I have trouble figuring out this problem:
Prove that the function $f: [0,\infty)\rightarrow[0,\infty)$ defined by $f(x)=\frac{x^2}{2x+1}$ is a bijection.
Work: First, I tried to show that $f$ is injective. $\frac{a^2}{2a+1}=\frac{b^2}{2b+1}$ I got $a^2(2b+1)=b^2(2a+1)$. However, I get stuck here and cannot simplify the equation to get $a=b$, which would prove that the function is injective.
Notice $f(x) = \frac{x^2}{2x+1} = \frac{x}{2 + \frac{1}{x}}$. Suppose $f(a) = f(b) $, then
$$ \frac{a}{2 + \frac{1}{a} } = \frac{b}{2 + \frac{1}{b}} \iff \frac{a}{b} = \frac{ 2 + \frac{1}{a}}{2 + \frac{1}{b}}$$
We want to show that $a = b$. Suppose not. then either $a > b $ or $b >a $.
Say $a > b $. Then $\frac{1}{b} > \frac{1}{a}$. and so
$$ \frac{a}{b} = \frac{ 2 + \frac{1}{a}}{2 + \frac{1}{b}} < \frac{ 2 + \frac{1}{b}}{2 + \frac{1}{b}} = 1 \implies a < b \; \; \text{contradiction} $$
Now, say $b > a $. then $\frac{1}{a} > \frac{1}{b} \implies \frac{1}{a} + 2 > \frac{1}{b} + 2 \implies \frac{ 2 + \frac{1}{a}}{2 + \frac{1}{b}} > 1 \implies \frac{a}{b} > 1 \implies a > b $ contradiction. Hence
$a= b$ and $f$ must be injective.
You have $$ a^2(2b+1)=b^2(2a+1) $$ that can be written as $$ 2a^2b-2ab^2=b^2-a^2 $$ or $$ 2ab(a-b)=-(a+b)(a-b). $$ If $a-b\ne0$, you get $$ a+b+2ab=0 $$ Is this possible?
For surjectivity, given $y\ge 0$, you have to solve the equation $$ x^2=2xy+y $$ or $$x^2-2xy-y=0$$ with respect to $x$. It has one and only one non negative solution. Why?
One way to prove injectiveness in this case, is to prove that the function is strictly increasing. This can be done via derivative.
Incresing: $f'(x)=\frac{2x(2x+1)-2x^2}{(2x+1)^2}=\frac{2x^2+2x}{(2x+1)^2}\gt 0$, if $x\ne0$ thus f is strictly increasing. and thus is injective, now we need to show surjectivity.
Pick $y\in[0,\infty)$, if $f(x)=y\Rightarrow \frac{x^2}{2x+1}=y\Rightarrow x^2-2xy-y=0$
Thus by the quadratic formula $x=\frac{2y\pm\sqrt{4y^2+4y}}{2}\in [0,\infty)$, thus $f$ is also surjective.
(note that we only take the "+" root)
