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The $f(x)=\sqrt{x}$ has a domain of $[0,\infty)$. As of in our textbook it is written that range is also $[0,\infty)$. How? $\sqrt{x}=-\sqrt x\space or \space \sqrt x.$ According to me range should be $(-\infty, \infty)$ then.

for example:

  1. f(1) = -1 or 1
  2. f(4) = -2 or 2
  3. f(9) = -3 or 3
    .
    .
    .
  4. $f(\infty)=-\infty \space or \space \infty$
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    $\begingroup$ $\sqrt x$ is by definition the positive square root of $x$ $\endgroup$
    – trisct
    Commented Mar 30, 2021 at 23:52
  • $\begingroup$ The function $x \mapsto \sqrt{x}$ for non-negative real $x$ is defined, by convention, to take non-negative values. $\endgroup$
    – Rob Arthan
    Commented Mar 30, 2021 at 23:54
  • $\begingroup$ @trisct: $0$ doesn't have a positive square root, but $\sqrt{0} = 0$. $\endgroup$
    – Rob Arthan
    Commented Mar 30, 2021 at 23:55
  • $\begingroup$ @RobArthan Nonnegative then. $\endgroup$
    – trisct
    Commented Mar 30, 2021 at 23:57
  • $\begingroup$ $f(1)=1$ and only $1$. $f(4)=2$ and only two, not negative two. Also, $\infty$ is not a real number so $f(\infty)$ is invalid. $\endgroup$
    – JMoravitz
    Commented Mar 30, 2021 at 23:57

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