surjectivity and inverse of a function 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: $f^{-1} (x) = \sqrt{x+xf^{-1}(x)^2}$.
 A: $f$ 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,
$$f^{-1}(x)=\sqrt{\frac{x}{1-x}}$$
A: 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 $f(x) = x^2/(1+x^2)$.
