Is the equation $y=\sqrt x$ a function?

Question

Is the equation $y=\sqrt{x}$ a function?

A function has one only one $y$ value for a $x$ value.

An equation is an expression.

For the equation $y=\sqrt{x}$, $y=\pm\sqrt{x}$ Therefore, for a given $x$, there are 2 $y$ values.

However when I plot this equation in the calculator it only shows me the top half.

Please explain.

Thanks!

Answer 1

Notice $\sqrt x$ is only the principal square root function, not "a thing that gives you every square root". There is only one principal square root for every non-negative real number.

This differentiates with the concept of square roots. While there are two square roots for every positive number, $\sqrt x$ gives the positive square root for positive $x$. Like when you say $x^2 = 25$, you have to denote the two square roots of $25$ to be $\sqrt {25}$ and $-\sqrt {25}$. And your calculator is following this notation.

Answer 2

Thought I'd elaborate some on peterwhy's answer, since this seems to be a common point of confusion for students in the algebra - calculus sequence.

"For the equation $y = \sqrt{x}$, $y = \pm\sqrt{x}$ $\ldots$"

For any non-negative real number $x$, there is one and only one non-negative real number $s^2$ such that $r^2 = x$. (For example: If $x = 4$, then $s = 2$.) This is sometimes called the principal square root of $x$.

The symbol $\sqrt{x}$ is by definition equal to $s$, for each $x \geq 0$. That's all there is to it. We humans agreed a long time ago that it would be useful to have a symbol for "the non-negative $s$ that, when squared, equals $x$", and putting $x$ beneath a "radix" $\sqrt{}$ to form $\sqrt{x}$" became the winner. It's a nice choice, too, since it's easy to write and has which $x$ we're referring to built into it. (It's worth noting that the radix symbol has been used in this way since the Dark Ages!)

With this in mind, there is no ambiguity in the definition of $y = \sqrt{x}$ for $x \geq 0$; you set $x = 0$, and the output is $y = 0$; you set $x = 4$, the output is $y = 2$; you set $x = \pi$, the output is $\sqrt{\pi}$; and so on.

Your confusion probably stems at least partially from your experience solving for $x$ in equations like $x^2 = 3$. If the correct solution to $x^2 = 3$ is $x = \pm\sqrt{3}$, and I know that the function $y = \sqrt{x}$ satisfies the equation $y^2 = x$, then why isn't it true that $y = \pm\sqrt{x}$?

Can you see the flaw in the reasoning that leads you to conclude $y = \pm\sqrt{x}$? When we start with $x^2 = 3$ and conclude $x = \pm\sqrt{3}$, we don't know what $x$ is, and so we list all possible values of $x$ for which the equation $x^2 = 3$ is true. The conclusion $x = \pm3$ says "if $x^2 = 3$ then $x$ could be $\sqrt{3}$ or $-\sqrt{3}$" (not that $x$ is both $\sqrt3$ and $-\sqrt3$ -- that's impossible!).

On the other hand, if $y = \sqrt{x}$ the reason why the solution to $y^2 = x$ is $y = \sqrt{x}$ and not $y = \pm\sqrt{x}$ is the I've already told you that $y = \sqrt{x}$, and $\sqrt{x} \neq -\sqrt{x}$ (for $x\neq0$). Saying that $y$ could equal $-\sqrt{x}$ for all $x \geq 0$ after I've already told you that $y = \sqrt{x}$ is akin to me telling you that $y = 2$ and you concluding that $y = -2$.