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When we write $x^2=4$. It means $x=+2$ or $x=-2$.

So then why is the range of the function $f(x) = \sqrt{x^2+4} \quad$ $[2,\infty)$ and not $(-\infty, - 2] \cup [2,\infty)$?

Please give me a hint, where am I going wrong?

Paul Frost
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Intermechanic
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  • Notice that you write "function". You have to take either the negative value or the positive value consistently to make this so. And the standard is to take the positive value. – Richard Jensen Feb 14 '22 at 09:02
  • Find the inverse function. Its domain is the range of your current function. – Bumblebee Feb 14 '22 at 09:02
  • Okay, I get it now, thank you everyone! – Intermechanic Feb 14 '22 at 09:03
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    This has been asked frequently: https://math.stackexchange.com/questions/linked/26363 – Martin R Feb 14 '22 at 09:14
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    $\sqrt{x}$ means the non-negative square root of $x.$ It is defined this way so that $f:x\to \sqrt{x}$ is a function. $f:x\to\pm \sqrt{x}$ is not a function. So it doesn't really make sense to write $f:x\to\pm \sqrt{x}.$ – Adam Rubinson Feb 14 '22 at 09:52
  • See this answer https://math.stackexchange.com/questions/380359/why-is-sqrt4-2-and-not-pm-2 – coudy Feb 14 '22 at 09:52

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