Range of Square Root of X

Asked Mar 30 '21 at 23:48

Active Mar 31 '21 at 00:03

Viewed 28 times

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:

f(1) = -1 or 1
f(4) = -2 or 2
f(9) = -3 or 3
.
.
.
$f(\infty)=-\infty \space or \space \infty$

edited Mar 31 '21 at 00:03

Greg Martin

78,820

asked Mar 30 '21 at 23:48

Aagam Sheth

1

$\sqrt x$ is by definition the positive square root of $x$ – trisct Mar 30 '21 at 23:52
The function $x \mapsto \sqrt{x}$ for non-negative real $x$ is defined, by convention, to take non-negative values. – Rob Arthan Mar 30 '21 at 23:54
@trisct: $0$ doesn't have a positive square root, but $\sqrt{0} = 0$. – Rob Arthan Mar 30 '21 at 23:55
@RobArthan Nonnegative then. – trisct Mar 30 '21 at 23:57
$f(1)=1$ and only $1$. $f(4)=2$ and only two, not negative two. Also, $\infty$ is not a real number so $f(\infty)$ is invalid. – JMoravitz Mar 30 '21 at 23:57
@trisct: that fixes it. $\ddot{\smile}$. – Rob Arthan Mar 30 '21 at 23:58

Range of Square Root of X

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