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I am looking for a function $f$ having the following characteristics:

  • $f$ defined on $[0,1]$
  • $f(0)=0$
  • $f(1)=1$
  • $ \forall x \in ]0,1[, x

  • $f$ differentiable on $]0,1]$

  • f'>0
  • f'(1)=1
  • \lim\limits_{x\to0} f'(x)=+\infty

Finally, I will also need an analytical expression of the inverse function $f^{-1}$.

Do you know such function?

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    Thanks andré, you are right: the last condition is incompatible. I replace it with: $\lim\limits_{x\to0} f'(x)=+\infty$2012-01-01

1 Answers 1

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A solution is the function $f:[0,1]\to[0,1]$ defined by $ f(x)=\tfrac12(1+x)\sqrt{x}, $ whose inverse function $g$ is defined by $ g(y)=\left(\sqrt{y^2+\tfrac1{27}}+y\right)^{2/3}+\left(\sqrt{y^2+\tfrac1{27}}-y\right)^{2/3}-\tfrac23 $

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    Jewel-li$k$e example!2012-01-07