If $f(x)=\frac{\sqrt{x}-x}{\sqrt{x}+2}$ and $x\in\mathbb D=[0,\infty]$ find $$f(\mathbb D)$$. I've tried to solve equation $y=f(x)$ and stopped to $x+(y-1)\sqrt{x}+2y=0$.
2026-04-17 13:41:17.1776433277
Find $f(\mathbb D)$
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$f(x)=\frac{\sqrt{x}-x}{\sqrt{x}+2}=\frac{\sqrt{x}+2-2-x}{\sqrt{x}+2}=1-\frac{{x}+2}{\sqrt(x)+2}$ $f(x)\leq1$ gives the upper bound
Now can you handle it