Is $x^{\frac{1}{2}}$ equal to $\sqrt{x}$ or $\pm\sqrt{x}$?

Question

I have seen that when graphing $f\left(x\right)=x^{\frac{1}{2}}$ the graph only outputs positive and zero values (the range is greater or equal to 0), but according to what I know about algebra (correct me if I'm wrong), $x^{\frac{1}{2}}$ is equal to $\pm\sqrt{x}$.

Is it because $x^{\frac{1}{2}}$ actually equals $\sqrt{x}$ or because $f\left(x\right)=x^{\frac{1}{2}}$ can only be a function if we ignore negative outputs?

Unless the context explicitly demands something else, most of the times it is assumed that $x\mapsto x^{1/2}$ is the function $g:[0,\infty)\to[0,\infty)$ such that $(g(x))^2=x$ for all $x$. — , Apr 13 '19 at 16:37

score 1 · Answer 1 · answered Apr 13 '19 at 16:37

There is a difference between the square root $$\sqrt x=x^{1/2},$$ which is defined to be the non-negative solution for $y$ of the equation $y^2=x$ for $x\geq0$, and the solutions of the equation $$x^2=y,$$ which are actually $x=\pm\sqrt y$.

If you view this as a complex function, then there would be two branches of the function; and you would have to decide if you take the one or the other.

Note that a function, however, only ever takes one $y$ value for each $x$ value, and never two. Then, this would not be a function anymore.

Is $x^{\frac{1}{2}}$ equal to $\sqrt{x}$ or $\pm\sqrt{x}$?

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