for $1 \geq x \geq 0: {2x^2\over{(2+x)}} \leq y \Rightarrow x \leq \left(\frac{3}{2} y \right)^{1/2}$

Question

So what I did is prove that $f(x) := {2x^2\over{2+x}}$ is increasing and then invert $f$ on $[0,\infty]$ this yields $(f\restriction_{[0,\infty[})^{-1}(y) = \frac{1}{4}(y+\sqrt{y}\sqrt{y+16})$ and this gives us that: $$ x \leq \frac{1}{4}(y+\sqrt{y}\sqrt{y+16}) $$

Answer 1

$0\le x\le1$

So we have

$ {2x^2\over{(2+1)}} \leq {2x^2\over{(2+x)}} \leq y\implies x\le (\frac{3}{2}y)^\frac{1}{2}$