Is $y=0^x$ really a function?

Question

I know that there are three main options when it comes to dealing with $0^0$: $$0^0=0$$ $$0^0=1$$ $$\nexists x(x=0^0)$$My question is, do we have to assert that one of these conditions is true in order to declare $y=0^x$ a function? I ask because Desmos plots both $(0,0)$ AND $(0,1)$ (to show the controversy I suppose). Thanks for your time!

Answer 1

If a rule is such that $r(k)$ is not defined then you simply can not define a function as being $f(x) = r(x)$ is $k$ is in the domain. Period.

If the domain of the function is $[0, \infty)$ then the function can not be $f(x) = 0^x$. It just plain can not be.

If you want to a function to be $f:[0,\infty) \to \mathbb R$ and $f(x)=0^x$ when $x\ne 0$ then you (must/only need to) extend the definition of the function to include a value for $f(0)$. So for example:

$f(x) = \begin{cases}0^x & \text{if }x \ne 0\\\frac 1{\sqrt \pi}+27.3&\text{if }x = 0\end{cases}$.

What you set $f(0)$ to doesn't matter. You can set it to $0$ if you want (that's the only value that will make the function continuous FWIW) or you can set it to $1$ or you can set it to $\frac 1{\sqrt \pi} + 27.3$. Whatever you set it to the function will not be $f(x) = 0^x$. That is simply not an option if $0\in $ the domain of the function. Period.

$0^0$ will remain undefined.