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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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How did you get a square root in your formula for $f^{-1}$? –  user108903 Dec 16 '12 at 15:22
    
i forgot to put ^2 to x in the original equation –  random guy Dec 16 '12 at 15:24
    
Take $y\in[0,1)$ and show that the equation $y={x^2\over1+x^2}$ has a solution $x\ge0$ (just solve for $x$). This will also give you a formula for the inverse. –  David Mitra Dec 16 '12 at 15:34
    
$x = \pm \sqrt { y\over 1-y} $? –  random guy Dec 16 '12 at 15:39
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2 Answers

up vote 1 down vote accepted

$f$ 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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thanks but i dont understand how is derivative important for surjectivity –  random guy Dec 16 '12 at 15:30
    
@randomguy I added some more detail –  Nameless Dec 16 '12 at 15:31
    
i see it now ;) umm do you have any idea how to build its inverse? –  random guy Dec 16 '12 at 15:33
    
@randomguy Sure. –  Nameless Dec 16 '12 at 15:34
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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 $f(x) = x^2/(1+x^2)$.

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