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I'd like to know if the following statement is true ?

If $f : (0,1) \to \mathbb{R}$ is a strictly monotonically increasing function and $f$ is differentiable at some $x \in (0,1)$ then $f^{-1}(y)$ is differentiable at $y = f(x)$ ?

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Yup, the answers down below cover it. Just to be sure you're not missing out on anything; the inverse $f^{-1}$ of a real to real function can aquired from the graph of $f$ by just flipping the xy plane along the diagonal. If you haven't already you might find it worthwhile to connect the algebraic picture below with this geometric picture. –  Eivind Dahl Apr 25 '11 at 10:09
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2 Answers

up vote 3 down vote accepted

Yes, if $f'(x)>0$; then $(f^{-1})'(y)=1/f'(x)$. But not if $f'(x)=0$.

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By strictly monotonous means does it imply $f'(x) \neq 0 \forall x $ –  Rajesh D Apr 25 '11 at 10:11
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No. For example, $f(x)=x^3$ is strictly increasing, but $f'(0)=0$ nevertheless. –  Hans Lundmark Apr 25 '11 at 10:21
    
It might be added that if you replace "strictly increasing" by just "bijective", then the statement is not true without additional assumptions. (For example, that $f^{-1}$ is continuous at $y$; this is not automatic even if $f$ is bijective and differentiable at $x$, even though it requires a bit of work to come up with a counterexample.) –  Hans Lundmark Apr 25 '11 at 10:24
    
Come to think of it, maybe you need to be a little careful here as well... For $f^{-1}$ to be differentiable at $y$, it needs first of all to be defined in a neighbourhood of $y$, so the range of $f$ must contain such an interval, which it doesn't necessarily do (say if $f$ has jump discontinuities accumulating at $x$). But if we add the assumption that $f$ is continuous in a neighbourhood of $x$, then we should be fine. –  Hans Lundmark Apr 25 '11 at 11:54
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I am afraid that is not true $f= (x-1/2)^2$.

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Maybe you meant cubed instead of squared? –  Hans Lundmark Apr 25 '11 at 9:58
    
sqrt is not differentiable in 0 either. cube is correct too. –  El Moro Apr 25 '11 at 10:05
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The point was that your function is not strictly increasing on $(0,1)$. –  Hans Lundmark Apr 25 '11 at 10:25
    
rightig sorry :) –  El Moro Apr 25 '11 at 12:09
    
so you should edit your answer, rightig? –  draks ... Jul 14 '12 at 20:27
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