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Bonsoir j'ai un petit problème avec un exercice.

On considère la function $f:[-1,1] \to [-\frac{1}{2},\frac{1}{2}]$ définie par $f(x) = \frac{x}{1+x²}$.

Je dois montrer que f admet une fonction réciproque et définir cette fonction. Merci pour votre aide.

"Hello, I have a little problem with an exercise. Consider the function $f\colon[−1,1]\to[−\tfrac12,\tfrac12]$ defined by $f(x)=\frac x{1+x^2}$. I have to show that $f$ admits an inverse function and define that function. Thanks for your help." (Translation by Hagen von Eitzen as in the comments.)

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  • $\begingroup$ English, please $\endgroup$ – John Dvorak Feb 6 '14 at 21:41
  • $\begingroup$ He is looking for an inverse function $\endgroup$ – imranfat Feb 6 '14 at 21:43
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    $\begingroup$ "Hello, I have a little problem with an exercise. Consider the function $f\colon[-1,1]\to [-\tfrac12,\tfrac12]$ defined by $f(x)=\frac x{1+x^2}$. Show that $f$ admits an inverse function and define that function. Thanks for your help" (I guess, je suis desole that I don't speak a siungle word of French) $\endgroup$ – Hagen von Eitzen Feb 6 '14 at 21:43
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    $\begingroup$ @imranfat "nous ne parlons pas". "parle" is the singular form. $\endgroup$ – John Dvorak Feb 6 '14 at 21:50
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    $\begingroup$ @JanDvorak This is not an English-only site; adding a translation is fine and good, but please do not remove the original question text. $\endgroup$ – user61527 Feb 7 '14 at 0:56
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La fonction étant dérivable et sa dérivée est $$f'(x)=\frac{1-x^2}{(1+x^2)^2}>0,\;\forall x\in]-1,1[$$ donc $f$ est continue et strictement croissante sur son domaine de définition donc elle admet une fonction réciproque. Pour déterminer $f^{-1}$ on exprime $x$ en fonction de $y$ dans l'égalité: $$y=f(x)$$ en résolvant une équation de second degré.

Translation

The function is differentiable and its derivative is $$ f '(x) = \frac {1-x ^ 2} {(1 + x ^ 2) ^ 2} >0, \, \forall x \in( -1,1 )$$ so $ f $ is continuous and strictly increasing on its domain of definition so it admits an inverse function. To determine $f ^ {-1}$ we express $ x $ in terms of $ y $ in the equation: $$ y = f (x) $$ by solving a quadratic equation.

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    $\begingroup$ En réalité, la fonction $f$ est strictement croissante. (Pourquoi voter +1 pour cette réponse ?) // Actually, the function $f$ is increasing. (Why the upvote?) $\endgroup$ – Did Feb 6 '14 at 23:03
  • $\begingroup$ @Did J'ai édité ma réponse. Merci. // I edited my answer. Thanks. $\endgroup$ – user63181 Feb 7 '14 at 6:51
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Hint : Solve the following equation in $x$
$y=\frac{x}{1+x^2}$
You will get 2 solutions. Because of the domain, only one makes sense

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  • $\begingroup$ Can u give me the full correction, plz. because when I tried to searched for a solution, I has a problem at that: yx² + y - x = 0. and thank u for you help. $\endgroup$ – Khalid Puerto Feb 6 '14 at 22:02
  • $\begingroup$ Using the discriminant (what you got is an equation of degree 2), you get two solutions: $\frac{1+\sqrt{1-4y^2}}{2y}$ and $\frac{1-\sqrt{1-4y^2}}{2y}$. $\endgroup$ – Lepanais Feb 6 '14 at 22:06
  • $\begingroup$ Thank u so much, That's what I reached by. $\endgroup$ – Khalid Puerto Feb 6 '14 at 22:11

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