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How to show that $f(x)=x^2/(1+x^2)$ is surjective on the codomain $[0,1)$ from $(0,\infty)$?

What would be its inverse? I already proved injectivity, so it must be bijective if $f(x)$ is also surjective.

I get stuck with recursive definition: $f^{-1} (x) = \sqrt{x+xf^{-1}(x)^2}$.

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    $x = \pm \sqrt { y\over 1-y} $?2012-12-16

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$f$ is continuous on $(0,+\infty)$ and $f^{\prime}(x)=\frac{(x^2)^{\prime}(1+x^2)-x^2(1+x^2)^{\prime}}{(1+x^2)^2}=\frac{2x+2x^3-2x^3}{(1+x^2)^2}=\frac{2x}{(1+x^2)^2}>0$ so $f$ is increasing in $(0,\infty)$. $\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{x^2}{1+x^2}=\lim_{x\to \infty}\frac{1}{\frac{1}{x^2}+1}=1 $ Therefore, $f((0,+\infty))=(f(0),\lim_{x\to \infty}f(x))=(0,1)$ $f$ is thus surjective.

For the inverse: We know $f$ is injective in $(0,+\infty)$ and $f((0,+\infty))=(0,1)$. $f(x)=\frac{x^2}{1+x^2}\Leftrightarrow x^2f(x)+f(x)=x^2\Leftrightarrow x^2(1-f(x))=f(x)$ Since $f(x)\neq 1$, $x^2=\frac{f(x)}{1-f(x)}$ Since $x>0$, $x=\sqrt{\frac{f(x)}{1-f(x)}}$ The inverse of $f$ is therefore, $f^{-1}(x)=\sqrt{\frac{x}{1-x}}$

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    @randomguy Sure.2012-12-16
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Observe first that $f(x) = 0 \Leftrightarrow x = 0$ and then your function is not surjective with that domain. You can prove easily that $f$ is increasing and therefore as $f(0) = 0$ and $\lim_{x\to \infty} f(x) = 1$ then you have that $f$ is surjective. To find its inverse you have to isolate $x$ form $f(x) = x^2/(1+x^2)$.