0,1]$ or $x^3$ on $[-1,1]$ will give examples of non-differentiable inverse functions.

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