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Let us take the function $$ y= \sqrt x. $$

Of course we must say that $x$ must be $\ge 0$ (domain $[0, +\infty]$). Let us suppose that we should determine its image.

I extract $x$ and obtain $$ x= y^2. $$

Now I observe that $y$ can assume any value in $\mathbb R$. BUT it is not true. $y$ is $\ge0$.

Can anyone explain me why? Where is the mistake I do?

Nick

Piyo
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Nick
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    Welcome to Maths SX! That IS simply because, of the two square roots of a positive number, $\sqrt x$ denotes, by convention, the positive one. – Bernard Oct 01 '21 at 10:38
  • Also: https://math.stackexchange.com/q/1569629/42969. – Martin R Oct 01 '21 at 11:06
  • More related questions: https://math.stackexchange.com/questions/linked/1448885? – Martin R Oct 01 '21 at 11:09
  • @MartinR The first two you have proposed as duplicate, surely related, are more on precalculus issues. Maybe there exists some othe duplicate more suitable as a target. – user Oct 01 '21 at 11:11
  • Another one: https://math.stackexchange.com/q/26363/42969 – Martin R Oct 01 '21 at 11:13
  • @MartinR I didn't find among the previous question given a good target duplicate for this question. Indeed, here the OP is aware that $y=\sqrt x \ge 0$, the doubt is about the steps used to obtain the image. Could you please me indicate what is the current target duplicate you have selected. Thanks – user Oct 01 '21 at 19:06

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The function $y=\sqrt x$ is indeed, by definition, the inverse function of the function $y=x^2$, defined in the domain $x\in [0,\infty)$ indeed assuming $x\in \mathbb R$ the latter doesn’t admit an inverse.

See also the related

user
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    And this has indeed been asked and answered many times before: https://math.stackexchange.com/questions/linked/1448885. – Martin R Oct 01 '21 at 11:07
  • @MartinR I supposed the subject is not new, I'll take a look to the target duplicates you have posted. Thanks – user Oct 01 '21 at 11:08