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Can $\sqrt{x}$ be considered a function for x (set of real numbers greater than or equal to 0)?

Moreover, why is the graph of $\sqrt{x}$ only maps to positive y values if for example $\sqrt{9}$ = 3 and -3 shouldn’t it be symmetrical over the x axis?

Ishraaq Parvez
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Anay Chadha
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  • By convention $\sqrt{x}$ is the non-negative real number which square is $x$. In fact, $x^2$ is not invertible on $\mathbb R$, we have to choose one of the branches. Usually, the branch with non-negative $x$'s is used and then $\sqrt{x}$ together with the convention is the inverse. – Peter Apr 05 '21 at 08:24
  • Plot imaginary if you want to see the graph on negative $y$ – Aderinsola Joshua Apr 05 '21 at 08:30

1 Answers1

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The domain and range of sqrt(x) is always non negative number. 1.) Domain must be non negative otherwise the number would become imaginary. E.g sqrt(-1) = iota. 2.) As square root of non negative number is also non negative, range must be toward positive side.

Gurjot
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