Is $y=\sqrt{6x}$ a function? How can you be sure just by looking at the equation alone?

Question

I thought it wouldn't be a function since the answer could be + or -, but I'm probably wrong.

Answer 1

Functions are a rule mapping a domain set $X$ to a range set $Y$, where every element $x$ in $X$ is associated with a unique element $y$ in $Y$.

So I can't tell if $y = \sqrt{6x}$ is a function until you tell me $X$ and $Y$. If you said $X$ and $Y$ are both all the real numbers, that includes negative numbers, and $\sqrt{6(-2)}$ doesn't exist. On the other hand, if you said the domain and range are all the non-negative real numbers, or the domain and range are all the complex numbers, then you might be onto something.

Answer 2

$x^2 = 2$ is not the same thing as $x = \sqrt 2$

The radical $\sqrt{}$ sign points us to the "principal root." Or in the case of square roots of positive real numbers, the positive root.

e.g. $|x| = \sqrt {x^2}$

If we want both roots, we must say so.

In an equation like $x^2 = 2$ the solution set is $\pm \sqrt 2,$ including both roots.

Looking at the problem above. $y = \sqrt {6x}$

This function is defined in the real numbers if $x\ge 0$ is the domain. The root is the positive root. So, the range is $y\ge 0$

Answer 3

$y = \sqrt{6x}$ is just $y = \sqrt{x}$ multiplied by a factor of $\sqrt{6}$. Use the vertical line test to check if $y = \sqrt{x}$ is a function, which it is. In other words, it's a function (and also monotonically increasing).

-FruDe